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ECS 152A/EEC 173A –Winter 2017 Midterm Solutions 1.Consider a packet of length Lwhich 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 diand Ridenote the length and transmission rate of link i, for i= 1, 2, 3. Let sdenote the propagation speed on each link. Each packet switch idelays a packet by dproc,i. a) Draw a diagram showing the above network configuration. b) What is the total end-to-end delay Tfor the packet (in terms of di, Ri, s, dproc,i, and L)? c) Suppose L= 20,000 bits; s = 2 x 108meters/sec; dproc,i= 1 msec; R1= 1 Mbps, R2= 0.5 Mbps; R3= 0.5 Mbps; d1= 100 km, d2= 200 km, d3= 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 R1= 0.5 Mbps, R2= 1 Mbps; R3= 2 Mbps, with everything else being the same. Answer: (Sabid) a) b) The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R2to transmit the packet onto the second link; the packet propagates over the second link in d2/s2. Similarly, we can find the delay caused by the second switch and the third link: L/R3, dproc, and d3/s3. Adding these eight delays gives dend-end= L/R1+ d1/s1+ dproc,1+ L/R2+ d2/s2+ dproc,2+ L/R3+ d3/s3c) dend-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 PS1 dend-end = (L/2)*2/R1+ d1/s1 + dproc,1 + (L/2)/R2 + d2/s2 + dproc,2+ (L/2)/R3 + d3/s3= 20 ms + 0.5 ms + 1 ms + 20 ms + 1 ms + 1 ms + 20 ms + 1.5 ms