The Conference on Computer Communications

NA

NETWORKING

Question 1
1.
The 9 leading bits will be;
class A:0, class B:10, class C:110, class D: 1110, class E:1111, class F:111110, class
G:1111110, class H:11111110, class I:111111110.
2.
a) In this class, the number of addresses on the network is
2^ (32-20) = 4096
To find the first address in the block we keep the first 20 bits and change the rest of the bits to
0s as shown.
Address 200.107.16.17/20 10100111 11000111 10101010 01010010
First address 200.107 .16.0/20 10100111 11000111 10100000 00000000
To find the first address in the block we keep the first 20 bits and change the rest of the bits to
1s as shown.
Address 200.107.16.17/20 10100111 11000111 10101010 01010010
Last address 200.107 .31.255/20 10100111 11000111 10101111 11111111
The first address is: 200.107 .16.0/20
The first address is: 200.107 .31.255/20
b) The mask is 255.255.240.0
This is obtained from last address in the block.

Question 2
1.
a) Distance vector for node A is:
A
A 0
B 1
C 3
D 3
E 4
b) Distance vector for node B is:
B
A 1
B 0
C 4
D 2
E 4
c) Distance vector for node C is:
C
A 3
B 4
C 0
D 6
E 1
d) Distance vector for node D is:
D
A 3
B 2
C 6
D 0
E 6
e) Distance vector for node E is:

E
A 4
B 4
C 1
D 6
E 0
2. No, the maximum distance of the network cannot be lowered by reducing the distance
between nodes B and E from 4 to 2. This is because it will affect the maximum
distance between B and D (Choi, Han, Cho, Kwon, & Choi,2011).
3. The maximum distance of the network after the link between nodes A and C is
removed.
a) maximum distance vector for node A is:
A
A 0
B 1
C 6
D 3
E 4
b) maximum distance vector for node B is:
B
A 1
B 0
C 5
D 3
E 4
c) maximum distance vector for node A is:
C
A 6
B 5
C 0
D 7
E 1

d) maximum distance vector for node D is:
A
A 3
B 3
C 7
D 0
E 6
e) maximum distance vector for node A is:
E
A 5
B 4
C 1
D 6
E 0
Losing the link will lead to reduced quality of service. This will be seen through data
congestion and queueing in the network because the network links between the nodes will be
subjected to carry more data than they can handle (Cohe, Postel, & Rom, 2012).
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 False

transport layer, while the network layer supports
connectionless communication.
[2].
a) The sequence numbers are 0,1,2,3,4,5,6,0,1,2,3,4,5,6,0,1,2…………
b) The acknowledgement numbers of the ACK packets are 31.
C) There is no effect on the Sn and Sf values since the acknowledgement numbers is equal to
Sf thus they are discarded.
[3].
a) Destination port number- 2 bytes.
Each hex = 4 bits.
93 E2 B0 17 0A B2 00 00
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.