HW 4 Solution

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Unformatted text preview: 3600 q 0.65 400 Ln ( e q 400 ) = Ln (0.65) = - 0.43078 ► q = 172 veh /hr n qt qt 3600 e 3600 P(n ) n! 4 172(30) e 3600 P ( 4) 4! 172 ( 30 ) 3600 0.042 ► P (n = 4) = 4.2% 3 CEE 362 - Homework #4 Solution Manual Problem # 4: Given: Queue start: 8 A.M, service start: 8:10 A.M Uniform deterministic distribution Average arrive rate (λ) = 8 veh /min Average departure rate (μ)= 10 veh /min D/D/1 ► Arrival pattern: Deterministic Service pattern: Deterministic Number of sever channels: 1 L = 100 Required: time needed for the initial queue to clear W=? L=? Wait time for 100th vehicle Solution: 8:10 – 8:00 = 10 min 8t = 10 (t – 10) 8t = 10 t - 100 100 = 2t ►t =50 min (the time needed for the queue to clear) The diagram helps understanding more clearly, go on to the next page 4 CEE 362 - Homework #4 Solution Manual μ = 10t L λ = 8t 400 80 t 10 min 50 (min) No queue Total delay = W The area between the arrival and departure lines: = 400 * 10 / 2 = 2000 veh-min ►Average delay = 2000 / 400 = 5 min Lmax (longest queue) happens when t =10 λ (t) = 8 * 10 = 80 veh ► maximum queue length = 80 veh When does the 100th car arrive: 100 / λ =100/8 =...
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This document was uploaded on 02/11/2014.

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