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GATE CS 2005 Q25 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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The maximum window size for data transmission using the selective reject protocol with n-bitn\text{-bit} frame sequence numbers is:

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Answer: B

In the Selective Reject (Selective Repeat) sliding window protocol, to avoid any ambiguity between new frames and retransmitted frames, the sum of the sender's window size (WTW_T) and the receiver's window size (WRW_R) must not exceed the total number of available sequence numbers (NN).

For Selective Repeat, the sender and receiver window sizes are typically equal:

WT=WR=WW_T = W_R = W

Thus, the condition is:

WT+WR≤N  ⟹  2W≤N  ⟹  W≤N2W_T + W_R \le N \implies 2W \le N \implies W \le \frac{N}{2}

Given that the sequence numbers are represented using nn bits, the total number of available sequence numbers is:

N=2nN = 2^n

Substituting N=2nN = 2^n into the inequality:

W≤2n2=2n−1W \le \frac{2^n}{2} = 2^{n-1}

Therefore, the maximum window size is 2n−12^{n-1}, which corresponds to Option B.

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