qs up questions 9

Quantitative Analysis BA 452 Supplemental Questions 9

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This document contains practice questions that supplement review questions for Lessons II-7 and II-8. This document first identifies the learning objectives of solving supplemental questions. The document then lists 35 questions and answers. All questions can be helpful. Questions marked with an asterisk * are similar to review questions.

Tip: Supplemental questions are grouped into sets of similar type. Once you have mastered the questions in a set, you can skip the rest of the questions in that set.


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Objectives By working through the homework questions and the supplemental questions, you will:

1. Be able to identify where waiting line problems occur and realize

why it is important to study these problems. 2. Know the difference between single-channel and multiple-channel

waiting lines. 3. Understand how the Poisson distribution is used to describe

arrivals and how the exponential distribution is used to describe services times.

4. Learn how to use formulas to identify operating characteristics of

the following waiting line models: a. Single-channel model with Poisson arrivals and exponential

service times b. Multiple-channel model with Poisson arrivals and exponential

service times 5. Know how to incorporate economic considerations to arrive at

decisions concerning the operation of a waiting line. 6. Understand the following terms: queuing theory steady state queue utilization factor single-channel operating characteristics multiple-channel arrival rate service rate queue discipline


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Supplemental Questions 9

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?

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 customer per minute. Use the exponential probability distribution to answer the following questions:

a. What is the probability the service time is one minute or less? b. What is the probability the service time is two minutes or less? c. What is the probability the service time is more than two minutes?

3. Use the single-channel 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

4. Use the single-channel 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?


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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?

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 check out system provides

an acceptable level of service?

7. Speedy Oil provides a single-channel 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?

8. For the Burger Dome single-channel 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.


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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?

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?

11. Agan Interior Design provides home and office decorating assistance to its customers. In normal operation, averages 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 arrival 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?

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 check out

process to no more than five minutes, what recommendations would you provide regarding the current checkout system?


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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 single channel 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-channel operation would have a service rate of 20 customers per hour for each channel.

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 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 waits 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 minute s 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?

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 with 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.


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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?

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 channel 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 you system will be providing? Why or why not?

18. All airplane passengers at the Lake City Regional Airport must pass through a security screening are 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 operation 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?


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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?

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 work stations. 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 post 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.

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 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.


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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 them drive to the service window t5o pay for and receive their orders. The following three service alternatives are being considered:

a. A single-channel 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.

b. A single-channel 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.

c. A two-channel operation with two service windows and two employees. The employee stationed at each window fills the order and takes the money from customers arriving at the window. The average service time for this alternative is 2 minutes for each channel.

Answer the following questions and recommend an alternative design for the fast-food franchise:

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?

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

channel.

What is the lowest-cost design for the fast-food business? 24. Patients arrive at a dentist’s office with an arrival rate of 2.8 patients per hour. The dentist can

treat patients at a service rate of 3 patients per hour. A study of patient waiting times shows that a patient waits o average of 30 minutes before seeing the dentist.

a. What are arrival and service rates in terms of patients per minute? b. What is the average number of patients in the waiting room? c. If as patient arrives at 10:10A.M., at what time is the patient expected to leave the

office?


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25. A study of the multiple-channel 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 channel 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?

26. Manning Auto operates an automotive service counter. While completing the repair work,

Manning mechanics arrive at the company’s parts department counter with an arrival rate of four per hour. The parts coordinator spends an average of 6 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 wait is 4 minutes before the parts coordinator is available to answer questions or retrieve parts from inventory. Find𝐿𝐿𝑞𝑞 , W, and L for this single-channel parts operation.

b. A trial period with a second parts coordinator showed that, on average, each mechanic waited only 1 minute before a parts coordinator is available. Find𝐿𝐿𝑞𝑞 , W, and L for this two-channel parts operation.

c. If the cost of each mechanic is $20 per hour and the cost of each parts coordinator is $12 per hours, is the one-channel or the two-channel system more economical?

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?


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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:

Service Time Design Mean Standard Deviation

A 6.0 3.0

B 6.25 0.6

a. What is the service rate in jobs per hours 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?

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-channel 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 said no to 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 work week?


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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 center. 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 past (b). How many access lines should this system have?

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 call 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?

32. City Can 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 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?


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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 machine per hours for each machine, and the service rate for each to individual assigned to the repair operation is 0.50 machine 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.

