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Two systems are connected with the link and link is having a bandwidth of 20 Mbps. Assume the distance between them is 2400 km and the size of the packet is 2 KB  with propagation speed is the same as the speed of light. What is the minimum number of bits required for the window to achieve 100% utilization in selective repeat protocol?
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BW = 20 Mbps or 20 * $10^{6}$ bits per second

distance = 2400 km or 2400 * $10^{3}$ m

packet size = 2 KB or 2 * $10^{3}$ * 8 bits

speed of light = 3 * $10^{8}$ m/sec

Window size to achieve 100% Link utilization can be calculated through a simple formula:

$w \geq$ 1 + 2a

where a = propagation delay / transmission time

propagation delay =  distance / speed of light = (2400 * $10^{3}$) / (3 * $10^{8}$) = 8 ms

Transmission time = packet size / BW = (2 * $10^{3}$ * 8) / (20 * $10^{6}$) = 0.8 ms

$w \geq$  1 + 2 * (8/0.8) = 1 + 20 = 21.

Sequence numbers = 2 * W = 2 * 21 = 42

To represent 42 sequence numbers in binary, $log_{2}(42)$, 6 bits are needed.

So a minimum of 6 bits is needed for 100% utilization.
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First of all W is the window size, not the number of bits as you have taken, and second, your calculation is incorrect. The value of W would be 21.

 

Now, for selective repeat, we know the window size is:-

 

W = $2^{m-1}$

$21$ =  $2^{m-1}$

$log_{2}21$ = $m-1$

$5$ =  $m-1$

$m$ = $6$ $bits$
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