LFCH14

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EACHINGEACHING

S

UGGESTIONSUGGESTIONS

Teaching Suggestion 14.1:

Teaching Suggestion 14.1: Topic of Queuing.Topic of Queuing.

Here is a chapter that all students can relate to. Ask about student Here is a chapter that all students can relate to. Ask about student experiences in lines. Stress that queues are a part of our everyday experiences in lines. Stress that queues are a part of our everyday lives and how things have changed at banks, post offices, and lives and how things have changed at banks, post offices, and air-ports in just the past decade. (We now wait in a common line for ports in just the past decade. (We now wait in a common line for the first available server.)

the ﬁrst available server.)

Teaching Suggestion 14.2: Cost of Waiting Time from anCost of Waiting Time from an Organizatio

Organizational nal Perspective.Perspective.

Students should realize that different organizations place different Students should realize that different organizations place different values on customer waiting time. Ask students to consider values on customer waiting time. Ask students to consider differ-ent scenarios, from a drive-through restaurant to a doctor’s office ent scenarios, from a drive-through restaurant to a doctor’s office to a registration line in their college or motor vehicle office. It to a registration line in their college or motor vehicle office. It be-comes clear that organizations place different values on their comes clear that organizations place different values on their cus-tomers’ time (with most colleges and DMVs unfortunately placing tomers’ time (with most colleges and DMVs unfortunately placing minimal cost on waiting time).

minimal cost on waiting time).

Teaching Suggestion 14.3: Use of Poisson and ExponentialUse of Poisson and Exponential Probabilit

Probability Distributions to Describe Arrival and y Distributions to Describe Arrival and Service Rates.Service Rates.

These two distributions are very common in basic models, but These two distributions are very common in basic models, but stu-dents should not take their appropriateness for granted. As a dents should not take their appropriateness for granted. As a pro- ject,

ject, ask ask students students to to visit visit a a bank bank or or drive-through drive-through restaurant restaurant andand time arrivals to see if they indeed are Poisson distributed. Note time arrivals to see if they indeed are Poisson distributed. Note that other distributions (such as exponential, normal, or Erlang) that other distributions (such as exponential, normal, or Erlang) are often more valid.

are often more valid.

Teaching Suggestion 14.4: Balking and Balking and Reneging AssumpReneging Assumptions.tions.

Note that most queuing models assume that balking and reneging Note that most queuing models assume that balking and reneging are not permitted. Since we know they do occur in supermarkets, are not permitted. Since we know they do occur in supermarkets, what can be done? This is one of many places to prepare students what can be done? This is one of many places to prepare students for the need for simulation, the topic of the next chapter.

for the need for simulation, the topic of the next chapter.

Teaching Suggestion 14.5: Use of Queuing Software.Use of Queuing Software.

The Excel QM and QM for Windows queuing software modules The Excel QM and QM for Windows queuing software modules are among the easiest models in the program to use since there are are among the easiest models in the program to use since there are so few inputs. Yet students should be reminded of how long it so few inputs. Yet students should be reminded of how long it would take to produce the programs in Chapter 14 by hand. would take to produce the programs in Chapter 14 by hand.

Teaching Suggestion 14.6: Importance of Importance of LLqqand and WWqqinin Economic Ana

Economic Analysis.lysis.

Although many parameters are computed for a queuing study, the Although many parameters are computed for a queuing study, the two most important ones are L

two most important ones are Lqqand Wand Wqqwhen it comes to an actualwhen it comes to an actual

cost analysis. cost analysis.

Teaching Suggestion 14.7: Teaching the New Teaching the New England England Foundry Case.

Foundry Case.

Here is a tip for this very teachable case. About half the students Here is a tip for this very teachable case. About half the students who tackle the case forget that time walking to the counter must who tackle the case forget that time walking to the counter must be noted

be notedand and that the return time also needs to be added.that the return time also needs to be added.

A

LTERNATIVELTERNATIVE

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XAMPLESXAMPLES

Alternative Example 14.1:

Alternative Example 14.1: A new shopping mall is consideringA new shopping mall is considering setting up an information desk manned by one employee. Based setting up an information desk manned by one employee. Based on information obtained from similar information desks, it is on information obtained from similar information desks, it is be-lieved that people will arrive at the desk at the rate of 20 per hour. lieved that people will arrive at the desk at the rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is It takes an average of 2 minutes to answer a question. It is as-sumed that arrivals are Poisson and answer times are sumed that arrivals are Poisson and answer times are exponen-tially distributed.

tially distributed. a.

a. Find Find the prthe probabilobability tity that thhat the emple employee ioyee is idls idle.e. b.

b. Find the propFind the proportion ortion of the of the time ttime that thhat the emple employee is oyee is busy.busy. c.

c. Find tFind the averhe average numage number of pber of people reople receivieceiving and wng and waitinaitingg to receive information.

to receive information. d.

d. Find the averFind the average numage number of ber of people people waitiwaiting in lng in line to ine to getget information.

information. e.

