Chapter-11
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1. Willow Brook National Bank operates a drive-up teller window that allows customers to
complete bank transactions without getting out of their cars. On weekday mornings, arrivals to
the drive-up teller window occur at random, with an arrival rate of 24 customers per hour or 0.4
customers per minute.
a. What is the mean or expected number of customers that will arrive in a five-minute period?
b. Assume that the Poisson probability distribution can be used to describe the arrival process.
Use the arrival rate in part (a) and compute the probabilities that exactly 0, 1, 2, and 3 customers
will arrive during a five-minute period.
c. Delays are expected if more than three customers arrive during any five-minute period. What is
the probability that delays will occur?
:: Solution ::
step: 1 of 9
(a)
step: 2 of 9
Mean or expected number of customers (Five-minute period)
step: 3 of 9
It is given that on weekday mornings, arrivals to the drive-up teller window occur at the arrival rate
of 24 customers per hour or 0.4 customers per minute.
Now, calculate the mean or expected number of customers in a five-minute period as follows:
Let
be the mean or expected number of customers in a five-minute period.
…… (1)
Hence, the expected number of customers in a time period is 2 per five-minute period.
step: 4 of 9
(b)
Poisson distribution
Poisson distribution is used to determine the customer arrivals. Using the Poisson distribution,
calculate the probability that x customers arrive within a particular time period. The formula for
calculating the probability of x customer arrivals during the five-minute period is as follows:
The calculated average number of customers during a five-minute period
(1)).
is 2 (refer Equation
The following table shows the Poisson probabilities of 0,1,2, and 3 customer arrivals during a fiveminute period
:
Number of arrivals, x Probability
0
0.1353 (refer to equation (2))
1
0.2707 (refer to equation (3))
2
0.2707 (refer to equation (4))
3
0.1804 (refer to equation (5))
step: 5 of 9
Working notes:
Probability of 0 customer arrival during a five-minute period
It is given that the arrival rate is 2 customers per five-minute period and the number of arrivals is 0
(x).
Now, calculate probability by using Poisson distribution as follows:
…… (2)
step: 6 of 9
Probability of 1 customer arrival during a five-minute period
It is given that the arrival rate is 2 customers per five-minute period and the number of arrival is 1
(x).
Now, calculate probability by using Poisson distribution as follows:
…… (3)
step: 7 of 9
Probability of 2 customer arrivals during a five-minute period
It is given that the arrival rate is 2 customers per five-minute period and the number of arrival is 2
(x).
Now, calculate probability by using Poisson distribution as follows:
…… (4)
step: 8 of 9
Probability of 3 customer arrivals during a five-minute period
It is given that the arrival rate is 2 customers per five-minute period and the number of arrivals is
3 (x).
Now, calculate probability by using Poisson distribution as follows:
…… (5)
step: 9 of 9
(c)
Calculate probability if delay occurs
It is given that the delays are expected if more than 3 customers arrive during a five-minute period.
Then the probability is calculated as follows:
Hence, the probability of delay is 0.1429.
Likes: 5
2. In the Willow Brook National Bank waiting line system (see Problem 1), assume that the service
times for the drive-up teller follow an exponential probability distribution with a service rate of 36
customers per hour, or 0.6 customers per minute. Use the exponential probability distribution to
answer the following questions:
a. What is the probability that the service time is one minute or less?
b. What is the probability that the service time is two minutes or less?
c. What is the probability that the service time is more than two minutes?
:: Solution ::
step: 1 of 14
Service time:
step: 2 of 14
Service times are described by a mean time and a probability distribution. The formula for
calculating the probability of completing service within a specified time period is as follows:
step: 3 of 14
Note:
Where,
e = 2.71828 (natural log base)
µ = the mean number of units that can be served per time period
step: 4 of 14
(a)
step: 5 of 14
Probability if the service time is one minute or less :
step: 6 of 14
It is given that the average service rate is 36 customers per hour or 0.6 (µ) customers per minute.
step: 7 of 14
Now, calculate the probability if actual service time is one minute or less as follows:
step: 8 of 14
Hence, the probability if the service time is one minute or less is 0.4512.
step: 9 of 14
(b)
step: 10 of 14
Probability if the service time is two minutes or less :
step: 11 of 14
It is given that the average service rate is 36 customers per hour or 0.6 (µ) customers per minute.
step: 12 of 14
Now, calculate the probability if actual service time is two minutes or less as follows:
step: 13 of 14
Hence, the probability if the service time is one minute or less is 0.6988.
step: 14 of 14
(c)
Probability if the service time is more than two minutes :
It is given that the average service rate is 36 customers per hour or 0.6 (µ) customers per minute.
Now, calculate the probability if actual service time is two minutes or less as follows:
Hence, the probability if the service time is more than two minutes is 0.3012.
Likes: 8
3. Use the single-server drive-up bank teller operation referred to in Problems 1 and 2 to
determine the following operating characteristics for the system:
a. The probability that no customers are in the system
b. The average number of customers waiting
c. The average number of customers in the system
d. The average time a customer spends waiting
e. The average time a customer spends in the system
f. The probability that arriving customers will have to wait for service
:: Solution ::
step: 1 of 6
Single-channel waiting line model with Poisson arrivals and exponential service times
Determine the operating characteristics for the system.
(a)
The probability that no customers are in the system
:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 2 of 6
(b)
The average number of customers waiting
:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 3 of 6
(c)
The average number of customers in the system
It is given that the average arrival rate of customer
:
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (2)].
Now, calculate the average number of customers in the system as follows:
is 1.333 [refer to
…… (3)
step: 4 of 6
(d)
The average time a customer spends waiting
:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (2)].
is 1.333 [refer to
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 5 of 6
(e)
The average time a customer spends waiting
It is given that the average arrival rate of customer
:
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (4)].
Now, calculate the average time a customer spends waiting as follows:
…… (5)
step: 6 of 6
is 3.333 [refer to
(f)
The probability that arriving customers will have to wait for service
:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute.
Now, calculate the probability that arriving customers will have to wait for service as follows:
…… (6)
Likes: 9
4. Use the single-server drive-up bank teller operation referred to in Problems 1–3 to determine
the probabilities of 0, 1, 2, and 3 customers in the system. What is the probability that more than
three customers will be in the drive-up teller system at the same time?
:: Solution ::
step: 1 of 6
Determine the probabilities of 0, 1, 2, and 3 customers in the system
The probability that no customers are in the system
:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 2 of 6
Probability of 0 customers in the system:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (1)].
Now, calculate probability of 0 customers in the system as follows:
is 0.3333 [refer to
…… (2)
step: 3 of 6
Probability of 1 customer in the system:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (1)].
is 0.3333 [refer to
Now, calculate probability of 1 customer in the system as follows:
…… (3)
step: 4 of 6
Probability of 2 customers in the system:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (1)].
is 0.3333 [refer to
Now, calculate probability of 2 customers in the system as follows:
…… (4)
step: 5 of 6
Probability of 3 customers in the system:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute, and the calculated
Equation (1)).
is 0.3333 (refer to
Now, calculate probability of 3 customers in the system as follows:
…… (5)
The following table shows the probability of n customers in the system:
n Probability
0 0.3333[refer to equation (2)]
1 0.2222 [refer to equation (3)]
2 0.1481 [refer to equation (4)]
3 0.0988 [refer to equation (5)]
step: 6 of 6
Probability that more than 3 customers in the system at the same time:
It is given that the average arrival rate of customer
is 0.4 customers per minute and the
average service rate of customer (µ) is 0.6 per minute.
Hence, the probability of more than 3 customers in the system at the same time is 0.1967.
Likes: 3
5. The reference desk of a university library receives requests for assistance. Assume that a
Poisson probability distribution with an arrival rate of 10 requests per hour can be used to
describe the arrival pattern and that service times follow an exponential probability distribution
with a service rate of 12 requests per hour.
a. What is the probability that no requests for assistance are in the system?
b. What is the average number of requests that will be waiting for service?
c. What is the average waiting time in minutes before service begins?
d. What is the average time at the reference desk in minutes (waiting time plus service time)?
e. What is the probability that a new arrival has to wait for service?
:: Solution ::
step: 1 of 6
Single-channel waiting line model with Poisson arrivals and exponential service times
Determine the operating characteristics for the system.
(a)
The probability that no requests for assistance are in the system (
):
It is assumed that the Poisson probability distribution has the arrival rate of customer ( ) is 10
requests per hour and the average service rate of request (µ) is 12 per hour.
Now, calculate the probability that no requests are in the system as follows:
…… (1)
step: 2 of 6
(b)
step: 3 of 6
The average number of requests that will be waiting for service (
):
It is assumed that the Poisson probability distribution has the arrival rate of customer ( ) is 10
requests per hour and the average service rate of request (µ) is 12 per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 4 of 6
(c)
The average time a customer spends waiting (
):
It is assumed that the Poisson probability distribution has the arrival rate of customer ( ) is 10
requests per hour and the average service rate of request (µ) is 12 per hour, and the calculated
4.6667 [refer to Equation (2)].
is
Now, calculate the average time a customer spends waiting as follows:
…… (3)
step: 5 of 6
(d)
The average time at the reference desk in minutes (
):
It is assumed that the Poisson probability distribution has the arrival rate of customer ( ) is 10
requests per hour and the average service rate of request (µ) is 12 per hour, and the calculated
0.4167 [refer to Equation (3)].
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 6 of 6
(e)
The probability that a new arrival has to wait for service
:
It is assumed that the Poisson probability distribution has the arrival rate of customer ( ) is 10
requests per hour and the average service rate of request (µ) is 12 per hour.
Now, calculate the probability that arriving customers will have to wait for service as follows:
…… (5)
is
Likes: 9
6. Movies Tonight is a typical video and DVD movie rental outlet for home-viewing customers.
During the weeknight evenings, customers arrive at Movies Tonight with an arrival rate of 1.25
customers per minute. The checkout clerk has a service rate of 2 customers per minute. Assume
Poisson arrivals and exponential service times.
a. What is the probability that no customers are in the system?
b. What is the average number of customers waiting for service?
c. What is the average time a customer waits for service to begin?
d. What is the probability that an arriving customer will have to wait for service?
e. Do the operating characteristics indicate that the one-clerk checkout system provides an
acceptable level of service?
:: Solution ::
step: 1 of 7
Single-channel waiting line model with Poisson arrivals and exponential service times
Determine the operating characteristics for the system.
(a)
The probability that no customers are in the system
:
It is given that the average arrival rate of customer ( ) is 1.25 customers per minute and the
average service rate of customer (µ) is 2 per minute.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 2 of 7
(b)
step: 3 of 7
The average number of customers waiting for service
:
step: 4 of 7
It is given that the average arrival rate of customer ( ) is 1.25 customers per minute and the
average service rate of customer (µ) is 2 per minute.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 5 of 7
(c)
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 1.25 customers per minute and the
average service rate of customer (µ) is 2per minute, and the calculated
Equation (2)].
is 1.0417 [(refer to
Now, calculate the average time a customer spends waiting as follows:
…… (3)
step: 6 of 7
(d)
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 1.25 customers per minute and the
average service rate of customer (µ) is 2 per minute.
Now, calculate the probability that arriving customers will have to wait for service as follows:
…… (4)
step: 7 of 7
(e)
Level of service:
The above operating characteristics for the one-clerk checkout system provide the reasonable level
of service to the customers. The average time a customer spends in a line is 50 seconds. Hence, it is
considered as an acceptable level of service.
Likes: 2
7. Speedy Oil provides a single-server automobile oil change and lubrication service. Customers
provide an arrival rate of 2.5 cars per hour. The service rate is 5 cars per hour. Assume that arrivals
follow a Poisson probability distribution and that service times follow an exponential probability
distribution.
a. What is the average number of cars in the system?
b. What is the average time that a car waits for the oil and lubrication service to begin?
c. What is the average time a car spends in the system?
d. What is the probability that an arrival has to wait for service?
:: Solution ::
step: 1 of 6
(a)
step: 2 of 6
Calculation of average number of cars in the system:
step: 3 of 6
The average number of customers waiting for service
It is given that the average arrival rate of customer
rate of customer (µ) is 5 cars per hour.
:
is 2.5 cars per hour, and the average service
Now, calculate the average number of customers in the waiting line as follows:
…… (1)
The average number of cars in the system
:
It is given that the average arrival rate of customer
is 2.5 cars per hour, and the average service
rate of customer (µ) is 5 cars per hour, and the calculated
is 0.5000 (refer to Equation (1)).
