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Consider a network connecting two systems located 4000 kilometers apart. The bandwidth of the network is 64 Mbps. The propagation speed of the media is 2/3 of the speed of light in vacuum. It is needed to design selective repeat sliding window protocol for this network. The average packet size is of 8 Kb. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be _________.
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Selective Repeat Sliding Window Protocol.

Bandwidth = 64 Mbps, Packet size = 8 Kb, Distance = 4000 kilometers, Speed = 2/3*speed of light, No of frames in sender window is n.
The network is to be used to its full capacity.
so TT = 2*PT

(n*Packet size) / Bandwidth = 2* Distance/ Speed
(n* 8 Kb) / 64 Mbps = 2*4000 kilometers / (2/3*Speed of light)

n = 320

SIZE OF SENDER WINDOW = 1 + 320 = 321
Size of sender window needs 321 different numbers to identify one window. Same for reciever . so we need total 321+321 different numbers.

No of bits required to represent 640 different numbers is log2(642) which is 10.

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your calculation is wrong.

check once again...

calculated window size= of 8 bits.

since it is selective repeat , so (1+8) = 9 bits
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Yes, The answer should be 10

Sender window size + receiver window size <= Available sequence numbers.

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