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Consider two stations communicates via 1 Mbps satellite link with propagation delay of 270 msec. The satellite serves merely to retransmit data received from one station to another, with negligible switch delay. If Ethernet frame size is 1024 bits with 3 bits sequence number, then the maximum possible data throughput is ____________

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Rtt is 2 x 270=540 ms. ( You can calculate yourself)

The RTT here in not of ACK but the time by which you, the sending station would get  the signal from your frame , if your frame got crashed right before me, the receiver station , that is the maximum time after which you can be sure that your frame is safely with me.

Transmission time is : 1ms ( You can clearly calculate)

Hence you could have transmitted 540 frames in this time.

But you have a 3 bit sequence number, meaning at most 7 frames, one of which is already the one sent.

So, basically you are sending 6 frames in the time you are supposed to send 540 ones .

Therefore throughput : 6/540 x 1 Mbps = 11.377 Kbps.
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Given Tp=270ms
Tt=1024bits/10^6 bps=1.024msec
Given maximum possible utilization, means that from transmission time  + propagation time + acknowledgement propagation time our link should be utilized, that is
we should keep sending frames for Tt+2tp time.
Totals frames that can be sent =Tt+2Tp
                                                =1+2a
                                                 =1+540/1.024
                                                 =1+540  (for simplicity)
                                                 =541
Means it can send 541 frames for maximum throughput but it is sending 7 frames only (because sequence number in gbn (best case) is 2^3-1 and 1 for window size)
therefore throughput is = data/time = 7*1024/541x10^-3=13.24 kbps

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