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HW_TN7_sol

# HW_TN7_sol - Therefore waiting cost = number in line times...

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Homework – Technical Note 7 1. Use model 1 (M/M/1 queue). = l 4/hour = m 6/hour a. 3333 . 6 4 1 1 1 = - = - = - m l r or 33.33% b. ) 4 6 ( 6 4 ) ( 2 2 - = - = l m m l q L =1.33, 4 33 . 1 = = l q q L W = 1/3 hour or 20 minutes c. ) 4 6 ( 6 4 ) ( 2 2 - = - = l m m l q L = 1.33 students d. At least one other student waiting in line is the same as at least two in the system. This probability is 1-(P 0 +P 1 ). n n P - = m l m l 1 0 0 6 4 6 4 1 - = P = .3333 1 1 6 4 6 4 1 - = P = .2222 Probability of at least one in line is 1-(.3333 + .2222) = .4444 2. Use model 1 (M/M/1 queue). a. = l 4/hour = m 10/hour (service rate 40% faster) ) 4 10 ( 10 4 ) ( 2 2 - = - = l m m l q L = .267 students Waiting cost = number in line times goodwill loss per hour times number of hours per day =.267(10)8 = \$21.33 per day Total cost = waiting cost + additional service cost =\$21.33 + \$99.50 = \$120.83 per day b. Use model 3 (M/M/s queue). M=2, 6 4 = m l =.667 Interpolating from Exhibit TN7.11, q L

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