CCNA Todd Lammle Ch 1 - 5
Textbook Notes by Michael Banks 2021-2023
(The jumping around between colors: a memory-hack i was trying for the exam!)
Chapter 1 Exam Essentials Internetworking
1.Identify the possible causes of LAN traffic congestion
too many hosts in a broadcast domain
broadcast storms
multicasting
low bandwidth
2.Describe the difference between collision and broadcast domains
COLLISION DOMAIN- collection of devices where if one devices sends a packet on a segment, all devices on that
segment must acknowledge the message
BROADCAST DOMAIN- all devices on a segment will hear a broadcaset
3.Describe the difference between MAC and IP address
MAC- hexadecimal number identifying the physical connection of a host. Operates at layer 2.
Hosts uses the MAC to find each other on the same physical segment.
IP- can be expressed in binary or hexadecimal. Operates at layer 3. Logical identifier.
Hosts use the IP to find each other on different physical segments.
4.Describe the difference between hub, bridge, switch, router
HUB creates one collision domain and one broadcast domain
BRIDGE breaks up collision domains (one per interface) but still only one broadcast domain
SWITCH breaks up collision domains (one per interface) but still
only one broadcast domain. Uses hardware addresses to filter the network.
ROUTER uses IP addresses to filter the netwrok.
Breaks up collision AND broadcast domains.
5. Functions and advantages of routers
Packet switching, filtering, and path selection. Reduce broadcast traffic.
6. Connection-oriented VS connectionless services
Connection-oriented uses acknowledgements and flow control to create a reliable connection.
Requires more overhead than connectionless.
7.Describe the OSI layers, functions of each, and how devices/protocols map to each layer
Application, Presentation, Session are 'upper layers' that communicate from the user interface.
Transport provides segmentation, sequencing, and virtual circuits
Network provides logical addressing and routing
Data Link provides framing and placing the data on the network medium
Physical encodes the 1's and 0's into a digital signal
Lab 1.1: OSI Questions
1. Which layer chooses and determines the availability of communicating partners and resources needed to make the connection,
coordinates partnering applications, and forms consensus on procedures for conrolling data inegrity and error recovery?
The Application layer is responsible for finding the network resources brodacast from a server and
adding flow/error control(if the developer chooses).
2. Which layer is responsible for converting data packets from the Data Link layer into electrical signals?
Physical
3. At which layer is routing implemented, enabling connections and path selection between two systems?
Network
4. Which layer defines how data is formatted, presented, encoded, and converted for use on the network?
Presentation layer ensures that data is in a readable format for the application layer
5. Which layer is responsible for creating, managing, and terminating sessions between applications?
Session layer sets up, maintains, and terminates sessions between applications
6. Which layer ensures the trustworthy transmission of data across a physical link and is primarily concerned with physical addressing,
line discipline, network topology, error notification, ordered delivery of frames, and flow control?
Data Link layer provides physical addressing and error notification but NOT error correction
7. Which layer is used for reliable communication between end nodes over the network and provides mechanisms for
establishing, maintaining, and terminating virtual circuits; transport-fault detection and recovery;
and controlling the flow of information?
Transport
8. Which layer provides logical addressing that routers will use for path determination?
Network
9. Which layer specifies voltage, wire speed, and pinout cables and moves bits between devices?
Physical
10. Which layer combines bits into bytes and then into frames, uses MAC addressing, and provides error detection?
Data Link
11. Which layer is responsible for keeping the data from different applications seperate on the network?
Session
12. Which layer is represented by frames?
Data Link
13. Which layer is represented by segments?
Transport
14. Which layer is represented by packets?
Network
15. Which layer is represented by bits?
Physical
16. Put these in order of encapuslation: packet, frame, bit, segment
- segment
- packet
- frame
- bit
17. Which layer segments and reassembles data into a data stream?
Transport
18. Which layer provides the physical transmission of the data and handles error notification, network topology and flow control?
Data Link
19. Which layer manages device addressing, tracks the location of devices on the network,
and determines the best way to move data on the network?
Network
20. What is the bit length and expression form of a mac address?
48 bit expressed as a 6 byte hexadecimal
Lab 1.2: Defining OSI Layers and Devices
1. This device sends and receives information about the Network layer
router
2. This layer creates a virtual circuit before transmitting between two end stations
transport
3. This device uses hardware addresses to filter a network
switch or bridge
4. Ethernet is defined at these layers
data link and physical
5. This device supports flow control, sequencing, and acknowledgements
transport
6. This device can measure the distance to a remote network
router
7. Logical addressing is used at this layer
network
8. Hardware addresses are defined at this layer
9. This device creates one big collision domain and one big broadcast domain
hub
10. This device creates many smaller collision domains, but the network is still one large broadcast domain
switch or bridge
11. This device can never run full duplex
hub
12. This device breaks up collision domains and broadcast domains
Router
Chapter 1 Review Questions
1. A receiving host has failed to receive all of the segments that it should acknowledge. What can the host do to improve
the reliability of this communication session?
a receiving host can control a transmitter by using flow control. By decreasing the window size, the receving host
can slow down the transmitting host to avoid buffer overlflow.
TCP uses windowing by default
Transport Layer = flow control
2. When a station sends a transmission to the MAC address ff:ff:ff:ff:ff:ff what type of transmission is it?
broadcast
3. Which layer 1 devices can be used to enlarge the area covered by a single LAN segment?
Switch or Hub
but the hub will enlarge a single collision domain
4. Segmentation of a data stream happens at which layer of the OSI model?
transport layer
receives large data streams from the upper layer and breaks it into segments
5. Describe the main router functions
packet switching, packet filtering, internetwork communication, path selection
note that routers do create/terminate collision domains, but that isnt their main function
6. At which layers do these devices operate? Rouer, LAN Switch, Ethernet hub, Word Processor
Router- layer 3
Switch- layer 2
Hub- layer 1
Word Processor- NONE- word processors communicate TO the application layer interface, but dont operate within it
7. When data is encapsulated, which is the correct order?
data > segment > packet > frame > bit
8. Why does the data communication industry use the OSI model?
OSI divides the process into smaller/simpler components which aids in development, design, and troubleshooting.
It allows develoers to changes aspects of a program at just one layer
9. What are two purposes of segmentation with a bridge?
add more broadcast domains
allow more broadcasts
10. What are causes of LAN congestion?
too many hosts in a broadcast domain, broadcast storms, low bandwidth
11. If a switch has three computers connected to it, with no VLANs present, how many broadcast/collision domains
is the switch creating?
one broadcast, three collision
12. Acknowledgements, sequencing, and flow control are characteristics of which OSI layer?
TRANSPORT layer 4
"a reliable connection" is defined by a virtual circuit that employs:
acknowledgements
sequencing
flow control
13. List types of flow control
buffering
windowing
congestion avoidance
14. If a hub has three computers connected to it, how many broadcast domains and collision domains is the hub creating?
one broadcast, one collision
15. Which is the purpose of flow control?
allows the receiving device to control the transmitter to avoid buffer overflow on the receiving end
16. Which three statements are true about the operation of a full-duplex network?
there are no collisions in full-duplex mode.
a dedicated switch port is required for each node in full-duplex.
the host network card and switch port must be capable of operating in full-duplex mode
Full duplex means you using using both wire pairs simultaneously to send and receive data. You will not have collisions
since you have a dedicated switch port for each node.
Both the host network card and the switch port must be capable and set to work in full duplex
17. Whare are benefits of OSI Model?
divides the network communication proces into smaller/simpler components
allows multiple-vendor development through standardization
allows various types of software and hardware to commnunicate
18. Which devices operate at ALL levels of the OSI model?
NMS- network management station
web and application servers
network hosts
19. When an HTTP document must be retrieved from a location other than the local machine,
what layer of the OSI model must be accessed first?
application
20. Which layer of the OSI model offers three different modes of communication: simplex, half duplex, and full duplex?
session
END CHAPTER 1
3 hours 12-11-21
original hours 3
3.5 hours recap 12-11-22
review hours 3
Chapter 2 Ethernet and Data Encapusulation Exam Essentials
1. Describe the operation of Carrier Sense Multiple Access with Collision Detection (CSMA/CD)
Helps devices share the bandwidth evenly without having two devices transmit at the same time. It does not eliminate
collisions, but it greatly reduces them, which reduces retransmissions.
How does it work? Pg 43
When a host wants to transmit, it first checks for signal on the wire. It all clear, proceed with transmission.
The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If another
signal is detected, send out a jam singal that causes all noted on the segment to stop transmitting (busy signal).
The other notes respond to that jam signal by waiting awhile before transmitting again.
Backoff algorithms determine when the colliding statinos can retransmit.
If collisions keep occuring after 15 tries, the nodes will timeout.
When a collision occurs, the following happens:
- A jam signal informs all devices that a collision occured
- The collision invokes a random backoff algorithm
- Each device on the segment stops transmitting until their backoff timers expire
- All hosts have equal priority to transmit after the timers have expired
A network sustaining heavy collisions will experience DELAY, LOW THROUGHPUT, CONGESTION
2. Differentiate half-duplex and full-duplex communications and define the requirements to utilize each method.
Full-duplex uses two paris of wires and allows for sending and receiving at the same time. Uses different wires
to eliminate collisions. For this to work, devices at each end must be capable and configured for full-duplex.
Half-duplex uses one wire pair can not send/recieve simlutaneously and can suffer collisions.
3. Describe the sections of a MAC address and the information contained in each section.
The first 24 bits (3 bytes) are the OUI- Organizationally Unique Identifier-
identifies the manufacturer
the remaining 24 bits identify the NIC
4. Identify the binary and hexadecimal equivalent of a decimal number
Any number expressed in one format can be also be expressed in the others.
5. Identify the fields in the data link portion of an ethernet frame
The fields in the Data Link portion of a frame include:
- Preamble
- Start Frame Delimiter
- Destination MAC Address (DA)
- Source MAC Address (SA)
- Length or Type
- Data
- Frame Check Sequence (FCS)
Note on the preamble and SFD:
6. Identify the IEEE physical standards for ethernet cabling
These standards describe the capabilities and physical characteristics of cable types including
10Base2, 10Base5, and 10BaseT
these three are the ORIGINAL IEEE 802.3 Standards
10Base2- 10Mbps, Baseband, 200 meters
"baseband"
Max meters: 185
thin coax and BNC connectors
Known as "thinnet" and can support up to 30 workstations on a single segment
connects everything via BUS
10Base5- 10Mbps, Baseband, 500 meters
uses THICK coax aka "thicknet"
AUI connectors- Attachment Unit Interface
Repeaters allow up to 2500 meters
Supports up to 1024 users
connects everything via BUS
10BaseT- 10Mbps 100 meters
Cat 3 UTP, RJ45 connector
No BUS anymore, each device must connect to a hub or switch
one host per segment/wire
Here are the EXPANDED 802.3 standards
100Base-TX 100 meters
802.3u
FAST ETHERNET Cat 5, 5e, or 6 UTP two-pair wiring
100Base-FX
FIBER
---------------------------------------------------
1000Base-CX
1000Base-T 100 meters
802.3ab
Cat 5 four-pair wiring
16 Gbps
1000Base-SX
FIBER
1000Base-LX
FIBER
1000Base-ZX
FIBER
---------------------------------------------------
10GBase-T 10Gbps, 100m, UTP
802.3an
10GBase-LR Long Range
FIBER
10GBase-ER Extended Range
FIBER
10GBase-SW Short Wavelength
FIBER
10GBase-EW Extra Long Wavelength
FIBER
7. Differentiate type of ethernet cabling and identify their proper application
Three types:
- straight-through
to connect a PC or router to a switch
- crossover
to connect hub/switch to hub/switch, or pc to pc
- rolled
console at PC to router or switch/span>
8. Describe the data encapsulation process and the role it plays in packet creation.
Information is added to the frame from each OSI Layer.
