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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 ?

In this case , what is the transmission time and propogation delay ?
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the R->B link transmits only 1 packet each second it means transmission time =1 sec .As we know in GBN the maximum window size can be given as 1+2*a where a=PT/TT and here 1+2*a=4 by using this formula propagation time is 1.5 sec .why this approach is wrong???
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i assumed pd =1 sec md trans delay negligible for link R-B

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Why can't we take transmission delay as 1 second here because it says that one packet is transmitted after every 1 sec ?
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ya if u assuming it as transmitting tym than p.d is negligible so.....soln  also same but here we need to consider  1 sec to transit the ack at receiver side.......becz nothing is clearly mentioned in question
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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 after at time 2, will still contain frames [0,1,2,3] as frame zero is not yet acknowledged.

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+1 vote
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