e. Find Find the avethe average trage time a ime a person person seekinseeking infog informatirmationon spends at the desk.

spends at the desk. f.

f. Find tFind the expeche expected tited time a peme a person sprson spends juends just waist waiting iting inn line to have a question answered.

line to have a question answered. ANSWER:

ANSWER: _20/hour_20/hour ___30/hour_30/hour

a. a. b. b. c. c. d. d. peoplepeople e. e. hourhour f. f. hourshours Alternative Example 14.2:

Alternative Example 14.2: In Alternative Example 14.1, the in-In Alternative Example 14.1, the in-formation desk employee earns $5/hour. The cost of waiting time, formation desk employee earns $5/hour. The cost of waiting time, in terms of customer unhappiness with the mall, is $12/hour of in terms of customer unhappiness with the mall, is $12/hour of time spent

time spentwaitingwaitingin line. Find the total expected costs over an 8-in line. Find the total expected costs over an 8-hour day.

hour day. a.

a. The avThe average erage person person waits waits 0.0667 0.0667 hour anhour and therd there are 1e are 16060 arrivals per day. So total waiting time

arrivals per day. So total waiting time_{(160)(0.0667)}_{(160)(0.0667)} 

10.67 hours @ $12/hour, implying a waiting cost of 10.67 hours @ $12/hour, implying a waiting cost of $128/day.

$128/day. b.

b. The The salsalary ary coscost it is $s $40/40/dayday.. c.

c. ToTotatal cl cososts ts arare $1e $12828_$40_$40_$168/day._$168/day.

W W _qq      ll ll   (( )) (( )) .. 20 20 30 30 30 30 2020 0 06670 0667 W W       1 1 11 330 0 2200 0 10 1    .. L L_qq             22 ₂₀₂₀22 30 30 30 30 2020 1 331 33 ( ( )) ( ( )) ( ( )) .. L L           20 20 330 0 2200 22 peoplepeople      0 660 66.. P P₀₀ 1 1 11 2020 30 30 333 3 3333               . . %%

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Waiting Line and Queuing Theory Models

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Alternative Example 14.3: A new shopping mall is considering setting up an information desk manned by two employees. Based on information obtained from similar information desks, it is be-lieved that people will arrive at the desk at the rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is as-sumed that arrivals are Poisson and answer times are exponen-tially distributed.

a. Find the proportion of the time that the employees are idle.

b. Find the average number of people waiting in the system. c. Find the expected time a person spends waiting in the system.

ANSWER: _20/hour, __30/hour, _M_{2 open channels}

(servers). a.

b.

c.

Alternative Example 14.4: Three students arrive per minute at a coffee machine that dispenses exactly 4 cups/minute at a con-stant rate. Describe the operating system parameters.

ANSWER: _3/minute __4/minute

_{1.125 people in queue on average}

_{0.375 minutes in the queue waiting}

_{1.87 people in the system}

_{0.625 minutes in the system}

W W    _q

1

  .375 1 4 L L   _q    1 125 3 4 . W _q        2 3 2 4 4 3 ( ) ( )( ) L_q        2 2 9 2 4 4 3 ( ) ( )( ) W L     3 4 20 3 80 0 0375 / . . hr  8 00 3      1 600 1 2 2 3 1 12 8 12 9 12 3 4 / ,

(

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peopple L   ( )( )( / ) ( )[( )( ) ] 20 3 0 2 0 30 1 2 30 20 1 2 2 2

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220₃₀      1 1 2 3 1 3 1 2 50%     1 1 2 3 1 2 4 9 60 60 20

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S

OLUTIONS TO

D

ISCUSSION

Q

UESTIONS AND

P

ROBLEMS

14-1. The waiting line problem concerns the question of ﬁnd-ing the ideal level of service that an organization should provide. The three components of a queuing system are arrivals, waiting line, and service facility.

14-2. The seven underlying assumptions are: 1. Arrivals are FIFO.

2. There is no balking or reneging. 3. Arrivals are independent. 4. Arrivals are Poisson.

5. Service times are independent. 6. Service times are negative exponential.

7. Average service rate exceeds average arrival rate.

14-3. The seven operating characteristics are:

1. Average number of customers in the system ( L) 2. Average time spent in the system (W )

3. Average number in the queue ( Lq)

4. Average time in the queue (W q)

5. Utilization factor (  ) 6. Percent idle time (Po)

7. Probability there are more thanK customers in the system

14-4. If the service rate is not greater than the arrival rate, an inﬁnite queue will eventually build up.

14-5. First-in, ﬁrst-out (FIFO) is often not applicable. Some ex-amples are (1) hospital emergency rooms, (2) an elevator, (3) an airplane trip, (4) a small store where the shopkeeper serves who-ever can get his or her attention ﬁrst, (5) a computer system set to accept priority runs, (6) a college registration system that allows juniors and seniors to register ahead of freshmen and sophomores, (7) a restaurant that may seat a party of 2 before a party of 4 even though the latter group arrived earlier, (8) a garage that repairs cars with minor problems before it works on major overhauls.

