217
T
EACHINGS
UGGESTIONSTeaching Suggestion 14.1:Topic of Queuing.
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 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 the first available server.)
Teaching Suggestion 14.2:Cost of Waiting Time from an Organizational Perspective.
Students should realize that different organizations place different values on customer waiting time. Ask students to consider differ-ent scenarios, from a drive-through restaurant to a doctor’s office to a registration line in their college or motor vehicle office. It be-comes clear that organizations place different values on their cus-tomers’ time (with most colleges and DMVs unfortunately placing minimal cost on waiting time).
Teaching Suggestion 14.3:Use of Poisson and Exponential Probability Distributions to Describe Arrival and Service Rates. These two distributions are very common in basic models, but stu-dents should not take their appropriateness for granted. As a pro-ject, ask students to visit a bank or drive-through restaurant and time arrivals to see if they indeed are Poisson distributed. Note that other distributions (such as exponential, normal, or Erlang) are often more valid.
Teaching Suggestion 14.4:Balking and Reneging Assumptions. Note that most queuing models assume that balking and reneging are not permitted. Since we know they do occur in supermarkets, what can be done? This is one of many places to prepare students for the need for simulation, the topic of the next chapter.
Teaching Suggestion 14.5:Use of Queuing Software.
The Excel QM and QM for Windows queuing software modules 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 would take to produce the programs in Chapter 14 by hand. Teaching Suggestion 14.6:Importance of Lqand Wqin Economic Analysis.
Although many parameters are computed for a queuing study, the two most important ones are Lqand Wqwhen it comes to an actual
cost analysis.
Teaching Suggestion 14.7:Teaching the New England Foundry Case.
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 be noted andthat the return time also needs to be added.
A
LTERNATIVEE
XAMPLESAlternative Example 14.1: A new shopping mall is considering setting up an information desk manned by one employee. 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 probability that the employee is idle.
b. Find the proportion of the time that the employee is busy.
c. Find the average number of people receiving and waiting
to receive information.
d. Find the average number of people waiting in line to get
information.
e. Find the average time a person seeking information
spends at the desk.
f. Find the expected time a person spends just waiting in
line to have a question answered.
ANSWER: 20/hour 30/hour
a.
b. c.
d. people
e. hour
f. hours
Alternative Example 14.2: In Alternative Example 14.1, the in-formation desk employee earns $5/hour. The cost of waiting time, in terms of customer unhappiness with the mall, is $12/hour of time spent waitingin line. Find the total expected costs over an 8-hour day.
a. The average person waits 0.0667 hour and there are 160
arrivals per day. So total waiting time (160)(0.0667)
10.67 hours @ $12/hour, implying a waiting cost of $128/day.
b. The salary cost is $40/day.
c. Total costs are $128 $40 $168/day.
Wq
l l
( ) ( ) .
20
30 30 20 0 0667 W
1 1
30 20 0 1. Lq
2 2
20
30 30 20 1 33
( )
( )
( ) .
L
20
30 20 2people
0 66. P0 1 1 20
30 33 33
. %
14
C H A P T E R
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, M2 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
WWq 1 .375
1 4 L L q
1 125 3 4 . Wq
2
3 2 4 4 3
( ) ( )( )
Lq
2
2
9 2 4 4 3
( ) ( )( )
WL
3 4 20
3
80 0 0375 /
. . hr
800 3
1 600 1 2
2 3
1 12
8 12
9 12
3 4 /
,
(
)
⎛ ⎝⎜⎞
⎠⎟ peopple
L
( )( )( / ) ( )[( )( ) ] 20 30 20 30 1 2 30 20
1 2
2
2
⎛ ⎝⎜
⎞ ⎠⎟
220 30
1
1 2 3
1 3
1 2 50%
1
1 2 3
1 2
4 9
60 60 20
⎛ ⎝⎜
⎞ ⎠⎟ ⎡ ⎣
⎢ ⎤
⎦ ⎥
( )
P0 0 1 1 1
0 20 30
1 1
20 30
1 2
20 30
! ! !