34. Five administrative assistants use an office copier. The average time between arrival for each assistant is 40 minutes, which is equivalent to an arrival rate of 1/40 = 0.025 arrival per minute. The mean time each assistant spends at the copier is 5 minutes, which is equivalent to a service of ½ = 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 as second copier? Explain.


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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 not trucks are at the truck dock b. The average number of truck 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-channel truck dock operation where the second channel should 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-channel truck dock economically feasible?

h. Should the company consider expanding to the two-channel truck dock? Explain.


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Wq =Lqλ

= 1.33330.4

= 3.3333 min.

P (x) = λ x e-λ

x!= 2x e-2

x!

P0 = 1 - λµ

= 1 - 0.40.6

= 0.3333

Lq = λ 2

µ (µ - λ)= (0.4)2

0.6 (0.6 - 0.4)= 1.3333

L = Lq + λµ

= 1.3333 + 0.40.6

= 2

W = Wq + 1µ

= 3.3333 + 10.6

= 5 min.

Pw = λµ

= 0.40.6

= 0.6667

Pn = λµ

nP0 = 0.4

0.6n

(0.3333)

Answers to Supplemental Questions 9 1. a. λ = 5(0.4) = 2 per five minute period b.

x P(x) 0 0.1353 1 0.2707 2 0.2707 3 0.1804

c. P(Delay Problems) = P(x > 3) = 1 - P(x ≤ 3) = 1 - 0.8571 = 0.1429 2. a. µ = 0.6 customers per minute P(service time ≤ 1) = 1 - e-(0.6)1 = 0.4512 b. P(service time ≤ 2) = 1 - e-(0.6)2 = 0.6988 c. P(service time > 2) = 1 - 0.6988 = 0.3012 3. a.

b. c. d.

e. f. 4.

n Pn

0

0.3333 1 0.2222 2 0.1481 3 0.0988

P(n > 3) = 1 - P(n ≤ 3) = 1 - 0.8024 = 0.1976


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Wq =Lqλ

= 0.4167 hours (25 minutes)

P0 = 1 - λµ

= 1 - 1012

= 0.1667

Lq = λ 2

µ (µ - λ)= 102

12 (12 - 10)= 4.1667

W = Wq + 1µ

= .5 hours (30 minutes)

5. a. b.

c. d.

e. 10 0.833312wP λ

µ= = =

6. a. 01.251 1 0.375

2P λ

µ= − = − =

b. 2 21.25 1.0417

( ) 2(2 1.25)qL λµ µ λ

= = =− −

c. 1.0417 0.83331.25

qq

LW

λ= = = minutes (50 seconds)

d. 1.25 0.6252wP λ

µ= = =

e. Average one customer in line with a 50 second average wait appears reasonable.

7. a. Lq =−

=−

=λ

µ µ λ

2 22 55 5 2 5

05000( )

( . )( . )

.

b. WL

qq= = =

λ05000

2 50 20.

.. hours (12 minutes)

c. W Wq= + = + =1 0 20 1

50 40

µ. . hours (24 minutes)

d. Pw = = =λµ

2 55

0 50. .

L Lq= + = + =λµ

05000 2 55

1. .


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8. λ = 1 and µ = 1.25

011 1 0.20

1.25P λ

µ= − = − =

2 1 3.2

( ) 1.25(0.25)qL λµ µ λ

= = =−

13.2 41.25qL L λ

µ= + = + =

3.2 3.21

qq

LW

λ= = = minutes

1 13.2 41.25qW W

µ= + = + = minutes

1 0.801.25wP λ

µ= = =

Even though the services rate is increased to µ = 1.25, this system provides slightly poorer service

due to the fact that arrivals are occurring at a higher rate. The average waiting times are identical, but there is a higher probability of waiting and the number waiting increases with the new system.

9. a. 02.21 1 0.565

P λµ

= − = − =

b. 1 02.2 (0.56) 0.24645

P Pλµ

= = =

c. 2 2

2 02.2 (0.56) 0.10845

P Pλµ

= = =

d. 3 3

3 02.2 (0.56) 0.04775

P Pλµ

= = =

e. P(More than 2 waiting) = P(More than 3 are in system) = 1 - (P0 + P1 + P2 + P3) = 1 - 0.9625 = 0.0375

f. 2 22.2 0.3457

( ) 5(5 2.2)qL λµ µ λ

= = =− −

0.157qq

LW

λ= = hours (9.43 minutes)