Now, calculate the average number of cars in the system as follows:
…… (2)
step: 4 of 6
(b)
The average time that a car waits for the oil and lubrication service to begin
It is given that the average arrival rate of customer
:
is 2.5 cars per hour, and the average service
rate of customer (µ) is 5 cars per hour, and the calculated
is 0.5000 (refer to Equation (1)).
Now, calculate the average time a car spends waiting for the service as follows:
…… (3)
step: 5 of 6
(c)
The average time a car spends in the system
It is given that the average arrival rate of customer
:
is 2.5 cars per hour, and the average service
rate of customer (µ) is 5 cars per hour, and the calculated
is 0.2000 (refer to Equation (3)).
Now, calculate the average time a car spends in the system as follows:
step: 6 of 6
(d)
The probability that a customer on arrival has to wait for service
It is given that the average arrival rate of customer
rate of customer (µ) is 5 cars per hour.
:
is 2.5 cars per hour, and the average service
Now, calculate the probability that arriving customers will have to wait for service as follows:
Likes: 5
8. For the Burger Dome single-server waiting line in Section 11.2, assume that the arrival rate is
increased to 1 customer per minute and that the service rate is increased to 1.25 customers per
minute. Compute the following operating characteristics for the new system: P0, Lq, L, Wq, W,
and Pw. Does this system provide better or poorer service compared to the original system?
Discuss any differences and the reason for these differences.
:: Solution ::
step: 1 of 9
Single-channel waiting line model with Poisson arrivals and exponential service times
Determine the operating characteristics for the system
(a)
The probability that no customers are in the system
:
It is given that the average arrival rate of customer
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 2 of 9
(b)
step: 3 of 9
The average number of customers waiting
:
It is given that the average arrival rate of customer
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 4 of 9
(c)
The average number of customers in the system
It is given that the average arrival rate of customer
:
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute, and the calculated
Equation (2)].
Now, calculate the average number of customers in the system as follows:
…… (3)
is 3.2 [refer to
step: 5 of 9
(d)
The average time a customer spends waiting
:
It is given that the average arrival rate of customer
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute, and the calculated
Equation (2)].
is 3.2 [refer to
Now, calculate the average time a customer spends waiting as follows:
…… (4)
(e)
The average time a customer spends waiting
It is given that the average arrival rate of customer
:
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute, and the calculated
Equation (4)].
is 3.2 [refer to
Now, calculate the average time a customer spends waiting as follows:
…… (5)
step: 6 of 9
(f)
The probability that arriving customers will have to wait for service
:
It is given that the average arrival rate of customer
is 1 customer per minute and the average
service rate of customer (µ) is 1.25 customers per minute.
Now, calculate the probability that arriving customers will have to wait for service as follows:
…… (6)
step: 7 of 9
Comparison between the original system and the new system:
step: 8 of 9
Justification:
step: 9 of 9
• The service rate (µ) is increased from 1 to 1.25 customers per minute.
• The arrival rate
of customer is increased from 0.75 to 1 customer per minute even if the
service rate is high. It shows the poor quality of service to the customers because of higher arrival
rate.
• The probability of average waiting time increases in the new system when compared to the
original system.
9. Marty’s Barber Shop has one barber. Customers have an arrival rate of 2.2 customers per hour,
and haircuts are given with a service rate of 5 per hour. Use the Poisson arrivals and exponential
service times model to answer the following questions:
a. What is the probability that no units are in the system?
b. What is the probability that one customer is receiving a haircut and no one is waiting?
c. What is the probability that one customer is receiving a haircut and one customer is waiting?
d. What is the probability that one customer is receiving a haircut and two customers are waiting?
e. What is the probability that more than two customers are waiting?
f. What is the average time a customer waits for service?
:: Solution ::
step: 1 of 8
Single-channel waiting line model with Poisson arrivals and exponential service times
Determine the operating characteristics for the system:
(a)
The probability that no units are in the system (
):
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 2 of 8
(b)
step: 3 of 8
The probability that one customer is receiving a haircut and no one is waiting (
):
step: 4 of 8
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour, and the calculated
(1)].
is 0.56 [refer to Equation
Now, calculate the probability that one customer is receiving and no one is waiting as follows:
…… (2)
step: 5 of 8
(c)
The probability that one customer is receiving a haircut and one customer is waiting (
):
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour, and the calculated
(1)].
is 0.56 [refer to Equation
Now, calculate the probability that one customer is receiving and one customer is waiting as follows:
…… (3)
step: 6 of 8
(d)
The probability that one customer is receiving a haircut and two customers are waiting (
):
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour, and the calculated
(1)].
is 0.56 [refer to Equation
Now, calculate the probability that one customer is receiving and two customers are waiting as
follows:
…… (4)
step: 7 of 8
(e)
The probability that more than two customers are waiting
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour, and the calculated
(1)], is 0.2464 [refer to Equation (2)],
Equation (4)].
is 0.56 [refer to Equation
is 0.1084 [refer to Equation (3)], and
is 0.0477 (refer to
Now, calculate the probability that more than two customers are waiting as follows:
step: 8 of 8
(f)
The average number of customers in the system (
)
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (5)
The average time a customer spends in the waiting line (
):
It is given that the average arrival rate of customer ( ) is 2.2 customer per hour and the average
service rate of customer (µ) is 5 customers per hour, and the calculated
equation (5)].
is 0.3457 [refer to
Now, calculate the average time a customer spends waiting as follows:
Likes: 4
10. Trosper Tire Company decided to hire a new mechanic to handle all tire changes for customers
ordering a new set of tires. Two mechanics applied for the job. One mechanic has limited
experience, can be hired for $14 per hour, and can service an average of three customers per hour.
The other mechanic has several years of experience, can service an average of four customers per
hour, but must be paid $20 per hour. Assume that customers arrive at the Trosper garage at the
rate of two customers per hour.
a. What are the waiting line operating characteristics using each mechanic, assuming Poisson
arrivals and exponential service times?
b. If the company assigns a customer waiting cost of $30 per hour, which mechanic provides the
lower operating cost?
:: Solution ::
step: 1 of 14
(a)
step: 2 of 14
Waiting line operating characteristics when average number of customer per hour (µ) is 3:
step: 3 of 14
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (1)
step: 4 of 14
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour, and the calculated
is 1.3333 [refer to Equation (1)].
Now, calculate the average number of customers in the system as follows:
…… (2)
step: 5 of 14
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour, and the calculated
is 1.3333 [refer to Equation (1)].
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 6 of 14
The average time a customer spends in the system (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour, and the calculated
is 0.6667 [refer to Equation (4)].
Now, calculate the average time a customer spends in the system as follows:
step: 7 of 14
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour.
Now, calculate the probability that arriving customers will have to wait for service as follows:
Waiting line operating characteristics when average number of customer per hour (µ) is 4:
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 4 per hour.
step: 8 of 14
Now, calculate the average number of customers in the waiting line as follows:
…… (5)
step: 9 of 14
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 3 per hour, and the calculated
is 0.5000 [refer to Equation (5)].
Now, calculate the average number of customers in the system as follows:
…… (6)
step: 10 of 14
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 4 per hour, and the calculated
is 0.5000 [refer to Equation (5)].
Now, calculate the average time a customer spends waiting as follows:
…… (7)
step: 11 of 14
The average time a customer spends in the system (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 4 per hour, and the calculated
is 0.2500 [refer to Equation (7)].
Now, calculate the average time a customer spends in the system as follows:
step: 12 of 14
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 2 customers per hour and the average
service rate of customer (µ) is 4 per hour.
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 13 of 14
(b)
Operating cost of hiring new mechanic:
It is given that the company assigns a customer waiting cost of $30 per hour, hiring cost of one new
mechanic is $14 per hour, and the calculated average number of customer
Equation (2)].
is 2 [refer to
Now, calculate the operating cost of new mechanic as follows:
step: 14 of 14
Operating cost of hiring experienced mechanic:
It is given that the company assigns a customer waiting cost of $30 per hour, hiring cost of one
experienced mechanic is $20 per hour, and the calculated average number of customer
[refer to Equation (6)].
is 1
Now, calculate the operating cost of new mechanic as follows:
Justification:
The operating cost of hiring new mechanic is higher than the cost of hiring experienced mechanic.
So, it is better to hire the experienced mechanic.
Likes: 0
Dislikes: 2
11. Agan Interior Design provides home and office decorating assistance to its customers. In
normal operation, an average of 2.5 customers arrive each hour. One design consultant is
available to answer customer questions and make product recommendations. The consultant
averages 10 minutes with each customer.
a. Compute the operating characteristics of the customer waiting line, assuming Poisson arrivals
and exponential service times.
b. Service goals dictate that an arriving customer should not wait for service more than an average
of 5 minutes. Is this goal being met? If not, what action do you recommend?
c. If the consultant can reduce the average time spent per customer to 8 minutes, what is the
mean service rate? Will the service goal be met?
:: Solution ::
step: 1 of 8
(a)
Operating characteristics of the customer waiting line:
It is given that the consultant averages 10 minutes with each customer.
Now, calculate the average service rate per minute as follows:
…… (1)
step: 2 of 8
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)].
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 3 of 8
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)], and the
calculated
is 0.2976 [refer to Equation (2)].
Now, calculate the average number of customers in the system as follows:
…… (3)
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)], and the
calculated
is 0.2976 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 4 of 8
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)], and the
calculated
is 0.1190 [refer to Equation (4)].
Now, calculate the average time a customer spends waiting as follows:
…… (5)
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)].
step: 5 of 8
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 6 of 8
(b)
Justification:
• The service goals are not met since the average time a customer spends waiting
minutes [refer to Equation (4)].
• AIG firm should increase the service rate (
consultant.
is 7.14
) for the consultants, otherwise they can hire a new
step: 7 of 8
(c)
Mean average rate:
It is given that the consultant averages 8 minutes with each customer.
Now, calculate the average service rate per minute as follows:
…… (6)
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 7.5 customers per hour [refer to Equation (6)].
Now, calculate the average number of customers in the waiting line as follows:
…… (7)
step: 8 of 8
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (1)], and the
calculated
is 0.1667 [refer to Equation (7)].
Now, calculate the average time a customer spends waiting as follows:
…… (8)
The service goals are met, when average service rate is 7.5 customers per hour.
Justification:
The average time a customer spends waiting (
than 5 minutes.
) is 4 minutes [refer to Equation (8)], which is less
Likes: 6
12. Pete’s Market is a small local grocery store with only one checkout counter. Assume that
shoppers arrive at the checkout lane according to a Poisson probability distribution, with an arrival
rate of 15 customers per hour. The checkout service times follow an exponential probability
distribution, with a service rate of 20 customers per hour.
a. Compute the operating characteristics for this waiting line.
b. If the manager’s service goal is to limit the waiting time prior to beginning the checkout process
to no more than five minutes, what recommendations would you provide regarding the current
checkout system?
:: Solution ::
step: 1 of 9
(a)
step: 2 of 9
Operating characteristics for the waiting line:
step: 3 of 9
The probability that no units are in the system (
):
It is given that the average arrival rate of customer ( ) is 15 customer per hour, and the average
service rate of customer (µ) is 20 customers per hour.
Now, calculate the probability that no customers are in the system as follows:
…… (1)
step: 4 of 9
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customer per hour, and the average
service rate of customer (µ) is 20 customers per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 5 of 9
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, the average service
rate of customer (µ) is 20 per hour, and the calculated
is 2.25 [refer to Equation (2)].
Now, calculate the average number of customers in the system as follows:
…… (2)
step: 6 of 9
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, the average service
rate of customer (µ) is 20 per hour, and the calculated
is 2.25 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 7 of 9
The average time a customer spends in the system (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, the average service
rate of customer (µ) is 20 per hour, and the calculated
is 0.15 [refer to Equation (4)].
Now, calculate the average time a customer spends in the system as follows:
step: 8 of 9
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, and the average
service rate of customer (µ) is 20 per hour.
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 9 of 9
(b)
Recommendation to the current checkout system:
The current checkout system needs to improve the service time to the customers, since the average
time a customer spends in the system is 9 minutes, which is more than 5 minutes. The
recommended average waiting time is 5 minutes or less than 5 minutes.
Likes: 1
13. After reviewing the waiting line analysis of Problem 12, the manager of Pete’s Market wants to
consider one of the following alternatives for improving service. What alternative would you
recommend? Justify your recommendation.
a. Hire a second person to bag the groceries while the cash register operator is entering the cost
data and collecting money from the customer. With this improved singleserver operation, the
service rate could be increased to 30 customers per hour.
b. Hire a second person to operate a second checkout counter. The two-server operation would
have a service rate of 20 customers per hour for each server.
