HW7_Sol - ECEN 646 Homework 7 Solutions 1 C inlar Exercise...

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Unformatted text preview: ECEN 646 Homework 7 Solutions 1 C inlar Exercise (8.6), Chapter 4 Each traffic flow process is Poisson described by the rate shown in the figure. 2 C inlar Exercise (8.8), Chapter 4 The rate of arrival of trucks at the restaurant is t = 0 . 1(0 . 3 80+0 . 2 60) = 3 . 6. The rate of arrival of cars at the restaurant is c = 0 . 1(0 . 7 80+ . 8 60) = 10 . 4. Therefore the rates of arrivals of cars with 1,2,3,4, and 5 passengers are 3.12, 3.12, 2.08, 1.04, and 1.04, respectively. (a) Then Z = 1 N t + 1 N 1 + 2 N 2 + 3 N 3 + 4 N 4 + 5 N 5 , where N t is the number of arrivals of trucks in one hour, and N i for i = 1 , 2 , 3 , 4 , 5 is the number of arrivals of cars with i passengers. So, E [ Z ] = t + 1 +2 2 +3 3 +4 4 +5 5 = 3 . 6+3 . 12+2 3 . 12+3 2 . 08+4 1 . 04+5 1 . 04 = 28 . 56 (b) E bracketleftbig Z bracketrightbig = E bracketleftbig N t + N 1 +2 N 2 +3 N 3 +4 N 4 +5 N 5 bracketrightbig = E bracketleftbig N t bracketrightbig E bracketleftbig N 1 bracketrightbig E bracketleftbig 2 N 2 bracketrightbig E bracketleftbig 3 N 3 bracketrightbig E bracketleftbig 4 N 4 bracketrightbig E bracketleftbig 4 N 4 bracketrightbig Each term involves the MGF of a Poisson process at time 1 hour. For example, E bracketleftbig 3 N 3 bracketrightbig = summationdisplay k =0 e- 3 k 3 k ! 3 k = e 3 3- 3 1 Thus, E bracketleftbig Z bracketrightbig = e 3 . 6 +3 . 12 +3 . 12 2 +2 . 08 3 +1 . 04 4 +1 . 04 5- 14 3 C inlar Exercise (8.9), Chapter 4 The overall Poisson process N with rate is divided into a device failure process with rate p , and a device doesnt fail process with rate q . Were given that T = T K , and K is the number of the shock that causes the failure....
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This note was uploaded on 01/19/2012 for the course ECEN 646 taught by Professor Serapsavari during the Fall '11 term at Texas A&M.

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HW7_Sol - ECEN 646 Homework 7 Solutions 1 C inlar Exercise...

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