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14 H
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Waiting Line and Queuing Theory Models
TEACHING SUGGESTIONS
ALTERNATIVE EXAMPLES
Teaching Suggestion 14.1: Topic of Queuing.
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 believed 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 assumed that arrivals are Poisson and answer times are exponentially distributed.
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 airports 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 Organizational Perspective.
Students should realize that different organizations place different values on customer waiting time. Ask students to consider different scenarios, from a drive-through restaurant to a doctor’s office to a registration line in their college or motor vehicle office. It becomes clear that organizations place different values on their customers’ time (with most colleges and DMVs unfortunately placing minimal cost on waiting time). Teaching Suggestion 14.3: Use of Poisson and Exponential Probability Probability Distributions to Describe Arrival and Service Rates.
These two distributions are very common in basic models, but students 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.
Alternative Example 14.1:
a. Find the probabil probability ity that that the the employee employee is is idle. idle. b. Find the the proportion proportion of the time that that the the employee employee is busy. c. Find the the average average number number of people people receivi receiving ng and waitin waiting g to receive information. d. Find the the average average number number of people waiting waiting in line line to get information. e. Find the average average time time a person seeking seeking informati information on spends at the desk. f. Find the the expected expected time time a person person spends spends just just waiting waiting in in line to have a question answered. ANSWER: 20/hour a.
b.
Teaching Suggestion 14.4: Balking 14.4: Balking and Reneging Assumptions. Assumptions.
c.
L
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.
d.
Lq
Teaching Suggestion 14.5: Use of Queuing Software.
e.
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.
f.
Teaching Suggestion 14.6: Importance 14.6: Importance of Lq and Wq in Economic Analysis. Analysis.
Although many parameters are computed for a queuing study, the two most important ones are Lq and Wq when 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 and that the return time also needs to be added.
30/hour
20 P0 1 1 .33 33% 30
0.66
20 2 people 30 20
(20)2 1.33 people ( ) 30( 30 20) 1 1 0.1 hour W 30 20 W q
2
l ( l)
20 0.0667 hours 30( 30 20)
In Alternative Example 14.1, the information desk employee earns $5/hour. The cost of waiting time, in terms of customer unhappiness with the mall, is $12/hour of time spent waiting in line. Find the total expected costs over an 8hour day. Alternative Example 14.2:
a. The average average person waits 0.0667 hour and and there there are 160 160 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 salary cost cost is is $40/ $40/day day.. c. Tota Totall cos costs ts are are $128 $128 $40 $168/day. 217
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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 believed 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 assumed that arrivals are Poisson and answer times are exponentially distributed. Alternative Example 14.3:
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: (servers). a. P
L
0
1 ⎛ 20 ⎞ 0 ! ⎜⎝ 30 ⎟ ⎠
1
1 ⎛ 20 ⎞ 1 ! ⎜⎝ 30 ⎟ ⎠
2
1 ⎛ 20 ⎞ 2 ! ⎜⎝ 30 ⎟ ⎠
The seven operating characteristics are:
4. Average time in the queue (W q)
3. Average number in the queue (Lq) 5. Utilization factor ( ) 6. Percent idle time (Po) 7. Probability there are more than K customers in the system
⎛ 1 ⎞ 20 ⎜ ⎟ (1)[( 2)( 30) 20]2 ⎝ 2 ⎠ 30 (20)(3 0)(2 0/ 30)2
(8 00 / 3) ⎛ 1 ⎞ L
2
1
8
9
3
peop ple ⎜⎝ 2 ⎟ ⎠ 3 12 12 12 4
1, 600
3/ 4 20
3 hr. 80
0.0375
2
4/minute
9 2( 4)( 4 3)
1.125 people in queue on average
2( )
3 2( 4)( 4 3)
0.375 minutes in the queue waiting
L Lq
14-3.
