Network Technology

Q1. A class less: Domain Routing (CIDR) address (8.0.0.0/7) has been

assigned to the network below:

London 60 host

Birmingham Manchester

256 hosts 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.

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

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

8.0.0.0/23 8.0.2.0/23 8.0.2.0/24

Advantages of VLSM

Standard subnetting benefits include:

1. Structure2. Performance3. Management4. Scalable

VLSM 1. Reduces address space wastage2. Reduce size of routing tables.

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

8.0.2.1/24(127 Total hosts)256 Total8.0.2.254/248.0.2.255/24

8.0.3.0/24

256 Total

8.0.3.1 8.0.3.62 BC.63

8.0.3.64/268.0.3.128/268.0.3.192/26

512

Total

8.0.1.254/23

8.0.1.255/23

512

Total

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.

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

----------------------

192.168.1.2/24 192.168.2.0/24Fast Ethernet 0/1 Fast Ethernet0/0

A DISTANCE PROTOCOL

120 RIP100 IGRP90 EIGRP110 OSPF115 ISIS

INPUT LAYAR DATA OUTPUT LAYAR DATA

R1 R2

Source IP = 192.168.10.3

Destination IP = 192.168.20.3

Source MAC = 000C. CF7C. 4B78

Destination MAC = 00ED. F75A.7601

LAYER 1 PORT Fast Ethernet 0/0

Source IP = 192.168.10.3

Destination IP = 192.168.20.3

Source MAC = 00E0.F75A.7604

Destination MAC = 00ED. F983.B5A6

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.

Answer.

Fast Ethernet 0/0 Fast Ethernet 0/1

00E0.F75A.7601 0de0.F75A.7604

X Over Cable

192.168.10.3 192.168.20.3

000C.CF7C.4B78 00ED.F983.B5A6

PC PC