Question 9137 – Sliding-Window-Protocol

Frames of 1000 bits are sent over a 10⁶ bps duplex link between two hosts. The propagation time is 25ms. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

Suppose that the sliding window protocol is used with the sender window size of 2^l, where l is the number of bits identified in the earlier part and acknowledgements are always piggy backed. After sending 2^l frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)

Correct Answer: C

From above diagram we can say that Total RTT will be
1ms (transmitting time for first frame)
+ 25ms (propagation delay from S to R)
+ 1ms (transmitting time for piggybacked ACK)
+ 25ms (propagation delay from R to S)
= 52 ms
Also total time for sender needed to transmit 32 frames (2² = 2⁵ = 32) is 32×1ms = 32ms
So sender has to wait for,
(52 – 32)ms
= 20ms