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Unformatted text preview: Questions and Answers, Tutorial No.1 COMP361, Spring 2006   Problem 1: Please check the Slide 128, show how to obtain the prob. = 0.0004. Answer: Suppose the status of a node (Active or not) is a random variable values: 1 or 0). P(X=1) = 0.1 and P(X=0)=0.9 For these 35 nodes, they are IID (Independent Identical Distribut Thus, suppose the number of simultaneously active nodes is a rand (0<=Y<=35). Y is binomial distributed with parameters 35 and 0.1. The probability can then be easily computed. l Problem 2: We are sending a 30 Mbit MP3 file from a source host to a destination the path between source and destination have a transmission rate of 1 the propagation speed is 2 * 10 8 meters/sec, and the distance between destination is 10,000 km. (20 points) a. Suppose there is only one link between source and destination. message switching is used, with the message consisting of the entir the transmission delay? (Hint: 3.05 seconds, 50 milliseconds, 3 se value?) (4 points) Answer: d tran = 30Mbits/10Mbps = 3 second b. Refer to the question above, calculate the endtoend delay (tr propagation delay). (Hint: 3 seconds, 3.05 seconds, 6 seconds, or points) Answer: d = d tran + d prop in which d prop = 10,000km/2 * 10 8 meters/sec = 0.05 seconds d = d tran + d prop = 3 + 0.05 = 3.05 seconds c. Refer to the above question, how many bits will the source have first bit arrives at the destination? (Hint: 50,000 bits, 30,000,00 some other value) (4 points) Answer: This is when the 1 st bit arrives at the destination, i.e., propagation del 10Mbps * d prop = 10Mps * 0.05 = 50,000 bits d. Now assume there are two links between source and destinat connecting the two links. Each link is 5,000 km long. Again suppose as one message (message switching). Assume that there is no congestas one message (message switching)....
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This note was uploaded on 01/24/2010 for the course WSE ECE taught by Professor Dognary during the Spring '10 term at ECE.
 Spring '10
 Dognary

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