HW 4 Solution

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Unformatted text preview: Lmax = + 2.3(10.45) = 12 veh 7 CEE 362 - Homework #4 Solution Manual Problem # 7: Given: Arrival rate = Arrivals: exponentially distributed Number of channels = m = 2 1 / μ = 12 sec /veh Required: Total time spent in the system: t Solution: μ = 60 / 12 = 5 ρ = 7 / 5 = 1.4 ρ<1 ρ / N = 1.4 / 2 = 0.7 <1 the probability of having no vehicles in the system with two booth open is: P0 P0 1 N n! n 0 N ! 1 N N 1 n = 0.176 The average queue length = 1.34 veh t= total time in the queue in 1 hour = 420 (0.391) = 164.22 min /hr 8 CEE 362 - Homework #4 Solution Manual Problem # 8: L B 2080 = 40t 1600 = 73.3t 800 µ 732 E = 36.6t F D A 20 min C 52 min 40 min λ= after 20 min : 40*20 = 800 veh in queue after 40 min: 40*40 = 1600 arrived 36.67* 20 = 733.4 departed 866.6+40t=73.33t 33.3t=866.6 min Total delay = + 9 t CEE 362 - Homework #4 Solution Manual Total delay = 35936.7 veh-min Average delay = 29888 / 2640 Average delay = 13.61 min 10...
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This document was uploaded on 02/11/2014.

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