Data-Link-Layer
Question 1 |
Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?
1 | |
2 | |
3 | |
4 |
Question 1 Explanation:
Let there be N such hosts. Then when one host is transmitting then others must be silent for successful transmission. So throughput per host,
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Question 2 |
In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as
01101011 | |
011010110 | |
011101100 | |
0110101100 |
Question 2 Explanation:
In the data link layer, bits stuffing is employed then bit stuffing is done using the flag delimiter. If there is a flag of n bits then we will compare the data sequence with the flag and for every n-1 bits matched found, a bit 0 is stuffed in the data sequence.
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Question 3 |
Which one of the following statements is FALSE?
Packet switching leads to better utilization of bandwidth resources than circuit switching. | |
Packet switching results in less variation in delay than circuit switching. | |
Packet switching requires more per packet processing than circuit switching. | |
Packet switching can lead to reordering unlike in circuit switching. |
There are 3 questions to complete.