Please answer all 4 parts to this question, and please give explanations. 5. The

Question

Please answer all 4 parts to this question, and please give explanations.

5. The following variation of the Stop and Wait protocol operates between nodes A and B which are c by a standard point-to-point link in which packets are received in order. Node A is sending data packets to node B which replies with ACK packets. Before transmission, a packet is divided to two sub-packets of equal length. The two sub-packets are transmitted one after the other and the receiver sends an ACK only if it receives both of them The distance between the two nodes is 100Km The propagation speed on the link is 2-10 m/sec The data packet length is 2000 bytes The ACK length is 100 bytes The transmission rate is 1Mbit/sec (10 bit/sec) A subpacket transmission fails with probability q. An ACK never fails. There is no processing or fragmentation delay. - - A. What is the propagation delay between the two nodes? B. What is the ideal timeout? C. What is the average time until 1 (full) packets are received by node B successfully? D. Assume that the probability that a (full) packet transmission fails is 0.1. What should be q in order for the described protocol to outperform the traditional Stop and Wait protocol?

Explanation / Answer

A)

Propagation delay = distance between two nodes/ propagation speed

                                                = 100km/2*(10^8)m/s

                                                = 100*1000m/(2*(10^8))m/s

                                                =1/(2*(10^3))s

                                                =0.5 ms

B)

Timeout= 2* propagation delay

                = 2*0.5ms

                =1ms

C)

For one sub packet:

Data packet Transimission delay = packet length/bandwidth

                                                =2000B/10^6bps

                                                =2000*8/10^6

                                                =16ms

Propagation delay=0.5 ms

Ack packet Transimission delay = packet length/bandwidth

                                                                = 100*8/10^6

                                                                =0.8ms

Total time For one sub packet = Data packet Transimission delay+ Ack packet Transimission delay+( Propagation delay)*2

                =16+0.8+1 = 17.8ms

Total time For 11 full packets = 11*(2*17.8) =195.8*2 = 391.6ms

D)

Probability of 1 packet transmission fails is 0.1

0.1= q+q

0.1=2q

q=0.05

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