1 1 50% 2 1 2 1 3 3
2 ( )
1.125
3 4
1.87 people in the system
W W q
7. Average service rate exceeds average arrival rate.
2. Average time spent in the system (W )
ANSWER: 3/minute
4. Arrivals are Poisson. 6. Service times are negative exponential.
⎡ 2(30) ⎤ ⎢ 2(30) 20 ⎥ ⎣ ⎦
Three students arrive per minute at a coffee machine that dispenses exactly 4 cups/minute at a constant rate. Describe the operating system parameters.
3. Arrivals are independent.
1. Average number of customers in the system L ()
Alternative Example 14.4:
W q
2. There is no balking or reneging.
2 1 ⎛ 4 ⎞ ⎡ 60 ⎤ 1 ⎜ ⎟ ⎢ ⎥ 3 2 ⎝ 9 ⎠ ⎣ (60 20) ⎦
W
The seven underlying assumptions are: 1. Arrivals are FIFO.
30/hour, M 2 open channels
1
Lq
14-2.
5. Service times are independent.
c.
14-1. The waiting line problem concerns the question of finding the ideal level of service that an organization should provide. The three components of a queuing system are arrivals, waiting line, and service facility.
1
0
b.
20/hour,
SOLUTIONS TO DISCUSSION QUESTIONS AND PROBLEMS
1
.375
1 4
0.625 minutes 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 examples are (1) hospital emergency rooms, (2) an elevator, (3) an airplane trip, (4) a small store where the shopkeeper serves whoever 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 airport 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 limited number of patients, and (5) a hospital ward with only 20 patients who need care. 14-7. a. Barbershop: usually a single-channel, multipleservice 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
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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
AND
219
Q UEUING THEORY MO D E L S
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 example 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 computer 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.
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 single-channel, singleserver 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
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 weekend, 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 arrive in batches 5 to 20 minutes before a show.
14-10. NUMBER OF CHECKOUT CLERKS 1
2
Number of customers Average waiting time
300 –16 hour
–1 hour 10
–1 hour 15
–1 hour 20
per customer Total customer waiting time Cost per waiting hour Total waiting costs Checkout clerk hourly salary Total pay of clerks for 8-hour shift Total expected cost
(10 minutes) 50 hours $10 $500 $8 $64
(6 minutes) 30 hours $10 $300 $8 $128
(4 minutes) 20 hours $10 $200 $8 $192
(3 minutes) 15 hours $10 $150 $8 $256
$564
$428
$392
$406
Optimal number of checkout clerks on duty
300
3
300
a
3
4
300
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a. The utilization rate, , is given by
d. The utilization rate, , is given by
=
3 8
0.375
0 1
b. The average down time, W, is the time the machine waits to be serviced plus the time taken to perform the service.
1
14-13.
1
W
d. Probability that more than one machine is in the system k 1
⎛ ⎞ ⎝ ⎟ ⎠
Probability that more than two machines are in the system: 3
27 0.053 512
2102 44, 100 280( 280 210) 280(70)
210 0.75 280
1
1 1 280 210 70
0.0143 hour in the line 0.857 minute 51.4 seconds
W q
9 0.141 64
⎛ 3 ⎞ Pn 2 ⎜ ⎟ ⎝ 8 ⎠
d. The average time spent by a patron waiting to get a ticket, W q, is given by
Pn k ⎜
2
( )
0.225 machine waiting
2
c. The average time a customer spends in the ticketdispensing system, W, is given by
32 8( 8 3)
⎛ 3 ⎞ Pn 1 ⎜ ⎟ ⎝ 8 ⎠
0. 8333 0. 1667
b. The average fraction of time the cashier is busy, , is given by
2 ( )
1
44, 100 2.25 patrons inline 19, 600
8 3
c. The number of machines waiting to be served, Lq, is, on average,
1
210 patrons/hour, 280 patrons/hour.