⎛ ⎝⎜
⎞ ⎠⎟
⎛ ⎝⎜
⎞ ⎠⎟
⎛ ⎝⎜⎜
⎞ ⎠⎟
⎡ ⎣
⎢ ⎤
⎦ ⎥
2
2 30 2 30 20
( ) ( )
S
OLUTIONS TOD
ISCUSSIONQ
UESTIONS ANDP
ROBLEMS14-1. 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 (Wq) 5. Utilization factor ()
6. Percent idle time (Po)
7. Probability there are more than K 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 one 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
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
Number of customers 300 300 300 300
Average waiting time
–
16hour 10–
1hour 15–
1hour 20–
1hour per customer (10 minutes) (6 minutes) (4 minutes) (3 minutes) Total customer waiting time 50 hours 30 hours 20 hours 15 hoursCost per waiting hour $10 $10 $10 $10
Total waiting costs $500 $300 $200 $150
Checkout clerk hourly salary $8 $8 $8 $8
Total pay of clerks for $64 $128 $192 $256
8-hour shift
Total expected cost $564 $428 $392 $406
Optimal number of checkout clerks on duty 3
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 minute51.4 seconds
d. The average time spent by a patron waiting to get a
ticket, Wq,is given by
38.6 seconds
e. The probability that there are more than two people in
the system, Pn2, 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, 6 0q w5 students/minute
a. The probability of more than two students in the system,
Pn 2, is given by
Pn2
3
4
5 0 512
⎛ ⎝⎜
⎞ ⎠⎟ . Pn k
k
⎛ ⎝⎜
⎞ ⎠⎟
1
Pn4 5
210
280 0 237
⎛ ⎝⎜
⎞ ⎠⎟ . Pn3
4
210
280 0 316
⎛ ⎝⎜
⎞ ⎠⎟ . Pn2
3
210
280 0 422
⎛ ⎝⎜
⎞ ⎠⎟ . Pn k
k
⎛ ⎝⎜
⎞ ⎠⎟
1
210
19 600, 0 011. hour 0 64. minute Wq
( ) ( ) ( )
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 in line Lq
2 2
210 280 280 210
44 100 280 70
( ) ( )
, ( ) P0 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, Wq,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 Wq
( ) ( ) ( )( ) 10
12 12 10 10 12 2 Lq
2 102 2
12 12 10 10 12 2
( ) ( ) ( )( )
Pn4 5
3 8
243
32 768 0 007
⎛ ⎝⎜
⎞
⎠⎟ , . Pn3
4
3 8
81
4 096 0 020
⎛ ⎝⎜
⎞
⎠⎟ , . Pn2
3
3 8
27 512 0 053
⎛ ⎝⎜
⎞
⎠⎟ .
Pn1 2
3 8
9 64 0 141
⎛ ⎝⎜
⎞
⎠⎟ .
Pn k k
⎛ ⎝⎜
⎞ ⎠⎟
1
3 8 8 3
2
( )
Lq
2
( )
1 8 3 W
1 3
The probability of more than three students in the system, Pn3, is
given by
The probability of more than four students in the system, Pn4, is
given by
b. The probability that the system is empty, P0, is given by
c. The average waiting time, Wq,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 m2
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 1 4 5 160 300 1 1 4 5 1 2 16 25 10 6 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ 1 1 4 5 1 2 4 5 2 5 10 4 2 ⎛ ⎝⎜ ⎞ ⎠⎟ ( ) 1 1 0 4 5 1 1 4 5 1 1 2 4 5 2
0 1 2
! ( ) ( ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ 55 2 5 4
) ( ) P n m n n n m 0 0 1 1 1 1 ! ! ⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎛ ⎝⎜ ⎞
∑
⎠⎠⎟ m m m 60 12 5 L 45 4 4 Lq
2 2
4
5 5 4 3 2
( ) ( ) . Wq ( ) ( ) . 4 5 5 4 0 8 P0 1 1 4
5 1 0 8 0 2
. .
Pn4
5
4
5 0 328
⎛ ⎝⎜
⎞ ⎠⎟ . Pn3
4
4
5 0 410
⎛ ⎝⎜
⎞ ⎠⎟ .
f (part c). The average waiting time, Wq, 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
Lq 0 153
4 5 0 95
. . student
L
m m P m ⎛ ⎝⎜ ⎞ ⎠⎟
( 1)!( )2 0
4 5 0 64 1 10 4 0 429
5 492 1 36 2 ( )( . ) ( ) ( . ) . ( ) Lq
4 5 4 5
2 1 2 5 4 0 429
2 2 ( ) ( )![ ( ) ] ( . ) ⎛ ⎝⎜ ⎞ ⎠⎟ L
m m P m ⎛ ⎝⎜ ⎞ ⎠⎟
( 1)!( )2 0 Lq L
1 373
1 36 0 038 2 3 .
( ) . minute . seconds
5 0 64 1 10 42 0 429
( . ) ( ) ( . ) Wq 5 4 5
2 1 2 5 4 0 429
2 2 ⎛ ⎝⎜ ⎞ ⎠⎟ ( )[ ( ) ] ( . ) W
m m P m ⎛ ⎝⎜ ⎞ ⎠⎟
( 1)!( ) 1
2 0
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, 6 0r 15 calls/hour.
a. The average time the catalog customer must wait, Wq, 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:
Ct/hourservice costwaiting cost
1012(0.267)(50)10160.2
$170.20/hour
10 12calls 0 267 50 hour hours call dollar . ⎛ ⎝⎜ ⎞ ⎠⎟ ss hour ⎛ ⎝⎜ ⎞ ⎠⎟ 12 15 15 12
144 15 3
144 45 3 2
2
( ) ( ) . customers
Lq 2 ( ) 12 15 15 12
12 15 3
12 45 0 267
( ) ( ) .