14-6. Examples of finite queuing situations include (1) a firm that has only 3 or 4 machines that need servicing, (2) a small air-port at which only 10 or 15 flights land each day, (3) a classroom that seats only 30 students for class, (4) a physician who has a lim-ited number of patients, and (5) a hospital ward with only 20 pa-tients who need care.

14-7. a. Barbershop: usually a single-channel, multiple-service system (if there is more than on e barber).

Arrivals_{customers wanting haircuts}

Waiting line_{seated customers who informally recognize}

who arrived ﬁrst among them

Service_{haircut, style, shampoo, and so forth; if}

service involves barber, then shampooist, then manicurist, it becomes a multiphase system

(3)

b. Car wash: usually either a single-channel, single-server system, or else a system with each service bay having its own queue.

Arrivals_{dirty cars or trucks}

Waiting time_{cars in one line (or more lines if there are}

service parallel wash systems); always FIFO Service_{either multiphase (if car ﬁrst vacuumed, then}

soaped, then sent through automatic cleaner, then dried by hand) or single-phase if all automatic or performed by one person c. Laundromat: basically a single-channel, multiserver, two-phase system.

Arrivals_{customers with dirty clothes}

Waiting line_{usually ﬁrst-come, ﬁrst-served in terms of}

selecting an available machine

Service_{ﬁrst phase consists of washing clothes in}

washing machines; second-phase is again queuing for the ﬁrst available drying machine d. Small grocery store: usually a channel, single-server system.

Arrivals_{customers buying food items}

Waiting line_{customers with carts or baskets of groceries}

who arrive ﬁrst at the cash register; sometimes not FIFO; grocer may care for regular customers ﬁrst or give priority to person making a small, quick, purchase Service_{ringing up sale on cash register, collecting}

money, and bagging groceries

14-8. The waiting time cost should be based on time in the queue in situations where the customer does not mind how long it takes to complete service once the service starts. The classic ex-ample of this is waiting in line for an amusement park ride. Waiting time cost should be based on the time in the system when the entire time is important to the customer. When a com-puter or an automobile is taken into the shop to be repaired, the customer is without use of the item until the service is ﬁnished. In such a situation, the time in the system is the relevant time.

14-9. The use of Poisson to describe arrivals:

a. Cafeteria: probably not. Most people arrive in groups and eat at the same time.

b. Barbershop: probably acceptable, especially on a week-end, in which case people arrive at the same rate all day long. c. Hardware store: okay.

d. Dentist’s ofﬁce: usually not. Patients are most likely scheduled at 15- to 30-minute intervals and do not arrive randomly.

e. College class: number of students come in groups at the beginning of class period; very few arrive during the class or very early before class.

f. Movie theater: probably not if only one movie is shown (if there are four or more auditoriums each playing a different movie simultaneously, it may be okay). Patrons all tend to ar-rive in batches 5 to 20 minutes before a show.

14-10.

NUMBER OFCHECKOUTCLERKS

1 2 3 4

Numberofcustomers 300 300 300 300

Average waiting time 1

–

₆hour ₁₀1

–

hour ₁₅1

–

hour ₂₀1

–

hour per customer (10 minutes) (6 minutes) (4 minutes) (3 minutes) Total customer waiting time 50 hours 30 hours 20 hours 15 hours

Costperwaitinghour $10 $10 $10 $10

Totalwaitingcosts $500 $300 $200 $150

Checkoutclerkhourlysalary $8 $8 $8 $8

Totalpayofclerksfor $64 $128 $192 $256

8-hour shift

Totalexpectedcost $564 $428 $392 $406

Optimal number of checkout clerks on duty3

(4)

d. The utilization rate,  , is given by

e. The probability that no cars are in the system,P0, is

given by:

14-13. _{210 patrons/hour,}__{280 patrons/hour.}

a. The average number of patrons waiting in line, Lq, is

given by

b. The average fraction of time the cashier is busy,  , is given by

c. The average time a customer spends in the ticket-dispensing system,W,is given by

_{0.0143 hour in the line} _{0.857 minute}_{51.4 seconds}

d. The average time spent by a patron waiting to get a ticket,W q,is given by

_{38.6 seconds}

e. The probability that there are more than two people in the system,Pn2, is given by

The probability that there are more than three people in the sys-tem,Pn3, is given by

The probability that there are more than four people in the system,

Pn₄, is given by

14-14. _{4 students/minute,}_ 60 q w_{5 students/minute}

a. The probability of more than two students in the system,

Pn₂, is given by P_n2  3 4 5 0 512

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1 P_n4  5 210 280 0 237

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. P_n3  4 210 280 0 316

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. Pn2  3 210 280 0 422

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1  210   19 600, 0 011 . hour 0 64. minute W q        ( ) ( ) ( ) 210 280 280 210 210 280 70 W      1 1 280 210 1 70        210 280 0 75.  44 100 19 600 2 25 , , . patrons inline L_q         2 ₂₁₀2 280 280 210 44 100 280 70 ( ) ( ) , ( ) 0      1 1 1 0 8333 0 1667    . .      10 12 0 8333. 14-11. a. The utilization rate,  , is given by

_0.375

b. The average down time,W, is the time the machine waits to be serviced plus the time taken to perform the service.