Lq
0.2 day, or 1.6 hours
a. The average number of patrons waiting in line, Lq, is given by
Lq
10 0.8333 12
e. The probability that no cars are in the system, P0, is given by:
W
( )
210 210 280( 280 210) 280(70)
210 0.011 hour 0. 64 minute 19, 600
38.6 seconds
e. The probability that there are more than two people in the system, Pn2, is given by k 1
4
⎛ 3 ⎞ ⎝ 8 ⎟ ⎠
Pn 3 ⎜
⎛ ⎞ ⎟
81 0.020 4, 096
Pn k ⎜
243 0.007 32, 768
Pn 2 ⎜
3
5
⎛ 3 ⎞ Pn 4 ⎜ ⎟ ⎝ 8 ⎠
14-12.
a.
10 cars/hour, 12 cars/hour.
The average number of cars in line, Lq, is given by
102 102 q ( ) 12(12 10) (12 )(2)
2
( )
10 10 12(12 10) (12)(2)
0.4167 hours
4
⎛ 210 ⎞ ⎝ 280 ⎟ ⎠
Pn 3 ⎜
1
0.316
The probability that there are more than four people in the system, Pn 4, is given by
1 12 10
⎛ 210 ⎞ ⎝ 280 ⎟ ⎠
Pn 4 ⎜ 14-14.
a.
0.237
4 students/minute, 60 q w 5 students/minute
The probability of more than two students in the system,
Pn 2, is given by k 1
c. The average time a car spends in the service system, W, is given by
W
0.422
5
b. The average time a car waits before it is washed, W q, is given by
W q
The probability that there are more than three people in the system, Pn 3, is given by
4.167 cars
⎛ 210 ⎞ ⎝ 280 ⎟ ⎠
1 0.5 hour 2
⎛ ⎞ ⎟
Pn k ⎜
3
⎛ 4 ⎞ ⎝ 5 ⎟ ⎠
Pn 2 ⎜
0.512
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The probability of more than three students in the system, Pn3, is given by
Q UEUING THEORY MO D E L S
AND
f (part c). The average waiting time, W q, for the twochannel system is given by
4
⎛ 4 ⎞ ⎝ 5 ⎟ ⎠
Pn 3 ⎜
b.
m
⎛ ⎞ ⎝ ⎟ ⎠
⎜
W
0.328
The probability that the system is empty, P0, is given by
where
5
⎛ 4 ⎞ ⎝ 5 ⎟ ⎠
Pn 4 ⎜
1
Wq W
0.410
The probability of more than four students in the system, Pn4, is given by
(m 1)!( m )
Then
4 P0 1 1 1 0.8 0. 2 5 ( )
2 ( )
W q
4 0.8 minute 5(5 4)
f.
42 3.2 students 5(5 4)
4 4 students 5 4
(2 1)[2( 5) 4]2
Lq L
where m
⎛ ⎞ ⎝ ⎟ ⎠
⎜
L
60 5 students/minute 12
(m 1)!( m )
P0
2
⎛ 4 ⎞ ⎝ 5 ⎟ ⎠
4( 5) ⎜
f (part b). The probability that the two-channel system is empty, P0, is given by
Lq
1
⎡ n m 1 1 ⎛ ⎞ ⎢ ∑ ⎠ ⎢⎣ n 0 n ! ⎜⎝ ⎟
n
⎤ 1 ⎛ ⎞ m ⎥ ⎜ ⎟ ⎥⎦ m ! ⎝ ⎠ m
0
1 ⎛ 4 ⎞ 0 ! ⎜⎝ 5 ⎟ ⎠
1
1 ⎛ 4 ⎞ 1 ⎜⎝ 5 ⎟ ⎠
(2 1)![2( 5) 4]2
m
1
4( 5)(0 .64) 1(10 4)
⎛ ⎞ ⎝ ⎟ ⎠
⎜
L
4 1 ⎛ 4 ⎞ 2( 5) 5 2 ⎜⎝ 5 ⎟ ⎠ 10 4
(m 1)!( m )
14-15.
4 1 ⎛ 16 ⎞ ⎛ 10 ⎞ 1 ⎜ ⎟ ⎜ ⎟ 5 2 ⎝ 25 ⎠ ⎝ 6 ⎠ 1
4 160 5 300
or
1 1 0.429 1 0.8 0. 53 2.33
Thus the probability of an empty system when using the second channel is 0.429.