Wq ( )
2weeks 7 1 728
year days week dollars day ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ , ⎜⎜ ⎞⎠⎟ annual waiting cost
CM16 30 0 2 hours day trucks hour hours truc ⎛ ⎝⎜ ⎞
⎠⎟ . kk
dollars hour ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ 18 Pn3
4
30
35 0 540
⎛ ⎝⎜ ⎞ ⎠⎟ . P k k n ⎛ ⎝⎜ ⎞ ⎠⎟ 1 30 35 6 7 0 857.
To determine total cost using the second clerk (a second channel):
or
Then
Cost with two clerks:
Ct/hourservice costwaiting cost
2012(0.0127)(50)207.62
$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. W0.167 hours
b. L4
c. Wq0.133
d. Lq3.2
e. P00.2
f. 0.8
g. P(n2)Pn 1Pn 20.6400.5120.128
20 12calls 0 0127 50 hour hours call dolla . ⎛ ⎝⎜ ⎞ ⎠⎟ rrs hour ⎛ ⎝⎜ ⎞ ⎠⎟ 4 12
1 324 0 0127 0 763 .
( ) . hour . seconds
15 0 64 1 30 122 0 429
( . ) ( ) ( . ) Wq 15 12 15
2 1 2 15 12 0 429
2 2 ⎛ ⎝⎜ ⎞ ⎠⎟ ( )[ ( ) ] ( . ) W
m m P q m ⎛ ⎝⎜ ⎞ ⎠⎟
( 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 ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ 1 1 4 5 1 2 4 5 2 15 30 12 2 ⎛ ⎝⎜ ⎞ ⎠⎟ ( ) 1 1 0 12 15 1 1 12 15 1 1 2 12 15 0 1 ! ( ) ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝⎜ ⎞⎞ ⎠⎟ 2 2 15 2 15 12
14-18. This is an M/M/2 system with 24 per hour and 30 per hour. Using QM for Windows we get the following:
a. W0.0397 hours
b. L0.9524
c. Wq0.0063
d. Lq0.1524
e. P00.4286
f. 0.4
g. P(n2)0.1371Pn1 Pn20.2286 0.0914
14-19and 14-20.
These results indicate that when only one loader is employed, the average truck must wait 3 r hour 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 the systemis 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
⎛ ⎝⎜
⎞ ⎠⎟
⎛ ⎝⎜
⎞ ⎠⎟
. P
n n n
0
0
1 2
1
1 3 4
1 2
3 4
2
! !
⎛ ⎝⎜
⎞ ⎠⎟ ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎛ ⎝⎜
⎞ ⎠⎟
∑
2 4( )(( )43hour
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 a single 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. Wq0.0643 for 2 channels; Wq0.0079 for 3 channels;
Wq0.0015 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.
14-24. This is an M/M/1 system with 10 per hour and 15 per hour.
a. Wq0.1333 hours
b. Lq1.333
c. W0.2 hours
d. L2
e. P00.333
14-25. This is an M/M/2 system with 10 per hour and 15 per hour.
a. Wq0.0083 hours
b. Lq0.083
c. W0.075
d. L0.75
e. P00.5
14-26. a. (8 hours per day)10(8)80 customers per day
b. Total time spent waitingWq(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 waiting0.0083(80)
0.664 hours. Wq0 123
3 0 041 .
. Lq0 873
3 4 0 123
. .
W0 873 3 0 291 .
. L
3 4 3 4
1 8 3 0 4545 3 4 0 873
2
2
( )
( )!( ) ( . ) .
⎛ ⎝⎜
⎞ ⎠⎟
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 (Wq)
3
–4hour 0.075 hour Utilization rate () 0.75 0.375 Probability system 0.25 0.625
empty (P0)
Probability of more K
than Ktrucks 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
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$16.602($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 M1:
But
Thus
This is the same formula. 14-29.
1/day
N5, n1
0.36
a. Number in queue
54.480.52 unit
b. Number in the system
LLq(1P0)0.52(10.36)
0.520.641.16 in system
c. Number running ok NL5 1.16
3.84
d. Average time in queue
days
e. Average wait in system
0.8171
1.817 days
14-30.
4/hour, N5, n1
60(0.1765)3120(0.1765)4120(0.1765)5
P0 1 2 1 5 0 1765 20 0 1765
( . ) ( . )
60
85 0 706. /hour, 0 1765. W Wq 1
W L
N L q
q
( )
.
( . )( . ) .