AKA Packet Creation
Each layer communicates only with it's peer layer
9. Understand how to connecte a console cable from a PC to a router and start HyperTerminal
Rolled cable connects to COM port of the PC to the console port of the router.
Start HyperTerminal and set bits per second to 9600
set flow control to none
10. Identify the layers in the Cisco three-layer model and describe the ideal function of each.
Pg.65 Note these are LOGICAL distinctions. A device can traverse layers, or vice versa
CORE aka Backbone
Transporting large amounts of traffic
DISTRIBUTION aka Routing
Routing, filtering, WAN access
AKA the workgroup layer. The communication point between Access layer and Core layer
Routing
Implementing access lists, packet filtering, queuing, security policies, address translation, firewalls
Redistributing between routing protocols/ static routing
Routing between VLANs
Definining broadcast and multicast domains
ACCESS aka Switching
AKA the desktop layer
client connectivity into the distribution layer
The network resources most users need will be avaialble locally.
Creation of seperate collision domains (Segmentation)
Lab 2.1: Binary/Decimal/Hexadecimal Conversion
1. convert from decimal IP address to binary format
192.168.10.15
11000000 10101000 00001010 00001111
2. convert from decimal IP address to binary format
172.16.20.55
10101100.0001000.00010100.00110111
3. convert from decimal IP address to binary format
10.11.12.99
00001010.00001011.00001100.01100011
4. convert from binary to decimal IP address
11001100.00110011.10101010.01010101
204.51.170.85
5. convert from binary to decimal IP address
11000110.11010011.00111001.11010001
198.211.57.209
6. convert from binary to decimal IP address
10000100.11010010.10111000.10100110
132.210.184.166
7. convert from binary to hexadecimal
11011000.00011011.00111101.01110110
D8 1B 3D 76
8. convert from binary to hexadecimal
11001010.11110101.10000011.11101011
CA F5 83 EB
9. convert from binary to hexadecimal
10000100.11010010.01000011.10110011
84 D2 43 B3
Lab 2.2: CSMA/CD Operations
1. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) helps to minimize collisions on the network, thereby
increasing data transmission efficiency.
List the correct order of the CSMA/CD steps.
A jam signal informs all devices that a collision occured
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
Lab 2.3: Cabling
For each situation which cable to use: straight-through, crossover, or rolled
1. host to host
crossover
2. host to switch or hub
straight
3. router direct to host
crossover
4. switch to switch
crossover
5. router to switch or hub
straight
6. hub to hub
crossover
7. hub to switch
crossover
8. host to a router console serial communication (COM) port
rolled
The straight cable only works when connecting UNLIKE devices
Lab 2.4: Encapsulation
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each
packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are
used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
Chapter 2 Review Questions
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired,
all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore.
Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1.Identify the possible causes of LAN traffic congestion
too many hosts in a broadcast domain
broadcast storms
multicasting
low bandwidth
broadcast storms
multicasting
low bandwidth
2.Describe the difference between collision and broadcast domains
COLLISION DOMAIN- collection of devices where if one devices sends a packet on a segment, all devices on that
segment must acknowledge the message
BROADCAST DOMAIN- all devices on a segment will hear a broadcaset
BROADCAST DOMAIN- all devices on a segment will hear a broadcaset
3.Describe the difference between MAC and IP address
MAC- hexadecimal number identifying the physical connection of a host. Operates at layer 2.
Hosts uses the MAC to find each other on the same physical segment.
IP- can be expressed in binary or hexadecimal. Operates at layer 3. Logical identifier. Hosts use the IP to find each other on different physical segments.
IP- can be expressed in binary or hexadecimal. Operates at layer 3. Logical identifier. Hosts use the IP to find each other on different physical segments.
4.Describe the difference between hub, bridge, switch, router
HUB creates one collision domain and one broadcast domain
BRIDGE breaks up collision domains (one per interface) but still only one broadcast domain
SWITCH breaks up collision domains (one per interface) but still only one broadcast domain. Uses hardware addresses to filter the network.
ROUTER uses IP addresses to filter the netwrok. Breaks up collision AND broadcast domains.
BRIDGE breaks up collision domains (one per interface) but still only one broadcast domain
SWITCH breaks up collision domains (one per interface) but still only one broadcast domain. Uses hardware addresses to filter the network.
ROUTER uses IP addresses to filter the netwrok. Breaks up collision AND broadcast domains.
5. Functions and advantages of routers
Packet switching, filtering, and path selection. Reduce broadcast traffic.
6. Connection-oriented VS connectionless services
Connection-oriented uses acknowledgements and flow control to create a reliable connection.
Requires more overhead than connectionless.
7.Describe the OSI layers, functions of each, and how devices/protocols map to each layer
Application, Presentation, Session are 'upper layers' that communicate from the user interface.
Transport provides segmentation, sequencing, and virtual circuits
Network provides logical addressing and routing
Data Link provides framing and placing the data on the network medium
Physical encodes the 1's and 0's into a digital signal
Transport provides segmentation, sequencing, and virtual circuits
Network provides logical addressing and routing
Data Link provides framing and placing the data on the network medium
Physical encodes the 1's and 0's into a digital signal
1. Which layer chooses and determines the availability of communicating partners and resources needed to make the connection,
coordinates partnering applications, and forms consensus on procedures for conrolling data inegrity and error recovery?
The Application layer is responsible for finding the network resources brodacast from a server and
adding flow/error control(if the developer chooses).
2. Which layer is responsible for converting data packets from the Data Link layer into electrical signals?
Physical
3. At which layer is routing implemented, enabling connections and path selection between two systems?
Network
4. Which layer defines how data is formatted, presented, encoded, and converted for use on the network?
Presentation layer ensures that data is in a readable format for the application layer
5. Which layer is responsible for creating, managing, and terminating sessions between applications?
Session layer sets up, maintains, and terminates sessions between applications
6. Which layer ensures the trustworthy transmission of data across a physical link and is primarily concerned with physical addressing,
line discipline, network topology, error notification, ordered delivery of frames, and flow control?
Data Link layer provides physical addressing and error notification but NOT error correction
7. Which layer is used for reliable communication between end nodes over the network and provides mechanisms for
establishing, maintaining, and terminating virtual circuits; transport-fault detection and recovery;
and controlling the flow of information?
Transport
8. Which layer provides logical addressing that routers will use for path determination?
Network
9. Which layer specifies voltage, wire speed, and pinout cables and moves bits between devices?
Physical
10. Which layer combines bits into bytes and then into frames, uses MAC addressing, and provides error detection?
Data Link
11. Which layer is responsible for keeping the data from different applications seperate on the network?
Session
12. Which layer is represented by frames?
Data Link
13. Which layer is represented by segments?
Transport
14. Which layer is represented by packets?
Network
15. Which layer is represented by bits?
Physical
16. Put these in order of encapuslation: packet, frame, bit, segment
- segment
- packet
- frame
- bit
17. Which layer segments and reassembles data into a data stream?
Transport
18. Which layer provides the physical transmission of the data and handles error notification, network topology and flow control?
Data Link
19. Which layer manages device addressing, tracks the location of devices on the network,
and determines the best way to move data on the network?
Network
20. What is the bit length and expression form of a mac address?
48 bit expressed as a 6 byte hexadecimal
Lab 1.2: Defining OSI Layers and Devices
1. This device sends and receives information about the Network layer
router
2. This layer creates a virtual circuit before transmitting between two end stations
transport
3. This device uses hardware addresses to filter a network
switch or bridge
4. Ethernet is defined at these layers
data link and physical
5. This device supports flow control, sequencing, and acknowledgements
transport
6. This device can measure the distance to a remote network
router
7. Logical addressing is used at this layer
network
8. Hardware addresses are defined at this layer
9. This device creates one big collision domain and one big broadcast domain
hub
10. This device creates many smaller collision domains, but the network is still one large broadcast domain
switch or bridge
11. This device can never run full duplex
hub
12. This device breaks up collision domains and broadcast domains
Router
Chapter 1 Review Questions
1. A receiving host has failed to receive all of the segments that it should acknowledge. What can the host do to improve
the reliability of this communication session?
a receiving host can control a transmitter by using flow control. By decreasing the window size, the receving host
can slow down the transmitting host to avoid buffer overlflow.
TCP uses windowing by default
Transport Layer = flow control
2. When a station sends a transmission to the MAC address ff:ff:ff:ff:ff:ff what type of transmission is it?
broadcast
3. Which layer 1 devices can be used to enlarge the area covered by a single LAN segment?
Switch or Hub
but the hub will enlarge a single collision domain
4. Segmentation of a data stream happens at which layer of the OSI model?
transport layer
receives large data streams from the upper layer and breaks it into segments
5. Describe the main router functions
packet switching, packet filtering, internetwork communication, path selection
note that routers do create/terminate collision domains, but that isnt their main function
6. At which layers do these devices operate? Rouer, LAN Switch, Ethernet hub, Word Processor
Router- layer 3
Switch- layer 2
Hub- layer 1
Word Processor- NONE- word processors communicate TO the application layer interface, but dont operate within it
7. When data is encapsulated, which is the correct order?
data > segment > packet > frame > bit
8. Why does the data communication industry use the OSI model?
OSI divides the process into smaller/simpler components which aids in development, design, and troubleshooting.
It allows develoers to changes aspects of a program at just one layer
9. What are two purposes of segmentation with a bridge?
add more broadcast domains
allow more broadcasts
10. What are causes of LAN congestion?
too many hosts in a broadcast domain, broadcast storms, low bandwidth
11. If a switch has three computers connected to it, with no VLANs present, how many broadcast/collision domains
is the switch creating?
one broadcast, three collision
12. Acknowledgements, sequencing, and flow control are characteristics of which OSI layer?
TRANSPORT layer 4
"a reliable connection" is defined by a virtual circuit that employs:
acknowledgements
sequencing
flow control
13. List types of flow control
buffering
windowing
congestion avoidance
14. If a hub has three computers connected to it, how many broadcast domains and collision domains is the hub creating?
one broadcast, one collision
15. Which is the purpose of flow control?
allows the receiving device to control the transmitter to avoid buffer overflow on the receiving end
16. Which three statements are true about the operation of a full-duplex network?
there are no collisions in full-duplex mode.
a dedicated switch port is required for each node in full-duplex.
the host network card and switch port must be capable of operating in full-duplex mode
Full duplex means you using using both wire pairs simultaneously to send and receive data. You will not have collisions
since you have a dedicated switch port for each node.
