GATE CSE 2015 Set 1 | Question: 53

Question

Suppose that the stop-and-wait protocol is used on a link with a bit rate of $64$ $\text{kilobits}$ per second and $20$ $\text{milliseconds}$ 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 $50$ $\text{%}$ is_________________.

Comments

That is not right. @Raj

Efficiency = etta(n)

Link utilization = etta * Bandwidth

 

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Was the guy who made the question was high on weed, or is there some concept behind the 160 in the official answer. Anyone know?

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haha

Answer 1

 % Link Utilization = T.T/(T.T + 2*P.T)*100

For 50% link utilization: 50 = T.T/(T.T + 2*P.T)*100 

Therefore the relation is T.T = 2*P.T

NOW T.T=FRAME SIZE/BANDWIDTH

P.T = 20 ms(given)

So frame size = 2* 20 *10^(-3) 64 * 10^3

frame size=2560 bits

frame size = 2560/8 bytes

frame size = 320 bytes(Ans)

 

 

 

 

Answer 2

Given B = 64 kbps
Tp = 20 ms
η ≥ 50%
For η≥50%⇒L≥BR
⇒ L=64×103×2×20×10-3
=2560bits
=320bytes