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

Why not 2?

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raja11sep

Here the size of frames to be send are already given, and we have calculated the window size for that only , which we got as $2.2$ , but can you really keep a window of size $2.2$?

if we keep it of size $2$ , then will the packet reside in it?

That’s why I approximated it to $3$.

anything wrong in this?

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if we keep it of size 2 , then will the packet reside in it?

Didn’t get this statement.

Window size means the maximum number of packets the sender can transmit without being acknowledged. 

Here, we are getting window size =2.2 means the maximum of 2.2 packets we can send, but sending a partial packet does not makes sense so we will transmit 2 packets.

please correct me if I'm wrong.

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@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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