5.492 1(36 )
m
2
1 ⎛ 4 ⎞ 2(5) 1(2 ) ⎜⎝ 5 ⎟ ⎠ 2(5) 4
1
(0.429)
0.15 student
2
1
2
(0.429)
f (part e). The average number of students in the twochannel system, L, is given by
1
P0
2
Then
m2
P0
(0.429)
f (part d). The average number of students in the queue for the two-channel system, Lq, is given by
Adding a second channel, we have 4 students/minute
5(0 .64) (0.429) 1(10 4)2 1.373 0. 038 minute 2. 3 sec onds 1(36 )
e. The average number of students in the system, L, is given as
L
1
2
d. The expected number of students in the queue, Lq, is given by
Lq
P0
5⎜
The average waiting time, W q, is given by
W q
2
⎛ 4 ⎞ ⎝ 5 ⎟ ⎠
c.
221
Lq
2
0.153
P0
4 0 .95 student 5
30 trucks/hour, 35 trucks/hour.
a. The average number of trucks in the system, L, is given by
L
30 30 6trucks in the system 35 30 5
b. The average time spent by a truck in the system, W, is given by
W
1
1 35 30
1 0.2 hour 12 minutes 5
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To determine total cost using the second clerk (a second channel):
The utilization rate for the bin area, , is given by
P0
30 6 0.857 35 7
k 1
⎛ ⎞ Pn k ⎜ ⎟ ⎝ ⎠ 4 ⎛ 30 ⎞ Pn 3 ⎜ ⎟ 0.540 ⎝ 35 ⎠
Thus the probability that there are more than three trucks in the system is 0.540.
hours ⎛
⎜⎝ 30
hour
hours ⎞ ⎛ dollars ⎞ 0. 2 18 ⎟ ⎜ ⎠ ⎝ truck ⎟ ⎠ ⎜⎝ hour ⎟ ⎠
2
weeks ⎛ days ⎞ ⎛ year
1, 728 ⎜⎝ 7 week ⎟ ⎠ ⎜⎝
day
⎟ ⎠
2
4 1 ⎛ 4 ⎞ 2(15) 1 ⎜ ⎟ 5 2 ⎝ 5 ⎠ 30 12 1
m
W q
16 minutes
⎛ 12 ⎞ ⎝ 15 ⎟ ⎠
W q
(2 1)[2( 15) 12]2
(0.429)
15(0 .64)
(0.429) 1( 30 12)2 4.12 0.0127 hour 0. 763 seconds 1(324)
Cost with two clerks: service cost waiting cost C /hour t
b.
The average number of callers waiting to place an order, Lq, is given by
P0
2
( )
12 12 12 0.267 15(15 12) 15(3) 45
(m 1)!( m )2
15 ⎜
⎛ ⎞ ⎝ ⎟ ⎠
Then
a. The average time the catalog customer must wait, W q, is given by
2(15) 1 ⎛ 12 ⎞ ⎜ ⎟ 1(2 ) ⎝ 15 ⎠ 2(15) 12
⎜
$24,192
12 calls/hour, 60 r 15 calls/hour.
Lq
2
1 1 0. 8 0. 53 1 0.429 2.33
dollars ⎞
Enlarging the bin will cut waiting costs by 50% next year, resulting in a savings of $12,096. Since the cost of enlarging the bin is only $9,000, the cooperative should proceed to enlarge the bin. The net savings is $3,096 ($12,096 $9,000).
W q
1 ⎛ 12 ⎞ ⎠ 1 ⎜⎝ 15 ⎟ 1
P0
f. Enlarging the bin will cut waiting costs by 50% next year. First, we must compute annual waiting costs:
14-16.
