This document discusses network addressing and routing. It presents a network with three segments in London, Birmingham, and Manchester of varying host sizes. It asks the reader to: a) Design an addressing scheme using VLSM for the network and populate a table with subnets and masks b) Label the network segments with addresses and broadcasts c) Explain benefits of VLSM It then asks the reader to determine the routing protocol and sketch the topology for a network routing between two interfaces. Finally, it provides input/output layer data to sketch a network diagram showing IP addresses, MAC addresses, and interfaces between devices.

1. Network Technology
Q1. A class less: Domain Routing (CIDR) address (8.0.0.0/7) has been
assigned to the network below:
London
Birmingham
256 hosts
60 host
Manchester
127 host
a. Employ VLSM to design a complete addressing scheme for the network.
Produce a table for your design and populate it with address and subnet
masks in forward slash notation format.
b. Label all network segments with Broadcast and full range of address.
c. Explain some benefits of employing VLSM in network design.

2. Bits
Total host Total Subnets
1
2
2
4
3
6
4
8
5
16
6
32
7
64
8
126
9
256
10
512
11
1024
12
2048

3. New Subnet mask
8.0.0.0/23
255.255.254.0
512 hosts
9 host bits
7 network bits
16 subnet bits
16 bits = 65536

4. 8.0.0.0/23
8.0.2.0/23
8.0.2.0/24
8.0.2.1/24
(127 Total hosts)
256 Total
8.0.2.254/24
8.0.2.255/24
8.0.3.0/24
256 Total
512
512
Total
Total
8.0.1.254/23
8.0.1.255/23
Advantages of VLSM
Standard subnetting benefits include:
1.
2.
3.
4.
Structure
Performance
Management
Scalable
VLSM
1. Reduces address space wastage
2. Reduce size of routing tables.
8.0.3.1
8.0.3.62 BC.63
8.0.3.64/26
8.0.3.128/26
8.0.3.192/26

5. Q2. Determine the route to a destination network given that packets to
the remote network pass through two fast Ethernet ports. The delay of a
fast Ethernet port is 100MS
256 X
[
107/105 + 200/10
]
256X120
= 30720
Show IP route
C
192.168.1.0/24 is a directly connected Fast Ethernet 0/0
C
192.168.3.0/24 is a directly connected Fast Ethernet 0/1
D
192.168.2.0/24 [90/30720] via is a directly connected Fast
Ethernet 0/0
EIGRP
AA Metric = Administrative distance
a. Determine the dynamic routing protocol active on the router.
b. Sketch the network topology using the information in the table.
Label all network segments with their network address. Indicate
suitable IP addresses for all interfaces on connected networks.

6. 192.168.1.0/24
192.168.3.1/24
192.168.1.1/24
192.168.2.1/24
192.168.3.0/24
R1
R2
----------------------
192.168.1.2/24
192.168.2.0/24
Fast Ethernet 0/1 Fast Ethernet0/0
A DISTANCE
120
100
90
110
115
PROTOCOL
RIP
IGRP
EIGRP
OSPF
ISIS

7. INPUT LAYAR DATA
OUTPUT LAYAR DATA
Source IP = 192.168.10.3
Source IP = 192.168.10.3
Destination IP = 192.168.20.3
Destination IP = 192.168.20.3
Source MAC = 000C. CF7C. 4B78
Source MAC = 00E0.F75A.7604
Destination MAC = 00ED. F75A.7601
Destination MAC = 00ED. F983.B5A6
LAYER 1 PORT Fast Ethernet 0/0
LAYER 1 PORT Fast Ethernet 0/1
(a) Sketch the network diagram using the data in the tables above. Assume
connected interfaces on the Layer device are configured with the first usable
IP address on their respective networks.
(b) Assume the source and destination devices are PC’s and show their IP
address on your diagram.
(c) Show the MAC address of all devices show in your diagram.

8. Answer.
Fast Ethernet 0/0
Fast Ethernet 0/1
00E0.F75A.7601
0de0.F75A.7604
X Over Cable
PC
PC
192.168.10.3
192.168.20.3
000C.CF7C.4B78
00ED.F983.B5A6