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Chapt 12

# Chapt 12 - Waiting Line Models 7 a Lq = 2(2.5 2 = = 0.5000...

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Waiting Line Models 7. a. L q = - = - = λ μ μ λ 2 2 2 5 5 5 2 5 05000 ( ) ( . ) ( . ) . b. W L q q = = = λ 05000 2 5 0 20 . . . hours (12 minutes) c. W W q = + = + = 1 0 20 1 5 0 40 μ . . hours (24 minutes) d. P w = = = λ μ 2 5 5 050 . . 10. a. λ = 2 μ = 3 μ = 4 Average number waiting ( L q ) 1.3333 0.5000 Average number in system ( L ) 2.0000 1.0000 Average time waiting ( W q ) 0.6667 0.2500 Average time in system ( W ) 1.0000 0.5000 Probability of waiting ( P w ) 0.6667 0.5000 b. New mechanic = \$30( L ) + \$14 = 30(2) + 14 = \$74 per hour Experienced mechanic = \$30( L ) + \$20 = 30(1) + 20 = \$50 per hour Hire the experienced mechanic 14. a. μ = = 60 7 5 8 . customers per hour b. 0 5 1 1 0.3750 8 P λ μ = - = - = c. L q = - = - = λ μ μ λ 2 2 5 8 8 5 10417 ( ) ( ) . d. W L q q = = = λ 10417 5 0 2083 . . hours (12.5 minutes) e. P w = = = λ μ 5 8 0 6250 . f. 62.5% of customers have to wait and the average waiting time is 12.5 minutes. Ocala needs to add more consultants to meet its service guidelines. L L q = + = + = λ μ 05000 2 5 5 1 . .

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L q = λ 2 σ 2 + ( λ / μ ) 2 2 (1 - λ / μ ) = (0.25) 2 (2) 2 + (0.25 / 0.3125) 2 2 (1 - 0.25 / 0.3125) = 2.225 W q = L q λ = 2.225 0.25 = 8.9 hours W = W q + 1 μ = 8.9 + 1 1.3125 = 12.1 hours Same at P w = λ μ = 0.25 0.3125 = 0.80 L q = λ 2 σ 2 + ( λ / μ ) 2 2 (1 - λ / μ ) = (.375) 2 (1.5) 2 + (.375/ .5) 2 2 (1 - .375/ .5) = 1.7578 Chapter 12 22. λ = 24 Characteristic System A ( k = 1, μ = 30) System B ( k = 1, μ = 48) System C ( k = 2, μ = 30) a. P 0 0.2000 0.5000 0.4286 b. L q 3.2000 0.5000 0.1524 c. W q 0.1333 0.0200 0.0063 d. W 0.1667 0.0417 0.0397 e. L 4.0000 1.0000 0.9524 f. P w 0.8000 0.5000 0.2286 System C provides the best service. 27. a. 2/8 hours = 0.25 per hour b. 1/3.2 hours = 0.3125 per hour c. d. e. f. 80% of the time the welder is busy. 29. a. λ = 3/8 = .375 μ = 1/2 = .5 b. L = L q + λ / μ = 1.7578 + .375 / .5 = 2.5078 TC = c w L + c s k = 35 (2.5078) + 28 (1) = \$115.71 c. Current System ( σ = 1.5) New System ( σ = 0) L q = 1.7578 L q = 1.125 L = 2.5078 L = 1.875 W q = 4.6875 W q = 3.00 W = 6.6875 W = 5.00 TC = \$115.77 TC = c w L + c s k = 35 (1.875) + 32 (1) = \$97.63 d. Yes; Savings = 40 (\$115.77 - \$97.63) = \$725.60
Waiting Line Models Note: Even with the advantages of the new system, W q = 3 shows an average waiting time of 3 hours. The company should consider a second channel or other ways of improving the emergency repair service. Chapter 12 Waiting Line Models Case Problem 1: Regional Airlines 1. Single-Channel Waiting Line Analysis The analysis that follows is based upon the assumptions of Poisson arrivals and exponential service times. With one call every 3.75 minutes, we have an average arrival rate of λ = 60/3.75 = 16 calls per hour Similarly, with an average service time of 3 minutes, we have a service rate of μ = 60/3 = 20 calls per hour The operating characteristics of a single channel system with λ = 16 and μ = 20 are as follows: 0 16 1 1 0.20 20 P λ μ = - = - = 2 2 16 3.2 ( ) 20(20 16) q L λ μ μ λ = = = - - 16 3.2 4 20 q L L λ μ = + = + = 3.2 0.20 16 q q L W λ + = = hours (12 minutes) 1 1 0.20 0.25 20 q W W μ = + = + = hours (15 minutes) 16 0.80 20 w P λ μ = = = Operating the telephone reservation service with only one ticket agent appears unacceptable.

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Chapt 12 - Waiting Line Models 7 a Lq = 2(2.5 2 = = 0.5000...

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