1 ⎛ 12 ⎞ ⎠ 0 ! ⎜⎝ 15 ⎟
1
or
$1,728/day or $12,096 per week.
annual waiting cost
1 0
4 1 ⎛ 16 ⎞ ⎛ 30 ⎞ 5 2 ⎜⎝ 25 ⎟ ⎠ ⎜⎝ 18 ⎟ ⎠ 1 4 480 1 5 900
trucks ⎞ ⎛
16(30)(0.2)(18)
1
e. Unloading cost: day
d. The probability that there are more than three trucks in the system, Pn 3, is given by
C M 16
1
⎡ n m 1 1 ⎛ ⎞ n ⎤ 1 ⎛ ⎞ m m ⎢ ∑ ⎥ ⎟ m ⎠ ⎥⎦ m ! ⎜⎝ ⎠ ⎢⎣ n 0 n ! ⎜⎝ ⎟
20 12
2
( )
12 2 144 144 3.2 customers 15(15 12) 15(3) 45
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: service cost waiting cost C /hour t
⎛ hours ⎞ ⎛ dollars ⎞ 0. 267 50 10 12 hour ⎜⎝ call ⎟ ⎠ ⎜⎝ hour ⎟ ⎠ calls
⎛ hours ⎞ ⎛ dollars ⎞ 0. 0127 50 ⎜ hour ⎝ call ⎟ ⎠ ⎜⎝ hour ⎟ ⎠ calls
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! This is an M/M/1 system with 24 per hour and 30 per hour. 14-17.
a.
W 0.167 hours
b. L 4 c.
W q 0.133
d. Lq 3.2 e.
P0 0.2
10 12(0.267)(50) 10 160.2
f.
0.8
$170.20/hour
g.
P(n 2) Pn 1 Pn 2 0.640 0.512 0.128
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2
This is an M/M/2 system with 24 per hour and 30 per hour. Using QM for Windows we get the following: 14-18.
a.
⎛ 3 ⎞ ⎝ 4 ⎟ ⎠
3( 4) ⎜
3 (0.4545) 0.873 4 (1)!(8 3)2 0.873 W 0.291 hour 3 3 Lq 0.873 0.123 4 0.123 W q 0.041 hour 3 L
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 Pn 1 Pn 2 0.2286 0.0914
14-19 and 14-20. NUMBER OF FRUIT LOADERS 1
Truck arrival rate ( ) Loading rate ( ) Average number in system ( L) Average time in system ( W ) Average number in queue ( Lq) Average time in queue ( W q) Utilization rate ( ) Probability system empty ( P 0) Probability of more K than K trucks in 0 system 1 2 3
2
3/hour 4/hour 3 trucks 1 hour 2.25 trucks –34 hour 0.75 0.25
3/hour 8/hour 0.6 truck 0.2 hour 0.225 truck 0.075 hour 0.375 0.625
0.75 0.56 0.42 0.32
0.375 0.141 0.053 0.020
These results indicate that when only one loader is employed, the average truck must wait 3r 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 decline in the queue when a second loader is employed. Referring to the data in Problems 14-19 and 14-20, we note that the average number of trucks in the system is 3 when there is only one loader and 0.6 when two loaders are employed. 14-21.
NUMBER OF LOADERS 1
Truck driver idle time costs ( average number trucks hourly rate) (3)($10) Loading costs Total expected cost per hour
2
223
Q UEUING THEORY MO D E L S
By looking back to Problems 14-19 and 14-20, we see that although length of the queue and average time in the queue are lowest 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 platform. Hence we would probably recommend not building a second gate. The queuing systems in this problem are the M/M/2, M/M/3, and the M/M/4 systems. 14-23.
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 channels, and 0.45 hours with 4 channels. c. The total daily waiting time cost is given in the table below: Service Service # cost per cost per Channels hour day
Total Total waiting time waiting Total cost Wq(10hr.) cost
2 3 4
19.29 2.37 0.45
$20 $30 $40
$200 $300 $400
$1929 237 45
$2129 $537 $445
The minimum daily cost is $445 with 4 channels. This is an M/M/1 system with 10 per hour and 15 per hour. 14-24.
a.
W q 0.1333 hours
b. Lq 1.333 c.
W 0.2 hours
d. L 2 e.
P0 0.333
This is an M/M/2 system with 10 per hour and 15 per hour. 14-25.
