Computer Networks Trimester

This document contains questions and solutions related to Computer Networks Trimester. It covers topics such as classful addressing, subnet masks, distance vectors, flow control, and TCP headers.

Computer Networks Trimester

Table of Contents
Question-1 (Chapter 18)..................................................................................................................3
Question-2 (Chapter 20)..................................................................................................................6
Question-3 (Chapters 23 and 24).....................................................................................................7

Question-1 (Chapter 18)
1) In current classful addressing, there are 5 classes as shown in Fig. 18.18 (p. 531 of the
textbook). There are 5 classes with variable-length leading bits as follows: Class A: 0; Class B:
10; Class C: 110; Class D: 1110; Class E: 1111 Suppose that we wish to modify to accommodate
9 classes. What are 9 (variable-length) leading bits for them?
The 9 leading bits of accommodating 9 classes is 1001010010
2) A classless address is given as 200.107.16.17/20
• Find the first and last addresses in the block (explain in detail how you can find them).
The given IP address is 200.107.16.17
It is a Class C network type
The prefix length is 20
Hence n= 20
The Subnet mask of the class C network is 255.255.255.0
The first address is calculated through the AND operation and will involve the Subnet mask and
the IP address
Step 1
The decimal number of IP address is 200 and needs to be converted into corresponding binary
number that is 11001000
Step 2
The rest of the numbers are 107, 16 and 17. These parts comprise of 8 digits in binary form.
Hence,
107=01101011
16=00010000
17=00010001
Therefore, the binary number is
11001000.01101011.00010000.00010001

Step 3
The Subnet mask of the class C network is 255.255.255.0. The address is to be designed with 32
digits and the netmask Is 20. Hence the it will comprise of twenty digits as one and other twelve
as zero. The outcome of network is:
11111111.11111111.11110000.00000000
Step 4
AND operation is to be conducted

Ip addess: 11001000.01101011.00010000.00010001
Subnet mask: 11111111.11111111.11110000.00000000
(AND operator)
First IP address 11001000.01101011.00010000.00000000
First IP address 200.107.16.0
Last Address
The above mentioned steps are to be followed and the Step 4 is to be changed and OR operator is
going to be implemented
Ip addess: 11001000.01101011.00010000.00010001
Subnet mask: 11111111.11111111.11110000.00000000
Mask complement: 00000000.00000000.00001111.11111111
(OR operator)
Last IP address 11001000.01101011.00011111.11111111
200.107.31.255
• Find the mask.
The subnet mask is calculated by performed by understanding the value of ‘n’
The value is in this case is 20
Thus, subnet mask address will comprise of twenty digits as one and other twelve as zero. The
result will be;
11111111.11111111.11110000.00000000
255.255.240.0

Question-2 (Chapter 20)
Consider the network consisting of 5 nodes.
1) Find the distance vectors for all the nodes.
Information at
each node
Distance to each node
A B C D E
A
B 1
C 3
D 2
E 1 4
Table 1: Initial distance vectors
The distance vectors for all the nodes are calculated through the distance vector routing
algorithm. Let dq(r) be the cost of the lead-cost trace from the node q to r. it is devised in
accordance to the Belman Ford equation
Dq(r) = minr{c(q,r) + dq(r)}
The distance vector q, Dq = [ Dq(r) : r in N] comprising of cost to every destination
Information at
each node
Final Distance vector to each node
A B C D E
A 0 4 2 4 3
B 1 2 1 1 2
C 3 1 3 3 1
D 2 2 2 1 1
E 1 3 1 2 4
Table 2: Final distance vector to each node
2) Can the maximum distance of the network (the maximum distance between any pair of
different nodes, i.e., {A,C}, {A,D}, {B,C}, etc) be lowered if the distance between nodes
B and E becomes 2? Please explain how you come to the answer (Yes or No). [4 marks]

Yes, the maximum distance of the network can be lowered if the distance between nodes B and E
is reduced to 2. It is because of the fact that the parameters of distance vector to the neighbours
can be manipulated through tweaking in the equation, Dv = [ Dv(y) : y in N ]
3) Find the maximum distance of the network if the link between nodes A and C is
removed. In addition, discuss what is the performance loss after losing the link between
nodes A and C.
The maximum distance is to be calculated through the help of the Belman Ford equation which
is stated by Dq(r) = maxr{c(q,r) + dq(r). The performance loss between nodes A and C is
significant in terms of routing loss and the buffer loss The routing loss is higher as it reflects on
the presence of the node link.
Question-3 (Chapters 23 and 24)
1) Choose “True” or “False” for each item
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
True
2) Suppose that the Go-Back-N protocol with = 3 and the sending window of size 7 is used to�
send packets. Let Sf = 31, Sn = 35, and Rn = 33.
a. What are the sequence numbers of data packets in transit?
The sending window size is 7 and is formulated by 2n-1

Thus, 23-1=7.
Hence 3-bit sequence number is chosen
The sequence numbers are: 000, 001,011,111,100,110,101,010
b. What are the acknowledgement numbers of ACK packets in transmit?
The acknowledgement numbers of the ACK packets are 2^32= 4,294,967,296
It is determined by the addition of the Sequence Number A with the number of octet that
comprises of the data in the segment, represented by (x)
c. If the sender’s process sends two more data packets to the sender’s transport layer and one
ACK packet is received from the receiver’s transport layer, what are the updated values for Sf
and Sn?
The updated values are: Sf = 40, Sn=42
3) The following is part of a TCP header in hexadecimal format:
93E2 B017 0AB2 0000 …
a. What is the destination port number?
Destination port number = B01716= 45079
b. What is the sequence number?
The sequence number field is 32 bits