MGM 315 - 4-12-16 Classwork - 238 =0)=(5(± ^ ² ^...

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σ queing systems Σ 3 part code A/B/k A = arrival distribution B = servce (departure) distribution μ k = the number of channels for the system stock transactions arrive at a mean rate of 20/hr 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 λμ ? =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 ?
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Unformatted text preview: 238 ? =0)=(5 (± ^ ² ^( −5 ) )/0!= .0067 ? = (± 1 )=(5^ 1 ^(−5))/ ² 1 != ° (°= ( 2 )=(5^ 2 °^(−5))/ 2 != ° (°=3)=(5^ ( 3 °^(−5))/3!=125(.0067)/6= .1396 ° (°= ( 4 )=(5^ 4 °^(−5))/ 4 != ° (°= ( 5 )=(5^ 5 °^(−5))/ 5 != ° (°= ( 6 )=(5^ 6 °^(−5))/ 6 != Answer average tme per order = 2 minuTes = 2/60(hr) = 1 / mean service tme =60/2 = 30 / hr Q: WhaT percenTage of The orders will Take less Than one minuTe To process A P(±<=1) = P(±<=1/60 hr) Q: whaT percenTage of The orders will be processed in exacTly 3 minuTes? A: since The exponental disTributon is a contnuous disTributon The probabiliTy a service tme exacTly equals any Q: whaT percenTage of The orders will require more Than 3 minuTes To process? (slide 24) A: ° ? ≤1)=1 − ( (± ² ° ^(−30/(1/60)) = 1 - .6065 = .3935 or 39.35% °° (°≥3/60)=° ( = ° ^(−30/(3/60)) = = .3935 or 39.35% specifc value is 0...
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