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Waiting Line

# Waiting Line - 1[Infinite source single channel = 3/day =...

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1. [Infinite source, single channel] λ = 3/day μ = 4/day a. customers 25 . 2 ) 3 4 ( 4 3 ) ( L 2 2 q = = λ μ μ λ = [Alternate solution, using infinite queuing table:] , 75 . r = μ λ = M = 1. From table 18–4, L q = 2.25 customers b. % 75 or , 75 . 4 3 = = ρ c. Percentage of time idle is 1 – p = 25%. Thus 25%(8) = 2 hours. d. P(L s 2) = 1 – [P o + P 1 ] = 1 – [.25 + .25(.75)] = .5625 3. λ = 30/hr. Single Channel μ = 40/hr. a. line in customers 25 . 2 ) 30 40 ( 40 30 ) ( L 2 2 q = = λ μ μ λ = minutes 5 . 4 . hrs 075 . 30 25 . 2 L W q q = = + λ = W S = W q +1/ μ = 4.5 + 1.5 = 6 minutes b. 25 . 40 30 1 1 P 0 = = μ λ = c. L q = 2.25 customers (see part a.)

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Solutions (continued) 4. [Infinite source, multiple channel] λ = .45 requests/hr. μ = .50 request/hr. M = 2 ambulances a. % 45 45 . ) 50 (. 2 45 . or M = = = μ λ ρ b. . 80 . 50 , 45 . = = = μ λ r From infinite queuing table with M = 2, L q = .229 c. minutes 4 5 . 30 . 509 . hr. customers/ ..45 customers 229 . or hr L W q q = = = λ 5. a. Period λ μ M r = λ / μ L q W q = L q P w λ morning 1.8/min. 1.5/min. 2 1.2 .675 .375 .450 afternoon 2.2/min. 1.0/min. 3 2.2 1.491 .678 0.542 evening 1.4/min. .7/min. 3 2.0 .889 .635 .444 b.
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