Testbook Test Series: Computer Networks - Stop And Wait

Question

$Que-$ A sender uses a Stop-and-Wait protocol for transmission of $8000 \ K-bits$ size frames on a $1Gbps$ satellite channel with a propagation delay of $400 \ ms$. What will be the link utilization (%) if a probability of single frame error is $0.001?$
 

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$\text{Note – Here Frame size is 8000 K- bits i.e 8}*10^6 \ bits$

Comments

I'm getting 0.9891% 😐

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You are getting answer in the range- $ .98 \ to \ 1$
Can you please comment/answer the explanation?

Answer 1

I took an example.

Prob = 0.001 = 0.1% which means 1 out of 1000 frames result in error.

I wish to send 1000 frames. So, total time to transmit it will be 1000*$T_t$ which is my useful time.

To send 1000 frames without any error, I actually need to transmit 1001 frames.

Thus, total time required will be 1001*$T_t$ + 1001*2*$T_p$

Thus, link utilization = $\frac{useful\:time}{total\:time}$=$\frac{1000*T_t}{1001*T_t + 1001*2*T_p}$

Put $T_t$ = 8ms and $T_p$ = 400ms

Link utilization = 0.009891 = 0.9891%