SIT202 Computer Networks: Problem Solving Report 1 - Trimester 2, 2019

This document presents a comprehensive solution to a Computer Networks problem-solving report (SIT202). The report addresses three key questions. The first question explores classless addressing, netmasks, wildcard masks, and IP address ranges. The second question delves into distance vector routing, analyzing distance vectors for multiple nodes and examining the impact of network topology changes on maximum distances and latency. The final question covers transport layer concepts, including true/false statements about port numbers and flow control, analysis of the Go-Back-N protocol with sequence and acknowledgement numbers, and interpretation of a TCP header in hexadecimal format, identifying port and sequence numbers. The solution provides detailed calculations, explanations, and answers to all the questions.

Running head: NETWORK AND COMMUNICATION
NETWORK AND COMMUNICATION
Name of the Student
Name of the University
Authors Note

1NETWORK AND COMMUNICATION
Question 1:
1)
The 9 (variable-length) leading bits for the 9 classes are the followings:
A: 0000 0000
B: 0000 0010
C: 0000 0110
D: 0000 1110
E: 0000 1111
F: 0010 1111
G: 0110 1111
H: 1110 1111
I: 1111 1111
2) A classless address is given as 200.107.16.17/20
The mask bit is 20
Address: 200.107.16.17 11001000.01101011.0001 0000.00010001
Netmask: 255.255.240.0 = 20 11111111.11111111.1111 0000.00000000
Wildcard: 0.0.15.255 00000000.00000000.0000 1111.11111111
=>
Network: 200.107.16.0/20 11001000.01101011.0001 0000.00000000 (Class C)
Broadcast: 200.107.31.255 11001000.01101011.0001 1111.11111111
HostMin: 200.107.16.1 11001000.01101011.0001 0000.00000001

2NETWORK AND COMMUNICATION
HostMax: 200.107.31.254 11001000.01101011.0001 1111.11111110
Hosts/Net: 4094
Subnet Name: A 4094
Needed Size: 4094
Allocated Size: 4094
200.107.16.0 /20
Address Mask: 200.107.16.0 /20
Dec Mask: 255.255.240.0
Assignable Range: 200.107.16.1 - 200.107.31.254
Broadcast: 200.107.31.255

3NETWORK AND COMMUNICATION

First assignable IP address - 200.107.16.1 and
Last assignable IP address - 200.107.31.254
 The mask address is 255.255.240.0
Question-2:
1) Distance vectors of the node:
Distance vector for node A:
Destination Next hop
B C
B 1 8
C 5 3
D 3 10
E 5 4
So following are the distance vectors of A:
A-B is 1
A-C is 3
A-D is 3
A-E is 4

4NETWORK AND COMMUNICATION
Distance vector for node B:
Destination Next hop
A D E
A 1 ∞ 8
C 4 ∞ 5
D ∞ 2 ∞
E 5 ∞ 4
So following are the distance vectors of A:
B-A is 1
B-C is 4
B-D is 2
B-E is 4
Distance vector for node C:
Destination Next hop
A E
A 3 6
B 4 5
D 6 7

5NETWORK AND COMMUNICATION
E 8 1
So following are the distance vectors of A:
C-A is 3
C-B is 4
C-D is 6
C-E is 1
Distance vector for node D:
Destination Next hop
B
A 3
B 2
C 6
E 6
So following are the distance vectors of A:
D-A is 3
D-B is 2
D-C is 6

6NETWORK AND COMMUNICATION
D-E is 6
Distance vector for node E:
Destination Next hop
B C
A 5 4
B 4 5
C 8 1
D 6 7
So following are the distance vectors of E:
E-A is 4
E-B is 4
E-C is 1
E-D is 6
2) The maximum distance between the nodes:
If the distance between nodes B and E becomes 2, the maximum distance between the nodes
is decreased which includes E as next hop. For example, the maximum distance between the
node B and node A through node E is 8. If the distance between nodes B and E is reduced to
2, then this distance is reduced to 6 and therefore the maximum distance between these two
nodes are lowered in this context.
3) The maximum distance between the nodes if nodes between the nodes A and C is
removed:

7NETWORK AND COMMUNICATION
The distance between the nodes A and C or nodes C and A is 6 and this is the maximum
distance between two nodes in the network design provided in this context
As the link between nodes A and C is removed, the distance between these two nodes is
increased from 3 to 6. Along with that the distance between nodes A and E is also increased
from 4 to 5. Due to this, time required to send data packets between these nodes will increase
and therefore, the latency of the network will also increase. Therefore, it will affect the
network speed as well which means that the performance of the network will be affected as
well in this context
Question 3:
1)
True or False
In the transport layer, port numbers are used to identify local
computers.
False
For flow control, we need two buffers: one at the sending transport
layer and the other at the receiving transport layer.
True
There are only connection-oriented protocols in the transport layer,
while the network layer supports connectionless communication.
False
2) For the Go-Back-N protocol, m=3 and the sending window of size 7. It is provided that Sf
= 31, Sn= 35, and Rn = 33
a) The sequence number of data packets that are in transit are 33, 34 and 35.
b) The acknowledgement numbers of ACK packets that are in transmit are 31 and 32.
c) The updated values for Sf and Sn are 36 and 40 respectively

8NETWORK AND COMMUNICATION
3) The TCP header in hexadecimal format: 93E2 B017 0AB2 0000
a) Port number: 45079
b) Sequence number: 179437568