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Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 4  packets each. Data packets (sent only from A to B) are all 1500 bytes long and the transmission time for such a packet is 60 μsμs. Acknowledgment packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 170 μs. What is the maximum achievable throughput in this communication?

  1. 3.75×10^6 Bps

  2. 7.5×10^6 Bps

  3. 10×10^6 Bps

  4. 12.75×10^6Bps

asked in Computer Networks by (125 points) | 16 views

1 Answer

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First of calculate the Maximum packets that we can send : 1+2*a

                                                                                                                 = 1 + 2*170/60

                                                                                                                 =  1+ 5.6

                                                                                                                  = 6.6   = 6

It means  if the window size is 6 then we get 100% throughput

but here we only send 4 packets

therefore efficiency = 4/6 =  0.67

Tt = length of the packet / bandwidth

Bandwidth = 1500 / 60 x 10 -6

                       = 25000000 bytes per sec

 If efficiency is not equal to 1 then it doesn't achieve the 100% then Throughput should decrease but how much ??

Throughput = $\eta$ *Bandwidth

                         = 0.67 * 25000000

                          = 16750000 Bytes per sec

 

answered by Loyal (5.9k points)
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