Question 6987 – UGC-NET June CS 2019 Paper 1
November 29, 2023JAVA
November 29, 2023Computer-Networks
| Question 3 |
- The time taken for processing the data frame by the receiver is negligible.
- The time taken for processing the acknowledgement frame by the sender is negligible.
- The sender has an infinite number of frames available for transmission.
- The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
- The link data rate in each direction is 1 Mbps (=106bits per second).
- One way propagation delay of the link is 100 milliseconds.
The minimum value of the sender’s window size in terms of the number of frames, (rounded to the nearest integer) needed to achieve a link utilization of 50% is ______.
| 51 |
Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2 x 10^-3 sec = 2 millisec
Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec
Tp = 100 millisec
Total time = Tt(packet) + 2 x Tp + Tt(Ack)
=> 2 + 2 x 100 + 0.01 = 202.01 millisec
Efficiency = 50 % = ½
Efficiency = Useful Time / Total time
½ = n x Tt / Total time
=> 2 x n x Tt = Total time
=>2 x n x 2 = 202.01
=> n = 202.01 / 4 => 50.50
For minimum, we have to take ceil, Hence size of window = 51
Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2 x 10^-3 sec = 2 millisec
Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec
Tp = 100 millisec
Total time = Tt(packet) + 2 x Tp + Tt(Ack)
=> 2 + 2 x 100 + 0.01 = 202.01 millisec
Efficiency = 50 % = ½
Efficiency = Useful Time / Total time
½ = n x Tt / Total time
=> 2 x n x Tt = Total time
=>2 x n x 2 = 202.01
=> n = 202.01 / 4 => 50.50
For minimum, we have to take ceil, Hence size of window = 51