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A Go - Back - N scheme has a frame with a 6 Byte header and is using the optimal frame size with a probability

A Go-Back-N scheme has a frame with a 6 Byte header and is using the optimal frame size with a probability of bit error of 12 in a million, or 0.000012. The transmission rate is 2Mb/s on the link . The propagation delay in both directions is 88us (micro-seconds) and the processing time is the same at the sender and receiver and is 400us (micro-seconds) for each.
The ACK is 6 Bytes and is being sent back on its own. The Transmission time of this must be considered. However the assumption is that there is no error in the ACK and the optimal Timeout is chosen.
What is the percentage efficiency of this system in percentage? Round your answer.
Based on the following information, can you answer the above question?
assume optimum choice of Timeout = TA +2*(Prop + Proc)
assume Window is large enough that Sender can transmit continuously if there
are no transmission errors. Can show average packet delay is
D =[ TF + r * Timeout ]/(1 r )
where r = p +(1 p )* q as before
average packet throughput =(1 r )/[ TF + r * Timeout ]
efficiency =[(1 r ) TF ]/[ TF + r * Timeout ]
note that as r ->0, efficiency of Go-back-N ->1(or 100%): this shows that Goback-N is capable of continuously delivering packets in the absence of errors
under the same conditions we showed that the efficiency of Stop-and-wait
ARQ is (1 r ) TF
/ TS =(1 r ) TF
/[ TF + Timeout ]
as r ->0 Stop-and-wait ARQ efficiency -> TF
/[ TF + Timeout ]
this shows the built-in inefficiency of the Stop-and-wait approach

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