:: Solution ::
step: 1 of 5
(a)
step: 2 of 5
One checkout system (2 employees)
step: 3 of 5
It is given that the average arrival rate of customer ( ) is 15 customers per hour, and the average
service rate of customer (µ) is 30 customers per hour.
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, and the average
service rate of customer (µ) is 30 customers per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (1)
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, and the average
service rate of customer (µ) is 30 per hour, and the calculated
is 0.5 [refer to Equation (1)].
Now, calculate the average time a customer spends waiting as follows:
step: 4 of 5
(b)
Two-channel checkout system
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, and the average
service rate of customer (µ) is 20 customers per hour.
Note: The table value of
with 0.75 for multiple-channel waiting lines is 0.4545.
Now, calculate the average number of customers waiting as follows:
…… (2)
step: 5 of 5
The average time a customer spends waiting (
):
It is given that the average arrival rate of customer ( ) is 15 customers per hour, the average service
rate of customer (µ) is 20 per hour, and the calculated
is 0.1227 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
Justification:
• Even though the two channel counter system provides better service, the cost of providing service
is high.
• The one checkout counter (2 employees) is recommended since it meets the service goal with the
average time a customer spends waiting (
) of 2.
Likes: 0
Dislikes: 2
14. Ocala Software Systems operates a technical support center for its software customers. If
customers have installation or use problems with Ocala software products, they may telephone
the technical support center and obtain free consultation. Currently, Ocala operates its support
center with one consultant. If the consultant is busy when a new customer call arrives, the
customer hears a recorded message stating that all consultants are currently busy with other
customers. The customer is then asked to hold and is told that a consultant will provide assistance
as soon as possible. The customer calls follow a Poisson probability distribution, with an arrival
rate of five calls per hour. On average, it takes 7.5 minutes for a consultant to answer a customer’s
questions. The service time follows an exponential probability distribution.
a. What is the service rate in terms of customers per hour?
b. What is the probability that no customers are in the system and the consultant is idle?
c. What is the average number of customers waiting for a consultant?
d. What is the average time a customer waits for a consultant?
e. What is the probability that a customer will have to wait for a consultant?
f. Ocala’s customer service department recently received several letters from customers
complaining about the difficulty in obtaining technical support. If Ocala’s customer service
guidelines state that no more than 35% of all customers should have to wait for technical support
and that the average waiting time should be two minutes or less, does your waiting line analysis
indicate that Ocala is or is not meeting its customer service guidelines? What action, if any, would
you recommend?
:: Solution ::
step: 1 of 7
Operating characteristics for the system
step: 2 of 7
(a)
The service rate in terms of customers per hour:
It is given that the consultant takes 7.5 minutes in average with each customer.
Now, calculate the average service rate per minute as follows:
…… (1)
step: 3 of 7
(b)
The probability that no customers are in the system (
):
It is given that the average arrival rate ( ) is 5 calls per hour, and the calculated average service rate
of customer (µ) is 8 calls per hour [refer to Equation (1)].
Now, calculate the probability that no customers are in the system as follows:
…… (2)
step: 4 of 7
(c)
The average number of customers waiting (
):
It is given that the average arrival rate ( ) is 5 calls per hour, and the calculated average service rate
of customer (µ) is 8 calls per hour [refer to Equation (1)].
Now, calculate the average number of customers in the waiting line as follows:
…… (3)
step: 5 of 7
(d)
The average time a customer spends waiting (
):
It is given that the average arrival rate ( ) is 5 calls per hour, the calculated average service rate of
customer (µ) is 8 calls per hour [refer to Equation (1)], and the calculated
Equation (3)].
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 6 of 7
(e)
is 1.0417 [refer to
The probability that arriving customers will have to wait for service (
)
It is given that the average arrival rate ( ) is 5 calls per hour, and the calculated average service rate
of customer (µ) is 8 calls per hour (refer to Equation (1)).
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 7 of 7
(f)
Waiting line analysis:
O’s customer service guidelines:
More than 35% of the customers should not wait and the average waiting time should be 2 minutes
or less.
Justification:
• Waiting line analysis shows 62.5% of customers should wait which is more than 35%.
• The average time a customer spends waiting is 12.5 minutes. But the service guidelines states that
the average waiting time should be 2 minutes or less.
• To improve the service goals of O, it should adjoin one or more consultants.
Likes: 0
15. To improve customer service, Ocala Software Systems (see Problem 14) wants to investigate
the effect of using a second consultant at its technical support center. What effect would the
additional consultant have on customer service? Would two technical consultants enable Ocala to
meet its service guidelines (no more than 35% of all customers having to wait for technical support
and an average customer waiting time of two minutes or less)? Discuss.
:: Solution ::
step: 1 of 4
Analysis of waiting line by using two consultants:
The average number of customers waiting (
):
It is given that the average arrival rate ( ) is 5 calls per hour, the calculated average service rate of
customer (µ) is 8 calls per hour, and the number of channels (k) is 2.
Note: The table value of
with 0.625
for multiple-channel waiting lines is 0.5238.
Now, calculate the average number of calls in the waiting line as follows:
…… (1)
step: 2 of 4
The average time a customer spends waiting (
):
It is given that the average arrival rate ( ) is 5 calls per hour, the calculated average service rate of
customer (µ) is 8 calls per hour, and the calculated
is 0.0676 [refer to Equation (1)].
Now, calculate the average time a customer spends waiting as follows:
step: 3 of 4
Probability of 0 customers in the system:
It is given that the average arrival rate ( ) is 5 calls per hour, and the calculated average service rate
of customer (µ) is 8 calls per hour.
Now, calculate the probability of 1 customer in the system as follows:
…… (2)
step: 4 of 4
Probability that 2 or less customers being in the system at the same time
The table value of
with 0.625 is 0.5238 and the calculated
is 0.3274 [refer to Equation (2)].
Now, calculate the probability of 2 or less customers being in the system at the same time as
follows:
Hence, the probability of 2 or less customers being in the system at the same time is 0.1488.
Justification:
O system met its service guidelines with 14.88% of customers and the average time that a customer
spends in the waiting line is 0.81 minutes, which is less than 2 minutes.
Likes: 1
16. The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume
that Fore and Aft decides to build a docking facility where one boat at a time can stop for gas and
servicing. Assume that arrivals follow a Poisson probability distribution, with an arrival rate of 5
boats per hour, and that service times follow an exponential probability distribution, with a
service rate of 10 boats per hour. Answer the following questions:
a. What is the probability that no boats are in the system?
b. What is the average number of boats that will be waiting for service?
c. What is the average time a boat will spend waiting for service?
d. What is the average time a boat will spend at the dock?
e. If you were the manager of Fore and Aft Marina, would you be satisfied with the service level
your system will be providing? Why or why not?
:: Solution ::
step: 1 of 7
Operating characteristics for the system
step: 2 of 7
(a)
step: 3 of 7
The probability that no boats are in the system (
):
It is given that the average arrival rate ( ) is 5 boats per hour, and the average service rate of
customer (µ) is 10 boats per hour.
Now, calculate the probability that no calls are in the system as follows:
…… (1)
step: 4 of 7
(b)
The average number of boats that will be waiting for service (
):
It is given that the average arrival rate ( ) is 5 boats per hour, and the average service rate of
customer (µ) is 10 boats per hour.
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 5 of 7
(c)
The average time a boat will spend waiting for service (
):
It is given that the average arrival rate ( ) is 5 boats per hour, and the average service rate of
customer (µ) is 10 boats per hour, and the calculated
is 0.50 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
…… (3)
step: 6 of 7
(d)
The average time a boat will spend at the dock (
)
It is given that the average arrival rate ( ) is 5 boats per hour, and the average service rate of
customer (µ) is 10 boats per hour, and the calculated
is 0.1 [refer to Equation (3)].
Now, calculate the average time a customer spends waiting as follows:
step: 7 of 7
(e)
Level of service:
The level of service is said to be satisfied, since the average time a boat will spend waiting for service
(
) is only 6 minutes, which is acceptable.
Likes: 3
17. The manager of the Fore and Aft Marina in Problem 16 wants to investigate the possibility of
enlarging the docking facility so that two boats can stop for gas and servicing simultaneously.
Assume that the arrival rate is 5 boats per hour and that the service rate for each server is 10
boats per hour.
a. What is the probability that the boat dock will be idle?
b. What is the average number of boats that will be waiting for service?
c. What is the average time a boat will spend waiting for service?
d. What is the average time a boat will spend at the dock?
e. If you were the manager of Fore and Aft Marina, would you be satisfied with the service level
your system will be providing? Why or why not?
:: Solution ::
step: 1 of 5
Analysis of waiting line by using two consultants:
(a)
The probability that the boat dock will be idle:
It is given that the average arrival rate ( ) is 5 boats per hour, the average service rate of customer
(µ) is 10 boats per hour.
Now, calculate the probability that the boat dock will be idle as follows:
From the table, using the ratio of 0.50 for 2 numbers of channels (k),
probability that the boat dock will be idle is 0.60.
is 0.60. Thus, the
step: 2 of 5
(b)
The average number of boats waiting for service (
):
It is given that the average arrival rate ( ) of 5 boats per hour, the average service rate of customer
(µ) is 10 boats per hour, and the number of channels (k) is 2.
Note: The table value of
with 0.50
for multiple-channel waiting lines is 0.60.
Now, calculate the average number of boats in the waiting line as follows:
…… (1)
step: 3 of 5
(c)
The average time a boat will spend waiting for service (
):
It is given that the average arrival rate ( ) of 5 boats per hour, the average service rate of customer
(µ) is 10 boats per hour, and the calculated
is 0.0333 [refer to Equation (1)].
Now, calculate the average time a boat spends waiting as follows:
…… (2)
step: 4 of 5
(d)
The average time a boat spends will spend at the dock (
)
It is given that the average arrival rate ( ) of 5 boats per hour, the average service rate of customer
(µ) is 10 boats per hour, and the calculated
is 0.0067 [refer to Equation (2)].
Now, calculate the average time a boat spends waiting as follows:
step: 5 of 5
(e)
Analysis of waiting line:
• The services which provide to the customer is much better, since the average time a customer
spends waiting for service is only 24.12 seconds (
) [refer to Equation (2)].
• 60% will be idle in both the channels.
Likes: 3
18. All airplane passengers at the Lake City Regional Airport must pass through a security
screening area before proceeding to the boarding area. The airport has three screening stations
available, and the facility manager must decide how many to have open at any particular time.
The service rate for processing passengers at each screening station is 3 passengers per minute. On
Monday morning the arrival rate is 5.4 passengers per minute. Assume that processing times at
each screening station follow an exponential distribution and that arrivals follow a Poisson
distribution.
a. Suppose two of the three screening stations are open on Monday morning. Compute the
operating characteristics for the screening facility.
b. Because of space considerations, the facility manager’s goal is to limit the average number of
passengers waiting in line to 10 or fewer. Will the two-screening-station system be able to meet
the manager’s goal?
c. What is the average time required for a passenger to pass through security screening?
:: Solution ::
step: 1 of 13
(a)
step: 2 of 13
Operating characteristics for the screening facility:
step: 3 of 13
The probability that no customers in the system:
step: 4 of 13
It is given that the average arrival rate ( ) is 5.4 passengers per minute and the average service rate
(µ) is 3 passengers per minute.
Now, calculate the ratio to find out the value of
as follows:
From the table, using the ratio of 1.80 for 2 numbers of channels (k), the table value of
is 0.0526.
step: 5 of 13
The average number of passengers waiting for service (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the number of channels (k) is 2.
Note: The table value of
with 1.80
for multiple-channel waiting lines is 0.0526.
Now, calculate the average number of passengers in the waiting line as follows:
…… (1)
step: 6 of 13
The average number of passengers in the system (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the calculated
is 7.67 [refer to Equation (1)].
Now, calculate the average number of passengers in the system as follows:
step: 7 of 13
The average time a passenger spends waiting (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the calculated
is 7.67 [refer to Equation (1)].
Now, calculate the average time a passenger spends waiting as follows:
…… (2)
step: 8 of 13
The average time a passenger spends in the system (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the calculated
is 1.42 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
step: 9 of 13
step: 10 of 13
The probability that a passenger has to wait for service:
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the number of channel (k) is 2.
Now, calculate the probability that a passenger has to wait for service as follows:
step: 11 of 13
(b)
step: 12 of 13
The two-screening station is able to meet the service goal since the average number of passengers
in the waiting line is 7.67, which is less than 10.
step: 13 of 13
(c)
The average time a passenger spends in the system (
)
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 3 passengers per minute, and the calculated
is 1.42 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
Hence, the average time for a passenger to pass through security screening is 1.75 minutes.
