hw1_sols - d 2/s 2 Similarly we can find the delay caused...

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Problem 6 Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires sec=7msec. The time required to transmit the packet is sec=sec. Propagation delay = 10 msec. The delay until decoding is 7msec +sec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec. Problem 7 a) 20 users can be supported. b) . c) . d) . We use the central limit theorem to approximate this probability. Let be independent random variables such that .
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“21 or more users” when is a standard normal r.v. Thus “21 or more users”. Problem 8 a) 10,000 b) Problem 9 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
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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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This note was uploaded on 12/11/2009 for the course TELECOM 2310 taught by Professor Rezgui during the Fall '09 term at Pittsburgh.

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hw1_sols - d 2/s 2 Similarly we can find the delay caused...

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