_{0.2 day, or 1.6 hours}

c. The number of machines waiting to be served, Lq,is,

on average,

_{0.225 machine waiting}

d. Probability that more than one machine is in the system

Probability that more than two machines are in the system:

14-12. _{10 cars/hour,}__{12 cars/hour.}

a. The average number of cars in line, Lq,is given by

_{4.167 cars}

b. The average time a car waits before it is washed,W q,is

given by

_{0.4167 hours}

c. The average time a car spends in the service system,W,

is given by W       1 1 12 10 1 2 0 5   . hour W q        ( ) ( ) ( )( ) 10 12 12 10 10 12 2 q _  _      2 ₁₀2 2 12 12 10 10 12 2 ( ) ( ) ( )( ) P_n4   5 3 8 243 32 768 0 007

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_, . P_n_₃   4 3 8 81 4 096 0 020

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_, . P_n2   3 3 8 27 512 0 053

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. P_n1   2 3 8 9 64 0 141

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1   3 8 8 3 2 ( ) Lq      2 ( )   1 8 3 W   1   3 8   

=

(5)

The probability of more than three students in the system,Pn3, is

given by

The probability of more than four students in the system,Pn4, is

given by

b. The probability that the system is empty,P0, is given by

c. The average waiting time,W q,is given by

minute

d. The expected number of students in the queue, Lq, is

given by

students

e. The average number of students in the system, L,is given as

students f. Adding a second channel, we have

_{4 students/minute}

students/minute

m₂

f (part b). The probability that the two-channel system is empty,P0, is given by

or

Thus the probability of an empty system when using the second channel is 0.429. P₀ 1 1 0 8 0 53 1 2 33 0 429   .  .  .  .   1 4 5 160 300   1 1 4 5 1 2 16 25 10 6

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   1 1 4 5 1 2 4 5 2 5 10 4 2

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( )   1 1 0 4 5 1 1 4 5 1 1 2 4 5 2 0 1 2 ! ( ) (

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_{2 5}_{( )}55_)₄ P n m n n n m 0 0 1 1 1 1      ! !    

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m m m     60 12 5 L        4 5 4 4 L_q         2 ₄2 5 5 4 3 2 ( ) ( ) . W q        ( ) ( ) . 4 5 5 4 0 8 P₀ 1 1 4 5 1 0 8 0 2          . . P_n4  5 4 5 0 328

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. P_n3  4 4 5 0 410

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.

f (part c). The average waiting time, W q, for the

two-channel system is given by

where

Then

f (part d). The average number of students in the queue for the two-channel system, Lq,is given by

where

Then

_{0.15 student}

f (part e). The average number of students in the two-channel system, L,is given by

14-15. _{30 trucks/hour,}__{35 trucks/hour.}

a. The average number of trucks in the system, L,is given by

b. The average time spent by a truck in the system,W,is given by      1 35 30 1 5 0 2 . hour 12minutes W   1       30 35 30 30

5 6trucks in the system

L        L_q     0 153 4 5 0 95 . . student L m m P m           

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( 1)!( )2 0    4 5 0 64 1 10 4 0 429 5 492 1 36 2 ( )( . ) ( ) ( . ) . ( ) L_q   4 5 4 5 2 1 2 5 4 0 429 2 2 ( ) ( )![ ( ) ] ( . )

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L m m P m           

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( 1)!( )2 0 L _q L    1 373  1 36 0 038 2 3 .

( ) . minute . seconds  5 0 64 1 10 42 0 429 ( . ) ( ) ( . ) W q   5 4 5 2 1 2 5 4 0 429 2 2

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( )[ ( ) ] ( . ) W m m P m          

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( 1)!( ) 1 2 0 W q W  1 

(6)

c. The utilization rate for the bin area,  , is given by

d. The probability that there are more than three trucks in the system,Pn₃, is given by

Thus the probability that there are more than three trucks in the system is 0.540.

e. Unloading cost:

_{16(30)(0.2)(18)}

_{$1,728/day or $12,096 per week.}

f. Enlarging the bin will cut waiting costs by 50% next year. First, we must compute annual waiting costs:

_$24,192

Enlarging the bin will cut waiting costs by 50% next year, result-ing in a savresult-ings of $12,096. Since the cost of enlargresult-ing the bin is only $9,000, the cooperative should proceed to enlarge the bin. The net savings is $3,096 ($12,096_$9,000).