Likes: 2
19. Refer again to the Lake City Regional Airport described in Problem 18. When the security level
is raised to high, the service rate for processing passengers is reduced to 2 passengers per minute
at each screening station. Suppose the security level is raised to high on Monday morning. The
arrival rate is 5.4 passengers per minute.
a. The facility manager’s goal is to limit the average number of passengers waiting in line to 10 or
fewer. How many screening stations must be open in order to satisfy the manager’s goal?
b. What is the average time required for a passenger to pass through security screening?
:: Solution ::
step: 1 of 9
(a)
step: 2 of 9
Operating characteristics for the screening facility:
step: 3 of 9
The probability that no passengers in the system:
step: 4 of 9
It is given that the average arrival rate ( ) is 5.4 passengers per minute and the average service rate
(µ) is 2 passengers per minute.
Now, calculate the ratio to find out the value of
as follows:
From the table, using the ratio of 2.7 for 3 numbers of channels (k), the table value of
is 0.02525.
step: 5 of 9
The average number of passengers waiting for service (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 2 passengers per minute, and the number of channels (k) is 3.
Note: The table value of
with 2.7
for multiple-channel waiting lines is 0.02525.
Now, calculate the average number of passengers in the waiting line as follows:
…… (1)
step: 6 of 9
Justification:
step: 7 of 9
The average number of passengers waiting for service is 7.45 customers, which is less than10.
Hence, the service goal of manager is being met with three-screening station.
step: 8 of 9
(b)
The average time a passenger spends waiting (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 2 passengers per minute, and the calculated
is 7.45 [refer to Equation (1)].
Now, calculate the average time a customer spends waiting as follows:
…… (2)
step: 9 of 9
The average time a passenger spends in the system (
):
It is given that the average arrival rate ( ) is 5.4 passengers per minute, the average service rate (µ)
is 2 passengers per minute, and the calculated
is 1.38 [refer to Equation (2)].
Now, calculate the average time a customer spends waiting as follows:
Hence, the average time required for the passenger to pass through security screening is 1.88
minutes.
Likes: 2
20. A Florida coastal community experiences a population increase during the winter months, with
seasonal residents arriving from northern states and Canada. Staffing at a local post office is often
in a state of change due to the relatively low volume of customers in the summer months and the
relatively high volume of customers in the winter months. The service rate of a postal clerk is 0.75
customers per minute. The post office counter has a maximum of three workstations. The target
maximum time a customer waits in the system is five minutes.
a. For a particular Monday morning in November, the anticipated arrival rate is 1.2 customers per
minute. What is the recommended staffing for this Monday morning? Show the operating
characteristics of the waiting line.
b. A new population growth study suggests that over the next two years the arrival rate at the
postal office during the busy winter months can be expected to be 2.1 customers per minute. Use
a waiting line analysis to make a recommendation to the post office manager.
:: Solution ::
step: 1 of 16
(a)
step: 2 of 16
Operating characteristics of the waiting line, when k = 2:
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute.
Now, calculate the ratio to find out the value of
as follows:
The arrival rate cannot be handled by one postal clerk since it is less than 1. Hence, there is no value
for
.
step: 3 of 16
Operating characteristics of the waiting line, when k = 2:
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute.
Now, calculate the ratio to find out the value of
as follows:
The arrival rate cannot be handled by one postal clerk since it is less than 1. Hence, there is no value
for
.
From the table, using the ratio of 1.6 for 2 postal clerks (k), the table value of
is 0.1111.
step: 4 of 16
The average number of customers waiting for service (
):
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute.
Note: The table value of
with 1.80
for multiple-channel waiting lines is 0.1111.
Now, calculate the average number of customers in the waiting line as follows:
…… (1)
step: 5 of 16
The average number of customers in the system (
):
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
2.8444 [refer to Equation (1)].
is
Now, calculate the average number of passengers in the system as follows:
step: 6 of 16
The average time a customer spends waiting (
):
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
2.8444 [refer to Equation (1)].
Now, calculate the average time a customer spends waiting as follows:
is
…… (2)
step: 7 of 16
The average time a customer spends in the system (
):
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
2.3704 [refer to Equation (2)].
is
Now, calculate the average time a customer spends waiting as follows:
step: 8 of 16
step: 9 of 16
The probability that a customer has to wait for service:
It is given that the average arrival rate of a postal clerk ( ) is 1.2 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the number of channel (k)
is 2.
Now, calculate the probability that a customer has to wait for service as follows:
step: 10 of 16
Hence, the average time that a customer spends waiting is 3.7037 minutes.
step: 11 of 16
(b)
Operating characteristics of the waiting line, when k = 3:
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute.
Now, calculate the ratio to find out the value of
as follows:
The arrival rate cannot be handled by one postal clerk since it is less than 1. Hence, there is value for
.
From the table, using the ratio of 2.8 for 3 postal clerks (k), the table value of
is 0.0160.
step: 12 of 16
The average number of customers waiting for service (
):
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute.
Note: The table value of
with 2.8
for multiple-channel waiting lines is 0.0160.
step: 13 of 16
Now, calculate the average number of customers in the waiting line as follows:
…… (3)
The average number of customers in the system (
):
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
[refer to Equation (3)].
is 12.3
Now, calculate the average number of passengers in the system as follows:
step: 14 of 16
The average time a customer spends waiting (
):
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
[refer to Equation (3)].
is 12.3
Now, calculate the average time of a customer spends waiting as follows:
…… (4)
step: 15 of 16
The average time a customer spends in the system (
):
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the calculated
[refer to Equation (4)].
Now, calculate the average time a customer spends waiting as follows:
step: 16 of 16
is 5.8
The probability that a customer has to wait for service:
It is given that the average arrival rate of a postal clerk ( ) is 2.1 customers per minute and the
average service rate (µ) of a postal clerk is 0.75 customer per minute, and the number of channel (k)
is 3.
Now, calculate the probability that a customer has to wait for service as follows:
Justification:
The average time that a customer spends in the system is 7.1 minutes and the average number of
customers in the system is 15.1, which is not the acceptable level of service. Hence, the post office
should expand more than three postal clerks.
Likes: 1
21. Refer to the Agan Interior Design situation in Problem 11. Agan’s management would like to
evaluate two alternatives:
• Use one consultant with an average service time of 8 minutes per customer.
• Expand to two consultants, each of whom has an average service time of 10 minutes per
customer.
If the consultants are paid $16 per hour and the customer waiting time is valued at $25 per hour
for waiting time prior to service, should Agan expand to the two-consultant system? Explain.
:: Solution ::
step: 1 of 8
Operating characteristics of the customer waiting line (one consultant)
It is given that one consultant’s average service time is 8 minutes per customer.
Now, calculate the average service rate per minute as follows:
…… (1)
step: 2 of 8
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 7.5 customers per hour [refer to Equation (1)].
Now, calculate the average number of customers in the waiting line as follows:
…… (2)
step: 3 of 8
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 7.5 customers per hour [refer to Equation (1)] and
0.1667 [refer to equation (2)].
is
Now, calculate the average number of customers in the system as follows:
…… (3)
step: 4 of 8
Total cost of one consultant:
It is given that the consultants are paid $16 per hour and the waiting time cost is $25 per hour, and
the calculated average number of customer
is 0.50 [refer to Equation (2)].
Now, calculate the total cost of one consultant as follows:
…… (4)
step: 5 of 8
Operating characteristics of the customer waiting line (two consultants)
It is given that one consultant’s average service time is 10 minutes per customer.
Now, calculate the average service rate per minute as follows:
…… (5)
The average number of customers waiting (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (5)].
Note: The table value of
with 0.4167
for multiple-channel waiting lines is 0.6552.
step: 6 of 8
Now, calculate the average number of passengers in the waiting line as follows:
…… (6)
step: 7 of 8
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 2.5 customers per hour, and the calculated
average service rate of customer (µ) is 6 customers per hour [refer to Equation (5)] and
[refer to equation (6)].
Now, calculate the average number of customers in the system as follows:
is 0.0189
…… (7)
step: 8 of 8
Total cost of two consultants:
It is given that the consultants are paid $16 per hour and the waiting time cost is $25 per hour, and
the calculated average number of customer
is 0.4356 [refer to Equation (5)].
Now, calculate the total cost of one consultant as follows:
…… (8)
Justification:
The total cost of one consultant is $28.50 [refer to Equation (5)] and the total cost of two
consultants is $42.80 [refer to Equation (8)]. Hence, use of one consultant is recommended with the
service time of 8 minutes.
Likes: 1
Dislikes: 2
22. A fast-food franchise is considering operating a drive-up window food-service operation.
Assume that customer arrivals follow a Poisson probability distribution, with an arrival rate of 24
cars per hour, and that service times follow an exponential probability distribution. Arriving
customers place orders at an intercom station at the back of the parking lot and then drive to the
service window to pay for and receive their orders. The following three service alternatives are
being considered:
• A single-server operation in which one employee fills the order and takes the money from the
customer. The average service time for this alternative is 2 minutes.
• A single-server operation in which one employee fills the order while a second employee takes
the money from the customer. The average service time for this alternative is 1.25 minutes.
• A two-server operation with two service windows and two employees. The employee stationed
at each window fills the order and takes the money for customers arriving at the window. The
average service time for this alternative is 2 minutes for each server.
Answer the following questions and recommend one of the design options.
a. What is the probability that no cars are in the system?
b. What is the average number of cars waiting for service?
c. What is the average number of cars in the system?
d. What is the average time a car waits for service?
e. What is the average time in the system?
f. What is the probability that an arriving car will have to wait for service?
:: Solution ::
step: 1 of 23
The service rate in terms of customers per hour:
It is given that the service time is 2 minutes on an average with each customer.
Now, calculate the average service rate per minute as follows:
…… (1)
step: 2 of 23
Operating characteristics for the system when k=1:
(a)
The probability that no cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 30 per hour [refer to Equation (1)].
Now, calculate the probability that no customers are in the system as follows:
…… (2)
step: 3 of 23
(b)
The average number of cars waiting for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 30 per hour [refer to Equation (1)].
Now, calculate the average number of cars in the waiting line as follows:
…… (3)
step: 4 of 23
(c)
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (1)] and the calculated
(3)].
is 3.2000 [refer to Equation
Now, calculate the average number of customers in the system as follows:
…… (4)
step: 5 of 23
(d)
The average time a cars spends for waiting (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (1)] and the calculated
(3)].
Now, calculate the average time a customer spends waiting as follows:
…… (5)
step: 6 of 23
is 3.2000 [refer to Equation
(e)
The average time in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (1)] and the calculated
(5)].
is 0.1333 [refer to Equation
Now, calculate the average time a customer spends waiting as follows:
step: 7 of 23
(f)
The probability that arriving cars will have to wait for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 30 per hour [refer to Equation (1)].
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 8 of 23
Operating characteristics for the system when k=1:
The service rate in terms of customers per hour:
It is given that the service time is 1.25 minutes in average with each customer.
Now, calculate the average service rate per minute as follows:
…… (6)
step: 9 of 23
The probability that no cars are in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 48 per hour [refer to Equation (6)].
Now, calculate the probability that no cars are in the system as follows:
…… (7)
step: 10 of 23
The average number of cars waiting (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 48 per hour [refer to Equation (6)].
Now, calculate the average number of cars in the waiting line as follows:
…… (8)
step: 11 of 23
(c)
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (1)] and the calculated
(8)].
step: 12 of 23
Now, calculate the average number of cars in the system as follows:
is 0.5000 [refer to Equation
…… (9)
step: 13 of 23
(d)
The average time a car waits for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 48 per hour [refer to Equation (6)] and the calculated
(8)].
is 0.5000 [refer to Equation
Now, calculate the average time a car spends waiting as follows:
…… (9)
step: 14 of 23
(e)
The average time in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 48 per hour [refer to Equation (1)] and the calculated
(9)].
Now, calculate the average time a car spends waiting as follows:
step: 15 of 23
(f)
is 0.0208 [refer to Equation
The probability that arriving customers will have to wait for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 48 per hour [refer to Equation (6)].
Now, calculate the probability that arriving customers will have to wait for service as follows:
step: 16 of 23
The service rate in terms of customers per hour:
It is given that the service time is 2 minutes on an average with each customer.
Now, calculate the average service rate per minute as follows:
…… (11)
step: 17 of 23
Operating characteristics for the system when k=2:
(a)
The probability that no cars in the system
It is given that the average arrival rate ( ) is 24 cars per hour, and the calculated average service
rate of customer (µ) is 30 per hour [refer to Equation (11)].
