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Correct Option (D)

in GBN Protocol Max No of packet that can be send at a particular time is equal to No of Sequence no possible which is Windows size + 1.

so, here max no of packet that can be send = 127 +1 = 128 

Now, Given Packet size is 512 B = 0.5 KB = 2 Kb  

So, the max data that can be send is 128 * 2 Kb = 256 Kb

therefore efficiency is already 100%, thus maximum throughput is equal to bandwidth * 1 = bandwidth which is 64 Kbps.   


Reference :https://gateoverflow.in/186305/ace-test-series

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