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Unformatted text preview: 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 five delays gives d end-end = L/R 1 + L/R 2 + L/R 3 + d 1 /s 1 + d 2 /s 2 + d 3 /s 3 + d proc + d proc To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec. Problem 13 It takes seconds to transmit the packets. Thus, the buffer is empty when a batch of packets arrive. The first of the packets has no queuing delay. The 2nd packet has a queuing delay of seconds. The th packet has a delay of seconds. The average delay is . Problem 26 a) 80,000,000 bits b) 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits. c) .25 meters Problem 28 Recall geostationary satellite is 36,000 kilometers away from earth surface. a) 150 msec b) 1,500,000 bits c) 600,000,000 bits...
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- Fall '09