Now, calculate the ratio to find out the value of
as follows:
…… (12)
step: 18 of 23
From the table, using the ratio of 0.8 for 2 numbers of channels (k), the table value of
step: 19 of 23
is 0.4286.
(b)
The average number of cars waiting for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (11)] and the number of channels (k) is 2.
Note: The table value of
with 0.8
for multiple-channel waiting lines is 0.4286.
Now, calculate the average number of cars in the waiting line as follows:
…… (13)
step: 20 of 23
(c)
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (11)] and the calculated
Equation (13)].
Now, calculate the average number of cars in the system as follows:
…… (14)
step: 21 of 23
(d)
is 0.1524 [refer to
The average time a car waits for service (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (11)] and the calculated
Equation (13)].
is 0.1524 [refer to
Now, calculate the average time a car spends waiting as follows:
…… (15)
step: 22 of 23
(e)
The average time in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (11)] and the calculated
Equation (15)].
is 0.0063 [refer to
Now, calculate the average time in the system as follows:
step: 23 of 23
(f)
The probability that a car has to wait for service:
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour [refer to Equation (11)] and the number of channel (k) is 2.
Now, calculate the probability that a car has to wait for service as follows:
The average time a customer spends waiting is 0.0063 in system C, which is less than system A and
system B. Hence, system C provides good service to the customers.
Likes: 3
23. The following cost information is available for the fast-food franchise in Problem 22:
• Customer waiting time is valued at $25 per hour to reflect the fact that waiting time is costly to
the fast-food business.
• The cost of each employee is $6.50 per hour.
• To account for equipment and space, an additional cost of $20 per hour is attributable to each
server.
What is the lowest-cost design for the fast-food business?
:: Solution ::
step: 1 of 17
Lowest-cost design for the fast-food business
step: 2 of 17
System A:
step: 3 of 17
Cost of service per hour for system A:
step: 4 of 17
It is given that the cost of each employee is $6.50 and the additional cost is $20.00.
step: 5 of 17
Now, calculate service cost per channel when k=1 as follows:
step: 6 of 17
…… (1)
step: 7 of 17
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour and the calculated
is 3.2000.
Now, calculate the average number of customers in the system as follows:
…… (2)
step: 8 of 17
Total cost of system A:
It is given that the waiting time is valued at $25 per hour. The calculated service cost per channel is
$26.50 [refer to Equation (1)] and the calculated
is 4.000 [refer to Equation (2)].
Now, calculate the total cost of system A as follows:
step: 9 of 17
System B:
step: 10 of 17
Cost of service per hour for system B:
step: 11 of 17
It is given that the cost of each employee is $6.50 and the additional cost is $20.00.
step: 12 of 17
Now, calculate the service cost per channel when k=2 as follows:
step: 13 of 17
…… (3)
step: 14 of 17
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour and the calculated
is 0.5000.
Now, calculate the average number of cars in the system as follows:
…… (4)
Total cost of system B:
It is given that the waiting time is valued at $25 per hour. The calculated service cost per channel is
$33.00 [refer to Equation (3)] and the calculated is 1.000 [refer to Equation (4)].
Now, calculate the total cost of system B as follows:
step: 15 of 17
System C:
Cost of service per hour for system C:
It is given that the cost of each employee is $6.50 and the additional cost is $20.00.
Now, calculate the service cost per channel when k=1 as follows:
…… (5)
step: 16 of 17
The average number of cars in the system (
):
It is given that the average arrival rate ( ) is 24 cars per hour. The calculated average service rate of
customer (µ) is 30 per hour and the calculated
is 0.152.
Now, calculate the average number of cars in the system as follows:
…… (6)
step: 17 of 17
Total cost of system C:
It is given that the waiting time is valued at $25 per hour. The calculated service cost per channel is
$26.50 [refer to Equation (5)] and the calculated is 0.9524 [refer to Equation (6)].
Now, calculate the total cost of system C as follows:
Likes: 1
24. A study of the multiple-server food-service operation at the Red Birds baseball park shows that
the average time between the arrival of a customer at the food-service counter and his or her
departure with a filled order is 10 minutes. During the game, customers arrive at the rate of four
per minute. The food-service operation requires an average of 2 minutes per customer order.
a. What is the service rate per server in terms of customers per minute?
b. What is the average waiting time in the line prior to placing an order?
c. On average, how many customers are in the food-service system?
:: Solution ::
step: 1 of 6
Operating characteristics of the customer waiting line:
(a)
The service rate per channel in terms of customers per minute:
It is given that the service operation requires an average of 2 minutes per customer order.
Now, calculate the average service rate per minute as follows:
…… (1)
Hence, the service rate per channel is 0.5 customers per minute.
step: 2 of 6
(b)
step: 3 of 6
The average waiting time in the line prior to placing an order:
step: 4 of 6
It is given that the average arrival rate of customer ( ) is 4 customers per minute. The calculated
average service rate of customer (µ) is 0.5 customers per hour [refer to Equation (1)]. The given
average time a customer spends waiting (
) is 10 minutes.
Now, calculate the average time a customer spends waiting as follows:
step: 5 of 6
Hence, the average waiting time in the line prior to placing an order is 8 minutes.
step: 6 of 6
(c)
The average number of customers in the system (
):
It is given that the average arrival rate of customer ( ) is 4 customers per minute. The calculated
average service rate of customer (µ) is 0.5 customers per hour [refer to Equation (1)]. The given
average time a customer spends waiting (
) is 10 minutes.
Now, calculate the average number of customers in the system as follows:
Hence, average number of customers in the food-service system is 40 customers.
Likes: 3
25. To understand how a multiple-server waiting line system with a shared queue compares to a
multiple-server waiting line system with a dedicated queue for each server, reconsider the Burger
Dome example. Suppose Burger Dome establishes two servers but arranges the restaurant layout
so that an arriving customer must decide which server’s queue to join. Assume that this system
equally splits the customer arrivals so that each server sees half of the customers. How does this
system compare with the two-server waiting line system with a shared queue from Section 11.3?
Compare the average number of customers waiting, average number of customers in the system,
average waiting time, and average time in the system.
:: Solution ::
step: 1 of 2
CASE1
When the system has 2 servers with dedicated waiting lines. It looks like
Various performance measures are
1. Expected number of customers in the queue is
Where
is the number of customers at the time t=0
2. Expected number of customers in the system is
3. Waiting time of customers in the queue is
4. Waiting time of customers in the system is
step: 2 of 2
CASE 2
When the system has 2 servers, but they had a single shared queue. It looks like
We can see that it looks like two M/M/1 system.
Various performance measures for each of the system are
1. Expected number of customers in the queue is
Where
is the number of customers at the time t=0
2. Expected number of customers in the system is
3. Waiting time of customers in the queue is
4. Waiting time of customers in the system is
Therefore, the combined results of the performance measure of case 2 are
1. Expected number of customers in the queue is
Where
is the number of customers at the time t=0
2. Expected number of customers in the system is
3. Waiting time of customers in the queue is
4. Waiting time of customers in the system is
By comparing the performance measure of both the cases, we get
Therefore, we can say that multiple server system is typically better when a single shared queue is
used, rather than a multiple dedicated waiting line.
Likes: 0
26. Manning Autos operates an automotive service. To complete their repair work, Manning
mechanics often need to retrieve parts from the company’s parts department counter. Mechanics
arrive at the parts counter at a rate of four per hour. The parts coordinator spends an average of
six minutes with each mechanic, discussing the parts the mechanic needs and retrieving the parts
from inventory.
a. Currently, Manning has one parts coordinator. On average, each mechanic waits four minutes
before the parts coordinator is available to answer questions or retrieve parts from inventory.
Find Lq, W, and L for this single-server parts operation.
b. A trial period with a second parts coordinator showed that, on average, each mechanic waited
only one minute before a parts coordinator was available. Find Lq, W, and L for this two-server
parts operation.
c. If the cost of each mechanic is $20 per hour and the cost of each parts coordinator is $12 per
hour, is the one-server or the two-server system more economical?
:: Solution ::
step: 1 of 22
(a)
step: 2 of 22
Single-channel parts operation:
step: 3 of 22
Arrival rate in terms of mechanics per minute:
step: 4 of 22
It is given that the arrival rate of mechanics per hour is 4.
step: 5 of 22
Now, calculate the arrival rate of patients per minute as follows:
step: 6 of 22
…… (1)
step: 7 of 22
The service rate in terms of customers per hour:
It is given that the service rate of mechanics per hour is 1.
Now, calculate the service rate of mechanics per minute as follows:
…… (2)
step: 8 of 22
The average number of mechanics in the waiting room
:
step: 9 of 22
It is given that a mechanic waits for 4 minutes on an average, and the calculated arrival rate is
0.0667 per minute [refer to Equation (1)].
Now, calculate the average number of mechanics in the waiting room as follows:
step: 10 of 22
The average time a mechanic spends in the system:
It is given that a mechanic waits for 4 minutes on an average, and the calculated service rate is
0.1667 per minute [refer to Equation (2)].
Now, calculate the average time a mechanic spends in the system as follows:
…… (3)
step: 11 of 22
The average number of mechanics in the system (
):
The calculated arrival rate is 0.0667 per minute [refer to Equation (1)], and the average time a
customer spends in the system is 10 minutes [refer to Equation (3)].
Now, calculate the average number of mechanics in the system as follows:
…… (4)
step: 12 of 22
(b)
step: 13 of 22
Two-channel parts operation
step: 14 of 22
The average number of mechanics in the waiting room
:
step: 15 of 22
It is given that a mechanic waits for 1 minute on an average, and the calculated arrival rate is 0.0667
per minute [refer to Equation (1)].
Now, calculate the average number of mechanics in the waiting room as follows:
The average time a mechanic spends in the system:
It is given that a mechanic waits for 1 minute on an average, and the calculated service rate is 0.1667
per minute [refer to Equation (2)].
step: 16 of 22
Now, calculate the average time a mechanic spends in the system as follows:
step: 17 of 22
The average number of mechanics in the system (
):
The calculated arrival rate is 0.0667 per minute [refer to Equation (1)], and the average time a
customer spends in the system is 7 minutes [refer to Equation (3)].
Now, calculate the average number of mechanics in the system as follows:
…… (5)
step: 18 of 22
(c)
step: 19 of 22
Calculation of cost of mechanic:
step: 20 of 22
One-channel system:
step: 21 of 22
It is given that the waiting time is valued at $20 per hour, the service cost per channel is $12, and the
calculated is 0.6667 [refer to Equation (2)].
Now, calculate the total cost of system A as follows:
step: 22 of 22
Two-channel system:
It is given that the waiting time is valued at $20 per hour, the service cost per channel is $12, and the
calculated is 0.4669 [refer to Equation (5)].
Now, calculate the total cost of system A as follows:
Likes: 0
27. Gubser Welding, Inc., operates a welding service for construction and automotive repair jobs.
Assume that the arrival of jobs at the company’s office can be described by a Poisson probability
distribution with an arrival rate of two jobs per 8-hour day. The time required to complete the jobs
follows a normal probability distribution, with a mean time of 3.2 hours and a standard deviation of
2 hours. Answer the following questions, assuming that Gubser uses one welder to complete all jobs:
a. What is the mean arrival rate in jobs per hour?
b. What is the mean service rate in jobs per hour?
c. What is the average number of jobs waiting for service?
d. What is the average time a job waits before the welder can begin working on it?
e. What is the average number of hours between when a job is received and when it is completed?
f. What percentage of the time is Gubser’s welder busy?
:: Solution ::
step: 1 of 14
(a)
step: 2 of 14
Mean arrival rate in jobs per hour:
step: 3 of 14
It is given that the arrival rate of two jobs per 8-hour day.
step: 4 of 14
Now, calculate the arrival rate in jobs per hour as follows:
step: 5 of 14
…… (1)
step: 6 of 14
(b)
step: 7 of 14
Mean service rate in jobs per hour:
step: 8 of 14
It is given that the time required for completing the jobs with a mean time of 3.2.
step: 9 of 14
Now, calculate the mean service rate in jobs as follows:
step: 10 of 14
…… (2)
step: 11 of 14
(c)
The average number of jobs waiting for service (
):
It is given that the time required for completing the jobs with a standard deviation of 2 hours, the
calculated arrival rate is 0.25 per hour [refer to Equation (1)], and the service rate is 0.3125 per hour
[refer to Equation (2)].
