Homework Assignment 3 Summer 2018 Question 2: Poisson processes Part 2 A toll st
ID: 3366357 • Letter: H
Question
Homework Assignment 3 Summer 2018 Question 2: Poisson processes Part 2 A toll station charges passing vehicles from 6am until 6pm. The other 12 hours, the toll station is closed. Vehicles pass a toll station with a rate of 50 per hour from 6am to 8am. The vehicles increase from 8am until 10am finally reaching 250 vehicles per hour. The rate stays at 250 vehicles per hour until 2pm, at which point they start decreasing until they hit 50 vehicles per hour at 6pm. Answer the following questions. (a) How many vehicles should they expect in the toll station between 6am and 6pm? (b) What is the probability no cars pass through the toll station in the first ten minutes of the toll station's operation (between 6am and 6.10am)? (c) What is the probability no cars pass through the toll station between 9am and 9.10am (d) Every vehicle passing from the toll station between 6am and 6pm needs to pay a toll de- pending on the size of the vehicle. With probability 0.3, vehicles pay $1, with probability 0.6 they need to pay $2, and with probability 0.1, they need to pay $3. How much money should the toll station expect to make from 6am to 6pm? Answer:Explanation / Answer
Solution
If a random variable X ~ Poisson(?), i.e., X has Poisson Distribution with mean ? then
probability mass function (pmf) of X is given by P(X = x) = e – ?.?x/(x!) …………..(1)
where x = 0, 1, 2, ……. , ?
Values of p(x) for various values of ? and x can be obtained by using Excel Function.(1a)
If X = number of times an event occurs during period t, Y = number of times the same event occurs during period kt, and X ~ Poisson(?), then Y ~ Poisson (k?) ……….. (2)
Mean = ?
Variance = ?
Part (a)
Number of vehicles expected to pass through the toll station from 6 am to 6 pm.
Time Interval
# of hours
Rate per hour
Total
6 am – 8 am
2
50
100
8 am – 10 am
2
(50 + 250)/2 = 150
300
10 am – 2 pm
4
250
1000
2 pm – 6 pm
4
(250 + 50)/2 = 150
600
Total
12
-
2500 ANSWER
Part (b)
Let X = Number of cars that through the toll station from 6 am to 7 am. Then, we are given X ~ Poisson (50) [vide above table Row 1]
So, by vide (2) under Back-up Theory, if Y = Number of cars that through the toll station from 6 am to 6:10 am, then Y ~ Poisson (50/6) [10 minutes = 1/6 hour]
So, required probability = P(Y = 0/? = 8.3333)
= 0.00024 [vide (1a) under Back-up Theory] ANSWER
Part (c)
Identical to the above analysis, required probability = P(W = 0/ ? = 150/6 = 25),
where W = Number of cars that through the toll station from 9 am to 9:10 am
= 1.39E – 11 [vide (1a) under Back-up Theory] ANSWER
Part (d)
Amount made by the toll station from 6am to 6 pm
= 2500 x {(1 x 0.3) + (2 x 0.6) + (3 x 0.1)} ANSWER
Time Interval
# of hours
Rate per hour
Total
6 am – 8 am
2
50
100
8 am – 10 am
2
(50 + 250)/2 = 150
300
10 am – 2 pm
4
250
1000
2 pm – 6 pm
4
(250 + 50)/2 = 150
600
Total
12
-
2500 ANSWER
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