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Stat Problems

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Question 8-1 a) the possible paths include; 1-3-6-7-9-8 Distance= 1+3+3+6+3 = 16 Path; 1-4-5 Distance= 2+4 =6 Path: 1-2 Distance= 4 Path: 9-10-11-13 Distance=4+3+3 =10 Path: 9-12-14 Distance= 5+4 =9 Total distance= 16+ 6 + 4 + 10 + 9 = 45 hudreds feet b. Recommended route for lines= 1-3-6-7-9-10-11-13-14. c. Question 8-2 a. ρ=lambda/mu = 3/8 = .375 rate = 1-.375 = .625 b. W_q = lambda/mu(mu-lambda) = 3/8(8-3) = .075 c. L_q= lambda^2/mu(mu-lambda) = 9/40 = .225 d. P_n>k = (lambda/mu)^k+1 = .141 k=1 e. P_n>k = .0527 k = 2 f. P_n>k = .0198 k = 3 g. P_n>k = .007 k = 4 Question 8-3 a. L_q = lambda^2/mu(mu-lambda) = 210^2/225(225-210) = 13.07 ~14 moviegoers b. Rho = lambda/mu = .9333*100 = 93.33% c. W_q = lambda/mu(mu-lambda) =.0622 d. W = 1/(mu – lambda) = .0667 e. P_n>k = .813 k=2 f. P_n>k = .708 k=4 g. P_o = 1 – lambda/mu = .0667 h. Since the waiting time to get a ticket is already high, Mike could opt to get another cashier. The cashier would assist with the customers minimizing the waiting time. Additionally, Mike could add another auditorium. Question 8-4 find the: a. L_s = 15 b. W_s = 0.5 hours c. Utilization rate = .9375 d. Probability = .7725 e. Total Cost = 58*16*30*.9375*0.5 = $13,050 f. When increased the cost in saving is $82,35 hence, it is significant to add another bin. Question 8-5 Discuss the results lambda = 3.25 mu = 4 The information above is accurate and the waiting time for a truck will be 1hr 5 minutes. (a) the hourly cost savings Current ...
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