Now, calculate the average number of jobs waiting for service as follows:
…… (3)
step: 12 of 14
(d)
The average time a customer spends waiting (
):
The calculated arrival rate is 0.25 per hour [refer to Equation (1)], and the average number of jobs
waiting for service (
) is 2.225.
Now, calculate the average time a customer spends waiting as follows:
…… (4)
step: 13 of 14
(e)
The average time a customer spends in the system:
The calculated service rate is 0.3125 per hour [refer to Equation (2)], and the average time a
customer spends waiting (
) is 8.9 hours [refer to Equation (4)].
Now, calculate the average time a customer spends in the system as follows:
step: 14 of 14
(f)
The probability that arriving customers will have to wait for service (
):
The calculated arrival rate is 0.25 per hour [refer to Equation (1)], and the service rate is 0.3125 per
hour [refer to Equation (2)].
Now, calculate the probability that arriving customers will have to wait for service as follows:
Hence, 80% of the time is considered to be as busy.
Likes: 6
28. Jobs arrive randomly at a particular assembly plant; assume that the arrival rate is five jobs per
hour. Service times (in minutes per job) do not follow the exponential probability distribution.
Two proposed designs for the plant’s assembly operation are shown.
a. What is the service rate in jobs per hour for each design?
b. For the service rates in part (a), what design appears to provide the best or fastest service rate?
c. What are the standard deviations of the service times in hours?
d. Use the M/G/1 model to compute the operating characteristics for each design.
e. Which design provides the best operating characteristics? Why?
:: Solution ::
step: 1 of 17
(a)
step: 2 of 17
The service rate in jobs per hour for each design
Service rate of design A:
It is given that the mean service time for design A is 6.0.
Now, calculate the service rate of design A:
…… (1)
step: 3 of 17
Service rate of design B:
It is given that the mean service time for design B is 6.25.
Now, calculate the service rate of design B:
…… (2)
step: 4 of 17
(b)
Level of service:
Design A appears to provide the best service rate since the service rate of A is higher than the
service rate of design B.
step: 5 of 17
(c)
The standard deviation of the service times in hours:
Design A:
It is given that the standard deviation of design A is 3.0 per minute.
Now, calculate standard deviation of the service time in hours as follows:
…… (3)
step: 6 of 17
Design B:
It is given that the standard deviation of design A is 0.6 per minute.
Now, calculate the standard deviation of the service time in hours as follows:
…… (4)
step: 7 of 17
(d)
Operating characteristics for each design by using
model:
For Design A:
The probability that no units are in the system (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated average
service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)].
Now, calculate the probability that no jobs are in the system as follows:
step: 8 of 17
The average number of jobs waiting for service (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour. The calculated average
service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)] and the calculated standard
deviation is 0.05 per hour [refer to Equation (3)].
Now, calculate the average number of jobs waiting for service as follows:
…… (5)
step: 9 of 17
The average number of jobs in the system (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour. The calculated average
service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)] and the calculated
0.3125 [refer to Equation (3)].
Now, calculate the average number of jobs in the system as follows:
step: 10 of 17
The average time a jobs spends waiting (
):
is
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated
0.3125 [refer to Equation (3)].
is
Now, calculate the average time a job spends waiting as follows:
…… (6)
step: 11 of 17
The average time a job spends in the system (
):
The calculated average service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)], and
the calculated
is 0.0625 [refer to Equation (6)].
Now, calculate the average time a customer spends in the system as follows:
step: 12 of 17
The probability that arriving jobs will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated average
service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)].
Now, calculate the probability that arriving customer will have to wait for service as follows:
step: 13 of 17
For Design B:
The probability that no units are in the system (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated average
service rate of customer (µ) is 9.6 jobs per hour [refer to Equation (2)].
Now, calculate the probability that no jobs are in the system as follows:
The average number of jobs waiting for service (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour. The calculated average
service rate of customer (µ) is 9.6 jobs per hour [refer to Equation (2)] and the calculated standard
deviation is 0.01 per hour [refer to Equation (4)].
Now, calculate the average number of jobs waiting for service as follows:
…… (7)
step: 14 of 17
The average number of jobs in the system (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour. The calculated average
service rate of customer (µ) is 9.6 jobs per hour [refer to Equation (2)] and the calculated
0.2857 [refer to Equation (7)].
Now, calculate the average number of jobs in the system as follows:
is
step: 15 of 17
The average time a jobs spends waiting (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated
0.2857 [refer to Equation (7)].
is
Now, calculate the average time a job spends waiting as follows:
…… (8)
step: 16 of 17
The average time a job spends in the system (
):
The calculated average service rate of customer (µ) is 10 jobs per hour [refer to Equation (1)], and
the calculated
is 0.0571 [refer to Equation (8)].
Now, calculate the average time a customer spends in the system as follows:
step: 17 of 17
The probability that arriving jobs will have to wait for service (
):
It is given that the average arrival rate of customer ( ) is 5 jobs per hour, and the calculated average
service rate of customer (µ) is 9.6 jobs per hour [refer to Equation (2)].
Now, calculate the probability that arriving job will have to wait for service as follows:
Likes: 0
29. The Robotics Manufacturing Company operates an equipment repair business where
emergency jobs arrive randomly at the rate of three jobs per 8-hour day. The company’s repair
facility is a single-server system operated by a repair technician. The service time varies, with a
mean repair time of 2 hours and a standard deviation of 1.5 hours. The company’s cost of the
repair operation is $28 per hour. In the economic analysis of the waiting line system, Robotics uses
$35 per hour cost for customers waiting during the repair process.
a. What are the arrival rate and service rate in jobs per hour?
b. Show the operating characteristics, including the total cost per hour.
c. The company is considering purchasing a computer-based equipment repair system that would
enable a constant repair time of 2 hours. For practical purposes, the standard deviation is 0.
Because of the computer-based system, the company’s cost of the new operation would be $32
per hour. The firm’s director of operations rejected the request for the new system because the
hourly cost is $4 higher and the mean repair time is the same. Do you agree? What effect will the
new system have on the waiting line characteristics of the repair service?
d. Does paying for the computer-based system to reduce the variation in service time make
economic sense? How much will the new system save the company during a 40-hour workweek?
:: Solution ::
step: 1 of 17
(a)
step: 2 of 17
Arrival rate and service rate in jobs per hour:
step: 3 of 17
Mean arrival rate in jobs per hour:
step: 4 of 17
It is given that the arrival rate is three jobs per 8-hour day.
step: 5 of 17
Now, calculate the arrival rate in jobs per hour as follows:
…… (1)
step: 6 of 17
Mean service rate in jobs per hour:
It is given that the time required to complete the jobs is a mean time of 2.
Now, calculate mean service rate in jobs as follows:
…… (2)
step: 7 of 17
(b)
Operating characteristics including the total cost per hour:
The average number of jobs waiting for service
:
It is given that the time required to complete the jobs is a standard deviation of 1.5 hours, the
calculated arrival rate is 0.375per hour [refer to Equation (1)], and the service rate is 0.5 per hour
[refer to Equation (2)].
Now, calculate the average number of jobs waiting for service as follows:
…… (3)
step: 8 of 17
The average number of jobs in the system
:
The calculated arrival rate is 0.375per hour [refer to Equation (1)], the service rate is 0.5 per hour
[refer to Equation (2)], and the average number of jobs waiting for service
Equation (3)].
Now, calculate average number of jobs in the system as follows:
…… (4)
step: 9 of 17
The average time a job spends waiting
:
is 1.7578 [refer to
The calculated arrival rate is 0.375per hour [refer to Equation (1)], and the average number of jobs
waiting for service
is 1.7578 [refer to Equation (3)].
Now, calculate the average time a customer spends waiting as follows:
…… (5)
step: 10 of 17
The average time a job spends in the system:
The calculated service rate is 0.50 per hour [refer to Equation (2)], and the average time a customer
spends waiting
is 4.6875 hours [refer to Equation (5)].
Now, calculate the average time a customer spends in the system as follows:
…… (5)
step: 11 of 17
Calculation of total cost when k =1:
It is given that the waiting time is valued at $35 per hour, the service cost per channel is $28, and the
calculated is 2.5078 [refer to Equation (4)].
Now, calculate the total cost of system A as follows:
…… (6)
step: 12 of 17
(c)
Operating characteristics of the new system:
The average number of jobs waiting for service
:
It is given that the time required to complete the jobs is a standard deviation of 0 hours, the
calculated arrival rate is 0.375per hour [refer to Equation (1)], and the service rate is 0.5 per hour
[refer to Equation (2)].
Now, calculate average number of jobs waiting for service as follows:
…… (7)
step: 13 of 17
The average number of jobs in the system
:
The calculated arrival rate is 0.375per hour [refer to Equation (1)], the service rate is 0.5 per hour
[refer to Equation (2)], and the average number of jobs waiting for service
Equation (7)].
is 1.125 [refer to
…… (8)
step: 14 of 17
The average time a job spends waiting
:
The calculated arrival rate is 0.375per hour [refer to Equation (1)], and the average number of jobs
waiting for service
is 1.125 [refer to Equation (7)].
Now, calculate the average time a customer spends waiting as follows:
…… (9)
step: 15 of 17
The average time a job spends in the system:
The calculated service rate is 0.50 per hour [refer to Equation (2)], and the average time a customer
spends waiting
is 3 hours [refer to Equation (9)].
Now, calculate the average time a customer spends in the system as follows:
…… (10)
step: 16 of 17
Calculation of total cost when k = 1:
It is given that the waiting time is valued at $35 per hour, the service cost of new operation is $32,
and the calculated L is 1.875 [refer to Equation (8)].
Now, calculate the total cost of system A as follows:
…… (11)
Hence, the total cost of the new operation is $97.63, which is lesser than the current system.
step: 17 of 17
(d)
Savings during a 4-hour work week:
The calculated total cost of current system is $115.77 [refer to Equation (6)], and the total cost of
new system is $97.63 [refer to Equation (11)].
Now, calculate the savings during a 4-hour work week as follows:
Justification:
Even though the average waiting time of the new system is lesser than the current system, the
company should also consider multiple channel waiting lines to improve the quality of service.
Likes: 2
30. A large insurance company maintains a central computing system that contains a variety of
information about customer accounts. Insurance agents in a six-state area use telephone lines to
access the customer information database. Currently, the company’s central computer system
allows three users to access the central computer simultaneously. Agents who attempt to use the
system when it is full are denied access; no waiting is allowed. Management realizes that with its
expanding business, more requests will be made to the central information system. Being denied
access to the system is inefficient as well as annoying for agents. Access requests follow a Poisson
probability distribution, with a mean of 42 calls per hour. The service rate per line is 20 calls per
hour.
a. What is the probability that 0, 1, 2, and 3 access lines will be in use?
b. What is the probability that an agent will be denied access to the system?
c. What is the average number of access lines in use?
d. In planning for the future, management wants to be able to handle λ = 50 calls per hour; in
addition, the probability that an agent will be denied access to the system should be no greater
than the value computed in part (b). How many access lines should this system have?
:: Solution ::
step: 1 of 9
Operating characteristics for the M/G/k model:
step: 2 of 9
The probabilities are calculated by using the formula given below:
step: 3 of 9
(a)
The probability of 0, 1, 2, and 3 access lines:
The probability of 0 access line:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability of 0 access line as follows:
step: 4 of 9
The probability of 1 access line:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability of 1 access line as follows:
step: 5 of 9
The probability of 2 access lines:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability of 2 access lines as follows:
step: 6 of 9
The probability of 3 access lines:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability of 3 access lines as follows:
step: 7 of 9
(b)
The probability that an agent will be denied access to the system:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability that an agent will be denied access to the system as follows:
step: 8 of 9
(c)
The average number of access lines in use:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
step: 9 of 9
(d)
The probability that an agent will be denied access to the system:
It is given that the arrival rate is 50 calls per hour, and the service rate per line is 20 calls per hour.
Now, calculate the probability that an agent will be denied access to the system as follows:
Justification:
The probability that an agent will be denied access to the system is 0.1499. This system should
have 4 access lines.
Likes: 1
31. Mid-West Publishing Company publishes college textbooks. The company operates an 800
telephone number whereby potential adopters can ask questions about forthcoming texts,
request examination copies of texts, and place orders. Currently, two extension lines are used,
with two representatives handling the telephone inquiries. Calls occurring when both extension
lines are being used receive a busy signal; no waiting is allowed. Each representative can
accommodate an average of 12 calls per hour. The arrival rate is 20 calls per hour.
a. How many extension lines should be used if the company wants to handle 90% of the calls
immediately?
b. What is the average number of extension lines that will be busy if your recommendation in part
(a) is used?
c. What percentage of calls receive a busy signal for the current telephone system with two
extension lines?