Both the host network card and the switch port must be capable and set to work in full duplex
17. Whare are benefits of OSI Model?
divides the network communication proces into smaller/simpler components
allows multiple-vendor development through standardization
allows various types of software and hardware to commnunicate
18. Which devices operate at ALL levels of the OSI model?
NMS- network management station
web and application servers
network hosts
19. When an HTTP document must be retrieved from a location other than the local machine,
what layer of the OSI model must be accessed first?
application
20. Which layer of the OSI model offers three different modes of communication: simplex, half duplex, and full duplex?
session
END CHAPTER 1
3 hours 12-11-21
original hours 3
3.5 hours recap 12-11-22
review hours 3
Chapter 2 Ethernet and Data Encapusulation Exam Essentials
1. Describe the operation of Carrier Sense Multiple Access with Collision Detection (CSMA/CD)
Helps devices share the bandwidth evenly without having two devices transmit at the same time. It does not eliminate
collisions, but it greatly reduces them, which reduces retransmissions.
How does it work? Pg 43
When a host wants to transmit, it first checks for signal on the wire. It all clear, proceed with transmission.
The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If another
signal is detected, send out a jam singal that causes all noted on the segment to stop transmitting (busy signal).
The other notes respond to that jam signal by waiting awhile before transmitting again.
Backoff algorithms determine when the colliding statinos can retransmit.
If collisions keep occuring after 15 tries, the nodes will timeout.
When a collision occurs, the following happens:
- A jam signal informs all devices that a collision occured
- The collision invokes a random backoff algorithm
- Each device on the segment stops transmitting until their backoff timers expire
- All hosts have equal priority to transmit after the timers have expired
A network sustaining heavy collisions will experience DELAY, LOW THROUGHPUT, CONGESTION
2. Differentiate half-duplex and full-duplex communications and define the requirements to utilize each method.
Full-duplex uses two paris of wires and allows for sending and receiving at the same time. Uses different wires
to eliminate collisions. For this to work, devices at each end must be capable and configured for full-duplex.
Half-duplex uses one wire pair can not send/recieve simlutaneously and can suffer collisions.
3. Describe the sections of a MAC address and the information contained in each section.
The first 24 bits (3 bytes) are the OUI- Organizationally Unique Identifier-
identifies the manufacturer
the remaining 24 bits identify the NIC
4. Identify the binary and hexadecimal equivalent of a decimal number
Any number expressed in one format can be also be expressed in the others.
5. Identify the fields in the data link portion of an ethernet frame
The fields in the Data Link portion of a frame include:
- Preamble
- Start Frame Delimiter
- Destination MAC Address (DA)
- Source MAC Address (SA)
- Length or Type
- Data
- Frame Check Sequence (FCS)
Note on the preamble and SFD:
6. Identify the IEEE physical standards for ethernet cabling
These standards describe the capabilities and physical characteristics of cable types including
10Base2, 10Base5, and 10BaseT
these three are the ORIGINAL IEEE 802.3 Standards
10Base2- 10Mbps, Baseband, 200 meters
"baseband"
Max meters: 185
thin coax and BNC connectors
Known as "thinnet" and can support up to 30 workstations on a single segment
connects everything via BUS
10Base5- 10Mbps, Baseband, 500 meters
uses THICK coax aka "thicknet"
AUI connectors- Attachment Unit Interface
Repeaters allow up to 2500 meters
Supports up to 1024 users
connects everything via BUS
10BaseT- 10Mbps 100 meters
Cat 3 UTP, RJ45 connector
No BUS anymore, each device must connect to a hub or switch
one host per segment/wire
Here are the EXPANDED 802.3 standards
100Base-TX 100 meters
802.3u
FAST ETHERNET Cat 5, 5e, or 6 UTP two-pair wiring
100Base-FX
FIBER
---------------------------------------------------
1000Base-CX
1000Base-T 100 meters
802.3ab
Cat 5 four-pair wiring
16 Gbps
1000Base-SX
FIBER
1000Base-LX
FIBER
1000Base-ZX
FIBER
---------------------------------------------------
10GBase-T 10Gbps, 100m, UTP
802.3an
10GBase-LR Long Range
FIBER
10GBase-ER Extended Range
FIBER
10GBase-SW Short Wavelength
FIBER
10GBase-EW Extra Long Wavelength
FIBER
7. Differentiate type of ethernet cabling and identify their proper application
Three types:
- straight-through
to connect a PC or router to a switch
- crossover
to connect hub/switch to hub/switch, or pc to pc
- rolled
console at PC to router or switch/span>
8. Describe the data encapsulation process and the role it plays in packet creation.
Information is added to the frame from each OSI Layer.
AKA Packet Creation
Each layer communicates only with it's peer layer
9. Understand how to connecte a console cable from a PC to a router and start HyperTerminal
Rolled cable connects to COM port of the PC to the console port of the router.
Start HyperTerminal and set bits per second to 9600
set flow control to none
10. Identify the layers in the Cisco three-layer model and describe the ideal function of each.
Pg.65 Note these are LOGICAL distinctions. A device can traverse layers, or vice versa
CORE aka Backbone
Transporting large amounts of traffic
DISTRIBUTION aka Routing
Routing, filtering, WAN access
AKA the workgroup layer. The communication point between Access layer and Core layer
Routing
Implementing access lists, packet filtering, queuing, security policies, address translation, firewalls
Redistributing between routing protocols/ static routing
Routing between VLANs
Definining broadcast and multicast domains
ACCESS aka Switching
AKA the desktop layer
client connectivity into the distribution layer
The network resources most users need will be avaialble locally.
Creation of seperate collision domains (Segmentation)
Lab 2.1: Binary/Decimal/Hexadecimal Conversion
1. convert from decimal IP address to binary format
192.168.10.15
11000000 10101000 00001010 00001111
2. convert from decimal IP address to binary format
172.16.20.55
10101100.0001000.00010100.00110111
3. convert from decimal IP address to binary format
10.11.12.99
00001010.00001011.00001100.01100011
4. convert from binary to decimal IP address
11001100.00110011.10101010.01010101
204.51.170.85
5. convert from binary to decimal IP address
11000110.11010011.00111001.11010001
198.211.57.209
6. convert from binary to decimal IP address
10000100.11010010.10111000.10100110
132.210.184.166
7. convert from binary to hexadecimal
11011000.00011011.00111101.01110110
D8 1B 3D 76
8. convert from binary to hexadecimal
11001010.11110101.10000011.11101011
CA F5 83 EB
9. convert from binary to hexadecimal
10000100.11010010.01000011.10110011
84 D2 43 B3
Lab 2.2: CSMA/CD Operations
1. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) helps to minimize collisions on the network, thereby
increasing data transmission efficiency.
List the correct order of the CSMA/CD steps.
A jam signal informs all devices that a collision occured
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
Lab 2.3: Cabling
For each situation which cable to use: straight-through, crossover, or rolled
1. host to host
crossover
2. host to switch or hub
straight
3. router direct to host
crossover
4. switch to switch
crossover
5. router to switch or hub
straight
6. hub to hub
crossover
7. hub to switch
crossover
8. host to a router console serial communication (COM) port
rolled
The straight cable only works when connecting UNLIKE devices
Lab 2.4: Encapsulation
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each
packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are
used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
Chapter 2 Review Questions
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired,
all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore.
Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. This device sends and receives information about the Network layer
router
2. This layer creates a virtual circuit before transmitting between two end stations
transport
3. This device uses hardware addresses to filter a network
switch or bridge
4. Ethernet is defined at these layers
data link and physical
5. This device supports flow control, sequencing, and acknowledgements
transport
6. This device can measure the distance to a remote network
router
7. Logical addressing is used at this layer
network
8. Hardware addresses are defined at this layer
9. This device creates one big collision domain and one big broadcast domain
hub
10. This device creates many smaller collision domains, but the network is still one large broadcast domain
switch or bridge
11. This device can never run full duplex
hub
12. This device breaks up collision domains and broadcast domains
Router
1. A receiving host has failed to receive all of the segments that it should acknowledge. What can the host do to improve
the reliability of this communication session?
a receiving host can control a transmitter by using flow control. By decreasing the window size, the receving host
can slow down the transmitting host to avoid buffer overlflow.
TCP uses windowing by default
Transport Layer = flow control
TCP uses windowing by default
Transport Layer = flow control
2. When a station sends a transmission to the MAC address ff:ff:ff:ff:ff:ff what type of transmission is it?
broadcast
3. Which layer 1 devices can be used to enlarge the area covered by a single LAN segment?
Switch or Hub
but the hub will enlarge a single collision domain
but the hub will enlarge a single collision domain
4. Segmentation of a data stream happens at which layer of the OSI model?
transport layer
receives large data streams from the upper layer and breaks it into segments
receives large data streams from the upper layer and breaks it into segments
5. Describe the main router functions
packet switching, packet filtering, internetwork communication, path selection
note that routers do create/terminate collision domains, but that isnt their main function
note that routers do create/terminate collision domains, but that isnt their main function
6. At which layers do these devices operate? Rouer, LAN Switch, Ethernet hub, Word Processor
Router- layer 3
Switch- layer 2
Hub- layer 1
Word Processor- NONE- word processors communicate TO the application layer interface, but dont operate within it
Switch- layer 2
Hub- layer 1
Word Processor- NONE- word processors communicate TO the application layer interface, but dont operate within it
7. When data is encapsulated, which is the correct order?
data > segment > packet > frame > bit
8. Why does the data communication industry use the OSI model?
OSI divides the process into smaller/simpler components which aids in development, design, and troubleshooting.
It allows develoers to changes aspects of a program at just one layer
It allows develoers to changes aspects of a program at just one layer
9. What are two purposes of segmentation with a bridge?
add more broadcast domains
allow more broadcasts
allow more broadcasts
10. What are causes of LAN congestion?
too many hosts in a broadcast domain, broadcast storms, low bandwidth
11. If a switch has three computers connected to it, with no VLANs present, how many broadcast/collision domains
is the switch creating?
one broadcast, three collision
12. Acknowledgements, sequencing, and flow control are characteristics of which OSI layer?
TRANSPORT layer 4
"a reliable connection" is defined by a virtual circuit that employs:
acknowledgements
sequencing
flow control
"a reliable connection" is defined by a virtual circuit that employs:
acknowledgements
sequencing
flow control
13. List types of flow control
buffering
windowing
congestion avoidance
windowing
congestion avoidance
14. If a hub has three computers connected to it, how many broadcast domains and collision domains is the hub creating?
one broadcast, one collision
15. Which is the purpose of flow control?
allows the receiving device to control the transmitter to avoid buffer overflow on the receiving end
16. Which three statements are true about the operation of a full-duplex network?
there are no collisions in full-duplex mode.
a dedicated switch port is required for each node in full-duplex.
the host network card and switch port must be capable of operating in full-duplex mode
Full duplex means you using using both wire pairs simultaneously to send and receive data. You will not have collisions since you have a dedicated switch port for each node.
Both the host network card and the switch port must be capable and set to work in full duplex
a dedicated switch port is required for each node in full-duplex.
the host network card and switch port must be capable of operating in full-duplex mode
Full duplex means you using using both wire pairs simultaneously to send and receive data. You will not have collisions since you have a dedicated switch port for each node.