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10. a. λ = 2 µ = 3 µ = 4

Average number waiting (Lq) 1.3333 0.5000 Average number in system (L) 2.0000 1.0000 Average time waiting (Wq) 0.6667 0.2500 Average time in system (W) 1.0000 0.5000 Probability of waiting (Pw) 0.6667 0.5000

b. New mechanic = $30(L) + $14 = 30(2) + 14 = $74 per hour Experienced mechanic = $30(L) + $20 = 30(1) + 20 = $50 per hour ∴ Hire the experienced mechanic 11. a. λ = 2.5 µ = 60/10 = 6 customers per hour

Lq = λ2

µ (µ - λ)= 2.5 2

6 (6 - 2.5)= 0.2976

L = Lq + λµ

= 0.7143

Wq =Lqλ

= 0.1190 hours (7.14 minutes)

W = Wq + 1µ

= 0.2857 hours

Pw = λµ

= 2.56

= 0.4167

b. No; Wq = 7.14 minutes. Firm should increase the mean service rate (µ) for the consultant or hire a

second consultant. c. µ = 60/8 = 7.5 customers per hour

Lq = λ2

µ (µ - λ)= 2.5 2

7.5 (7.5 - 2.5)= 0.1667

Wq =Lqλ


The service goal is being met.


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12. P0 = 1 - λ

µ= 1 - 15

20= 0.25

Lq = λ 2

µ (µ - λ)= 152

20 (20 - 15)= 2.25

L = L + λµ

= 3

Wq =Lqλ


W = Wq + 1µ


Pw = λµ

= 1520

= 0.75

With Wq = 9 minutes, the checkout service needs improvements. 13. Average waiting time goal: 5 minutes or less. a. One checkout counter with 2 employees λ = 15 µ = 30 per hour

Lq = λ 2

µ (µ - λ)= 152

30 (30 - 15)= 0.50

Wq =Lqλ


b. Two channel-two counter system λ = 15 µ = 20 per hour for each From Table, P0 = 0.4545

Lq = (λ / µ)2 λ µ1! (2 (20) - 15)2

P0 = (15 / 20)2 (15) (20)(40 - 15)2

(0.4545) = 0.1227

Wq =Lqλ

= 0.0082 hours (0.492 minutes)

Recommend one checkout counter with two people. This meets the service goal with Wq = 2

minutes. The two counter system has better service, but has the added cost of installing a new counter.


20

14. a. µ = =607 5

8.

customers per hour

b. 051 1 0.37508

P λµ

= − = − =

c. Lq =−

=−

=λ

µ µ λ

2 258 8 5

10417( ) ( )

.

d. WL

qq= = =

λ10417

50 2083. . hours (12.5 minutes)

e. Pw = = =λµ

58

0 6250.

f. 62.5% of customers have to wait and the average waiting time is 12.5 minutes. Ocala needs to add

more consultants to meet its service guidelines. 15. k = 2, λ = 5, µ = 8 Using the equation for P0, P0 = 0.5238

Lk

Pq =−

=λ µ λµ

µ λ/

!( ).b g2

2 010 0676

WL

qq= = =

λ0 0676

50 0135. . hours (0.81 minutes)

P P P0 1

1

00 52381

58

05238 0 3274= = = =./!

( . ) . λ µb g

Pw = P(n ≥ 2) = 1 - P(n ≤ 1) = 1 - 0.5238 - 0.3274 = 0.1488 Two consultants meet service goals with only 14.88% of customers waiting with an average waiting

time of 0.81 minutes (49 seconds).


21

P0 = 1 - λµ

= 1 - 510

= 0.50

Lq = (λ / µ)2 λ µ1! (k µ - λ)2

P0 = 0.0333

Wq =Lqλ

= 0.0067 hours (24.12 seconds)

W = Wq + 1µ

= 0.1067 (6.4 minutes)

16. a.

b. 2 25 0.50

( ) 10(10 5)qL λµ µ λ

= = =− −

c. 0.1qq

LW

λ= = hours (6 minutes)

d. 1 0.2qW Wµ

= + = hours (12 minutes)

e. Yes, unless Wq = 6 minutes is considered too long. 17. a. From Table, P0 = 0.60 b. c. d. e. This service is probably much better than necessary with average waiting time only 24 seconds.

Both channels will be idle 60% of the time.


