Project ForgeSuppose that the sliding window protocol is used with the sender window size of 2^l (l=5 from the previous part) on the same 1000-bit-frame, 10⁶bps, 25ms-propagation-time link, 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.)
Explanation
The sender finishes transmitting all 32 frames at t=32ms (32×1ms). The ACK for frame 1 arrives back at the sender at t = Ttrans(frame1) + Tprop + Ttrans(piggybacked return frame) + Tprop = 1+25+1+25 = 52ms. The sender must wait from t=32ms until t=52ms to free a window slot: a wait of 52−32=20ms.
Source: GATE CS 2009 (2009)
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