Computer Networks Assignment: Problem Solving and Solutions Report

This report presents solutions to a computer networks assignment, addressing various concepts within the field. The assignment begins with an exploration of IPv4 classful addressing, including the design of fixed prefixes for different network sizes and the division of addresses into classes A through E. The first question delves into classful addressing, subnetting, calculating usable IP addresses, and determining subnet masks. The second question focuses on distance vector routing, analyzing distance vectors in different nodes, the impact of link changes, and calculating maximum distances. Finally, the report examines transport layer concepts, including port numbers, flow control, sequence numbers, and acknowledgement numbers. It concludes with a bibliography of relevant sources.

TEMAOS
COMPUTER NETWORKING

Question 1
1)
This is ipv4 Classful addressing. For small and large networks to be accommodated, three fixed
prefixes were designed instead of one this is (n = 8, n= 16, and n = 24).
The whole addresses are then divided into five classes these are class A, B, C, D, and E.
• IP was initially calculated with fixed length prefixes.
• Diverse lengths were reinforced to support different network sizes, known as classes
• For class-A, the first bit (0) defines the class, we have 2? =128 networks
• For Class B- (10) first two bots define class, 214 =16384 networks Address space:
4,294,967,296 addresses C 110 Prefix Suffix Class D 1110 Multicast addresses
Class E 1111 Reserved for future use Class Prefixes A n = 8 bits B n = 16 bits C n =24 bits
D Not applicable
E Not applicable First byte 0 to 127 128 to 191 192 to 223 224 to 239 240 to 255 2,097,152
networks 28-2 hosts per network (254 hosts)
2)
In this case, the number of usable IP addresses will be as follows.
2^ (32-20) = 4096-2
So the number of usable IP addresses will be 4094
To find the first IP address the first 20 bits are kept and the other bits are changed to one and 0s.
As shown below
Address 200.107.16.17/20 10100111 11000111 10101010 01010010
First address 200.107 .16.1/20 10100111 11000111 10100000 00000000
The first IP address is found by changing the first 20 bits while maintaining the ones and zeros.
Address 200.107.16.17/20 10100111 11000111 10101010 01010010
Last address 200.107 .31.255/20 10100111 11000111 10101111 11111111
The first address will be : 200.107 .16.0/20

The first address is: 200.107 .31.255/20
The first usable IP address is 200.107 .16.0/20 .
The last usable IP address is: 200.107.31.254/20
The subnet mask, in this case, will be: 255.255.240.0
This calculated as shown below.
We first divide the mask into 4 bits as shown.
128 192 224 240 248 252 254 255
128 64 32 16 8 4 2 1
00000000. 00000000. 00000000. 00000000 =32 bits
8 16 24
We get that 20 is in the range of 24 bits.
24-20 = 4
2^4= 16
16 goes with 240.
So the subnet mask is 255.255.240.0

Question 2
1)
Distances vectors in node A is:
A
A 0
B 1
C 3
D 3
E 4
Distances vectors in node B is:
B
A 1
B 0
C 4
D 2
E 4
Distances vectors in C is:
C
A 3
B 4
C 0
D 6
E 1
Distances vectors in node D is:
D
A 3
B 2
C 6
D 0
E 6

Distances vectors in node E is:
E
A 4
B 4
C 1
D 6
E 0
2)
No, the most extreme distance of the network can't be brought down by decreasing the
separation between node B and E from 4 to 2. This is on the grounds that it will influence
the maximum separation among B and D
3) The maximum distance between the nodes will be as follows after the link nodes are
removed.
The following tables shows the maximum distances from one node to another.
maximum distance vector for node A is:
A
A 0
B 1
C 6
D 3
E 4
maximum distance vector for node B is:
B
A 1
B 0
C 5
D 3
E 4
maximum distance vector for node A is:

C
A 6
B 5
C 0
D 7
E 1
maximum distance vector for node D is:
A
A 3
B 3
C 7
D 0
E 6
maximum distance vector for node A is:
E
A 5
B 4
C 1
D 6
E 0

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.
a. The sequence numbers of the packet in transit will be
0,1,2,3,4,5,6,0,1,2,3,4,5,6,0,1,2….
b. 31 will be the acknowledgement number of the ACK packets.
c. There will be no effect on the Sn and Sf values since the acknowledgment
number is equal to the b value of the Sn hence the values are cast-off.
3.
a.
The destination port No: is B017==23
b. Sequence number- 4 bytes.
Thus the sequence No: is 0AB20000==1

Bibliography
Cohe, D., Postel, J. B., & Rom, R. (2012, May). IP addressing and routing in a local wireless
network. In [Proceedings] IEEE INFOCOM'92: The Conference on Computer
Communications(pp. 626-632). IEEE.
Choi, J., Han, J., Cho, E., Kwon, T., & Choi, Y. (2011). A survey on content-oriented
networking for efficient content delivery. IEEE Communications Magazine, 49(3), 121-127.