HW4-solution

HW4-solution - Probability and Stochastic Processes: A...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 2.6.5 2.7.2 2.7.4 2.7.6 2.7.7 2.8.5 2.8.9 2.8.11 2.9.7 and 2.10.5 Problem 2.6.5 Solution (a) The source continues to transmit packets until one is received correctly. Hence, the total number of times that a packet is transmitted is X = x if the first x- 1 trans- missions were in error. Therefore the PMF of X is P X ( x ) = q x- 1 (1- q ) x = 1 , 2 , . . . otherwise (1) (b) The time required to send a packet is a millisecond and the time required to send an acknowledgment back to the source takes another millisecond. Thus, if X transmis- sions of a packet are needed to send the packet correctly, then the packet is correctly received after T = 2 X- 1 milliseconds. Therefore, for an odd integer t > 0, T = t iff X = ( t + 1) / 2. Thus, P T ( t ) = P X (( t + 1) / 2) = q ( t- 1) / 2 (1- q ) t = 1 , 3 , 5 , . . . otherwise (2) Problem 2.7.2 Solution Whether a lottery ticket is a winner is a Bernoulli trial with a success probability of 0 . 001. If we buy one every day for 50 years for a total of 50 365 = 18250 tickets, then the number of winning tickets T is a binomial random variable with mean E [ T ] = 18250(0 . 001) = 18 . 25 (1) Since each winning ticket grosses $1000, the revenue we collect over 50 years is R = 1000 T dollars. The expected revenue is E [ R ] = 1000 E [ T ] = 18250 (2) But buying a lottery ticket everyday for 50 years, at $2.00 a pop isnt cheap and will cost us a total of 18250 2 = $36500. Our net profit is then Q = R- 36500 and the result of our loyal 50 year patronage of the lottery system, is disappointing expected loss of E [ Q ] = E [ R ]- 36500 =- 18250 (3) 1 Problem 2.7.4 Solution Given the distributions of D , the waiting time in days and the resulting cost, C , we can answer the following questions. (a) The expected waiting time is simply the expected value of D . E [ D ] = 4 X d =1 d P D ( d ) = 1(0 . 2) + 2(0 . 4) + 3(0 . 3) + 4(0 . 1) = 2 . 3 (1) (b) The expected deviation from the waiting time is E [ D- D ] = E [ D ]- E [ d ] = D- D = 0 (2) (c) C can be expressed as a function of D in the following manner. C ( D ) = 90 D = 1 70 D = 2 40 D = 3 40 D = 4 (3) (d) The expected service charge is E [ C ] = 90(0 . 2) + 70(0 . 4) + 40(0 . 3) + 40(0 . 1) = 62 dollars (4) Problem 2.7.6 Solution Since our phone use is a geometric random variable M with mean value 1 /p , P M ( m ) = (1- p ) m- 1 p m = 1 , 2 , . . ....
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This note was uploaded on 08/25/2008 for the course EEE 350 taught by Professor Duman during the Spring '08 term at ASU.

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HW4-solution - Probability and Stochastic Processes: A...

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