22

18. Arrival rate: λ = 5.4 per minute Service rate: µ = 3 per minute for each station a. Using the table of values of P0, λ/µ = 1.8, k = 2, and P0 = 0.0526

( ) 2

02 2

/ (1.8) (5.4)(3) (0.0526) 7.67( 1)!( ) (2 1)!(6 5.4)

k

qL Pk k

λ µ λµµ λ

= = =− − − −

7.67 1.8 9.47qL L λµ

= + = + =

7.67 1.425.4

qq

LW

λ= = = minutes

1 1.42 0.33 1.75qW Wµ

= + = + = minutes

2

01 1 5.4 2(3) 0.0526 0.8526! 2! 3 2(3) 5.4

k

wkP P

k kλ µµ µ λ

= = = − −

b. The average number of passengers in the waiting line is 7.67. Two screening stations will be able to

meet the manager’s goal. c. The average time for a passenger to move through security screening is 1.75 minutes.


23

19. a. For the system to be able to handle the arrivals, we must have kµ > λ, where k is the number of channels. With µ = 2 and λ = 5.4, we must have at least k = 3 channels.

To see if 3 screening stations are adequate, we must compute Lq. Using the table of values of P0, λ/µ = 2.7, k = 3, and P0 = 0.02525 (halfway between λ/µ = 2.6 and

λ/µ = 2.8)

( ) 3

02 2

/ (2.7) (5.4)(2) (0.02525) 7.45( 1)!( ) 2!(6 5.4)

k

qL Pk k

λ µ λµµ λ

= = =− − −

Having 3 stations open satisfies the manager’s goal to limit the average number of passengers in the

waiting line to at most 10. b. The average time required for a passenger to pass through security screening is

7.45 1.385.4

qq

LW

λ= = =

1 1.38 0.5 1.88qW Wµ

= + = + = minutes

Note: The above results are based on using the tables of P0 and an approximate value for P0. If a

computer program is used, we obtain exact results as follows: P0 = 0.0249 Lq = 7.35 W = 1.86 minutes


24

20. a. Note 1.2 1.60 1.0.75

λµ

= = > Thus, one postal clerk cannot handle the arrival rate.

Try k = 2 postal clerks

From Table with 1.60λµ

= and k = 2, P0 = 0.1111

2

02

( / ) 2.84441!(2 )qL Pλ µ λµ

µ λ= =

−

4.4444qL L λµ

= + =

2.3704qq

LW

λ= = minutes

1 3.7037qW Wµ

= + = minutes

Pw = 0.7111 Use 2 postal clerks with average time in system 3.7037 minutes. No need to consider k = 3. b. Try k = 3 postal clerks.

From Table with 2.1 2.80.75

λµ

= = and k = 3, P0 = 0.0160

3

02

( / ) 12.27352(3 )qL Pλ µ λµ

µ λ= =

−

15.0735qL L λµ

= + =

5.8445qq

LW

λ= = minutes

1 7.1778qW Wµ

= + = minutes

Pw = 0.8767 Three postal clerks will not be enough in two years. Average time in system of 7.1778 minutes and

an average of 15.0735 customers in the system are unacceptable levels of service. Post office expansion to allow at least four postal clerks should be considered.


25

21. From question 11, a service time of 8 minutes has µ = 60/8 = 7.5

Lq = λ2

µ (µ - λ)= (2.5)2

7.5 (7.5 - 2.5)= 0.1667

L = Lq + λµ

= 0.50

Total Cost = $25L + $16 = 25(0.50) + 16 = $28.50 Two channels: λ = 2.5 µ = 60/10 = 6 Using equation, P0 = 0.6552

Lq = (λ / µ)2 λ µ1! (2 µ - λ)2

P0 = 0.0189

L = Lq + λµ

= 0.4356

Total Cost = 25(0.4356) + 2(16) = $42.89 Use the one consultant with an 8 minute service time. 22. λ = 24

Characteristic

System A (k = 1, µ = 30)

System B (k = 1, µ = 48)

System C (k = 2, µ = 30)

a. P0

0.2000

0.5000

0.4286

b. Lq 3.2000 0.5000 0.1524 c. Wq 0.1333 0.0200 0.0063 d. W 0.1667 0.0417 0.0397 e. L 4.0000 1.0000 0.9524 f. Pw 0.8000 0.5000 0.2286

System C provides the best service. 23. Service Cost per Channel

System A: 6.50 + 20.00 = $26.50/hour System B: 2(6.50) + 20.00 = $33.00/hour System C: 6.50 + 20.00 = $26.50/hour

Total Cost = cwL + csk System A: 25(4) + 26.50(1) = $126.50 System B: 25(1) + 33.00(1) = $ 58.00 System C: 25(0.9524) + 26.50(2) = $ 76.81

System B is the most economical.


