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 $0\mathrm{.}000012.$ The transmission rate is $2$Mb$/$s on the link $.$ The propagation delay in both directions is $88$us $($micro$-$seconds$)$ and the processing time is the same at the sender and receiver and is $400$us $($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