Both the host network card and the switch port must be capable and set to work in full duplex
17. Whare are benefits of OSI Model?
divides the network communication proces into smaller/simpler components
allows multiple-vendor development through standardization
allows various types of software and hardware to commnunicate
allows multiple-vendor development through standardization
allows various types of software and hardware to commnunicate
18. Which devices operate at ALL levels of the OSI model?
NMS- network management station
web and application servers
network hosts
web and application servers
network hosts
19. When an HTTP document must be retrieved from a location other than the local machine,
what layer of the OSI model must be accessed first?
application
20. Which layer of the OSI model offers three different modes of communication: simplex, half duplex, and full duplex?
session
END CHAPTER 1
3 hours 12-11-21
original hours 3
3.5 hours recap 12-11-22
review hours 3
Chapter 2 Ethernet and Data Encapusulation Exam Essentials
1. Describe the operation of Carrier Sense Multiple Access with Collision Detection (CSMA/CD)
Helps devices share the bandwidth evenly without having two devices transmit at the same time. It does not eliminate
collisions, but it greatly reduces them, which reduces retransmissions.
How does it work? Pg 43
When a host wants to transmit, it first checks for signal on the wire. It all clear, proceed with transmission.
The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If another
signal is detected, send out a jam singal that causes all noted on the segment to stop transmitting (busy signal).
The other notes respond to that jam signal by waiting awhile before transmitting again.
Backoff algorithms determine when the colliding statinos can retransmit.
If collisions keep occuring after 15 tries, the nodes will timeout.
When a collision occurs, the following happens:
- A jam signal informs all devices that a collision occured
- The collision invokes a random backoff algorithm
- Each device on the segment stops transmitting until their backoff timers expire
- All hosts have equal priority to transmit after the timers have expired
A network sustaining heavy collisions will experience DELAY, LOW THROUGHPUT, CONGESTION
2. Differentiate half-duplex and full-duplex communications and define the requirements to utilize each method.
Full-duplex uses two paris of wires and allows for sending and receiving at the same time. Uses different wires
to eliminate collisions. For this to work, devices at each end must be capable and configured for full-duplex.
Half-duplex uses one wire pair can not send/recieve simlutaneously and can suffer collisions.
3. Describe the sections of a MAC address and the information contained in each section.
The first 24 bits (3 bytes) are the OUI- Organizationally Unique Identifier-
identifies the manufacturer
the remaining 24 bits identify the NIC
4. Identify the binary and hexadecimal equivalent of a decimal number
Any number expressed in one format can be also be expressed in the others.
5. Identify the fields in the data link portion of an ethernet frame
The fields in the Data Link portion of a frame include:
- Preamble
- Start Frame Delimiter
- Destination MAC Address (DA)
- Source MAC Address (SA)
- Length or Type
- Data
- Frame Check Sequence (FCS)
Note on the preamble and SFD:
6. Identify the IEEE physical standards for ethernet cabling
These standards describe the capabilities and physical characteristics of cable types including
10Base2, 10Base5, and 10BaseT
these three are the ORIGINAL IEEE 802.3 Standards
10Base2- 10Mbps, Baseband, 200 meters
"baseband"
Max meters: 185
thin coax and BNC connectors
Known as "thinnet" and can support up to 30 workstations on a single segment
connects everything via BUS
10Base5- 10Mbps, Baseband, 500 meters
uses THICK coax aka "thicknet"
AUI connectors- Attachment Unit Interface
Repeaters allow up to 2500 meters
Supports up to 1024 users
connects everything via BUS
10BaseT- 10Mbps 100 meters
Cat 3 UTP, RJ45 connector
No BUS anymore, each device must connect to a hub or switch
one host per segment/wire
Here are the EXPANDED 802.3 standards
100Base-TX 100 meters
802.3u
FAST ETHERNET Cat 5, 5e, or 6 UTP two-pair wiring
100Base-FX
FIBER
---------------------------------------------------
1000Base-CX
1000Base-T 100 meters
802.3ab
Cat 5 four-pair wiring
16 Gbps
1000Base-SX
FIBER
1000Base-LX
FIBER
1000Base-ZX
FIBER
---------------------------------------------------
10GBase-T 10Gbps, 100m, UTP
802.3an
10GBase-LR Long Range
FIBER
10GBase-ER Extended Range
FIBER
10GBase-SW Short Wavelength
FIBER
10GBase-EW Extra Long Wavelength
FIBER
7. Differentiate type of ethernet cabling and identify their proper application
Three types:
- straight-through
to connect a PC or router to a switch
- crossover
to connect hub/switch to hub/switch, or pc to pc
- rolled
console at PC to router or switch/span>
8. Describe the data encapsulation process and the role it plays in packet creation.
Information is added to the frame from each OSI Layer.
AKA Packet Creation
Each layer communicates only with it's peer layer
9. Understand how to connecte a console cable from a PC to a router and start HyperTerminal
Rolled cable connects to COM port of the PC to the console port of the router.
Start HyperTerminal and set bits per second to 9600
set flow control to none
10. Identify the layers in the Cisco three-layer model and describe the ideal function of each.
Pg.65 Note these are LOGICAL distinctions. A device can traverse layers, or vice versa
CORE aka Backbone
Transporting large amounts of traffic
DISTRIBUTION aka Routing
Routing, filtering, WAN access
AKA the workgroup layer. The communication point between Access layer and Core layer
Routing
Implementing access lists, packet filtering, queuing, security policies, address translation, firewalls
Redistributing between routing protocols/ static routing
Routing between VLANs
Definining broadcast and multicast domains
ACCESS aka Switching
AKA the desktop layer
client connectivity into the distribution layer
The network resources most users need will be avaialble locally.
Creation of seperate collision domains (Segmentation)
Lab 2.1: Binary/Decimal/Hexadecimal Conversion
1. convert from decimal IP address to binary format
192.168.10.15
11000000 10101000 00001010 00001111
2. convert from decimal IP address to binary format
172.16.20.55
10101100.0001000.00010100.00110111
3. convert from decimal IP address to binary format
10.11.12.99
00001010.00001011.00001100.01100011
4. convert from binary to decimal IP address
11001100.00110011.10101010.01010101
204.51.170.85
5. convert from binary to decimal IP address
11000110.11010011.00111001.11010001
198.211.57.209
6. convert from binary to decimal IP address
10000100.11010010.10111000.10100110
132.210.184.166
7. convert from binary to hexadecimal
11011000.00011011.00111101.01110110
D8 1B 3D 76
8. convert from binary to hexadecimal
11001010.11110101.10000011.11101011
CA F5 83 EB
9. convert from binary to hexadecimal
10000100.11010010.01000011.10110011
84 D2 43 B3
Lab 2.2: CSMA/CD Operations
1. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) helps to minimize collisions on the network, thereby
increasing data transmission efficiency.
List the correct order of the CSMA/CD steps.
A jam signal informs all devices that a collision occured
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
Lab 2.3: Cabling
For each situation which cable to use: straight-through, crossover, or rolled
1. host to host
crossover
2. host to switch or hub
straight
3. router direct to host
crossover
4. switch to switch
crossover
5. router to switch or hub
straight
6. hub to hub
crossover
7. hub to switch
crossover
8. host to a router console serial communication (COM) port
rolled
The straight cable only works when connecting UNLIKE devices
Lab 2.4: Encapsulation
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each
packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are
used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
Chapter 2 Review Questions
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired,
all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore.
Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. Describe the operation of Carrier Sense Multiple Access with Collision Detection (CSMA/CD)
Helps devices share the bandwidth evenly without having two devices transmit at the same time. It does not eliminate
collisions, but it greatly reduces them, which reduces retransmissions.
How does it work? Pg 43
When a host wants to transmit, it first checks for signal on the wire. It all clear, proceed with transmission. The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If another signal is detected, send out a jam singal that causes all noted on the segment to stop transmitting (busy signal). The other notes respond to that jam signal by waiting awhile before transmitting again. Backoff algorithms determine when the colliding statinos can retransmit. If collisions keep occuring after 15 tries, the nodes will timeout.
When a collision occurs, the following happens:
A network sustaining heavy collisions will experience DELAY, LOW THROUGHPUT, CONGESTION
How does it work? Pg 43
When a host wants to transmit, it first checks for signal on the wire. It all clear, proceed with transmission. The transmitting host constantly monitors the wire to make sure no other hosts begin transmitting. If another signal is detected, send out a jam singal that causes all noted on the segment to stop transmitting (busy signal). The other notes respond to that jam signal by waiting awhile before transmitting again. Backoff algorithms determine when the colliding statinos can retransmit. If collisions keep occuring after 15 tries, the nodes will timeout.
When a collision occurs, the following happens:
- A jam signal informs all devices that a collision occured
- The collision invokes a random backoff algorithm
- Each device on the segment stops transmitting until their backoff timers expire
- All hosts have equal priority to transmit after the timers have expired
A network sustaining heavy collisions will experience DELAY, LOW THROUGHPUT, CONGESTION
2. Differentiate half-duplex and full-duplex communications and define the requirements to utilize each method.
Full-duplex uses two paris of wires and allows for sending and receiving at the same time. Uses different wires
to eliminate collisions. For this to work, devices at each end must be capable and configured for full-duplex.
Half-duplex uses one wire pair can not send/recieve simlutaneously and can suffer collisions.
Half-duplex uses one wire pair can not send/recieve simlutaneously and can suffer collisions.
3. Describe the sections of a MAC address and the information contained in each section.
The first 24 bits (3 bytes) are the OUI- Organizationally Unique Identifier-
identifies the manufacturer
the remaining 24 bits identify the NIC
the remaining 24 bits identify the NIC
4. Identify the binary and hexadecimal equivalent of a decimal number
Any number expressed in one format can be also be expressed in the others.
5. Identify the fields in the data link portion of an ethernet frame
The fields in the Data Link portion of a frame include:
Note on the preamble and SFD:
- Preamble
- Start Frame Delimiter
- Destination MAC Address (DA)
- Source MAC Address (SA)
- Length or Type
- Data
- Frame Check Sequence (FCS)
Note on the preamble and SFD:
6. Identify the IEEE physical standards for ethernet cabling
These standards describe the capabilities and physical characteristics of cable types including
10Base2, 10Base5, and 10BaseT
these three are the ORIGINAL IEEE 802.3 Standards
10Base2- 10Mbps, Baseband, 200 meters
"baseband"
Max meters: 185
thin coax and BNC connectors
Known as "thinnet" and can support up to 30 workstations on a single segment
connects everything via BUS
10Base5- 10Mbps, Baseband, 500 meters
uses THICK coax aka "thicknet"
AUI connectors- Attachment Unit Interface
Repeaters allow up to 2500 meters
Supports up to 1024 users
connects everything via BUS
10BaseT- 10Mbps 100 meters
Cat 3 UTP, RJ45 connector
No BUS anymore, each device must connect to a hub or switch
one host per segment/wire
Here are the EXPANDED 802.3 standards
100Base-TX 100 meters
802.3u
FAST ETHERNET Cat 5, 5e, or 6 UTP two-pair wiring
100Base-FX
FIBER
---------------------------------------------------
1000Base-CX
1000Base-T 100 meters
802.3ab
Cat 5 four-pair wiring
16 Gbps
1000Base-SX
FIBER
1000Base-LX
FIBER
1000Base-ZX
FIBER
---------------------------------------------------
10GBase-T 10Gbps, 100m, UTP
802.3an
10GBase-LR Long Range
FIBER
10GBase-ER Extended Range
FIBER
10GBase-SW Short Wavelength
FIBER
10GBase-EW Extra Long Wavelength
FIBER
these three are the ORIGINAL IEEE 802.3 Standards
10Base2- 10Mbps, Baseband, 200 meters
"baseband"
Max meters: 185
thin coax and BNC connectors
Known as "thinnet" and can support up to 30 workstations on a single segment
connects everything via BUS
10Base5- 10Mbps, Baseband, 500 meters
uses THICK coax aka "thicknet"
AUI connectors- Attachment Unit Interface
Repeaters allow up to 2500 meters
Supports up to 1024 users
connects everything via BUS
10BaseT- 10Mbps 100 meters
Cat 3 UTP, RJ45 connector
No BUS anymore, each device must connect to a hub or switch
one host per segment/wire
Here are the EXPANDED 802.3 standards
100Base-TX 100 meters
802.3u
FAST ETHERNET Cat 5, 5e, or 6 UTP two-pair wiring
100Base-FX
FIBER
---------------------------------------------------
1000Base-CX
1000Base-T 100 meters
802.3ab
Cat 5 four-pair wiring
16 Gbps
1000Base-SX
FIBER
1000Base-LX
FIBER
1000Base-ZX
FIBER
---------------------------------------------------
10GBase-T 10Gbps, 100m, UTP
802.3an
10GBase-LR Long Range
FIBER
10GBase-ER Extended Range
FIBER
10GBase-SW Short Wavelength
FIBER
10GBase-EW Extra Long Wavelength
FIBER
7. Differentiate type of ethernet cabling and identify their proper application
Three types:
- straight-through to connect a PC or router to a switch
- crossover to connect hub/switch to hub/switch, or pc to pc
- rolled console at PC to router or switch/span>
8. Describe the data encapsulation process and the role it plays in packet creation.
Information is added to the frame from each OSI Layer.
