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Consider sliding window protocol for a 10 MBps point to point link with propagation delay of 2 sec. If frame size is of 4kB then what wil be the minimum number of bits number required for the sequence number?

Which approach to take and why? Notice the difference in frame numbers, why does this arise and which one is correct. The answer is not affected in this question.

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Question is incomplete- I suppose we want to maximize the "utilization".

Now, if you blindly follow the formula you will always have confusion. If we imagine the actual scenario- its all very easy. For maximum utilization, the sender must send frames continuously until the ACK of first packet send comes back. This time for arrival of ACK is called RTT.

RTT = Propagation time for frame  (time for the first bit of frame to reach destination) + Time for frame to transmit (Time difference between first and last bit of frame to reach destination) + Propagation time for ACK + Transmission time for ACK.

= 2s + 4kB/10MBps + 2s + 0 (ACK being small)

= 2s + .0004 + 2s = 4000.4 ms.

Now, no, of bytes that could be sent during this time = 10MBps * 4000.4ms = 40004 KB = 10001 frames.

Now, for sending 10001 frames we need $\lceil \lg 10001 \rceil = 15$ bits for the sequence number field.

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