Suppose you are designing a sliding window protocol for a 10

Suppose you are designing a sliding window protocol for a 10 Mbps point-to-point link to the moon, which has a one-way latency of 2.0 seconds. Assuming that each frame carries 5KB of data, what is the minimum number of bits you need for the sequence number if RWS = SWS?

Solution

Answer :

...>Two steps to get it.

       1.Find the window size based on delay bandwidth product

        2.Find the max sequence number

Delay bandwidth= rtt * bw=2.0 * 2*10^6

size of sender window =2.0 * 2*10^6 /8*10^3

                                 =4000000/8000

                                 =500

              if RWS=SWS

              No of sequence number=500*2=1000bits

             To Represent 1000 you will need at least 10bits are required.