HW3_solution_2009

HW3_solution_2009 - INFSCI 1004 TELCOM 2000 Homework 3...

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INFSCI 1004 & TELCOM 2000 Homework 3 Problem 12.3a a) Circuit switching: Latency = SetupTime + TransmitTime + PropagationTime + QueuingDelay = (S) + (MessageLength/Bandwidth) + (N*D) + 0 = S + L/B + N*D= 0.2 + 3200/9600 + 4*0.001 =537ms b) Virtual circuit packet switching Latency = SetupTime + TransmitTime + PropagationTime + QueuingDelay = (S) + (number- ofpackets)*( overhead)/Bandwidth + (MessageLength/Bandwidth) + (N*D) + 0 Number of packets: 3200/1024 = 3.125 => 4 packets Latency = S + 4*H/B + L/B + N*D =0.2 +4*(16)/9600+ 3200/9600+ 4*0.001=544 ms c) Datagram packet switching Latency = TransmitTime + PropagationTime + QueuingDelay = (number-ofpackets)*( overhead)/Bandwidth + (MessageLength/Bandwidth) + (N*D) + 0 =4*H/B + L/B +N*D =4*(16)/9600+ 3200/9600+4*0.001=344 ms Problem 12.4 One telephone call requires 4 kHz of bandwidth. The exact number is 3400 Hz (see Chapter 2.1) but 4 kHz number is usually used in calculations. We have a full-duplex trunk in the problem; it means that 2 * 4 kHz = 8 kHz of bandwidth is used simultaneously. Half duplex means that data

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HW3_solution_2009 - INFSCI 1004 TELCOM 2000 Homework 3...

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