wq17-mid-sol - ECS 152A\/EEC 173A Winter 2017 Midterm Solutions 1 Consider a packet of length L which begins at end system A and travels over a path

wq17-mid-sol - ECS 152A/EEC 173A Winter 2017 Midterm...

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ECS 152A/EEC 173A –Winter 2017 Midterm Solutions 1. Consider a packet of length L which begins at end system A and travels over a path consisting of two packet switches (PS1 and PS2), i.e., three links, to a destination end system B. Let d i and R i denote the length and transmission rate of link i , for i = 1, 2, 3. Let s denote the propagation speed on each link. Each packet switch i delays a packet by d proc,i . a) Draw a diagram showing the above network configuration. b) What is the total end-to-end delay T for the packet (in terms of d i , R i , s, d proc,i , and L )? c) Suppose L = 20,000 bits; s = 2 x 10 8 meters/sec; d proc,i = 1 msec; R 1 = 1 Mbps, R 2 = 0.5 Mbps; R 3 = 0.5 Mbps; d 1 = 100 km, d 2 = 200 km, d 3 = 300 km. For these values, what is the end-to-end delay T ? d) Now suppose the data is sent in two packets, each of which is of length L/2. Repeat part (b) to determine what will now be the end-to-end delay T . e) Extra credit: Repeat part d) where R 1 = 0.5 Mbps, R 2 = 1 Mbps; R 3 = 2 Mbps, with everything else being the same. Answer: (Sabid) a) b) The first end system requires L/R 1 to transmit the packet onto the first link; the packet propagates over the first link in d 1 /s 1 ; the packet switch adds a processing delay of d proc ; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R 2 to transmit the packet onto the second link; the packet propagates over the second link in d 2 /s 2 . Similarly, we can find the delay caused by the second switch and the third link: L/R 3 , d proc , and d 3 /s 3 . Adding these eight delays gives d end-end = L/R 1 + d 1 /s 1 + d proc,1 + L/R 2 + d 2 /s 2 + d proc,2 + L/R 3 + d 3 /s 3 c) d end-end = 20000/10^6 + 100*10^3/2*10^8 + 10^-3 +20000/0.5*10^6 + 200*10^3/2*10^8 + 10^-3 + 20000/0.5*10^6 + 300*10^3/2*10^8 = (0.02 + 0.0005 + 0.001 + 0.04 + 0.001 + 0.001 + 0.04 + 0.0015) sec = 105 ms d) Packet-2 can be processed simultaneously when packet-1 is being transmitted at PS 1 d end-end = (L/2)*2/R 1 + d 1 /s 1 + d proc,1 + (L/2)/R 2 + d 2 /s 2 + d proc,2 + (L/2)/R 3 + d 3 /s 3 = 20 ms + 0.5 ms + 1 ms + 20 ms + 1 ms + 1 ms + 20 ms + 1.5 ms
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