0 52
5 1 16 0 1667 0 817 Lq 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
( ) ( )( )
⎛ ⎝⎜
⎞ ⎠⎟ P0 1
1
1
1
⎛ ⎝⎜
⎞ ⎠⎟ L P
⎛
⎝⎜ ⎞ ⎠⎟ ⎡
⎣ ⎢ ⎢ ⎢ ⎢ ⎢
⎤
⎦ ⎥ ⎥ ⎥ ⎥ ⎥
1 2 0
2
( ) (
)2 P0
0.34
a. Average number waiting Lq
b. Average number broken L
Lq(1P0)
0.576(10.34)1.24
c. P00.34, as seen above.
d. Average time in queue
hour
e. Average time in system
0.2170.25
0.467 hour
14-31. a. Entering: 84/minute, 30/minute, 2.8
Exiting: 48/minute, 30/minute, 1.6
The manager desires that Wq0.1 minute 6 seconds and that
Lq8 customers in queue.
Entering:
If M3, Lq12.27 and Wq0.14 minute (too high) If M4, Lq1.00 and Wq0.01 minute (this is okay) If M5, Lq0.24 and Wq0.003 minute (this is also
okay)
So the manager must open M4 or more entrances.
Exiting:
If M2, Lq2.8, Wq0.06 minute (this is okay) If M3, Lq0.31, Wq0.006 minute (also okay)
So the manager must open M2 or more exits. Since there are
only 6 turnstiles, 4 must be used as entrances and 2 as exits. b. The students should recognize and questionall the limiting queuing assumptions that have been applied in solving the case. For example, it may be reasonable to assume that arrivals at the entranceturnstiles 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 are notformed at turnstiles and the problem
be-comes a series of single channel queues.
W Wq 1μ = 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 .
. ( . ) . .
⎛ ⎝⎜
⎞ ⎠⎟
N P
(1 0)
1
1 0 88 0 62 0 30 0 12 0 02 1 2 944
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. Pn3= (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 TOI
NTERNETH
OMEWORKP
ROBLEMS 14-34. 12/hour; 4/hour/barber; M4 channelsa. P00.0377 3.8% (from formula)
b. L⯝4.528
c. W⯝0.377 hour 22.6 minutes
d. Wq0.127 hour 7.6 minutes
e. Lq1.5282 (from formula)
f. 0.75 75%
g. with m5 barbers drops to 60%
14-35. a. 9 A.M.–3 P.M.; 6 patients/hour; 5 patients/ hour/doctor
Want Wqto be 5 minutes 0.0833 hour. Wq0.0833 implies
that
or Lq0.0833or Lq0.50
Thus m3 channels or doctors are needed (with m2, Lq
0.6748; with m3, Lq0.0904).
b. 3 P.M.–8 P.M.; 4 patients/hour; 5 patients/hour/doctor
Wq0.0833 hour implies that or Lq0.0833or
Lq0.03333. This means m2 doctors.
c. 8 P.M.–midnight; 12 patients/hour; 5 patients/
hour/doctor
Want Wq0.0833 hour or or Lq0.0833or Lq
1.00. m4 doctors are needed.
14-36. a. 1 per minute and 2 per minute
b. M/M/1
c. 1 w 0.5
d. P01 1 w 0.5. The cashier is idle 50% of the time.
e. Lq0.5
Lq
0 0833. 12
5 2 4.
Lq
0 0833. 4
5 0 80. WqLq
0 0833. 6
5 1 20.
f. Wq0.5 minute
g. W1 minute
14-37. a. 3 per minute and 4 per minute
b. M/M/1
c. 3/40.75
d. P013/40.25. The cashier is idle 25% of the time.
e. Lq2.25
f. Wq0.75 minute
g. W1 minute
h. P(n1) 0.188
P(n2) 0.141 P(n3) 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. Lq0.1227
b. Wq0.0409 minute
c. W0.2909 minutes
d. P(n1)0.3409, P(n2)0.1278, P(n3)0.0479.
S
OLUTION TON
EWE
NGLANDF
OUNDRYC
ASE1. 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 (M2). 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
( ) ⎛ (
⎝⎜ ⎞ ⎠⎟ ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎛ ⎝⎜
⎞ ⎠⎟
55 5 2 7
0 18 )
( )
.
⎡
⎣
⎢ ⎤
⎦ ⎥ P
n M n
n M
0
0 1
1
1 1
! !
⎛
⎝⎜ ⎞ ⎠⎟ ⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
⎛ ⎝⎜
⎞ ⎠⎟
∑
M M
M M
W
M M P M
( / )
( 1)!( )
1
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 an increasein 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 a decreasein 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\v60 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 TOW
INTERP
ARKH
OTELC
ASE1. 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. Ws
1 W
1 7 3
1 4 W
1 1
6 4 1 2
average time in the system gives W7.6 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
NTERNETC
ASES
TUDYPantry 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 be slowerthan