14-16. _{12 calls/hour,}_ 60 r_{15 calls/hour.}

a. The average time the catalog customer must wait,W q, is

given by

_{16 minutes}

b. The average number of callers waiting to place an order,

Lq, is given by

c. To decide whether or not to add the second clerk, we must (a) compute present total cost, (b) compute total cost with the second clerk, and (c) compare the two. Present total cost:

C /hourt service costwaiting cost

₁₀_{12(0.267)(50)}₁₀_160.2 _$170.20/hour  10 12  calls 0 267 50 hour hours call dollar .

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     12 15 15 12 144 15 3 144 45 3 2 2 ( ) ( ) . customers L_q      2 ( )      12 15 15 12 12 15 3 12 45 0 267 ( ) ( ) . W _q     ( ) 2 weeks 7 1 728 year days week dollars day

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annual waiting cost C_M16 hours 30 0 2 day trucks hour hours truc

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18dollars_hour

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Pn3  4 30 35 0 540

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. P k k n    

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1   30 35 6 7 0 857.    

To determine total cost using the second clerk (a second channel):

or

Then

Cost with two clerks:

C /hourt service costwaiting cost

₂₀_{12(0.0127)(50)}₂₀_7.62 _$27.62/hour

There is a savings of 170.20_27.62_{142.5/hour. Thus a second}

clerk should certainly be added!

14-17. This is an M/M/1 system with _{24 per hour and}_

30 per hour. a. W _{0.167 hours} b. L₄ c. W q0.133 d. Lq3.2 e. P00.2 f.  _0.8 g. P(n₂₎_P_n1Pn20.6400.5120.128  20 12 calls 0 0127 50 hour hours call dolla .

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 4 12   1 324 0 0127 0 763 . ( ) . hour . seconds   15 0 64 1 30 122 0 429 ( . ) ( ) ( . ) W q   15 12 15 2 1 2 15 12 0 429 2 2

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( )[ ( ) ] ( . ) W m m P q m        

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( 1)!( )2 0  1  2 33. 0 429. P₀ 1 1 0 8 0 53   .  .    1 1 4 5 480 900    1 1 4 5 1 2 16 25 30 18

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    1 1 4 5 1 2 4 5 2 15 30 12 2

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( )    1 1 0 12 15 1 1 12 15 1 1 2 12 15 0 1 ! ( )

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2 2 15 2 15 12 ( ) ( ) P n m n n n m 0 0 1 1 1 1      ! !    

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m m m   

(7)

30 per hour. Using QM for Windows we get the following: a. W _{0.0397 hours} b. L_0.9524 c. W q0.0063 d. Lq0.1524 e. P00.4286 f.  _0.4 g. P(n₂₎_0.1371_P_n₁Pn₂0.22860.0914 14-19and14-20.

These results indicate that when only one loader is employed, the average truck must wait 3rhour before it is loaded. Furthermore, there are an average of 2.25 trucks waiting in line to be loaded. This situation may be unacceptable to management. Note the de-cline in the queue when a second loader is employed.

14-21. Referring to the data in Problems 14-19 and 14-20, we note that the average number of trucks in thesystemis 3 when there is only one loader and 0.6 when two loaders are employed.

The ﬁrm will save $18/hour by adding the second loader.

14-22.      1 1 3 4 1 2 3 4 8 8 3 0 454 2

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. P n n n 0 0 1 2 1 1 3 4 1 2 3 4 2    ! !

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∑

_{2 4}_{( )}(( )4 ₃  hour hour

By looking back to Problems 14-19 and 14-20, we see that al-though length of the queue and average time in the queue are low-est by opening the second platform, the average number of trucks in the system and average time spent waiting in the system are smallest when two workers are employed loading at asingle plat-form. Hence we would probably recommend not building a second gate.

14-23. The queuing systems in this problem are the M/M/2, M/M/3, and the M/M/4 systems.

a. Wq0.0643 for 2 channels; Wq0.0079 for 3 channels;

Wq0.0015 for 4 channels;

b. The total time spent waiting is Wq(10 hours per day).

This is 19.29 hours with 2 channels, 2.37 hours with 3 chan-nels, and 0.45 hours with 4 channels.

c. The total daily waiting time cost is given in the table below:

Service Service Total Total

# cost per cost per waiting time waiting Total Channels hour day  Wq(10hr.) cost cost

2 $20 $200 19.29 $1929 $2129

3 $30 $300 2.37 237 $537

4 $40 $400 0.45 45 $445

The minimum daily cost is $445 with 4 channels.

15 per hour. a. W q0.1333 hours b. Lq1.333 c. W _{0.2 hours} d. L₂ e. P00.333

15 per hour. a. W q0.0083 hours b. Lq0.083 c. W _0.075 d. L_0.75 e. P00.5

14-26. a. (8 hours per day)₁₀₍₈₎_{80 customers per day}

b. Total time spent waiting_W_q_{(number of customers)}

0.1333(80)_{10.66 hours.}

Total waiting time cost_$25(10.66)_$266.5

c. With 2 tellers, total time spent waiting_0.0083(80)

0.664 hours. W q   0 123 3 0 041 . . L_q 0 873 3 4 0 123 . . W  0 873 3 0 291 . . L    3 4 3 4 1 8 3 0 4545 3 4 0 873 2 2 ( ) ( )!( ) ( . ) .