:: Solution ::
step: 1 of 8
(a)
step: 2 of 8
The probability of 0, 1, 2, and 3 access lines:
The probability of 0 access line:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 0 access line as follows:
step: 3 of 8
The probability of 1 access line:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 1 access line as follows:
step: 4 of 8
The probability of 2 access lines:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 2 access lines as follows:
Justification:
If the company wants to handle 90% of the calls immediately, it should use 2 extension lines.
step: 5 of 8
(b)
The probability of 3 access lines:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 3 access lines as follows:
step: 6 of 8
The probability of 4 access lines:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 4 access lines as follows:
step: 7 of 8
The average number of access lines in use:
It is given that the arrival rate is 42 calls per hour, and the service rate per line is 20 calls per hour.
Hence, the company should select 4 access lines with k = 4.
step: 8 of 8
(c)
Percentage of calls that receives a busy signal for the current system with two extension lines:
The probability of 2 access lines:
It is given that the arrival rate is 20 calls per hour, and the service rate per line is 12 calls per hour.
Now, calculate the probability of 2 access lines as follows:
Hence, 34.25% of calls receive a busy signal for the current system with 2 extension lines.
Likes: 1
Dislikes: 5
32. City Cab, Inc., uses two dispatchers to handle requests for service and to dispatch the cabs. The
telephone calls that are made to City Cab use a common telephone number. When both
dispatchers are busy, the caller hears a busy signal; no waiting is allowed. Callers who receive a
busy signal can call back later or call another cab service. Assume that the arrival of calls follows a
Poisson probability distribution, with a mean of 40 calls per hour, and that each dispatcher can
handle a mean of 30 calls per hour.
a. What percentage of time are both dispatchers idle?
b. What percentage of time are both dispatchers busy?
c. What is the probability that callers will receive a busy signal if two, three, or four dispatchers
are used?
d. If management wants no more than 12% of the callers to receive a busy signal, how many
dispatchers should be used?
:: Solution ::
step: 1 of 10
(a)
step: 2 of 10
Percentage of time when both dispatchers are idle:
step: 3 of 10
The probability of 0 dispatchers:
It is given that the arrival rate is 40 calls per hour, and the service rate per line is 30 calls per hour.
Now, calculate the probability of 0 access line as follows:
step: 4 of 10
step: 5 of 10
Hence, both dispatchers are idle 31.04% of time.
step: 6 of 10
(b)
Percentage of time when both dispatchers are busy:
The probability of 2 dispatchers:
It is given that the arrival rate is 40 calls per hour, and the service rate per line is 30 calls per hour.
Now, calculate the probability of 2 access lines as follows:
Hence, both dispatchers are busy 27.58% of time.
step: 7 of 10
(c)
The probability that callers will receive a busy signal if 2, 3, or 4 dispatchers are used:
The probability of 2 dispatchers:
It is given that the arrival rate is 40 calls per hour, and the service rate per line is 30 calls per hour.
Now, calculate the probability that callers will receive a busy signal if 2 dispatchers are used as
follows:
step: 8 of 10
The probability of 3 dispatchers:
It is given that the arrival rate is 40 calls per hour, and the service rate per line is 30 calls per hour.
Now, calculate the probability that callers will receive a busy signal if 3 dispatchers are used as
follows:
step: 9 of 10
The probability of 4 dispatchers:
It is given that the arrival rate is 40 calls per hour, and the service rate per line is 30 calls per hour.
Now, calculate the probability that callers will receive a busy signal if 4 dispatchers are used as
follows:
step: 10 of 10
(d)
If the management wants less than 12% of the callers to receive a busy signal, they should
use 2 dispatchers at 10.92%.
Likes: 0
Dislikes: 4
33. Kolkmeyer Manufacturing Company (see Section 11.9) is considering adding two machines to
its manufacturing operation. This addition will bring the number of machines to eight. The
president of Kolkmeyer asked for a study of the need to add a second employee to the repair
operation. The arrival rate is 0.05 machines per hour for each machine, and the service rate for
each individual assigned to the repair operation is 0.50 machines per hour.
a. Compute the operating characteristics if the company retains the single-employee repair
operation.
b. Compute the operating characteristics if a second employee is added to the machine repair
operation.
c. Each employee is paid $20 per hour. Machine downtime is valued at $80 per hour. From an
economic point of view, should one or two employees handle the machine repair operation?
Explain.
:: Solution ::
step: 1 of 18
(a)
step: 2 of 18
Operating characteristics if the company retains the single-employee repair operation:
step: 3 of 18
The probability that no units are in the system:
step: 4 of 18
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8.
step: 5 of 18
Now, calculate the probability that no units are in the system as follows:
step: 6 of 18
Table 1:
step: 7 of 18
n
0
1.0000
1
0.8000
2
0.5600
3
0.3360
4
0.1680
5
0.0672
6
0.0202
7
0.0040
8
0.0004
Total 2.9558
step: 8 of 18
With the help of the above table, the following probability is calculated:
…… (1)
Hence, the probability that no units are in the system is 0.3383.
step: 9 of 18
The average number of units in the waiting line:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.3383 [refer to Equation (1)].
Now, calculate the average number of units waiting for service as follows:
…… (2)
step: 10 of 18
The average number of units in the system
:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.7215 [refer to Equation (2)].
Now, calculate the average number of units in the system as follows:
…… (3)
step: 11 of 18
The average time a unit spends waiting
:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.3383 [refer to Equation (1)], and
is 0.7215 [refer to Equation (2)].
Now, calculate the average time a unit spends waiting as follows:
…… (4)
step: 12 of 18
The average time a unit spends in the system:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 2.1808 [refer to Equation (4)].
Now, calculate the average time a unit spends in the system as follows:
…… (5)
step: 13 of 18
(b)
Operating characteristics if a second employee is added to the machine
The value of
is 0.4566 if a second employee is added to the machine.
The average number of units in the waiting line:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.4566.
Now, calculate the average number of jobs waiting for service as follows:
…… (6)
step: 14 of 18
The average number of units in the system
:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.0646 [refer to Equation (6)].
Now, calculate the average number of units in the system as follows:
…… (7)
step: 15 of 18
The average time a unit spends waiting
:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.0646 [refer to Equation (6)].
Now, calculate the average time a customer spends waiting as follows:
…… (8)
step: 16 of 18
The average time a unit spends in the system:
It is given that the arrival rate is 0.05 and the service rate is 0.50, and the size of the population (N) is
8, and the calculated
is 0.1791 [refer to Equation (8)].
Now, calculate the average time a customer spends in the system as follows:
…… (9)
step: 17 of 18
Calculation of total cost when one employee handles the machine repair operation:
It is given that the waiting time is valued at $80 per hour, the service cost is $20, and the
calculated
is 1.3832 [refer to Equation (3)].
Now, calculate the total cost of system A as follows:
…… (10)
step: 18 of 18
Calculation of total cost when two employees handle the machine repair operation:
It is given that the waiting time is valued at $80 per hour, the service cost is $20, and the
calculated
is 0.7860 [refer to Equation (7)].
Now, calculate the total cost of system A as follows:
…… (11)
The total cost of two employees ($102.88) is lesser than the cost of a single employee ($130.65).
Hence, it is recommended to use two employees.
Likes: 0
Dislikes: 1
34. Five administrative assistants use an office copier. The average time between arrivals for each
assistant is 40 minutes, which is equivalent to an arrival rate of
= 0.025 arrivals per minute.
The mean time each assistant spends at the copier is 5 minutes, which is equivalent to a service
rate of ⅕ = 0.20 per minute. Use the M/M/1 model with a finite calling population to determine
the following:
a. The probability that the copier is idle
b. The average number of administrative assistants in the waiting line
c. The average number of administrative assistants at the copier
d. The average time an assistant spends waiting for the copier
e. The average time an assistant spends at the copier
f. During an 8-hour day, how many minutes does an assistant spend at the copier? How much of
this time is waiting time?
g. Should management consider purchasing a second copier? Explain.
:: Solution ::
step: 1 of 15
(a)
step: 2 of 15
The probability that the copier is idle:
step: 3 of 15
It is given that the arrival rate is 0.025 per minute and the service rate is 0.20 per minute, and the
size of the population (N) is 5.
step: 4 of 15
Now, calculate the probability that the copier is idle as follows:
step: 5 of 15
Table 1:
step: 6 of 15
n
0
1.0000
1
0.6250
2
0.3125
3
0.1172
4
0.0293
5
0.0037
Total 2.0877
step: 7 of 15
With the help of the above table, the following probability is calculated:
…… (1)
Hence, the probability that the copier is idle is 0.4790.
step: 8 of 15
(b)
The average number of administrative assistants in the waiting line:
It is given that the arrival rate is 0.025 per minute, the service rate is 0.20 per minute, the size of the
population (N) is 5, and the calculated
is 0.4790 [refer to Equation (1)].
Now, calculate the average number of administrative assistants waiting for service as follows:
…… (2)
step: 9 of 15
(c)
The average number of administrative assistants at the copier
:
It is given that the arrival rate is 0.025 per minute, the service rate is 0.20 per minute, the size of the
population (N) is 5, and the calculated
is 0.3110 [refer to Equation (2)].
Now, calculate the average number of administrative assistants in the system as follows:
…… (3)
step: 10 of 15
(d)
The average time an assistant spends waiting for the copier
:
It is given that the arrival rate is 0.025 per minute, the service rate is 0.20 per minute, the size of the
population (N) is 5, and the calculated
Equation (2)].
is 0.4790 [refer to Equation (1)], and
is 0.8321 [refer to
Now, calculate the average time an assistant spends waiting for the copier as follows:
…… (4)
step: 11 of 15
(e)
The average time an assistant spends at the copier:
It is given that the arrival rate is 0.025 per minute, the service rate is 0.20 per minute, the size of the
population (N) is 5, and the calculated
is 2.9854 [refer to Equation (4)].
Now, calculate the average time an assistant spends at the copier as follows:
…… (5)
step: 12 of 15
(f)
The average minutes an assistant spends in the waiting line during an 8-hour day:
It is given that the arrival rate is 0.025 per minute and the service rate is 0.20 per minute.
Now, calculate the number of trips per day as follows:
…… (6)
step: 13 of 15
The average time an assistant spends at the copier per day:
The calculated average time an assistant spends at the copier (W) is 7.9854 [refer to Equation (5)],
and the number of trips is 12 per day [refer to Equation (6)].
Now, calculate the average time an assistant spends at the copier per day as follows:
step: 14 of 15
The average time an assistant spends waiting at the copier per day:
The calculated average time an assistant spends waiting is 2.9854 [refer to Equation (4)], and the
number of trips is 12 per day [refer to Equation (6)].
Now, calculate the average time an assistant spends waiting at the copier per day as follows:
step: 15 of 15
(g)
Management should consider purchasing a second copier .
Justification:
By using 5 administrative assistants, the loss of hours per day is calculated as follows:
Hence, it is recommended to purchase a second copier.
Likes: 0
35. Schips Department Store operates a fleet of 10 trucks. The trucks arrive at random times
throughout the day at the store’s truck dock to be loaded with new deliveries or to have incoming
shipments from the regional warehouse unloaded. Each truck returns to the truck dock for service
two times per 8-hour day. Thus, the arrival rate per truck is 0.25 trucks per hour. The service rate
is 4 trucks per hour. Using the Poisson arrivals and exponential service times model with a finite
calling population of 10 trucks, determine the following operating characteristics:
a. The probability that no trucks are at the truck dock
b. The average number of trucks waiting for loading/unloading
c. The average number of trucks in the truck dock area
d. The average waiting time before loading/unloading begins
e. The average waiting time in the system
f. What is the hourly cost of operation if the cost is $50 per hour for each truck and $30 per hour
for the truck dock?
g. Consider a two-server truck dock operation where the second server could be operated for an
additional $30 per hour. How much would the average number of trucks waiting for
loading/unloading have to be reduced to make the two-server truck dock economically feasible?
h. Should the company consider expanding to the two-server truck dock? Explain.
