HW8_Solutions - IE:3610 Stochastic Modeling(Fall 2015...

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IE:3610 Stochastic Modeling (Fall 2015) Homework 8 (due Thu, Oct 29) Solutions HW8 (Due Thu, OCT 29): 17.6-3; 17.6-10 (a,c,e); 17.6-12; 17.6-15. 17.6-3 Solution: M/M/1 model Proportion of time no one is waiting to be served = = P (number of customers in the system number of servers) = = P ( n s ) λ = 10 , μ = 15 = ) = 2 / 3 P 0 = 1 - = 1 / 3 , P 1 = (1 - ) = 1 / 3 2 / 3 = 2 / 9 P ( n s ) = 1 / 3 + 2 / 9 = 5 / 9 = 0 . 5556 17.6-10(a,c,e) Solution: (a) M/M/1 model with λ = 30 and μ = 40 L = λ μ - λ = 30 40 - 30 = 3 customers, W = 1 μ - λ = 1 40 - 30 = 0 . 1 hours, W q = λ μ ( μ - λ ) = 30 40(40 - 30) = 0 . 075 hours, L q = λ W q = 30 0 . 075 = 2 . 25 customers, P 0 = 1 - = 1 - 3 / 4 = 1 / 4 = 0 . 25, P 1 = (1 - ) = 0 . 1875, P 2 = 2 (1 - ) = 0 . 140625 P (more than 2 customers at the check-out stand) = P ( n > 2) = = 1 - ( P 0 + P 1 + P 2 ) P (more than 2 customers at the check-out stand) = 1 - 0 . 578125 = 0 . 421875 0 . 423 (c) M/M/1 model with λ = 30 and μ = 60 L = λ μ - λ = 30 60 - 30 = 1 customer, W = 1 μ - λ = 1 60 - 30 = 0 . 033 hours, W q = λ μ ( μ - λ ) = 30 60(60 - 30) = 0 . 017 hours, L q = λ W q = 30 0 . 017 = 0 . 51 customers, P 0 = 1 - = 1 - 0 . 5 = 0 . 5, P 1 = (1 - ) = 0 . 25, P 2 = 2 (1 - ) = 0 . 125 1
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P (more than 2 customers at the check-out stand) = P ( n > 2) = = 1 - ( P 0 + P 1 + P 2 ) P
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