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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan Solutions to Homework 7 1. You must be careful in this problem because λ is not the arrival rate in this problem. Because λ is being used in the service rate distribution, I will use η (read as “eta”) to denote the arrival rate. Because the arrival rate is given in terms of hours, I will first convert it into minutes to match the service distribution’s units. (You can also convert the service rate information into hours; both methods will work, so long as the units agree.) η = 30 customers 1 hour 1 hour 60 minutes = 1 customer 2 minutes Because the arrival distribution is exponential, the squared coefficient of variation is 1, so c 2 A = 1, which is always true for the exponential distribution. Now that the arrival information is calculated, I need to determine the mean and variance for the arrival distribution (which if you look carefully should see is not an exponential distribution). The first way to do this is to calculate the mean and variance directly from the definitions. Let S be a random variable representing the service time. E ( S ) = Z ∞ sf ( s ) d s = Z ∞ s 4 λ 2 se 2 λs d s = 1 λ = 3 2 minutes Var( S ) = E ( S 2 ) ( E ( S )) 2 = Z ∞ s 2 f ( s ) d s Z ∞ sf ( s ) d s 2 = Z ∞ s 2 4 λ 2 se 2 λs d s Z ∞ s 4 λ 2 se 2 λs d s 2 = 3 2 λ 2 1 λ 2 = 1 2 λ 2 = 9 8 An alternative method for calculating the mean and variance is possible that avoids the ne cessity of calculating the above integrals but requires knowledge of the Erlang k distribution. The p.d.f. for an Erlang k distribution with parameter α is, g ( s ) = α k s k e αs ( k 1)! for s ≥ otherwise By setting α = 2 λ and k = 2, and considering only s ≥ 0, I can rewrite the given p.d.f. as follows. f ( s ) = 4 λ 2 se 2 λs = (2 λ ) 2 se (2 λ ) s (2 1)! = α 2 se αs ( k 1)! I have now shown that the service times have an Erlang2 distribution with parameter α = 2 λ ....
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 Fall '07
 Billings

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