GATE CS 2015 Set 1 — Question 36
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Computer Networks → Data Link Layer → Efficiency Numericals
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Question
Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgement and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is ___________.
Correct answer
160 to 160
Solution
For Stop-and-Wait protocol, the efficiency (utilization) is given by:where is the transmission time and is the propagation delay.Given:
Bandwidth kbps bps.
Propagation delay ms s.
Target utilization .Substituting into the efficiency formula:Let be the frame size in bits. Then .Converting to bytes:Note on the Official Answer (160):
The official answer key provided in the document is 160. This result is obtained if the "20 milliseconds propagation delay" is interpreted as the total round-trip delay (or if the term in the denominator is taken as 20 ms). If ms, then ms. Following the same logic (), we would get ms, leading to . Given the answer key, this interpretation is likely intended.
Bandwidth kbps bps.
Propagation delay ms s.
Target utilization .Substituting into the efficiency formula:Let be the frame size in bits. Then .Converting to bytes:Note on the Official Answer (160):
The official answer key provided in the document is 160. This result is obtained if the "20 milliseconds propagation delay" is interpreted as the total round-trip delay (or if the term in the denominator is taken as 20 ms). If ms, then ms. Following the same logic (), we would get ms, leading to . Given the answer key, this interpretation is likely intended.
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