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NUMBER OFFRUIT LOADERS 1 2

Truck arrival rate ( ) 3/hour 3/hour Loading rate () 4/hour 8/hour Average number in system (L) 3 trucks 0.6 truck Average time in system (W ) 1 hour 0.2 hour Average number in queue (Lq) 2.25 trucks 0.225 truck

Average time in queue (W q) –3₄hour 0.075 hour

Utilization rate (  ) 0.75 0.375 Probability system 0.25 0.625 empty (P 0) Probability of more K thanK trucks in ₀ _0.75 _0.375 system ₁ _0.56 _0.141 2 0.42 0.053 3 0.32 0.020 NUMBER OF LOADERS 1 2

Truck driver idle time costs (average number truckshourly

rate) (3)($10)_$30 _$6_(0.6)($10) Loading costs 36 12(2)($6) Total expected cost per hour $36 $18

(8)

Total waiting time cost_$25(0.664)_$16.60

d. Total cost with 1 teller_$266.5_$96_$362.5

Total cost with 2 tellers_$16.60_2($96)_$208.60

14-27. a. Average number in line_0.666

b. Average number in system_1.333

c. Average wait in line_{0.1666 minute}_{10 seconds}

14-28. For M _1:

But

Thus

This is the same formula.

14-29. _1/day N _5,_n₁ _0.36 a. Number in queue ₅_4.48_{0.52 unit}

b. Number in the system

L_L_q₍₁_P₀₎_0.52₍₁_0.36) _0.52_0.64_{1.16 in system}

c. Number running ok _N_L₅_1.16 _3.84

d. Average time in queue

days

e. Average wait in system

_0.817₁ _{1.817 days} 14-30. _4/hour, _N_5,_n₁ _60(0.1765)3__120(0.1765)4__120(0.1765)5 P₀ 1 ₂ 1 5 0 1765 20 0 1765   ( . )  ( . )     60  85 0 706 . /hour, 0 1765.  W W  _q1  W L N L q q      ( ) . ( . )( . ) .  0 52 5 1 16 0 1667 0 817 L _q N      P  ( 1 0) 5 7 1 0 36( . ) P0 ₁ 6 1 6 2 1 6 3 1 6 4 1 6 1 1 5 20 60 120 120   ( )  ( )  ( )  ( )  ( ))5  0 1667 1 6 . / day /day                      2 ₍ ₎ ( )( ) ( )( ) L                     2 2 2 ( ) ( )( )

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P₀ 1 1 1 1                   

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L  P            

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1 2 0 2 ( ) (      )2 P0 _0.34

a. Average number waiting_L_q

b. Average number broken_L _L_q₍₁_P₀₎

_0.576₍₁_0.34)_1.24

c. P00.34, as seen above.

d. Average time in queue

hour e. Average time in system

_0.217_0.25 _{0.467 hour}

14-31. a. Entering: _84/minute,__30/minute,__2.8

Exiting: _48/minute,__30/minute,__1.6

The manager desires thatW q0.1 minute6 seconds and that Lq8 customers in queue.

Entering:

If M _3,_L_q_{12.27 and}_W_q_{0.14 minute (too high)}

If M _4,_L

q1.00 andW q0.01 minute (this is okay)

If M _5,_L_q_{0.24 and}_W_q_{0.003 minute (this is also}

okay)

So the manager must open M _{4 or more entrances.}

Exiting:

If M _2,_L_q_2.8,_W_q_{0.06 minute (this is okay)}

If M _3,_L_q_0.31,_W_q_{0.006 minute (also okay)}

So the manager must open M _{2 or more exits. Since there are}

only 6 turnstiles, 4 must be used as entrances and 2 as exits. b. The students should recognize andquestionall the limiting queuing assumptions that have been applied in solving the case. For example, it may be reasonable to assume that arrivals at theentranceturnstiles are independent and Poisson. But are

exitingpassengers independent? More realistically, they arrive in batches (as a train arrives), and unless trains unload every minute or two, this assumption may be unreasonable.

Other problems arise as well. If an exiting passenger’s card does not have the correct fare, the card is rejected and the passen-ger must leave the line, go to an “add fare” machine to correct the deﬁciency, and enter the queue again. This resembles the reneging customer.

Note:In the real-world subway station in Washington, D.C., common queues arenot formed at turnstiles and the problem be-comes a series of single channel queues.

 W  W _q 1 μ

=

0 576

=

2 65 0 217 . . .  0 576 5 1 24 0 706 . ( . )( . )    W L N L q q )     5 4 706 0 706 0 66 5 4 4 0 6 . . ( . ) . .

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 N   P  ( 1 0)        1 1 0 88 0 62 0 30 0 12 0 02 1 2 944 . . . .

