Homework Assignment 3 Summer 2018 Question 2: Poisson processes Part 2 sing vehi

Question

Homework Assignment 3 Summer 2018 Question 2: Poisson processes Part 2 sing vehicles from 6am until 6pm. The other 12 hours, the toll station Vehicles pass a toll station with a rate of 50 per hour from 6am to 8am. The vehicles A toll station charges pas is closed. 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 tol de- pending on the size of the vehicle. With probability 0.3, vehicles pay S1, with probability 0.0 they need to pay S2, 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

Calculation:

6 am – 8 am - - - Average no. of vehicles passing is 50/hour

   So total no. of vehicles passing from 6 am – 8 am = 50*2 = 100

8 am – 10 am - - - vehicles increases from 50 to 250 per hour

We can say that

8 am – 9 am vehicles increases from 50 to 150 per hour

So Average no. of vehicles passing = (50+150)/2 = 100 per hour

9 am – 10 am vehicles increases from 150 to 250 per hour

So Average no. of vehicles passing = (150+250)/2 = 200 per hour

10 am – 2 pm Average no. of vehicles passing is 250/hour

So total no. of vehicles passing from 10 am – 2 pm = 250*4 = 1000

2 pm – 6 pm vehicles decrease from 250 to 50 per hour

We can say that

2 pm – 3 pm Vehicles passing decrease from 250 to 200 per hour

So Average no. of vehicles passing = (250+200)/2 = 225 per hour

3 pm – 4 pm Vehicles passing decrease from 200 to 150 per hour

So Average no. of vehicles passing = (200+150)/2 = 175 per hour

4 pm – 5 pm Vehicles passing decrease from 150 to 100 per hour

So Average no. of vehicles passing = (150+100)/2 = 125 per hour

5 pm – 6 pm Vehicles passing decrease from 100 to 50 per hour

So Average no. of vehicles passing = (100+50)/2 = 75 per hour

Therefore expected no. of vehicles passing from 8 am – 6 pm = 100+100+200+1000+225+175+125+75 = 2000

b>

Mean no. of vehicles passing in 60 mins is 50

Mean no. of vehicles passing in 0 mins = 50/6 = 8.33

So lamda = 8.33

We know that inter arrival posses Poison’s distribution

So P(x) = {e^(-lamda)*(lamda)^x}/(x!)

For no. of zero vehicles passing, x=0

P(0) = {e^(-8.33)*(8.33)^0}/(0!) =0.00024

c>

9 am – 10 am Average no. of vehicles passing = 200

9 am – 9:10 am Average no. of vehicles passing = 200/6 = 33.33 ( fractional value is allowable during calculation of mean value )

So lamda = 33.33

For no. of zero vehicles passing, x=0

P(0) = {e^(-33.33)*(33.33)^0}/(0!) =3.35 *10^(-15)

d>

PROBABILITY OF VEHICLES

TAX IN DOLLARS

0.3

1

0.6

2

0.1

3

Expected tax paid by a vehicle = 0.3*1 + 0.6*2 + 0.1*3 = 1.8 in dollars

Tax paid by 2000 vehicles = 2000*1.8 = 3600 in dollars

PROBABILITY OF VEHICLES

TAX IN DOLLARS

0.3

1

0.6

2

0.1

3

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