HW9_Solutions_2014

HW9_Solutions_2014 - IE:3610 Stochastic Modeling(Fall 2014...

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Unformatted text preview: IE:3610 Stochastic Modeling (Fall 2014) Homework 9 (due Thu, Nov 6) Solutions HW9 (Due Thu, Nov 6): 17.6-3; 17.6-10 (a,c,e); 17.6-11; 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 =⇒ ρ = 2/3 P0 = 1 − ρ = 1/3, P1 = (1 − ρ) ∗ ρ = 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 λ 30 L = µ−λ = 40−30 = 3 customers, 1 1 W = µ−λ = 40−30 = 0.1 hours, 30 λ = 40(40−30) = 0.075 hours, Wq = µ(µ−λ) Lq = λWq = 30 ∗ 0.075 = 2.25 customers, P0 = 1 − ρ = 1 − 3/4 = 1/4 = 0.25, P1 = (1 − ρ)ρ = 0.1875, P2 = ρ2 (1 − ρ) = 0.140625 P (more than 2 customers at the check-out stand) = P (n > 2) = = 1 − (P0 + P1 + P2 ) 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 λ 30 L = µ−λ = 60−30 = 1 customer, 1 1 W = µ−λ = 60−30 = 0.033 hours, λ 30 Wq = µ(µ−λ) = 60(60−30) = 0.017 hours, Lq = λWq = 30 ∗ 0.017 = 0.51 customers, P0 = 1 − ρ = 1 − 0.5 = 0.5, P1 = (1 − ρ)ρ = 0.25, P2 = ρ2 (1 − ρ) = 0.125 P (more than 2 customers at the check-out stand) = P (n > 2) = 1 = 1 − (P0 + P1 + P2 ) P (more than 2 customers at the check-out stand) = 1 − 0.875 = 0.125 (e) The manager should hire another person to help the cashier by bagging the groceries 17.6-11 Solution: The three criteria are: 1) the average number of customers in the queue should not exceed 1: Lq ≤ 1 2) with 95% probability the number of customers in the queue (waiting for clearance) should not exceed 4, which means the total number of customers in the system should not exceed 5 with probability 95%: P4+s i=0 Pi ≥ 0.95 3) P (Wq ≤ 30 mins) ≥ 0.99 or P (Wq ≤ 0.5 hrs) ≥ 0.99 17.6-11.(a) M/M/s model with λ = 10, µ = 20, s = 1 Allcriteria the criteria are currently satisfied (a) All the are currently satisfied. s= Data 10 20 1 Results (mean arrival rate) (mean service rate) (# servers) Pr(W > t) = 0.006738 when t = L= Lq = 1 0.5 W= Wq = 0.05 0.1 0.5 0.5 Prob(W q > t) = 0.003369 when t = 0.5 n 0 1 2 3 4 5 model with λ = 15, µ = 20, s = 1 (b) None(b) of M/M/s the criteria are satisfied. None of the criteria are currently satisfied Data s= 15 20 1 (mean arrival rate) (mean service rate) (# servers) Pr(W > t) = 0.082085 when t = Pn cumulative 0.5 0.25 0.125 0.0625 0.03125 0.015625 0.5 0.75 0.875 0.9375 0.96875 0.984375 Results L= Lq = 3 2.25 W= Wq = 0.15 0.2 0.5 0.75 Prob(W q > t) = 0.061564 when t = 0.5 n 2 Pn (c) The first and third criteria are satisfied, but the second is not. Data 25 (mean arrival rate) cumulative 0 0.25 0.25 1 0.1875 0.4375 2 0.140625 0.578125 3 0.10546875 0.68359375 4 0.079101563 0.76269531 5 0.059326172 0.82202148 Results L = 2.051282051 Wq = Pr(W > t) = 0.006738 when t = 0.5 when t = 0.5 0.5 0.5 Prob(W q > t) = 0.003369 Prob(W q > t) = 0.003369 when t = 0.5 when t = 0.5 s= s= (mean arrival rate) (mean (mean arrival servicerate) rate) (mean service rate) (# servers) (# servers) Pr(W > t) = 0.082085 Pr(W > t) = 0.082085 when t = 0.5 when t = 0.5 Pn Pn n n 0 01 12 23 34 45 5 0.5 0.5 0.25 0.25 0.125 0.125 0.0625 0.0625 0.03125 0.03125 0.015625 0.015625 L= LLq == Lq = Results Results 3 3 2.25 2.25 W= W == W q Wq = 0.2 0.2 0.15 0.15 (b) None of the criteria are satisfied. (b) None of the criteria are satisfied. Data Data 15 15 20 20 1 1 0.05 cumulative cumulative 0.5 0.5 0.75 0.75 0.875 0.875 0.9375 0.9375 0.96875 0.96875 0.984375 0.984375 0.75 0.75 Prob(W q > t) = 0.061564 Prob(W q > t) = 0.061564 when t = 0.5 when t = 0.5 Pn Pn n n 0 01 12 23 34 45 5 0.25 0.25 0.1875 0.1875 0.140625 0.140625 0.10546875 0.10546875 0.079101563 0.079101563 0.059326172 0.059326172 L= LLq == Lq = Results 2.051282051 2.051282051 0.801282051 0.801282051 W= W= Wq = Wq = 0.082051282 0.082051282 0.032051282 0.032051282 cumulative cumulative 0.25 0.25 0.4375 0.4375 0.578125 0.578125 0.68359375 0.68359375 0.76269531 0.76269531 0.82202148 0.82202148 (c) M/M/s model with λ = 25, µ = 20, s = 2 (c) The first and third