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# homework1 - Dept. of Computer Science HKUST CSIT 561...

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Dept. of Computer Science HKUST CSIT 561 Computer Networks: An Internet Perspective Fall Semester 2009 Homework 1 Due Date: Sunday 27 Sep. 2009 11:59 pm Submission Instructions: Please submit your solutions electronically through the LMES website http://lmes2.ust.hk . Please go the CSIT 561 website and look under Assignments. You will see Homework 1 listed there. Please submit your work either in PDF format (preferred), or MS Word format. Problems: 1. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m metres, and suppose the propagation speed along the link is s metres/ sec. Host A is to send a packet of size L bits to host B. a) Express the propagation delay, d prop , in terms of m and s . d prop = length of physical link / propagation speed in medium = m/s sec b) Determine the transmission time of the packet, d trans , in terms of L and R . d tran = packet length / link bandwidth = L/R sec c) Ignoring processing and queuing delays, obtain an expression for the end-to-end delay. End-to-end delay = d prop + d tran = m/s + L/R sec d) Suppose host A begins to transmit the packet at time t = 0 . At time t = d trans , where is the last bit of the packet? The last bit of the packet is just leaving the host A. CSIT 561 (Fall 2009) Homework 1 1

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Dept. of Computer Science HKUST e) Suppose d prop is greater than d trans . At time t = d trans , where is the first bit of the packet? The first bit of the packet leaves the host A and does not reach the host B. It is on the physical link. f) Suppose d prop is less than d trans . At time t = d trans , where is the first bit of the packet? The first bit of the packet arrivals at host B. g) Suppose s = 2.5 × 10 8 , L = 100 bits, and R = 28 Kbps. Find the distance m so that d prop equals d trans . d prop = d tran => m/s = L/R => m = L/R*s => m = 100bit / 28Kbps * 2.5*108 => m = 892.86 km h) Use a program like Microsoft Excel to explore the interplay of transmission and propagation delay further. Consider three different packet sizes L = 100 bits, 1000 bits and 1,000,000 bits. Further consider three different link bandwidths R = 56 Kbps, 1.5 Mbps and 1.5 Gbps. Also, consider three different distances m = 2.5 metres, 25 Km, and 2500 Km. For each of these 27 combinations, compute d prop , d trans and the end-to-end delay. Explain your observations in a few sentences. L R m d prop d tran end-to-end 100 56000 2.5 1.00E-08 0.001786 0.00178572 100 56000 25000 0.0001 0.001786 0.00188571 100 56000 2500000 0.01 0.001786 0.01178571 100 150000 0 2.5 1.00E-08 6.67E-05 6.67E-05 100 150000 25000 0.0001 6.67E-05 0.00016667 CSIT 561 (Fall 2009) Homework 1 2
Dept. of Computer Science HKUST 0 100 150000 0 2500000 0.01 6.67E-05 0.01006667 100 1.5E+09 2.5 1.00E-08 6.67E-08 7.67E-08 100 1.5E+09 25000 0.0001 6.67E-08 0.00010007 100 1.5E+09 2500000 0.01 6.67E-08 0.01000007 1000 56000 2.5 1.00E-08 0.017857 0.01785715 1000 56000 25000 0.0001 0.017857 0.01795714 1000 56000 2500000 0.01 0.017857 0.02785714 1000 150000 0 2.5 1.00E-08 0.000667 0.00066668 1000 150000 0 25000 0.0001 0.000667 0.00076667 1000 150000 0 2500000 0.01 0.000667 0.01066667 1000 1.5E+09 2.5 1.00E-08 6.67E-07 6.77E-07 1000 1.5E+09 25000 0.0001 6.67E-07 0.00010067 1000 1.5E+09 2500000 0.01 6.67E-07 0.01000067 100000 0 56000 2.5 1.00E-08 17.85714 17.8571429 100000 0 56000 25000 0.0001 17.85714 17.8572429 100000 0 56000 2500000 0.01 17.85714 17.8671429 100000 0 150000 0 2.5 1.00E-08 0.666667 0.66666668 100000 0 150000 0 25000 0.0001 0.666667 0.66676667 100000 0 150000 0 2500000 0.01 0.666667 0.67666667 100000 0 1.5E+09 2.5 1.00E-08 0.000667 0.00066668 100000 0 1.5E+09 25000 0.0001 0.000667 0.00076667 100000 0 1.5E+09 2500000 0.01 0.000667 0.01066667

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## This note was uploaded on 11/21/2009 for the course CSE csit561 taught by Professor Zq during the Spring '09 term at HKUST.

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homework1 - Dept. of Computer Science HKUST CSIT 561...

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