GATE CSE 2021 Set 1 | Question: 49

Question

Consider the sliding window flow-control protocol operating between a sender and a receiver over a full-duplex error-free link. Assume the following:

• 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 infinite number of frames available for transmission.
• The size of the data frame is $2,000$ bits and the size of the acknowledgement frame is $10$ bits.
• The link data rate in each direction is $1$ Mbps ($=10^6$ bits 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_____________.

Comments

Instead of focusing on what A or B has given for a question if one ocus on what is the answer to the question based on the concept involved, GATE will become easy.

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I think they consider both 50 and 51 because if we find link utilization putting the value of N=50 so it is 49.5% and N=51 so it is 50.49 %. Also, they do not mention that “link utilization at least 50%” instead of they mention “exactly 50%”. Because of that, they consider 0.5 error.

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Can anyone tell me what is role of full duplex link here? Will the answer change or something that i am missing?

Answer 1

Let the sender window size be $N$.

$\text{One way propagation delay} =100\,ms=0.1\,s$

$\text{Transmission delay}_{packet}=\frac{\text{Size of data frame}}{\text{Link bit rate}}=\frac{2000}{10^{6}}=0.002\,s$

$\text{Transmission delay}_{ACK}=\frac{\text{Size of ACK frame}}{\text{Link bit rate}}=\frac{10}{10^{6}}=0.00001\,s$

$\text{Link Utilization}(\eta )=\frac{\text{Useful Data Transfer time}}{\text{Total time}}$

$\implies \eta = \frac{N.(T_{f})}{T_{f}+2(T_{p})+T_{ACK}}$

$\implies N = \left \lceil \frac{\eta.\left(T_{f}+2(T_{p})+T_{ACK}\right)}{T_{f}}\right \rceil = \left \lceil \frac{0.5(0.002+0.2+0.00001)}{0.002}\right \rceil =\lceil 50.5025 \rceil  =51$

Correct Answer: $51$

Comments

Not sure but we can say since this is an sliding window protocol.. We will wait for 2tp time until acknowledgement of first one comes then we move the window...

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@abhijeet, but it's full-duplex. @arjun sir, please help.

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Vandit Shah I don't understand how is piggybacking going to help us with anything in this case as the receiver is not sending any data.
Coming to your second point, let's assume even if the receiver is sending some data in return, along with the acknowledgment, it will end up taking an extra transmission time, for transmitting data+ ack on the line, an extra propagation time, for sending data+ack and another extra propagation time, for sender acknowledging receivers data. (Which is bizarre, as nothing about receiver sending data is mentioned in the question)

Do correct me if i am missing something.

Answer 2

let window size =N

a=Tp/Tt

a=100*(10^-3)*(10^6/2000)

a=100/2=50

so,to get 50% utilization

 

N/(1+2a)>=(1/2)

N>=(1+2a)(1/2)

N>=(1+100)(1/2)

N>=51

The minimum size of window size is 51