HW 4 Solution

# HW 4 Solution

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/11/2014.

Ask a homework question - tutors are online