In a sliding window protocol where 4 bits are used for sequence numbers, what is the maximum sender window size? (Assume Go-Back-N unless otherwise specified.)

Introduction / Context:
Sliding window protocols use a finite sequence number space. The chosen ARQ variant—Go-Back-N (GBN) or Selective Repeat (SR)—determines the maximum permissible sender window to avoid ambiguity when sequence numbers wrap around.

Given Data / Assumptions:

• Number of sequence bits m = 4 → sequence space size = 2^m = 16.
• Assume Go-Back-N unless otherwise specified.
• Acknowledgments are reliable and eventually arrive.

Concept / Approach:
For Go-Back-N, the maximum sender window W_max is 2^m − 1. This ensures that when sequence numbers wrap, outstanding frames cannot be confused with newly sent frames. For Selective Repeat, to avoid overlap between outstanding and future frames, W_max is 2^(m−1). The problem statement does not specify SR, so the common default is GBN.

Step-by-Step Solution:
Compute sequence space: 2^4 = 16 distinct numbers.Apply GBN rule: W_max = 2^m − 1 = 16 − 1 = 15.Validate no ambiguity at wrap-around with this bound.

Verification / Alternative check:
If it were Selective Repeat, W_max would be 2^(4−1) = 8. Many textbooks explicitly state the default unspecified case as Go-Back-N, yielding 15.

Why Other Options Are Wrong:
16: Equals the entire sequence space; would cause ambiguity in GBN.

8: Correct only for Selective Repeat, which was not specified.

4: Too small relative to the allowable window.

It depends on RTT only: RTT influences throughput, not the protocol limit.

Common Pitfalls:
Mixing up GBN and SR window bounds; forgetting that sequence wrap-around safety drives these limits, not bandwidth-delay product.

Final Answer:
15