Understanding acknowledgement mechanism in sliding window protocol

Networking , , , , ,


I came across following problem:

Node A sends packets to B through router R. The link A-R is instantaneous , but the R->B link transmits only 1 packet each second , one at a time . Assume A sends to B using sliding window protocol with window size =4 , For time t=2 , state what packets arrive at R and what are the packets present in the queue at R ?

The solution given was:

Since link A to B is instantaneous, at time 0, all frames 0,1,2,3,4… will be at router R. The window is of size 4. So, lets represent it as [0,1,2,3],4,…. Link R to B can transmit only 1 packet each second. So, at time 1, R will send frame 0 to B. At time 1, B will send ACK for packet 0 to R. So window at R will slide making it: 0,[1,2,3,4],…

The doubt

Go Back N uses cumulative acknowledgements, so what if at both time 0 and 1, router sends packet 0 and 1 to receiver? The router window at time 2, will still contain frames [0,1,2,3] as frame zero is not yet acknowledged. Is this wrong? Or I am wrong and sliding window protocol means Selective Repeat and it sends ACK for each frame?



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