ENC3250 Solutions pg 9

ENC3250 Solutions pg 9 - 0.108 T = 0.428 + 0.108 + 0.108 +...

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-9- C 2 = Message Delivery Time C 1 = S = 0.2 C 2 = Propagation Delay + Transmission Time = N × D + L/B = 4 × 0.001 + 3200/9600 = 0.337 T = 0.2 + 0.337 = 0.537 sec Datagram Packet Switching T = D 1 + D 2 + D 3 + D 4 where D 1 = Time to Transmit and Deliver all packets through first hop D 2 = Time to Deliver last packet across second hop D 3 = Time to Deliver last packet across third hop D 4 = Time to Deliver last packet across forth hop There are P – H = 1024 – 16 = 1008 data bits per packet. A message of 3200 bits requires four packets (3200 bits/1008 bits/packet = 3.17 packets which we round up to 4 packets). D 1 = 4 × t + p where t = transmission time for one packet p = propagation delay for one hop D 1 = 4 × (P/B) + D = 4 × (1024/9600) + 0.001 = 0.428 D 2 = D 3 = D 4 = t + p = (P/B) + D = (1024/9600) + 0.001 =
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Unformatted text preview: 0.108 T = 0.428 + 0.108 + 0.108 + 0.108 = 0.752 sec Virtual Circuit Packet Switching T = V 1 + V 2 where V 1 = Call Setup Time V 2 = Datagram Packet Switching Time T = S + 0.752 = 0.2 + 0.752 = 0.952 sec b. Circuit Switching vs. Diagram Packet Switching T c = End-to-End Delay, Circuit Switching T c = S + N D + L/B T d = End-to-End Delay, Datagram Packet Switching N p = Number of packets = L P H T d = D 1 + (N 1)D 2 D 1 = Time to Transmit and Deliver all packets through first hop D 2 = Time to Deliver last packet through a hop D 1 = N p (P/B) + D D 2 = P/B + D T = (N p + N 1)(P/B) + N x D...
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