26

24. λ = 2.8, µ = 3.0, Wq = 30 minutes a. λ = 2.8/60 = 0.0466 µ = 3/60 = 0.0500 b. Lq = λWq = (0.0466)(30) = 1.4 c. W = Wq + 1/µ = 30 + 1/0.05 = 50 minutes ∴ 11:00 a.m. 25. λ = 4, W = 10 minutes a. µ = 1/2 = 0.5 b. Wq = W - 1/µ = 10 - 1/0.5 = 8 minutes c. L = λW = 4(10) = 40 26. a. Express λ and µ in mechanics per minute λ = 4/60 = 0.0667 mechanics per minute µ = 1/6 = 0.1667 mechanics per minute Lq = λWq = 0.0667(4) = 0.2668 W = Wq + 1/µ = 4 + 1/0.1667 = 10 minutes L = λW = (0.0667)(10) = 0.6667 b. Lq = 0.0667(1) = 0.0667 W = 1 + 1/0.1667 = 7 minutes L = λW = (0.0667)(7) = 0.4669 c. One-Channel Total Cost = 20(0.6667) + 12(1) = $25.33 Two-Channel Total Cost = 20(0.4669) + 12(2) = $33.34 One-Channel is more economical.


27

Lq = λ 2 σ2 + (λ / µ)2

2 (1 - λ / µ)= (0.25)2 (2)2 + (0.25 / 0.3125)2

2 (1 - 0.25 / 0.3125)= 2.225

Wq =Lqλ

= 2.2250.25

= 8.9 hours

W = Wq + 1µ

= 8.9 + 11.3125

= 12.1 hours

Same at Pw = λµ

= 0.250.3125

= 0.80

27. a. 2/8 hours = 0.25 per hour b. 1/3.2 hours = 0.3125 per hour c. d. e. f. 80% of the time the welder is busy. 28. λ = 5 a.

Design µ A 60/6 = 10 B 60/6.25 = 9.6

b. Design A with µ = 10 jobs per hour. c. 3/60 = 0.05 for A 0.6/60 = 0.01 for B d.

Characteristic Design A Design B P0 0.5000 0.4792 Lq 0.3125 0.2857 L 0.8125 0.8065

Wq 0.0625 0.0571 W 0.1625 0.1613 Pw 0.5000 0.5208

e. Design B is slightly better due to the lower variability of service times.

System A: W = 0.1625 hrs (9.75 minutes) System B: W = 0.1613 hrs (9.68 minutes)


28

Lq = λ2 σ2 + (λ / µ)2

2 (1 - λ / µ)= (.375)2 (1.5)2 + (.375/ .5)2

2 (1 - .375 / .5)= 1.7578

29. a. λ = 3/8 = .375 µ = 1/2 = .5 b. L = Lq + λ / µ = 1.7578 + .375 / .5 = 2.5078 TC = cwL + csk = 35 (2.5078) + 28 (1) = $115.71 c.

Current System (σ = 1.5) New System (σ = 0) Lq = 1.7578 Lq = 1.125 L = 2.5078 L = 1.875

Wq = 4.6875 Wq = 3.00 W = 6.6875 W = 5.00

TC = $115.77 TC = cwL + csk = 35 (1.875) + 32 (1) = $97.63 d. Yes; Savings = 40 ($115.77 - $97.63) = $725.60 Note: Even with the advantages of the new system, Wq = 3 shows an average waiting time of 3

hours. The company should consider a second channel or other ways of improving the emergency repair service.

30. a. λ = 42 µ = 20

i (λ/µ)i / i ! 0 1.0000 1 2.1000 2 2.2050 3 1.5435

6.8485

j Pj 0 1/6.8485 = 0.1460 1 2.1/6.8485 = 0.3066 2 2.2050/6.8485 = 0.3220 3 1.5435/6.8485 = 0.2254

1.0000 b. 0.2254 c. L = λ/µ(1 - Pk) = 42/20 (1 - 0.2254) = 1.6267 d. Four lines will be necessary. The probability of denied access is 0.1499.


