Question

Customers arrive at a suburban ticket outlet at the rate of 4 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 12 minutes per customer, and varies exponentially. There is 1 ticket agent on duty on Mondays. Determine the System Utilization

Question 12 options:

	.8
	1.6
	.4
	.88
	None of the Above

Question 13 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 15 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 3 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. What is the Probability that an arriving customer will find nobody in the System?

Question 13 options:

	.94
	.235
	.47
	.517
	None of the Answers Listed is Correct

Question 14 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 5 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 30 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. Determine the System Utilization

Question 14 options:

	1.666
	.833
	.9163
	.4165
	None of the Answers Listed is Correct

Question 15 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 1 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 30 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. On Average, how much time does a customer spend Waiting in Line?

Question 15 options:

	.503
	.003
	.5533
	.0015
	None of the Answers Listed is Correct

Question 16 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 15 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 10 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. On Average, how much time does a customer spend Waiting in Line and getting tickets?

Question 16 options:

	.234
	.4
	.44
	.2
	None of the Answers Listed is Correct

Question 17 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 6 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 20 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. On Average, how much many customers are Waiting in Line?

Question 17 options:

	3.1768
	2.888
	.444
	.888
	None of the Answers Listed is Correct

Question 18 (1.5 points)

Customers arrive at a suburban ticket outlet at the rate of 2 per hour on Monday mornings. This can be described by a Poisson distribution. Selling the tickets and providing general information takes an average of 30 minutes per customer, and varies exponentially. There are 3 ticket agents on duty on Mondays. On Average, how much many customers are in the System?

Question 18 options:

	1.045
	.045
	.5225
	.0495
	None of the Answers Listed is Correct

Answer 1

1st Question

Arrival rate = 4 per hour

Service rate = 1/12 per minute = 5 per hour

Utilization = Arrival rate/Service rate = 4/5 = 0.8

Question 13

Arrival rate = 15 per hour

Service rate = 1/3 per minute = 20 per hour

Utilization = Arrival rate/(number of Servers* Service rate) = 15/(3*20) = 0.25

The probability that there is nobody in the system =((0.75^0/1)+(0.75^1/1)+(0.75^2/2)+((0.75^3/6)*(1/(1-0.25))))^-1 = 0.470588235 = 0.47(Ans)

Question 14

Arrival rate = 5 per hour

Service rate = 1/30 per minute = 2 per hour

Utilization = Arrival rate/(number of Servers* Service rate) = 5/(3*2) = 0.833 (Ans)

Question 15

Arrival rate = 1 per hour

Service rate = 1/30 per minute = 2 per hour

Utilization = Arrival rate/(number of Servers* Service rate) = 1/(3*2) = 0.16667

P0 = Probability of no customers in the system =((0.5^0/1)+(0.5^1/1)+(0.5^2/2)+((0.5^3/6)*(1/(1-0.16667))))^-1
= 0.606061

S = number of servers = 3

Number of customers in waiting line = (P0*(Arrival rate/service rate)^s*Utilization)/(s!*(1-utilization)^2)

= (0.606061*(1/2)^3*0.16667)/(3!*(1-0.16667)^2)

= 0.00303039

Time a student spends in waiting line = 0.00303039/1 hr = 0.00303039 = 0.003 (Ans)