AKA Packet Creation
Each layer communicates only with it's peer layer
AKA Packet Creation
Each layer communicates only with it's peer layer
9. Understand how to connecte a console cable from a PC to a router and start HyperTerminal
Rolled cable connects to COM port of the PC to the console port of the router.
Start HyperTerminal and set bits per second to 9600
set flow control to none
Start HyperTerminal and set bits per second to 9600
set flow control to none
10. Identify the layers in the Cisco three-layer model and describe the ideal function of each.
Pg.65 Note these are LOGICAL distinctions. A device can traverse layers, or vice versa
CORE aka Backbone
Transporting large amounts of traffic
DISTRIBUTION aka Routing Routing, filtering, WAN access
AKA the workgroup layer. The communication point between Access layer and Core layer
Routing
Implementing access lists, packet filtering, queuing, security policies, address translation, firewalls
Redistributing between routing protocols/ static routing
Routing between VLANs
Definining broadcast and multicast domains
ACCESS aka Switching
AKA the desktop layer
client connectivity into the distribution layer
The network resources most users need will be avaialble locally.
Creation of seperate collision domains (Segmentation)
CORE aka Backbone
Transporting large amounts of traffic
DISTRIBUTION aka Routing Routing, filtering, WAN access
AKA the workgroup layer. The communication point between Access layer and Core layer
Routing
Implementing access lists, packet filtering, queuing, security policies, address translation, firewalls
Redistributing between routing protocols/ static routing
Routing between VLANs
Definining broadcast and multicast domains
ACCESS aka Switching
AKA the desktop layer
client connectivity into the distribution layer
The network resources most users need will be avaialble locally.
Creation of seperate collision domains (Segmentation)
1. convert from decimal IP address to binary format
192.168.10.15
192.168.10.15
11000000 10101000 00001010 00001111
2. convert from decimal IP address to binary format
172.16.20.55
172.16.20.55
10101100.0001000.00010100.00110111
3. convert from decimal IP address to binary format
10.11.12.99
10.11.12.99
00001010.00001011.00001100.01100011
4. convert from binary to decimal IP address
11001100.00110011.10101010.01010101
11001100.00110011.10101010.01010101
204.51.170.85
5. convert from binary to decimal IP address
11000110.11010011.00111001.11010001
11000110.11010011.00111001.11010001
198.211.57.209
6. convert from binary to decimal IP address
10000100.11010010.10111000.10100110
10000100.11010010.10111000.10100110
132.210.184.166
7. convert from binary to hexadecimal
11011000.00011011.00111101.01110110
11011000.00011011.00111101.01110110
D8 1B 3D 76
8. convert from binary to hexadecimal
11001010.11110101.10000011.11101011
11001010.11110101.10000011.11101011
CA F5 83 EB
9. convert from binary to hexadecimal
10000100.11010010.01000011.10110011
10000100.11010010.01000011.10110011
84 D2 43 B3
Lab 2.2: CSMA/CD Operations
1. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) helps to minimize collisions on the network, thereby
increasing data transmission efficiency.
List the correct order of the CSMA/CD steps.
A jam signal informs all devices that a collision occured
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
Lab 2.3: Cabling
For each situation which cable to use: straight-through, crossover, or rolled
1. host to host
crossover
2. host to switch or hub
straight
3. router direct to host
crossover
4. switch to switch
crossover
5. router to switch or hub
straight
6. hub to hub
crossover
7. hub to switch
crossover
8. host to a router console serial communication (COM) port
rolled
The straight cable only works when connecting UNLIKE devices
Lab 2.4: Encapsulation
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each
packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are
used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
Chapter 2 Review Questions
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired,
all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore.
Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. Carrier Sense Multiple Access with Collision Detection (CSMA/CD) helps to minimize collisions on the network, thereby
increasing data transmission efficiency.
List the correct order of the CSMA/CD steps.
List the correct order of the CSMA/CD steps.
A jam signal informs all devices that a collision occured
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
The collision invokes a random backoff aglorithm
Each device on the segment stops transmitting until the timers expire
All hosts have equal priority to transmit after the timers have expired
1. host to host
crossover
2. host to switch or hub
straight
3. router direct to host
crossover
4. switch to switch
crossover
5. router to switch or hub
straight
6. hub to hub
crossover
7. hub to switch
crossover
8. host to a router console serial communication (COM) port
rolled
The straight cable only works when connecting UNLIKE devices
Lab 2.4: Encapsulation
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each
packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are
used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
Chapter 2 Review Questions
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired,
all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore.
Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. List the steps of the encapsulation process
encapsulation i.e. SENDING the data
encapsulation i.e. SENDING the data
User information is converted to data for transmission on the network.
Data is converted into segments, and a reliable connection is set up between the transmitting and receiving hosts.
Segments are converted to packets or datagrams, and a logical address is placed in the header so each packet can be routed.
Packets or datagrams are converted to frames for transmission on the local network. Hardware(ethernet) addresses are used to uniquely identify hosts on a local network segment.
Frames are convered to bits, and a digital encoding and clocking scheme is used.
1. Which fields are contained within an IEEE Ethernet frame?
source and destination MAC
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
FCS
pg.51- preamble(including SFD) + DA + SA + type + data + FCS
2. Which are unique characteristics of half-duplex Eternet when compared to full-duplex?
half-duplex operates in a shared collision domain and has lower effective throughput
3. You want to implement a network medium that is not susceptible to EFI. Which type of cabling should be used?
fiber
4. Which connections can use full duplex?
switch to switch
host to host
switch to host
any connections involving a hub are going to be half-duplex
host to host
switch to host
any connections involving a hub are going to be half-duplex
5. What type of RJ45 UTP cable is used between switches?
crossover
6. How does a host on an Ethernet LAN know when to transmit after a collision has occured?
in CSMA/CD, stations must wait until the media is not in use before transmitting
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired, all stations have equal priority to transmit.
after a collsion, all stations run a random backoff algorithm. When the backoff delay period has expired, all stations have equal priority to transmit.
7. What type of RJ45 UTP cable do you use to connect a PC's COM port to a router or switch console port?
rolled
8. You have the following binary number: 10110111. What are the decimal and hexadecimal equivalents?
dec: 183
hex: B7
hex: B7
9. Which of the following contention mechanisms is used by Ethernet?
CSMA/CD
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore. Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
Since switches have replaced hubs in most of today’s LANs, CSMA/CD is not often used anymore. Switches work in full-duplex mode and each port on a switch is in a seperate collision domain, so no collisions can occur.
geek-university.com/ccna/csma-cd-explained/
10. In the operation of CSMA/CD, which host(s) have priority after the expiration of the backoff algorithm?
all hosts have equal priority
11. Which is correct about full-duplex
full-duplex uses two wire pairs
12. Which are true about full-duplex?
there are no collisions in full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
a dedicated switch port is required for each node
the host NIC and switch port must be capable and configured for full-duplex
13. Which statement is correct about MAC addresses?
a MAC is a hardware harddress written as 48 byte hex
14. Which part of a MAC address is called the organizationally unique identifier (OUI)?
the first half, or 24 bits/3 bytes
15. Which layer of the OSI model is responsible for combining bits into bytes and bytes into frames?
Link
16. What is the specific term for the unwanted signal interference from adjacent pairs in the cable?
Crosstalk
17. Which of the following is part of the IEEE 802.3u standard?
100Base-TX
802.3u is "fast ethernet"
802.3u includes:
802.3u is "fast ethernet"
802.3u includes:
- 100 BaseTX
- 100 BaseFX (fiber)
- 100 BaseT4 (early implementation)
18. 10G-Base-Long Wavelength is known as which IEEE standard?
802.3ae
802.3ae is fiber and includes:
802.3ae is fiber and includes:
- 10Gbase-SR
- 10Gbase-LR
- 10Gbase-ER
- 10Gbase-SW
- 10Gbase-LW
- 10Gbase-E
19. 1000Base-T is which IEEE standard?
802.3ab
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
1000Base-T, "gigabit ethernet", 1000Mbps or 1Gbps
20. When making a HyperTerminal connection, what must the bit rate be set to?
9600bps
QUICK SUMMARY OF THE BASIC STANDARDS
802.3
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
"ethernet"
10 Base T
ten base t
802.3 U
"fast ethernet"
100 Base TX
one hundred base tx
802.3 AB
"gigabit ethernet"
1000 Base T
one thousand base t
802.3 Z
"gigabit fiber"
1000 Base LX
one thousand base LX
802.3 AN
"ten gigabit ethernet"
10G Base T
ten g base t
END CHAPTER 2
8 hours 1-2-22
original total hours 11
1 hour recap 12-18-22
review total hours 4
Chapter 3 Intro to TCP/IP Exam Essentials
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Lab 3.1: TCP/IP
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. Differentiate the DoD and OSI network models
The DoD models is a condensed version of OSI, composed of 4 layers instead of 7.
2. Identify Process/Application layer protocols
Telnet- terminal emulation program that allows you to login to a remote host
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
FTP- connection-oriented service for file transfer
TFTP- connectionless file transfer
SMTP- send mail
3. Identify Host-to-Host layer protocols
TCP- provides reliable network service by using acknowledgements and flow control
UDP- unreliable but low overhead
UDP- unreliable but low overhead
4. Identify Internet layer protocols
IP- connectionless protocol that network addressing and routing.
