Go-back n uses concept of pipelining, so efficiency of go-back n is given by

Efficiency = N/ (1 + 2a)

Here N = sender window size. a = Tp /Tt where Tp = propagation delay and Tt = transmission time.

Now to get 100% efficiency, N = 1+2a.

My doubt is can we have window size greater than 1+2a ? If we have window size greater than 1 + 2a then efficiency exceeds 100%. Is this possible? According to the formula it doesn't seem to be possible but I think we can increase the window size beyond 1 + 2a

  • Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide and accept your own answer. – Ron Maupin Feb 19 '18 at 20:40

You could indeed increase the window size beyond 1+2a, but it would indeed not increase the efficiency beyond 1 (or 100% if you like). I guess the correct formula would be

Efficiency = min[1 , N/(1 + 2a)]

If the window size exceeds 1 + 2a, and assuming the network delivers all acks, the sender would never be able to fill the window.

After sending 1+2a packets, the ack for the first packet will be received. So the number of outstanding packets will stay at 1+2a, even if the window size would allow for more.

  • Ahh, yes. The window would shift after receiving the ack. So number of packets inside the window could never exceed 1+2a . What if acks are delayed ? Then we will be able to fill the window with number of packets greater than (1+2a). – Zephyr Dec 21 '17 at 4:27

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