Unformatted text preview: 12.5 min
When does the 100th car departure: 10(t10) = 100 ► t = 20min
L μ = 10t λ = 8t 100 t
12.5 20 = 20 – 12.5 = 7.5 min = 7 min and 30 sec
5 CEE 362  Homework #4 Solution Manual Problem # 5:
Solution
proportion of numbers have not been considered in the diagram for better illustration L 200 E = 2(t30)
160
120 = 0(t20)
B C μ = 4t = 8t
D A 20 min F 30 (min) 50 (min) (y160) = (x30)*2
y= 2x +100
we also know y = 4x
x= 50, y = 200 total delay = sum of area between arrival lines and departure line
[0.5*20*160] + [160*10] + [160*2] + [0.5*20*40]  [0.5 *50*200]
= 1800 vehmin 6 t CEE 362  Homework #4 Solution Manual Problem # 6:
Given: Deterministic time varying rate Departure rate = μ = 2 veh / min = 2 veh /min
Required: Time needed for the queue to clear
Total delay =
Maximum queue length Solution:
Arrival:
Departure:
The time required to clear the queue with a departure rate of 2 veh/min will be:
=
►t = 21.05 min = arrivals – departures =  2t = + 2.3t + 2.3t = 0
t = 10.45 ►...
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 Fall '14
 µ, Min Farshaw

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