MGM 315 4-14-16 Classwork - queing systems 3 part code...

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σ queing systems μ>λ Σ 3 part code A/B/k A = arrival distribution λ B = servce (departure) distribution μ k = the number of channels for the system stock transactions arrive at a mean rate of 20/hr Formula S each order requires an average of 2 minutes to process orders = 20/hr λ = 20 - average arrival rate (1 hr) or μ = 30 - average service rate for each server (1 hr) 1 order / 3 minutes so in 15 minutes = 5 orders λ = 15/3 = 5 - average arrival rate (15 minutes) Arrival Rate Distribution Q: what is the probability that no (0) orders are received within a 15 minute period? Answer Q: what is the probability that exactly 3 orders are receieved within a 15 minute time period? Answer Q: what is the probability that more than 6 orders arrive within a 15 minute time period? Answer Q: What is the mean service rate per hour? Slide 21 ? =0)=(5 (? ^ 0 ? ^( −5 ) )/0! =? ^( −5 ) = .0067 𝑃 (𝑃=3)=(5 ( ^ 3 𝑃 ^( −5 ) )/3! =125(.0067)/6= .1396 𝑃 (𝑃>6)= ( 1-P(X=0)-P(X=1)-P(X=2)- P(X=3)-P(X=4)-P(X=5)-P(X=6) = 1-.762 = . 238 ? =0)=(5 (? ^ 0 ? ^( −5 ) )/0!= .0067 ? = (? 1 )=(5^ 1 ^(−5))/ ? 1 != 𝑃 (𝑃= ( 2 )=(5^ 2 𝑃^(−5))/ 2 != 𝑃 (𝑃=3)=(5^ ( 3 𝑃^(−5))/3!=125(.0067)/6= .1396 𝑃 (𝑃= ( 4 )=(5^ 4 𝑃^(−5))/ 4 != 𝑃 (𝑃= ( 5 )=(5^ 5 𝑃^(−5))/ 5 != 𝑃 (𝑃= ( 6 )=(5^ 6 𝑃^(−5))/ 6 != A B C 1 2 3 4 5 6 7 8 9 O p e r a t i n g C h P r o b a b i l i t y o f n o o r d A v e r a g e n u m b e r o f A v e r a g e n u m b e r o f A v e r a g e t i m e a n o r d A v e r a g e t i m e a n o r d P o i s s o n A r r i v a l R a t e E x p o n e n t i a l S e r v i c e P r o b a b i l i t y a n o r d e r
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Answer average time per order = 2 minutes = 2/60(hr) = 1 / mean service time =60/2 = 30 / hr Q: What percentage of the orders will take less than one minute to process A P(T<=1) = P(T<=1/60 hr) Q: what percentage of the orders will be processed in exactly 3 minutes? A: since the exponential distribution is a continuous distribution the probability a service time exactly equals any Q: what percentage of the orders will require more than 3 minutes to process? (slide 24) A: Q: What is the average time an order must wait from the time Joe receives the order unit it is finished being proce A: Q: What is the average number of orders Joe has waiting to be processed 𝑃 ? ≤1)=1 − ( (? ? 𝑃 ^(−30/(1/60)) = 1 - .6065 = .3935 or 39.35% 𝑃𝑃 (𝑃≥3/60)=𝑃 ( = 𝑃 ^(−30/(3/60)) = = .3935 or 39.35%
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Spreadsheet (slide 29 - CH 11) A B C D E F G H 1 2 0 2 3 0 3 4 P o = 1 - H 1 / H 2 5 L g = H 1 ^ 2 / ( H 2 * ( H 2 - H 1 ) ) 6 L = H 5 + H 1 / H 2 7 W q = H 5 / H 1 8 W = H 7 + 1 / H 2 9 P w = H 1 / H 2 O p e r a t i n g C h a r a c t e r i s t i c s P r o b a b i l i t y o f n o o r d e r s i n s y s t e m A v e r a g e n u m b e r o f o r d e r s w a i t i n g A v e r a g e n u m b e r o f o r d e r s i n s y s t e m A v e r a g e t i m e a n o r d e r w a i t s A v e r a g e t i m e a n o r d e r i s i n s y s t e m P o i s s o n A r r i v a l R a t e E x p o n e n t i a l S e r v i c e R a t e P r o b a b i l i t y a n o r d e r m u s t w a i t
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specific value is 0 essed
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σ Σ λ =1.5(20) =30 per hour < new arrival rate of orders μ Question: Will Joe Ferris alone not be able to handle the increase in orders?
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  • Spring '16
  • Dr. Chao
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