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
ARP- finds a hardware address from a known IP address
RARP- finds an IP from a known hardware address
ICMP- provides diagnostics and destination unreachable messages
5. Describe the functions of DNS and DHCP in the network
DHCP provides network config info
DNS resolves hostnames to IP addresses
DNS resolves hostnames to IP addresses
6. Identify what is contained in the TCP header of a connection-oriented transmission
source port, dest, port, sequence number, ack number, header length,
code bits, window size, checksum, urgent pointer, options, DATA
7. Identify what is contained in the UDP header of a connectionless transmission
source port, dest port, length, checksum, DATA
8. Identify what is contained in the IP header
version, header length, priority/type of service, total length, id, flags, frag offset, ttl, protocol, header checksum,
source IP, dest IP, options, DATA
9. Compare and contrast UDP and TCP characteristics and features
TCP- connection-oriented, acknowledged, sequenced, flow/error control
UDP- none of the above
UDP- none of the above
10. Explain the role of port numbers
identify the protocol or service that is being used
11. Identify the role of ICMP
Network layer, used by IP for manaagement and messaging
12. Define the Class A IP range
1-126
13. Define the Class B IP range
128-191
14. Define the Class C IP range
192-223
15. Identify the private IP ranges
A- 10.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
B- 172.16.x.x - 172.31.x.x
C- 192.x.x.x
16. Explain the difference between a broadcast, unicast, and multicast address
Broadcast- all devices in a subnet
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
Unicast- one device
mutlicast- to some but not all (subscribed to multicast group)
1. What is the Class C address range in decimal and in binary?
class c range
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
192-223
first 3 bits cant change
this first octet represents the range of 192.x.x.x - 223.x.x.x
110xxxxx
2. What layer of the DoD model is equivalent to the OSI Transport layer?
host to host
3. What is the valid range of a Class A network address?
1-126
4. What is the 127.0.0.1 used for?
loopback- test a NIC without generating traffic on the network
5. How do you find the network address from a listed IP address?
set all host bits to zero (you would need to know the subnet mask)
6. How do you find the broadcast address from a listed IP address?
set all host bits to one (you would need to know the subnet mask)
7. What is the Class A private address space?
10.x.x.x
8. What is the Class B private address space?
172.16.x.x - 172.31.x.x
9. What is the Class C private address space?
192.x.x.x
10. What are all the available characters that you can use in hexadecimal addressing?
1-9, A-F
largest value of 15, which is the max available from a 4 bit nibble
largest value of 15, which is the max available from a 4 bit nibble
Lab 3.2: Mapping Applications to the DoD Model
The four layers of the DoD model are:
- Process/Application
- Host to Host
- Internet
- Network Access
Identify the layer at which each of these protocols operates:
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
Chapter 3 Review Questions
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that
all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. Internet Protocol (IP)
internet
2. Telnet
application
3. FTP
application
4. SNMP
application
5. DNS
application
6. Address Resolution Protocol (ARP)
internet
7. DHCP/BootP
application
8. Transmission Control Protocol (TCP)
host
9. X Window
application
10. User Datagram Protocol (UDP)
host
11. NFS
application
12. Internet Control Message Protocol (ICMP)
internet
13. Reverse Address Resolution Protocol (RARP)
internet
14. Proxy ARP
internet
15. TFTP
application
16. SMTP
application
17. LPD
application
1. What must happen if a DHCP conflict occurs?
Admin must fix the conflict by hand at the DHCP server
2. Which allows a router to respond to an ARP request that is intended for a remote host?
Proxy ARP can help machines on a subnet reach remote subnets without configuring routing or a default gateway.
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
Proxy ARP allows the router to provide its own MAC in response to ARP queries attempting to resolve a distant IP to a MAC
the disadvantage is that hosts on the network think that all the other machines are reachable by an ARP request, then increase the amount of infomation in their ARP tables.
https://scientificsentence.net/Networking/ARP.html
3. What protocol can be used to implement a mechanism that automates the IP configuration, including IP address,
subnet mask, default gateways, and DNS information
DHCP
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
Client: Broadcast DHCP discover port 67
Server: Unicast DHCP offer
Client: Broadcast DHCP request
Server: Unicast DHCP ACK
4. What protocol is used to find the hardware address of a local device?
ARP
5. Which three layers of the TCP/IP aka DoD model also exist in the OSI model?
application
transport
internet
transport
internet
6. Which class of IP addresses provides a maximum of only 254 hosts per network?
C
8 host bits -- 2^8 = 256 -2 = 254
8 host bits -- 2^8 = 256 -2 = 254
7. Which describes the DHCP discover message?
Sends out a UDP broadcast at
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
layer 2- FF:FF:FF:FF:FF:FF
and
layer 3- 255.255.255.255 (any network and all hosts)
8. Which layer-4 protocol is used for Telnet?
TCP
9. How does a DHCP client ensure that no other computer has its assigned IP address?
Gratuitous ARP- broadcast an ARP request for it's own IP to see if another host responds
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
A Gratuitous ARP is an ARP Response that was not prompted by an ARP Request.
https://www.practicalnetworking.net/series/arp/gratuitous-arp/
10. Which services use TCP?
SMTP
FTP
HTTP
FTP
HTTP
11. Which services use UDP?
DHCP
SNMP
TFTP
SNMP
TFTP
12. Which TCP/IP protocols are used at the OSI Application layer?
Telnet
FTP
TFTP
FTP
TFTP
13. See illustration- protocol is this header from?
TCP
14. If you use either Telnet or FTP, what layer are you using to generate data?
Application
15. The DoD model (AKA TCP/IP Stack) has four layers. Which layer is equivalent to the Network layer of the OSI model?
Internet
16. Which are private IP addresses?
172.20.14.36
172.33.194.30
192.168.24.43
172.33.194.30
192.168.24.43
17. What layer in the TCP/IP stack is equivalent to OSI Transport layer?
host to host
18. Which are true statements regarding ICMP packets?
ICMP can provide hosts iwth info about network problems
ICMP is encapsulated within IP datagrams
ICMP is encapsulated within IP datagrams
19. What is the address range of a Class B network address in binary?
10xxxxxx
20. Which protocol uses both TCP and UDP?
DNS
TCP for zone exchanges between servers
UDP when client is resolving
TCP for zone exchanges between servers
UDP when client is resolving
END CHAPTER 3
7 hours 1-2-22
original total hours 18
0.5 hour recap 12-18-22
review total hours 4.5
Chapter 4 Notes - Subnetting
Concepts:
The "interesting octet" is the first octet of the Subnet Mask whose value is not 255.
Class A
8 net bits taken by default
22 potential subnet bits
mask cidr
255.0.0.0
/8
255.128.0.0
/9
255.192.0.0
/10
255.224.0.0
/11
255.240.0.0
/12
255.248.0.0
/13
255.252.0.0
/14
255.254.0.0
/15
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.248
/30
Class B
16 net bits taken by default
14 potential subnet bits
mask cidr
255.255.0.0
/16
255.255.128.0
/17
255.255.192.0
/18
255.255.224.0
/19
255.255.240.0
/20
255.255.248.0
/21
255.255.252.0
/22
255.255.254.0
/23
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
Class C
24 net bits taken by default
6 potential subnet bits
mask cidr
255.255.255.0
/24
255.255.255.128
/25
255.255.255.192
/26
255.255.255.224
/27
255.255.255.240
/28
255.255.255.248
/29
255.255.255.252
/30
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Chapter 4 Exercises - Class A Subnetting
Class A minimum netmask is 255.0.0.0 or /8
Ex 1A
255
.255
.0.0
(/16)
8 classful bits
+ 8 subnet bits
+ 16 host bits
Ex 2A
255
.255
.240
.0
(/20)
8 classful bits
+ 12 subnet bits
+ 12 host bits
Ex 3A
255
.255.240
.0
(/26)
8 classful bits
+ 18 subnet bits
+ 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
Class A
8 net bits taken by default 22 potential subnet bits| mask | cidr |
|---|---|
| 255.0.0.0 | /8 |
| 255.128.0.0 | /9 |
| 255.192.0.0 | /10 |
| 255.224.0.0 | /11 |
| 255.240.0.0 | /12 |
| 255.248.0.0 | /13 |
| 255.252.0.0 | /14 |
| 255.254.0.0 | /15 |
| 255.255.0.0 | /16 |
| 255.255.128.0 | /17 |
| 255.255.192.0 | /18 |
| 255.255.224.0 | /19 |
| 255.255.240.0 | /20 |
| 255.255.248.0 | /21 |
| 255.255.252.0 | /22 |
| 255.255.254.0 | /23 |
| 255.255.255.0 | /24 |
| 255.255.255.128 | /25 |
| 255.255.255.192 | /26 |
| 255.255.255.224 | /27 |
| 255.255.255.240 | /28 |
| 255.255.255.248 | /29 |
| 255.255.255.248 | /30 |
Class B
16 net bits taken by default 14 potential subnet bits| mask | cidr |
|---|---|
| 255.255.0.0 | /16 |
| 255.255.128.0 | /17 |
| 255.255.192.0 | /18 |
| 255.255.224.0 | /19 |
| 255.255.240.0 | /20 |
| 255.255.248.0 | /21 |
| 255.255.252.0 | /22 |
| 255.255.254.0 | /23 |
| 255.255.255.0 | /24 |
| 255.255.255.128 | /25 |
| 255.255.255.192 | /26 |
| 255.255.255.224 | /27 |
| 255.255.255.240 | /28 |
| 255.255.255.248 | /29 |
| 255.255.255.252 | /30 |
Class C
24 net bits taken by default 6 potential subnet bits| mask | cidr |
|---|---|
| 255.255.255.0 | /24 |
| 255.255.255.128 | /25 |
| 255.255.255.192 | /26 |
| 255.255.255.224 | /27 |
| 255.255.255.240 | /28 |
| 255.255.255.248 | /29 |
| 255.255.255.252 | /30 |
The CIDR SLASH notation is relative to the CLASSFUL STARTING POINT
For example-
Class A /30 indicates 8 classful bits + 22 subnet bits
Class B /30 indicates 16 classful bits + 14 subnet bits
Class C /30 indicates 24 classful bits + 6 subnet bits
Ex 1A
255 .255 .0.0 (/16)
8 classful bits + 8 subnet bits + 16 host bits
255 .255 .0.0 (/16)
8 classful bits + 8 subnet bits + 16 host bits
Ex 2A
255 .255 .240 .0 (/20)
8 classful bits + 12 subnet bits + 12 host bits
255 .255 .240 .0 (/20)
8 classful bits + 12 subnet bits + 12 host bits
Ex 3A
255 .255.240 .0 (/26)
8 classful bits + 18 subnet bits + 6 host bits
255 .255.240 .0 (/26)
8 classful bits + 18 subnet bits + 6 host bits
Concept:
When subnet bits expand into the neighboring octet, the subnets will iterate within one octet, then the other
Class A Summary
Chapter 4 Exercises - Class B Subnetting
Class B minimum netmask is 255.255.0.0 or /16
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255
.128
.0
(/17)
16 classful bits