(9)

14-32. This is an M/M/1 system with = 12 per hour and= 15 per hour. a. Wq= 0.27 hours b. Lq= 3.2 c. L = 4 d. W = 0.33 hours e. Pn3= (12/15)3+1= 0.4096

14-33. This is an M/M/2 system with = 12 per hour and= 15 per hour.

a. Wq= 0.013 hours

b. Lq= 0.152

c. L = 0.952 d. W = 0.079 hours

S

OLUTIONS TO

I

NTERNET

H

OMEWORK

P

ROBLEMS

14-34. _12/hour;__{4/hour/barber;}_M_{4 channels}

a. P00.03773.8% (from formula) b. L4.528 c. W 0.377 hour22.6 minutes d. W q0.127 hour7.6 minutes e. Lq1.5282 (from formula) f.  _0.75_75%

g.  withm_{5 barbers drops to 60%}

14-35. a. 9A.M.–3P.M.; _{6 patients/hour;}__{5 patients/}

hour/doctor

WantW qto be5 minutes0.0833 hour.W q0.0833 implies

that

or Lq0.0833or Lq0.50

Thusm _{3 channels or doctors are needed (with}_m_2,_L_q

0.6748; withm_3,_L_q_0.0904).

b. 3P.M.–8P.M.; _{4 patients/hour;}_₅

patients/hour/doctor

W q0.0833 hour implies that or Lq0.0833 or Lq0.03333. This meansm2 doctors.

c. 8P.M.–midnight; _{12 patients/hour;}__{5 patients/}

hour/doctor

WantW q0.0833 hour or or Lq0.0833 or Lq

1.00.m_{4 doctors are needed.}

14-36. a. _{1 per minute and}__{2 per minute}

b. M/M/1 c.   1w_0.5

d. P01 1w0.5. The cashier is idle 50% of the time.

e. Lq0.5 L_q  0 0833.   12 5 2 4. L_q  0 0833.   4 5 0 80. W _q Lq   0 0833.      6 5 1 20. f. W q0.5 minute g. W _{1 minute}

14-37. a. _{3 per minute and}__{4 per minute}

b. M/M/1 c.  _3/4_0.75

d. P013/40.25. The cashier is idle 25% of the time.

e. Lq2.25 f. W q0.75 minute g. W _{1 minute} h. P(n₁₎_0.188 P(n₂₎_0.141 P(n₃₎_0.106

14-38. This is an M/M/2 system with _{3 per minute and}_₄

per minute. Solving with QM for Windows we obtain the following: a. Lq0.1227

b. W q0.0409 minute

c. W _{0.2909 minutes}

d. P(n₁₎_0.3409,_P₍_n₂₎_0.1278,_P₍_n₃₎_0.0479.

S

OLUTION TO

N

EW

E

NGLAND

F

OUNDRY

C

ASE

1. To determine how much time the new layout would save, the present system must be compared to the new system. The amount of time that an employee spends traveling to the maintenance de-partment added to the time that he or she spends in the system being serviced and waiting for service presently, compared to this value under the proposed system, will give the savings in time.

Under the present system, there are two service channels with a single line ( M _{2). The number of arrivals per hour is 7 (}__{7). The}

number of employees that can be serviced in an hour by each channel is 5 (_{5). The average time that a person spends in the system is}

where

In this case

Therefore,

_{0.396 hour, or 23 minutes and 45 seconds}

Added to the travel times involved (6 minutes total for main-tenance personnel and 2 minutes total for molding personnel), the total trip takes:

For maintenance—29 minutes and 45 seconds For molding—25 minutes and 45 seconds

Under the new system, waiting lines are converted to single-channel, single-line operations. Bob will serve the maintenance

W    5 7 5 1 10 7 0 18 1 5 2 2 ( / ) ( ) ( . ) / P0 ₁ ₂ 1 1 1 1 1 1 7 5 1 2 7 5 2    ( )

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_{5 2}55 ₇ 0 18 ) ( ) .  

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P n M n n M 0 0 1 1 1 1     ! !    

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M M M M    W M M P M           ( / ) ( 1)!( ) 1 2 0

(10)

personnel and Pete will serve the molding personnel.

Bob can now service 6 people per hour (_{6). Four people}

arrive from the maintenance department every hour ( _{4). The}

time spent in Bob’s department is

hour, or 30 minutes

The reduced travel time is equal to 2 minutes, making the total trip time equal to 32 minutes. This is anincreasein time of 2 minutes and 15 seconds for the maintenance personnel.

Pete can now service 7 people per hour (_{7). Three}

peo-ple arrive from the molding department every hour ( _{3). The}

time in Pete’s department is

hour, or 15 minutes

The travel time is equal to 2 minutes, making the total trip time equal to 17 minutes. This is adecreasein time of 8 minutes and 45 seconds per trip for the molding personnel.

