Solution is 2 or 3?

Question

Should the solution for the above question be 2 or 3?

Answer 1

$\eta = \frac{N}{1+2a}$   , where $N =window$ $size$ , $a = \frac{T_{p}}{T_{x}}$ , $\eta = efficiency$

and also , $\eta = \frac{Throughput}{Bandwidth}$

so ,  $\frac{Throughput}{Bandwidth}$ = $\frac{N}{1+2a}$

$\frac{1 }{5} = \frac{N}{1+2*5}$

$N = \frac{11}{5} = 2.2 \approx 3$

Comments

@raja11sep

If we see in terms of bits required for sequence number field , then Here we will get

$2^{n} = 2.2$ , so $n$ will be 2 bits,

so sequence number = $2^2$ = 4 

moreover as here considering general sliding window protocol  $W_{s}$ must be 4.

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@raja11sep

by that i meant the packet will not be able to reside in it as we cannot send 2.2 packets.

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