criteria are satisfied, but the second is not. firstthird andData third criteria are satisfied, but theisResults second (c) The The first and criteria are satisfied, but the second not. is not s= s= Data 25 25 20 20 2 2 (mean arrival rate) (mean (mean arrival servicerate) rate) (mean service rate) (# servers) (# servers) Pr(W > t) = 0.001022 Pr(W > t) = 0.001022 when t = 0.5 when t = 0.5 0.625 0.625 Prob(W q > t) = 0.000266 Prob(W q > t) = 0.000266 when t = 0.5 when t = 0.5 n n 0 01 12 23 34 45 5 Pn cumulative Pn cumulative 0.230769231 0.23076923 0.230769231 0.288461538 0.23076923 0.51923077 0.288461538 0.180288462 0.51923077 0.69951923 0.180288462 0.112680288 0.69951923 0.81219952 0.112680288 0.07042518 0.81219952 0.8826247 0.07042518 0.92664044 0.8826247 0.044015738 0.044015738 0.92664044 Criteria: 1) Lq = 0.8013 ≤ 1 is satisfied; P P6 2) 4+s i=0 Pi = i=0 Pi = 0.027509836 + 0.926640437 = 0.954150273 ≥ 0.95 is satisfied; 3)P (Wq ≤ 0.5hr) = 1 − 0.00026617-16 ≥ 0.99 is satisfied 17-16 17.6-12 Solution: Guidlines (M/M/4 model): 1)The average number of customers waiting in line to begin service should not exceed 1: Lq ≤ 1 3 2)At least 95 percent of the time, the number of customers waiting in line should not exceed 5: P (n − s ≤ 5) ≥ 0.95 or P (n ≤ 9) ≥ 0.95 3)For at least 95 percent of the customers, the time spent in line waiting 17.6-12.to begin service should not exceed 5 minutes: P (Wq ≤ 5 mins) ≥ 0.95 (a) All the guidelines currently (a)M/M/4are model. All met. the guidelines are currently met. s= Data 2 1 4 (mean arrival rate) (mean service rate) (# servers) W = 1.086956522 W q = 0.086956522 Pr(W > t) = 0.007902 when t = Results L = 2.173913043 Lq = 0.173913043 5 0.5 Prob(W q > t) = when t = 7.9E-06 5 n Pn 0 1 2 3 4 5 6 7 8 9 0.130434783 0.260869565 0.260869565 0.173913043 0.086956522 0.043478261 0.02173913 0.010869565 0.005434783 0.002717391 cumulative 0.1304 0.3913 0.6522 0.8261 0.9130 0.9565 0.9783 0.9891 0.9946 0.9973 (b)The first two guidelines will not be satisfied in a year, but the third (b) The will firstbe. two guidelines will not be satisfied in a year, but the third will be. s= Data 3 1 4 (mean arrival rate) (mean service rate) (# servers) W = 1.509433962 W q = 0.509433962 Pr(W > t) = 0.023901 when t = Results L = 4.528301887 Lq = 1.528301887 5 0.75 Prob(W q > t) = 0.003433 when t = 5 4 (c) Five tellers are needed in a year. n Pn 0 1 2 3 4 5 6 7 8 9 0.037735849 0.113207547 0.169811321 0.169811321 0.127358491 0.095518868 0.071639151 0.053729363 0.040297022 0.030222767 cumulative 0.0377 0.1509 0.3208 0.4906 0.6179 0.7134 0.7851 0.8388 0.8791 0.9093 1 2 3 4 5 6 7 8 9 0.260869565 0.260869565 0.173913043 0.086956522 0.043478261 0.02173913 0.010869565 0.005434783 0.002717391 0.3913 0.6522 0.8261 0.9130 0.9565 0.9783 0.9891 0.9946 0.9973 (b) The first two guidelines will not be satisfied in a year, but the third will be. s= Data 3 1 4 (mean arrival rate) (mean service rate) (# servers) W = 1.509433962 W q = 0.509433962 Pr(W > t) = 0.023901 when t = Results L = 4.528301887 Lq = 1.528301887 5 0.75 Prob(W q > t) = 0.003433 when t = 5 n Pn 0 1 2 3 4 5 6 7 8 9 0.037735849 0.113207547 0.169811321 0.169811321 0.127358491 0.095518868 0.071639151 0.053729363 0.040297022 0.030222767 cumulative 0.0377 0.1509 0.3208 0.4906 0.6179 0.7134 0.7851 0.8388 0.8791 0.9093 Fiveare tellers areinneeded (c) Five(c) tellers needed a year. in a year: 1) Lq = 0.354227405 ≤ 1 is satisfied 2) P (n − s ≤ 5) = P (n ≤ 10) = 0.988982111 ≥ 0.95 is satisfied 3) P (Wq ≤ 5 mins) = 1 − (1.07213E − 05) ≥ 0.95 is satisfied 17.6-15 Solution: M/M/1 model with λ = 20 per hr, µ = 1/2 per min = 30 per hr 17-17 P ( An arriving customer does not have to wait before service) = P0 = 1 − λ/µ = 1/3 Expected price of gasoline per gallon: 4 ∗ 1/3 + 3.5 ∗ 2/3 = \$3.667 5 ...
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• Spring '08
• KROKHMAL
• Ode, mean arrival rate, mean service rate

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