+ 1 subnet bit
+ 15 host bits
Ex 2B
255.255
.192
.0
(/18)
16 classful bits
+ 2 subnet bits
+ 14 host bits
Ex 3B
255.255
.240
.0
(/20)
16 classful bits
+ 4 subnet bits
+ 12 host bits
Ex 4B
255.255
.254
.0
(/23)
16 classful bits
+ 7 subnet bits
+ 9 host bits
Ex 5B
255.255
.255
.0
(/24)
16 classful bits
+ 8 subnet bits
+ 8 host bits
Ex 6B
255.255
.255
.128
(/25)
16 classful bits
+ 9 subnet bits
+ 7 host bits
Ex 7B
255.255
.255
.192
(/26)
16 classful bits
+ 10 subnet bits
+ 6 host bits
Ex 8B
255.255
.255
.224
(/27)
16 classful bits
+ 11 subnet bits
+ 5 host bits
Class B Summary
Chapter 4 Exercises - Class C Subnetting
Class C minimum netmask is 255.255.255.0 or /24
Ex 1C
255.255.255
.128
(/25)
24 classful bits
+ 1 subnet bit
+ 7 host bits
Ex 2C
255.255.255
.192
(/26)
24 classful bits
+ 2 subnet bits
+ 6 host bits
Ex 3C
255.255.255
.224
(/27)
24 classful bits
+ 3 subnet bits
+ 5 host bits
Ex 4C
255.255.255
.240
(/28)
24 classful bits
+ 4 subnet bits
+ 4 host bits
Ex 5C
255.255.255
.248
(/29)
24 classful bits
+ 5 subnet bits
+ 3 host bits
ans
Ex 6C
255.255.255
.252
(/30)
24 classful bits
+ 6 subnet bits
+ 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
Ex 1B -- 12-18-22 Needs correction, broadcasts are wrong
255.255 .128 .0 (/17)
16 classful bits + 1 subnet bit + 15 host bits
255.255 .128 .0 (/17)
16 classful bits + 1 subnet bit + 15 host bits
Ex 2B
255.255 .192 .0 (/18)
16 classful bits + 2 subnet bits + 14 host bits
255.255 .192 .0 (/18)
16 classful bits + 2 subnet bits + 14 host bits
Ex 3B
255.255 .240 .0 (/20)
16 classful bits + 4 subnet bits + 12 host bits
255.255 .240 .0 (/20)
16 classful bits + 4 subnet bits + 12 host bits
Ex 4B
255.255 .254 .0 (/23)
16 classful bits + 7 subnet bits + 9 host bits
255.255 .254 .0 (/23)
16 classful bits + 7 subnet bits + 9 host bits
Ex 5B
255.255 .255 .0 (/24)
16 classful bits + 8 subnet bits + 8 host bits
255.255 .255 .0 (/24)
16 classful bits + 8 subnet bits + 8 host bits
Ex 6B
255.255 .255 .128 (/25)
16 classful bits + 9 subnet bits + 7 host bits
255.255 .255 .128 (/25)
16 classful bits + 9 subnet bits + 7 host bits
Ex 7B
255.255 .255 .192 (/26)
16 classful bits + 10 subnet bits + 6 host bits
255.255 .255 .192 (/26)
16 classful bits + 10 subnet bits + 6 host bits
Ex 8B
255.255 .255 .224 (/27)
16 classful bits + 11 subnet bits + 5 host bits
255.255 .255 .224 (/27)
16 classful bits + 11 subnet bits + 5 host bits
Class B Summary
Ex 1C
255.255.255 .128 (/25)
24 classful bits + 1 subnet bit + 7 host bits
255.255.255 .128 (/25)
24 classful bits + 1 subnet bit + 7 host bits
Ex 2C
255.255.255 .192 (/26)
24 classful bits + 2 subnet bits + 6 host bits
255.255.255 .192 (/26)
24 classful bits + 2 subnet bits + 6 host bits
Ex 3C
255.255.255 .224 (/27)
24 classful bits + 3 subnet bits + 5 host bits
255.255.255 .224 (/27)
24 classful bits + 3 subnet bits + 5 host bits
Ex 4C
255.255.255 .240 (/28)
24 classful bits + 4 subnet bits + 4 host bits
255.255.255 .240 (/28)
24 classful bits + 4 subnet bits + 4 host bits
Ex 5C
255.255.255 .248 (/29)
24 classful bits + 5 subnet bits + 3 host bits
255.255.255 .248 (/29)
24 classful bits + 5 subnet bits + 3 host bits
ans
Ex 6C
255.255.255 .252 (/30)
24 classful bits + 6 subnet bits + 2 host bits
255.255.255 .252 (/30)
24 classful bits + 6 subnet bits + 2 host bits
Chapter 4 Subnetting Exam Essentials
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Lab 4.1 Written Subnet Practice
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. Identify the advantages of subnetting
Reduced traffic on each segment, optimized performance
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
Simplified management- easier to isolate problems
Connecting smaller, multiple networks is more efficeint.
A single large network that spans long distances can be problematic; WAN links are slower and more expensive
2. Describe the effect of the ip subnet-zero command
allows you to use the first and last subnet
3. Identify the steps to subnet a classful network
determine block size by subtracing the interesting octet from 256
4. Determine possible block sizes
the valid blocks are always 2, 4, 8, 16, 32, 64, 128, etc
OR could be blocks of 1
OR could be blocks of 1
5. Describe the role of a subnet mask in IP addressing
32 bit value that allows a node to distinguish between network portion and host portion of an IP address
6. Understand and apply the (2^n) - 2 formula
N = HOST BITS
2^N = available hosts
subtract 2 to account for network id and broadcast address
2^N = available hosts
subtract 2 to account for network id and broadcast address
7. Explain the impact of Class Inter-Domain Routing (CIDR)
allows the creation of networks of a size other than allowed with classful.
Write the subnet, broadcast address, and valid host range for questions 1 - 6
1. 192.168.100.25/30
borrowed 6 ---> 256-252= 4 block at octet 4
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
our subnet is 192.168.100.24
next subnet is 192.168.100.28
our host range is .25 - .26
2. 192.168.10.37/28
borrowed 4 ---> 256-240= 16 block at octet 4
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
our subnet is 192.168.100.32
next subnet is 192.168.100.48
our host range is .33 - .46
3. 192.168.100.66/27
borrowed 3 ---> 256-224= 32 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
our subnet is 192.168.100.64
next subnet is 192.168.100.96
our host range is .65 - .94
4. 192.168.100.17/29
borrowed 5 ---> 256-248= 8 block at octet 4
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
our subnet is 192.168.100.16
next subnet is 192.168.100.24
our host range is .17 - .22
5. 192.168.100.99/26
borrowed 2 ---> 256-192- 64 block at octet 4
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
our subnet is 192.168.100.64
next subnet is 192.168.100.128
our host range is .65 - .126
6. 192.168.100.99/25
borrowed 1 ---> 256-128= 128 block at octet 4
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
our subnet is 192.168.100.0
next subnet is 192.168.100.128
our host range is .1 - .126
7. You have a Class B network and need 29 subnets. What is your mask?
25=32 subnets
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
borrow 5
default mask is /16.. add 5 = /21 aka 255.255.248.0
8. What is the broadcast address of 192.168.192.10/29?
class c = /24 with 5 borrowed at octet 4. 256-248 - 8 block at octet 4.
we are on subnet .8 and next subnet is .16
our broadcast is .15
we are on subnet .8 and next subnet is .16
our broadcast is .15
9. How many hosts are available with a Class C /29 mask?
23= 8 hosts - 2 reserved = 6
10. What is the subnet for host ID 10.16.3.65/23?
class A = /8 with 15 borrowed aka 8 borrowed at octet 2 and 7 borrowed at octet 3
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
256-254= 2 block at octet 3
27 borrowed=128 subnets in blocks of 2 at octet 3 FOR EACH of the 256 subnets at octet 2
29=512 hosts each
our network is 10.16.2.0
next network is 10.16.4.0
Lab 4.2 Written Subnet Practice
Given a Class B network and the following CIDR, complete the table to identify the mask and hosts available for each
Classful address
Subnet mask
number of hosts per subnet
/16
255.255.0.0
65534
/17
255.255.128.0
32764
/18
255.255.192.0
16382
/19
255.255.224.0
8190
/20
255.255.240.0
4094
/21
255.255.248.0
2046
/22
255.255.252.0
1022
/23
255.255.254.0
510
/24
255.255.255.0
254
/25
255.255.255.128
126
/26
255.255.255.192
62
/27
255.255.255.204
30
/28
255.255.255.240
14
/29
255.255.255.248
6
/30
255.255.255.252
2
Lab 4.3 Written Subnet Practice
Complete the table based on the decimal IP address:
Decimal IP
Address Class
Subnet bits
Host bits
subnet qty
host qty
10.25.66.154/23
A
15
9
32768
510
172.31.254.12/24
B
8
8
256
254
192.168.20.123/28
C
4
4
16
14
63.24.89.21/18
A
10
14
1024
16382
128.1.1.254/20
B
4
12
16
4094
208.100.54.209/30f
C
6
2
64
2
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
| Classful address | Subnet mask | number of hosts per subnet |
|---|---|---|
| /16 | 255.255.0.0 | 65534 |
| /17 | 255.255.128.0 | 32764 |
| /18 | 255.255.192.0 | 16382 |
| /19 | 255.255.224.0 | 8190 |
| /20 | 255.255.240.0 | 4094 |
| /21 | 255.255.248.0 | 2046 |
| /22 | 255.255.252.0 | 1022 |
| /23 | 255.255.254.0 | 510 |
| /24 | 255.255.255.0 | 254 |
| /25 | 255.255.255.128 | 126 |
| /26 | 255.255.255.192 | 62 |
| /27 | 255.255.255.204 | 30 |
| /28 | 255.255.255.240 | 14 |
| /29 | 255.255.255.248 | 6 |
| /30 | 255.255.255.252 | 2 |
Complete the table based on the decimal IP address:
| Decimal IP | Address Class | Subnet bits | Host bits | subnet qty | host qty |
|---|---|---|---|---|---|
| 10.25.66.154/23 | A | 15 | 9 | 32768 | 510 |
| 172.31.254.12/24 | B | 8 | 8 | 256 | 254 |
| 192.168.20.123/28 | C | 4 | 4 | 16 | 14 |
| 63.24.89.21/18 | A | 10 | 14 | 1024 | 16382 |
| 128.1.1.254/20 | B | 4 | 12 | 16 | 4094 |
| 208.100.54.209/30f | C | 6 | 2 | 64 | 2 |
Chapter 4 Review Questions
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0
then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
END CHAPTER 4
17 hours 1-3 thru 1-7-22
original total hours 35
3.5 hour review 12-21-22
review total hours 8
Chapter 5 Route Summarization Notes
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1. What is the max number of IP addresses that can be assigned to hosts on a local subnet that uses
the 255.255.255.224 mask?
/224 = 3 bits at 4th octet. Leaving 5 bits. 25 = 32 hosts
2. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet.
How many bits must you borrow from the host field to provide the correct subnet mask?
How many bits must you borrow from the host field to provide the correct subnet mask?
orignal class is not stated..
to get 29 subnets, borrow 5 bits.. 25 = 32
to get 29 subnets, borrow 5 bits.. 25 = 32
3. What is the subnetwork address for a host with the IP address 200.10.5.68/28 ?
orig class C.. /24
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
borrowed 4 = 240.. 256 - 240 = block 16 @ 4th octet
the host is on the .64 subnet
4. The network address of 172.16.0.0/19 provides how many subnets and hosts?
class B- /16... borrowed 3 = 224... 256-224= 32 block @ third octet
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
borrowed 3... 23 = 8 subnets
13 host bits... 213 = 8192 hosts each
5. Two true staments about the IP address 10.16.3.65/23
class A- /8... borrowed 15.. 8 at octet2 and 7 at octet 3.. 256-254= 2 block @ 3rd octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
9 host bits... 29 = 512 hosts each
10.16.2.0 is the network address
the broadcast is 10.16.3.255, because the next subnet starts at 10.16.4.0
hosts are iterating in the 4th octet
6. If a host on a network has the address 172.16.45.14/30, what is the subnetwork this host belongs to?
class B borrowed 8 at octet 2 and 6 at octet 4...