29

31. a. λ = 20 µ = 12

i (λ/µ)i / i ! 0 1.0000 1 1.6667 2 1.3889

4.0556

j Pj 0 1/4.0556 = 0.2466 1 1.6667/4.0556 = 0.4110 2 1.3889/4.0556 = 0.3425

P2 = 0.3425 34.25% b. k = 3 P3 = 0.1598 k = 4 P4 = 0.0624 Must go to k = 4. c. L = λ/µ(1 - P4) = 20/12(1 - 0.0624) = 1.5626 32. a. λ = 40 µ = 30

i (λ/µ)i / i ! 0 1.0000 1 1.3333 2 0.8888

3.2221 P0 = 1.0000/3.2221 = 0.3104 31.04% b. P2 = 0.8888/3.2221 = 0.2758 27.58% c.

i (λ/µ)i / i ! 3 0.3951 4 0.1317

P2 = 0.2758 P3 = 0.3951/(3.2221 + 0.3951) = 0.1092 P4 = 0.1317/(3.2221 + 0.3951 + 0.1317) = 0.0351 d. k = 3 with 10.92% of calls receiving a busy signal.


30

33. a. λ = 0.05 µ = 0.50 λ/µ = 0.10 N = 8

n

N !(N - n) !

λµ

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

2.9558

P0 = 1/2.9558 = 0.3383

L N Pq = −+FHG IKJ − = −FHG IKJ − =

λ µλ

( ) ..

( . ) .1 8 0550 05

1 0 3383 0 72150

L = Lq + (1 - P0) = 0.7213 + (1 - 0.3383) = 1.3832

WL

N Lqq=

−=

−=

( ).

( . )( . ).

λ0 7215

8 13832 0 0521808 hours

W Wq= + = + =1 21808 1

05041808

µ.

.. hours

b. P0 = 0.4566 Lq = 0.0646 L = 0.7860 Wq = 0.1791 hours W = 2.1791 hours c. One Employee Cost = 80L + 20 = 80(1.3832) + 20 = $130.65 Two Employees Cost = 80L + 20(2) = 80(0.7860) + 40 = $102.88 Use two employees.


31

34. N = 5 λ = 0.025 µ = 0.20 λ/µ = 0.125 a.

n

N !(N - n) !

λµ

n

0 1.0000 1 0.6250 2 0.3125 3 0.1172 4 0.0293 5 0.0037

2.0877

P0 = 1/2.0877 = 0.4790

b. 00.225(1 ) 5 (1 0.4790) 0.31100.025qL N Pλ µ

λ+ = − − = − − =

c. L = Lq + (1 - P0) = 0.3110 + (1 - 0.4790) = 0.8321

d. WL

N Lqq=

−=

−=

( ).

( . )( . ).

λ0 3110

5 08321 0 0252 9854 min

e. W Wq= + = + =1 2 9854 1

0 207 9854

µ.

.. min

f. Trips/Days = (8 hours)(60 min/hour) (λ) = (8)(60)(0.025) = 12 trips Time at Copier: 12 x 7.9854 = 95.8 minutes/day Wait Time at Copier: 12 x 2.9854 = 35.8 minutes/day g. Yes. Five administrative assistants x 35.8 = 179 min. (3 hours/day) 3 hours per day are lost to waiting. (35.8/480)(100) = 7.5% of each administrative assistant's day is spent waiting for the copier.


32

35. N = 10 λ = 0. 25 µ = 4 λ/µ = 0.0625 a.

n

N !(N - n) !

λµ

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 2.2698

P0 = 1/2.2698 = 0.4406

b. L N Pq = −+FHG IKJ − = −FHG IKJ − =

λ µλ

( ) ..

( . ) .1 10 4 250 25

1 0 4406 0 48950

c. L = Lq + (1 - P0) = 0.4895 + (1 - 0.4406) = 1.0490

d. WL

N Lqq=

−=

−=

( ).

( . )( . ).

λ0 4895

10 10490 0 250 2188

e. W Wq= + = + =1 0 2188 1

40 4688

µ. .

f. TC = cw L + cs k = 50 (1.0490) + 30 (1) = $82.45 g. k = 2 TC = cw L + cs k = 50L + 30(2) = $82.45 50L = 22.45 L = 0.4490 or less. h. The solution with k = 2 is, L = 0.6237 TC = cw L + cs k = 50 (1.6237) + 30 (2) = $91.18 The company should not expand to the two-channel truck dock.

qs up questions 9

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