Homework 3 Solutions - CS421/S03 Chapter 10(7/e 1 Problem...

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CS421/S03 Solutions to Homework Three Chapter 10 (7/e) 1. Problem 10.1 Each telephone makes 0.5 calls/hour at 6 minutes each. Thus a telephone occupies a circuit for 3 minutes per hour. Twenty telephones can share a circuit (although this 100% utilization implies long queuing delays). Since 10% of the calls are long distance, it takes 200 telephones to occupy a long distance (4 kHz) channel full time. The interoffice trunk has 10 6 /(4 × 10 3 ) = 250 channels. With 200 telephones per channel, an end office can support 200 × 250 = 50,000 telephones. 2. Problem 10.4 a) Circuit Switching T = C 1 + C 2 where C 1 = Call Setup Time 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 require 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
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= 0.428 D 2 = D 3 = D
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Homework 3 Solutions - CS421/S03 Chapter 10(7/e 1 Problem...

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