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
256-252= 4 block at 4th octet AND 256-255 = 1 block at 3rd octet
26=64 subnets at octet 4, for each of 256 subnets at octet 3...
172.16.45.12 is the network.. because the next network start at 172.16.45.16, then .20, then .24... etc
7. Which mask should you use on point-to-point WAN links in order to reduce waste of IP addresses?
/30
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
leaves 2 host bits.. 22 = 4 - 2 = 2 usable hosts
8. What is the subnetwork number of a host with an IP address of 172.16.66.0/21 ?
class B = /16 ... 5 borrowed bits at octet 3 = 248. 256-248 = 8 block. Subnets are 8.0, 16.0, 24.0, 32.0, 40.0, 48.0, 56.0, 64.0, 72.0
Our network is 172.16.64.0
Our network is 172.16.64.0
9. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts
can have IP addresses on the LAN attached to the router interface?
/29 leaves 3 host bits. 23 = 8 total hosts
network and broadcast address are n/a
6 available hosts including the router interface
network and broadcast address are n/a
6 available hosts including the router interface
10. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address.
Which should you assign to the server?
Which should you assign to the server?
class C = /24.. 5 borrowed at octet 4 = 248. 256-248= 8 block at octet 4
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
our network is 192.168.19.24
next network is 192.168.18.32
our broadcast is 19.168.19.31
our available hosts are .25 - .30
server can be at .26
11. You have an interface on a router with the IP address of 192.168.192.10/29.
What is the broadcast address the hosts will use on this LAN?
class C = /24 + 5 borrowed = 248 at octect 4. 256-248 = 8 block at octet 4.
we are on the 192.168.192.8 network
broadcast is .15
we are on the 192.168.192.8 network
broadcast is .15
12. You need to subnet a network that has 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
Class A = /8 with 24 host bits
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
Class B = /16 with 16 host bits
Class C = /24 with 8 host bits
255.255.255.224 provides 8 subnets..
.224 = 3 borrowed bits.. 23 = 8 subnets
leaving 5 host bits.. 25 = 32 hosts each
We are using CIDR here.. the question refers to 'classful' in terms of the starting point
13. You configure a router interface with the IP address 192.168.10.62 255.255.255.192 and receive this error:
Bad mask /26 for address 192.168.10.62
Why did you get the error?
Bad mask /26 for address 192.168.10.62
Why did you get the error?
192.168.10.62 / 26
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
class C = 24. 2 borrowed. 256-192 = 64 block at octet 4
networks are .0 , .64, .128, etc
our network is .0
broadcast is .63
hosts are .1 - .62
the error occurs because we are on the ZERO SUBNET
the router wont allow it by default. Must enable with the ip subnet-zero command
this command allows you to use the FIRST and LAST subnets
14. If an ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this interface?
class B = /16. borrowed 9. aka borrowed 8 at octet 3 and borrowed 1 at octet 4.
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
256-255 = 1 block at octet 3
256-128 = 128 block at octet 4
2 subnets at octet 4 for each of 256 subnets at octet 3
we are on the 172.16.112.0 subnet
next subnet is 172.16.112.128, then 113.0, then 113.128, etc
we are on the 172.16.112.0 subnet. Broadcast is 112.127
valid hosts 112.1 - 112.126
15. Using the following illustation, what would be the IP address of E0 if you were using the eighth subnet?
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
The network id id 192.168.10.0/28 and you need to use the last available IP address in the range.
The zero subnet should not be considered valid for this question.
network 192.168.10.0/28
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0 then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
class C = /24 with 4 borrowed---> 256-240 = 16 block at octet 4
first subnet is .0 then .16, .32, .48, .64, .80, .96, .112 is the eighth subnet, .128 is the ninth
host range is .113 - .126
broacst is .127
192.168.10.126 is the last available ip in the range
WRONG
Question stated the zero subnet should NOT be considered valid. Need to start counting at .16 instead of .0
the EIGHTH subnet is .128 and the ninth is .144
host range is .129-.142
last availabile host is 192.168.10.142
16. Using the illustration from the previous question, what would be the IP address of S0 if you were using the first subnet?
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range.
Again, the zero subnet should not be considered valid for this question.
the first subnet (excluding subnet zero) is 192.168.10.16, next is .32
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
host range is .17-.30 and broadcast is .31
last host is 192.168.10.30
17. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224 ?
borrowed 3 for 224... 23= 8 subnets
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
but you can only use six unless you run the command
ip subnet-zero
possible book typo...
we should be able to use SEVEN without the command...
ex: 192.168.10.0/27
networks
.0
.32
.64
.96
.128
.160
.192
.224
eight, including subnet zero. Books says you end up with six after excluding subnet zero. Typo?
18. You have a network with a subnet of 172.16.17.0/22. Which is the valid host address?
class B = /16 with 6 borrowed at octet 3---> 256-252= 4 block
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
26=64 subnets in blocks of 4 at octet 3
210 host bits = 1024 hosts per subnet
172.16.16.0 is our network
172.16.20.0 is the next network (note subnets are iterating at octet 3, NOT octet 4 which is for hosts only)
172.16.16.1 - 172.16.19.254 is our host range
172.16.16.255 is broadcast
19. Your router has the following IP address on Ethernet0: 172.16.2.1/23.
Which can be valid host IDs on the LAN interface attached to the router?
class B = /16 with 7 borrowed---> 256-254= 2 block at octet 3
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
29= 512 hosts per subnet
our network is 172.16.2.0
next network is 172.16.4.0
host range is 172.16.2.1 - 172.16.3.254
router is using the first valid address
20. To test the IP stack on your local host, which IP address would you ping?
127.0.0.1
loopback
loopback
pg 187 example
Summarize networks 172.16.32.0 through 172.16.50.0
Summarize networks 172.16.32.0 through 172.16.50.0
pg 188 fig 5.13
What is the range of addresses for summary 192.168.144.0/20 ?
What is the range of addresses for summary 192.168.144.0/20 ?
pg 188 fig 5.14
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Summarize these addresses:
172.1.4.128/25
172.1.7.0/24
172.1.6.0/24
172.1.5.0/24
172.1.4.0/25
Route Summarization Examples
Chapter 5 Troubelshooting IP Addressing
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2.
ping your own IP address
failure indicates a problem at the NIC
3.
ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4.
ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
Chapter 5 VLSM, Summarization, Troubleshooting Exam Essentials
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
FOUR TROUBLESHOOTING STEPS
scenario: User cannot log into a Window server
scenario: User cannot log into a Window server
1.
ping loopback address the affected client.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2. ping your own IP address
failure indicates a problem at the NIC
3. ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4. ping a remote server.
if all four steps pass, issue is probably name resolution.
An effective ping indicates your IP stack is initalized.
A FAILURE to ping 127.0.0.1 would require re-installation of TCP/IP on the host.
2. ping your own IP address
failure indicates a problem at the NIC
3. ping the default gateway
failure indicates an issue with the network somehwere between the NIC and the router
4. ping a remote server.
if all four steps pass, issue is probably name resolution.
Solve the IP address problems
1. Describe the benefits of VLSM
enable the creation of subnets of specific sizes that dont need to be equal, making the use of address space more efficient
2. Understand the relationship between the subnet mask and the resulting block size and the
allowable IP addresses in each resulting subnet
the relationship between the classful default network bits and the 'borrowed' subnet bits determines the block size.
Block size delineates networks and defines valid hosts within each network
3. Describe the process of summarization or route aggregation and its realtionship to subnetting
Summarization is the combining of subnets for the purpose of advertising a single route to
neighboring routers, reducing the size of routing tables and speeding the route process.
4. Calculate the summary mask that will advertise a single network representing all subnets.
the network address used to advertise the summary address is always the first network address in the block.
5. Remember the four diagnostic steps
ping loopback
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
ping your own ip aka ping your NIC
ping your default gateway
ping remote address
6. Identify and mitigate an IP address problem?
try the four troubleshooting steps. Then find out if the ip addresses in question are configured properly
7. Understand the troubleshooting tools that you can use from your host and a Cisco router.
ping 127.0.0.1 - test your local ip stack
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Windows
tracert track the packet
arp -a display IP to MAC mapping
ip config /all display your PC network configuration
Cisco IOS
traceroute track the packet
trace shorthand for traceroute
Lab 5.0
For each of the following sets of networks, determine the summary address and the mask to be used
that will summarize the subnets
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
Chapter 5 Review Questions
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11
1) 192.168.10/24 through 192.168.12.0/24
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
2) 172.144.0.0 through 172.159.0.0
3) 192.168.32.0 through 192.168.63.0
4) 192.168.96.0 through 192.168.111.0
5) 66.66.0.0 through 66.66.15.0
6) 192.168.1.0 through 192.168.120.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
7) 172.16.1.0 through 172.16.7.0
8) 192.168.128.0 through 192.168.190.0
9) 53.60.96.0 through 53.60.127.0
10) 172.16.10.0 through 172.16.63.0
1 On a VLSM network, which mask should you use on point-to-point WAN links in order
to reduce the waste of waste of IP addresses?
/30
2. To test the IP stack of your local host, which IP address would you ping?
127.0.0.1
3. What is the only connection type that supports the use of the /30 mask?
point to point
4. To use VLSM, what capability must the routing protocols possess?
transmission of subnet mask info
5. What is another term for route aggregation?
summarization
6. Which of the following is a result of route aggregation?
smaller routing tables
7. The network address used to advertise a summary address is always __________
the first network in the block
8. When a ping to the loopback address fails, what can you assume?
your IP stack has failed to initizalize
9. When a ping to the local host IP address fails, what can you assume?
NIC failure
10. When a ping to the local IP address succeeds but a ping to the default gateway fails, what can you rule out?
NIC or IP stack failure can be ruled out.
IP address of local host or IP address of default gateway could be incorrect
IP address of local host or IP address of default gateway could be incorrect
11. If a remote host can be pinged what problems can you rule out?
any issues described above; if you can ping a remote host, your IP stack, NIC, ip address must all be ok
12. What network service is the most likely problem if you can ping a computer by IP address but not by name?
DNS
13. When you issue the ping command, what protocol are you using?
ICMP
14. Which command displays the networks traversed on a path to a destination?
traceroute
15. Which command uses ICMP echo requests and replies?
ping
16. What command is the Windows version of the Cisco command that display the networks
traversed on a path to a network destination?
tracert
17. Which command displays IP to MAC address mappings on a Windows PC?
arp -a
18. What command displays the ARP table on a Cisco router?
show ip arp
19. What switch must be added to the ipconfig command on a PC to verify DNS configuration?
/all
20. What is the best summarization of the following networks:
192.168.128.0 thru 192.168.159.0
192.168.128.0 thru 192.168.159.0
need to cover 31 at octet 3, use 32 block. 256-32= 224 is our 3rd octet value
192.168.128.0 /19 or 255.255.224.0
192.168.128.0 /19 or 255.255.224.0
END CHAPTER 5
13 hours 1-9 thru 1-14-22
original total hours 48
3 hours review 12-21-22
review total hours 11