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GATE | GATE-CS-2015 (Set 2) | Question 18

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A link has a transmission speed of 106 bits/sec. It uses data packets of size 1000 bytes each. Assume that the acknowledgment has negligible transmission delay, and that its propagation delay is the same as the data propagation delay. Also assume that the processing delays at nodes are negligible. The efficiency of the stop-and-wait protocol in this setup is exactly 25%. The value of the one-way propagation delay (in milliseconds) is ___________.
(A) 4
(B) 8
(C) 12
(D) 16

Answer: (C)


In stop and wait, protocol next packet is sent only when 
acknowledgement of previous packet is received.  This 
causes poor link utilization. 

Transmission speed = 106

Time to send a packet = (1000 * 8) bits / 106
                      = 8 milliseconds
Since link utilization or efficiency is 25%, total time 
taken for 1 packet is 8 * 100/25 = 32 milliseconds.

Total time is twice the one way propagation delay plus 
transmission delay. Propagation delay has to be considered 
for packet and ack both.  
Transmission delay is considered only for packet as the 
question says that trans. time for ack is negligible.

Let propagation delay be x.
2x + 8 = 32.
x = 12.

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Last Updated : 09 Sep, 2021
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