2. To evaluate systemwide savings, the times must be mone-tized. For the maintenance personnel who are paid $9.50 per hour, the 2Z\vminutes lost per trip costs the company 36 cents per trip [2Z\v₆₀ _{0.0375 of an hour; 0.0375(9.50)}_{$0.36]. For the}

molding personnel who are paid $11.75 per hour, the 8 minutes and 45 seconds per trip saved saves in monetary terms $1.71 per trip. The net savings is $1.71_0.36_{$1.35 per trip. (Students}

may also ﬁnd the cost savings on an hourly or daily basis.) Because the net savings for the new layout is small, other fac-tors should be considered before a ﬁnal decision is made. For ex-ample, the cost of changing from the old layout to the new layout could completely eliminate the advantages of operating the new layout. In addition, there may be other factors, some noneco-nomic, that were not discussed in the case that could cause you to want to stay with the old layout. In general, when the cost savings of a new approach (a new layout in this case) is small, careful analysis should be made of other factors.

S

OLUTION TO

W

INTER

P

ARK

H

OTEL

C

ASE

1. Which of the two plans appears to be better? The current sys-tem has ﬁve clerks each with his or her own waiting line. This can be treated as ﬁve independent queues each with an arrival time of

_90/5_{18 per hour. The service rate is one every 3 minutes,}

or_{20 per hour. Assuming Poisson arrivals and exponential}

service times, the average amount of time that a guest spends wait-ing and checkwait-ing in is given by

hour, or 30 minutes

If 30% of the arrivals [that is, _0.3(90)_{27 per hour] are}

diverted to a quick-serve clerk who can register them in an aver-age of 2 minutes (_{30 per hour) their average time in the}

sys-tem will be 20 minutes. The remaining 63 arrivals per hour would distribute themselves equally among the four remaining clerks ( _63/4_{15.75 per hour), each of whose mean service time is}

3.4 minutes (or 0.5667 hour), so that_1/0.5667_{17.65 per}

hour. The average time in the system for these guests will be 0.53 hour or 31.8 minutes. The average time for all arrivals would be 0.3(20)_0.7(31.8)_{28.3 minutes.}

A single waiting line for the ﬁve clerks yields an M/M/5 queue with _{90 per hour,}__{20 per hour. The calculation of}

   1 20 18 0 5. W _s  1   W    1 7 3 1 4 W      1 1 6 4 1 2  

average time in the system givesW _{7.6 minutes. This plan is}

clearly faster.

Use of an ATM with the same service rate as the clerks (20 per hour) by 20 percent of the arrivals (18 per hour) gives the same average time for these guests as the current systems—30 minutes. The remaining _{72 per hour form an}_M/M/_{4 or}_M/M/₅

queuing system. With four servers, the average time in the system is 8.9 minutes, resulting in an overall average of:

0.2₃₀_0.8_8.9_{13.1 minutes}

With ﬁve servers, the average time is 3.9 minutes resulting in an overall average of:

0.2₃₀_0.8_3.9_{9.1 minutes}

I

NTERNET

C

ASE

S

TUDY

Pantry Shopper

Beth wants to get a general idea of the system behavior. She ﬁrst will need to decide whether she is interested in time waiting or time in system. Some students may use system time, but since most shoppers are relieved when it is their turn, we use waiting time as our measure. For all of our analyses, we use current ser-vice times, even though a UPC reader is going to be installed. This means that our waiting times are an upper bound for the new, bet-ter system (the M/M/smodel).

We begin with a rough analysis (one that is going to have a very interesting feature, by the way). We assume that there are no express lanes. Then, we want to ﬁnd the average service time and rate. The time is given by

t _{.2(2 min.)}_{.8(4 min.)} _.4_3.2

_{3.6 min.}

This means that the average service rate is 60/3.6_16.67

cus-tomers per hour. Notice that this is not the same as taking 20 per-cent of the rate of 30 and 80 perper-cent of the rate of 15, which would equal 18 and would be wrong.

Using an arrival rate of 100 and a service rate of 16.67, the min-imum number of servers is 6. (This is due to round off.) In reality, the minimum number is 7, and the average waiting time is 2.2 min-utes. Trying one more server leads to a waiting time of .64 minmin-utes.

Now we separate the express and regular. Assume that all ex-press customers go into the exex-press (even though they can go into any lane) and assume that all non-express customers go into the proper lanes (even though we all have seen people with twenty packages get into a ten-items-or-less line).

For the express lane, with an arrival rate of 20 and a service rate of 30, one server yields an average wait of 4 minutes, while two servers yield an average wait of .25 minutes.

For the regular lane, with an arrival rate of 80 and a service rate of 15, 6 servers yield an average wait of 4.28 minutes and 7 servers yield an average wait of .98 minutes.

If Beth uses 7 servers, they will be split this way: 6 in regular lanes and 1 in an express lane. If Beth uses 8 severs, a 6–2 split be-tween regular lanes and express lanes yields an average wait of

(.2)(.25)_(.8)(4.28)_.05_3.424_{3.47 min.}

A 7–1 split yields an average of

(.2)(4)_(.8)(.98)_.8_.784_{1.584 min.,}

which is better. However, the express lane would beslower than the regular lanes!