GATE | GATE-CS-2015 (Set 1) | Question 62

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 acknowledgment 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 _________.
(A) 160
(B) 320
(C) 640
(D) 220


Answer: (B)

Explanation:

Transmission or Link speed = 64 kb per sec
Propagation Delay = 20 milisec

Since stop and wait is used, a packet is sent only
when previous one is acknowledged.

Let x be size of packet, transmission time = x / 64 milisec

Since utilization is at least 50%, minimum possible total time
for one packet is twice of transmission delay, which means 
x/64 * 2 = x/32

x/32 > x/64 + 2*20
x/64 > 40
x > 2560 bits = 320 bytes

Answer in GATE keys says 160 bytes, but the answer keys seem incorrect. See question 36 here.

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