# HW7_Sol - ECEN 646 Homework 7 Solutions 1 C ¸inlar...

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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 λ t = 0 . 1(0 . 3 × 80+0 . 2 × 60) = 3 . 6. The rate of arrival of cars at the restaurant is λ c = 0 . 1(0 . 7 × 80+ 0 . 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 ] = λ t + λ 1 +2 λ 2 +3 λ 3 +4 λ 4 +5 λ 5 = 3 . 6+3 . 12+2 × 3 . 12+3 × 2 . 08+4 × 1 . 04+5 × 1 . 04 = 28 . 56 (b) E bracketleftbig α Z bracketrightbig = E bracketleftbig α N t + N 1 +2 N 2 +3 N 3 +4 N 4 +5 N 5 bracketrightbig = E bracketleftbig α N t bracketrightbig E bracketleftbig α N 1 bracketrightbig E bracketleftbig α 2 N 2 bracketrightbig E bracketleftbig α 3 N 3 bracketrightbig E bracketleftbig α 4 N 4 bracketrightbig E bracketleftbig α 4 N 4 bracketrightbig Each term involves the MGF of a Poisson process at time 1 hour. For example, E bracketleftbig α 3 N 3 bracketrightbig = summationdisplay k =0 e - λ 3 λ k 3 k ! α 3 k = e λ 3 α 3 - λ 3 1 This study resource was shared via CourseHero.com

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Thus, E bracketleftbig α Z bracketrightbig = e 3 . 6 α +3 . 12 α +3 . 12 α 2 +2 . 08 α 3 +1 . 04 α 4 +1 . 04 α 5 - 14 3 C ¸inlar Exercise (8.9), Chapter 4 The overall Poisson process N with rate λ is divided into a device failure process with rate , and a device doesn’t fail process with rate . We’re given that T = T K , and K is the number of the shock that causes the failure.
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