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GATE CSE 2021 Set 1 Q59: Sliding window size for 50% utilisation

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 (=106= 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 ________.

Try it: numerical answer: type the value

Numerical answer. No negative marking.
Show answer and explanation

Answer: 50 to 52 (official GATE 2021 final answer key)

Concept used: Utilisation =W⋅TframeTframe+2Tp+Tack= \dfrac{W \cdot T_{\text{frame}}}{T_{\text{frame}} + 2T_p + T_{\text{ack}}} when the window does not fill the cycle.

First step: Tframe=2T_{\text{frame}} = 2 ms, Tack=0.01T_{\text{ack}} = 0.01 ms and 2Tp=2002T_p = 200 ms; set the utilisation equal to 0.5 and solve for WW.

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Source: official GATE 2021 Set 1 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key

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