Computer Networks Doubt

Question

What will be the formula for the efficiency of a link in stop and wait protocol when acknowledgements are piggybacked???

My guess is  $2*T/(2*T+2*P)$

where T=Transmission delay P=propagation delay

And frames sent by sender and reciever are of equal size

Comments

yeah i think so

Answer 1

Normal S&W
efficiency $=\frac{useful\_time}{total\_time} = \frac{T}{T+2P}$ (2P caz P time to send packet and P time to receive ACK)

with piggbacked ACK is sent along with the next data packet
so it saves the P time to receive ACK caz it overlaps with the P time required to receive data packet from the receiver

so here efficiency will be $\eta = \frac{useful\_time}{total\_time} = \frac{T}{T+P}$

Comments

Thanks.I had another doubt.

The optimal window size in sliding window protocol is 1+2a where a=Propogation delay/transmission delay

But in this question https://gateoverflow.in/1820/gate2006-44 the answer which is chosen as the best answer uses a slightly different approach resulting in the answer to be 40 instead of 41.Can you please help me out?