:: Solution ::
step: 1 of 15
(a)
step: 2 of 15
The probability that no trucks are at the truck dock:
step: 3 of 15
It is given that the arrival rate is 0.25 trucks per hour and the service rate is 4 trucks per hour, and
the size of the population (N) is 10 trucks.
step: 4 of 15
Now, calculate the probability that no trucks are at the truck dock as follows:
step: 5 of 15
Table 1:
step: 6 of 15
n
0
1.0000
1
0.6250
2
0.3516
3
0.1758
4
0.0769
5
0.0288
6
0.0090
7
0.0023
8
0.0004
9
0.0001
10
0.0000
Total 2.2698
step: 7 of 15
With the help of the above table, the following probability is calculated:
…… (1)
Hence, the probability that the copier is idle is 0.4406.
step: 8 of 15
(b)
The average number of trucks waiting for loading/unloading:
It is given that the arrival rate is 0.25 trucks per hour, the service rate is 4 trucks per hour, the size of
the population (N) is 10 trucks, and the calculated
is 0.4406 [refer to Equation (1)].
Now, calculate the average number of administrative assistants waiting for service as follows:
…… (2)
step: 9 of 15
(c)
The average number of trucks in the truck dock area
:
It is given that the arrival rate is 0.25 trucks per hour, the service rate is 4 trucks per hour, the size of
the population (N) is 10 trucks, and the calculated
is 0.4895 [refer to Equation (2)].
Now, calculate the average number of trucks in the truck dock area as follows:
…… (3)
step: 10 of 15
(d)
The average waiting time before loading/unloading begins
:
It is given that the arrival rate is 0.25 trucks per hour, the service rate is 4 trucks per hour, the size of
the population (N) is 10 trucks, and the calculated
is 0.4895 [refer to Equation (2)].
step: 11 of 15
Now, calculate the average waiting time before loading/unloading begins as follows:
…… (4)
step: 12 of 15
(e)
The average waiting time in the system:
It is given that the arrival rate is 0.25 trucks per hour, the service rate is 4 trucks per hour, the size of
the population (N) is 10 trucks, and the calculated
is 0.2188 [refer to Equation (4)].
Now, calculate the average waiting time in the system as follows:
…… (5)
step: 13 of 15
(f)
Cost of operation:
It is given that the waiting time is valued at $50 per hour, the service cost is $30, and the
calculated is 1.0490 [refer to Equation (3)].
Now, calculate the total cost of system A as follows:
…… (10)
step: 14 of 15
(g)
The average number of trucks waiting for loading/unloading if two-channel truck dock operation
is used:
It is given that the waiting time is valued at $50 per hour, addition cost is $30, and the calculated
total cost is $82.45 [refer to Equation (10)].
Now, calculate the average number of trucks waiting by using the below equation:
…… (11)
step: 15 of 15
(h)
Total cost of two-channel operation:
It is given that the waiting time is valued at $50 per hour, the service cost is $30, and the value of L is
0.6237 when two-channel truck dock operation is used.
Now, calculate the total cost as follows:
…… (12)
Justification:
The total cost of two-channel operation is $91.18 [refer to Equation (11)], which is higher than the
cost of single-channel operation. Hence, it is recommended to the company not to expand their
operations.
Likes: 1
Dislikes: 1
Regional Airlines
Regional Airlines is establishing a new telephone system for handling flight reservations. During the
10:00 A.M. to 11:00 A.M. time period, calls to the reservation agent occur randomly at an average
of one call every 3.75 minutes. Historical service time data show that a reservation agent spends an
average of 3 minutes with each customer. The waiting line model assumptions of Poisson arrivals
and exponential service times appear reasonable for the telephone reservation system.
Regional Airlines’ management believes that offering an efficient telephone reservation system is
an important part of establishing an image as a service-oriented airline. If the system is properly
implemented, Regional Airlines will establish good customer relations, which in the long run will
increase business. However, if the telephone reservation system is frequently overloaded and
customers have difficulty contacting an agent, a negative customer reaction may lead to an eventual
loss of business. The cost of a ticket reservation agent is $20 per hour. Thus, management wants to
provide good service, but it does not want to incur the cost of overstaffing the telephone reservation
operation by using more agents than necessary.
At a planning meeting, Regional’s management team agreed that an acceptable customer service
goal is to answer at least 85% of the incoming calls immediately. During the planning meeting,
Regional’s vice president of administration pointed out that the data show that the average service
rate for an agent is faster than the average arrival rate of the telephone calls. The vice president’s
conclusion was that personnel costs could be minimized by using one agent and that the single agent
should be able to handle the telephone reservations and still have some idle time. The vice president
of marketing restated the importance of customer service and expressed support for at least two
reservation agents.
The current telephone reservation system design does not allow callers to wait. Callers who attempt
to reach a reservation agent when all agents are occupied receive a busy signal and are blocked
from the system. A representative from the telephone company suggested that Regional Airlines
consider an expanded system that accommodates waiting. In the expanded system, when a
customer calls and all agents are busy, a recorded message tells the customer that the call is being
held in the order received and that an agent will be available shortly. The customer can stay on the
line and listen to background music while waiting for an agent. Regional’s management will need
more information before switching to the expanded system.
Managerial Report
Prepare a managerial report for Regional Airlines analyzing the telephone reservation system.
Evaluate both the system that does not allow waiting and the expanded system that allows waiting.
Include the following information in your report:
1. A detailed analysis of the operating characteristics of the reservation system with one agent as
proposed by the vice president of administration. What is your recommendation concerning a
single-agent system?
2. A detailed analysis of the operating characteristics of the reservation system based on your
recommendation regarding the number of agents Regional should use.
3. A detailed analysis of the advantages or disadvantages of the expanded system. Discuss the
number of waiting callers the expanded system would need to accommodate.
4. This report represents a pilot study of the reservation system for the 10:00 A.M. to 11:00 A.M.
time period during which an average of one call arrives every 3.75 minutes; however, the arrival
rate of incoming calls is expected to change from hour to hour. Describe how your waiting line
analysis could be used to develop a ticket agent staffing plan that would enable the company to
provide different levels of staffing for the ticket reservation system at different times during the
day. Indicate the information that you would need to develop this staffing plan.
:: Solution ::
step: 1 of 34
Synopsis of Regional airlines:
• Regional airlines established a new system to handle flight reservation in order to provide efficient
service to customers.
• The reservation calls are done at an average of 1 call every 3.75 minutes, and it spends 3 minutes
with each customer on an average.
• The analysis of the waiting line is made based on the Poisson arrivals and exponential service time.
• The per hour cost of tick reservation is $20.
• The service goal of the new system is that 85% of incoming calls should be answered.
step: 2 of 34
1)
An analysis of current reservation system that does not allow callers to wait:
It is given that the reservation calls are done at an average of 1 call every 3.75 minutes, and spends 3
minutes with each customer on an average.
Operating characteristics with one-reservation system:
Arrival rate per hour:
It is calculated by dividing minutes per hour and the average minutes per call.
Hence, the arrival rate is
.
step: 3 of 34
Service rate per hour:
The service data shows that an agent spends 3 minutes with each customer on an average.
It is calculated by dividing minutes per hour and the average minutes that the agent spend on each
customer.
Hence, the service rate is
step: 4 of 34
The probability of 0 calls in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20.
step: 5 of 34
The probability of 1 call in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
Hence, the probability that a caller gets a busy signal with the value of
Operating characteristics with two-reservation system:
The probability of 0 calls in the system:
is
.
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
step: 6 of 34
The probability of 1 call in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
step: 7 of 34
The probability of 2 calls in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
Hence, the probability that a caller gets busy signal with the value of
step: 8 of 34
Justification:
is
.
As per the current reservation system, 85% of incoming calls will be answered immediately if tworeservation agents are used. 15.09% of callers will get a busy signal.
step: 9 of 34
2)
An analysis of the expanded system proposed by the telephone company:
An analysis with one-reservation agent:
The probability that no units are in the system
:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
It is calculated by subtracting the value attained by dividing arrival rate and service rate from 1.
Hence, the value of probability is
.
step: 10 of 34
The average number of customers waiting
:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
It is calculated by dividing the square of arrival rate and the value attained by multiplying service and
the value obtained by subtracting arrival rate from service rate.
Hence, the average number of customers waiting is
.
step: 11 of 34
The average number of customers in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the calculated
is 3.2.
It is calculated by adding the average number of customers waiting and the value attained by
dividing arrival rate and service rate.
step: 12 of 34
The average time a customer spends waiting
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the calculated
is 3.2.
It is calculated by dividing average number of customers waiting and the arrival rate.
step: 13 of 34
The average time a customer spends in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the calculated
is 3.2.
It is calculated by adding the average time customer spends waiting and the value attained by
dividing 1 and service rate.
step: 14 of 34
The probability that the arriving customers will have to wait for service
The average time a customer spends in the system
:
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the calculated
is 3.2.
It is calculated by dividing arrival rate and service rate.
The service goal of the company is to answer 85% of the incoming calls. However, in this case, 80%
of incoming calls will have to wait for service. Hence, it is not acceptable. It is recommended to
consider two or more agents.
step: 15 of 34
Two-reservation agents:
Operating characteristics of the waiting line, when k = 2:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the calculated
the calculated
is 3.2, and
is 12 minutes.
It is calculated by dividing arrival rate and service rate.
Note: The table value of
with 0.8 for multiple channel waiting line is
.
step: 16 of 34
The probability that a customer has to wait for service:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the calculated
calculated
is 12 minutes, and the number of channel (k) is 2.
step: 17 of 34
Three-reservation agents:
is 3.2, the
Operating characteristics of the waiting line, when k = 3:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the calculated
calculated
is 3.2, the
is 12 minutes, and the number of channel (k) is 2.
It is calculated by dividing arrival rate and service rate.
Note: The table value of
with 0.8 for multiple channel waiting line is
.
step: 18 of 34
The average number of customers waiting for service
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the calculated
calculated
is 3.2, the
is 12 minutes, and the number of channel (k) is 2.
Note: The table value of
with 0.8
for multiple-channel waiting lines is 0.4472.
Now, calculate the average number of customers in the waiting line as follows:
step: 19 of 34
The average number of customers in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the number of channel (k) is
3, and the calculated
is 0.0189.
It is calculated by adding the average number of customers waiting and the value attained by
dividing arrival rate and service rate.
step: 20 of 34
The average time a customer spends waiting
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the number of channel (k) is
3, and the calculated
is 0.0189.
It is calculated by dividing the average number of customers waiting with the arrival rate.
step: 21 of 34
The average time a customer spends in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the number of channel (k) is
3, and the calculated
is 0.0012.
It is calculated by adding the average time customer spends waiting and the value attained by
dividing 1 and service rate.
step: 22 of 34
The probability that a customer has to wait for service:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the number of channel (k)
is 3.
Now, calculate the probability that a customer has to wait for service as follows:
step: 23 of 34
Justification:
The above calculation states that three-reservation agents will be needed to meet the service goal
(85% of the incoming calls should be answered) of the company.
c)
Recommendation for the selection of a system:
The above two analyses are made in order to identify the best system for the customers. As a result,
three-reservation agents will be required to meet the service goal of the company.
The following are the further analyses which are required by the management to take a better
decision:
The probability of 0 units in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
Now, calculate the ratio to find out the value of
Note: The table value of
as follows:
with 0.8 for multiple channel waiting line is
step: 24 of 34
The probability of 1 unit in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
step: 25 of 34
.
The probability of 2 units in the system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
step: 26 of 34
The probability of 3 or more customers in the system:
The calculated
is 0.4472,
is 0.3577, and
is 0.1431.
Now, calculate the probability of 3 or more customers in the system as follows:
With the above analysis, the management should be informed that the three-reservation agents
would be underutilized. It is recommended to the management to reassess the two-reservation
agents.
step: 27 of 34
Operation characteristics of the two-reservation system:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
It is calculated by dividing arrival rate and service rate.
The table value of
with 0.8 for multiple channel waiting line is
.
step: 28 of 34
The average number of customers waiting for service
:
The calculated arrival rate is 16 calls per hour and service rate is 20 hours.
Note: The table value of
with 0.8
for multiple-channel waiting lines is 0.4286.
step: 29 of 34
The average number of customers in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the number of channel (k) is
2, and the calculated
is 0.1524.
Now, calculate the average number of passengers in the system as follows:
step: 30 of 34
The average time a customer spends waiting
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, the number of channel (k) is
2, and the calculated
is 0.1524.
step: 31 of 34
Now, calculate the average time a customer spends waiting as follows:
step: 32 of 34
The average time a customer spends in the system
:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the calculated
0.0095.
is
Now, calculate the average time a customer spends waiting as follows:
step: 33 of 34
The probability that a customer has to wait for service:
The calculated arrival rate is 16 calls per hour, service rate is 20 hours, and the number of channel (k)
is 2.
Now, calculate the probability that a customer has to wait for service as follows:
step: 34 of 34
Conclusion:
Hence, the selection of the best system depends on the methods and measures used by the analyst.
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