$30 36 $36
$6 (0.6)($10) 12 (2)($6) $18
a.
W q 0.0083 hours
b. Lq 0.083 The firm will save $18/hour by adding the second loader.
c.
W 0.075
d. L 0.75 14-22.
P0
1
⎡ 1 1 ⎛ 3 ⎞ n ⎤ 1 ⎛ 3 ⎞ 2 2( 4) ⎢ ∑ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢⎣ n 0 n ! ⎝ 4 ⎠ ⎥⎦ 2 ! ⎝ 4 ⎠ 2(4) 3
1 2
3 1 ⎛ 3 ⎞ 1 ⎜ ⎟ 4 2 ⎝ 4 ⎠
⎛ 8 ⎞ ⎜ 8 3 ⎟
0.454
e. 14-26.
P0 0.5
a.
(8 hours per day) 10(8) 80 customers per day
b. Total time spent waiting Wq(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.
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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.
1 1 1 0.88 0.62 0 .30 0. 12 0. 02 2 .944
0.34
a. Average number in line 0.666
a. Average number waiting Lq
b. Average number in system 1.333
c. Average wait in line 0.1666 minute 10 seconds 14-28.
But
Thus
L
1
1
1
(1 P0 )
5 ⎜
⎡ ⎛ ⎞ ⎤ ⎢ ⎜ ⎟ ⎥ ⎝ ⎠ ⎥ 2 L ⎢⎢ P P 0 2 ⎥ 2 0 ( ) ⎢ 1( ) ⎥ ⎢⎣ ⎥⎦ ⎛ ⎞ 1 ⎜ ⎟ ⎝ ⎠
⎛ 4.706 ⎞ (0.66) 5 4. 4 0. 6 ⎝ 0.706 ⎟ ⎠
For M 1:
P0
N
b. Average number broken L Lq
c.
N L) 0.576 (5 1.24)(0 . 706) 0.576 = = 0.217 hour 2.65
2
2
( ) ( )() ( )()
e. Average time in system W
This is the same formula. 14-29.
0.1667 / day 61 / day
14-31.
1/day
1
1 5( 61 ) 20( 61 )2 60( 61 )3 120( 61 ) 4 120( 61 )5 0.36
(1 P0 ) 5 7(1 0. 36)
5 4.48 0.52 unit
L Lq (1 P0) 0.52 (1 0.36)
0.52 0.64 1.16 in system
c. Number running ok N L 5 1.16
Lq ( N L)
3.84
0.52 0.817 days (5 1.16)(0 . 1667)
e. Average wait in system W W q
14-30.
60 0.706/hour , 85 4/hour,
P0
1
0.817 1
1.817 days
0.1765
N 5, n 1
1 1 5(0.1765) 20( 0. 1765)2
0.217 0.25
0.467 hour
1 μ
a. Entering: 84/minute, 30/minute, 2.8
The manager desires that W q Lq 8 customers in queue.
0.1 minute
6 seconds and that
Entering: If M 4, Lq 1.00 and W q 0.01 minute (this is okay) If M 5, Lq 0.24 and W q 0.003 minute (this is also okay) Exiting: If M 2, Lq 2.8, W q 0.06 minute (this is okay) If M 3, Lq 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.
d. Average time in queue
W q
So the manager must open M 4 or more entrances.
b. Number in the system
W q
If M 3, Lq 12.27 and W q 0.14 minute (too high)
a. Number in queue
Lq N
Exiting: 48/minute, 30/minute, 1.6
N 5, n 1
P0
Lq
d. Average time in queue W q
2
0.576 (1 0.34) 1.24
P0 0.34, as seen above.
2 ⎛ ⎞ ⎜ ⎟ ( )( ) ⎝ ⎠ ( )
(1 P0)
60(0.1765)3 120(0.1765)4 120(0.1765)5
b. The students should recognize andquestion all the limiting queuing assumptions that have been applied in solving the case. For example, it may be reasonable to assume that arrivals at the entrance turnstiles are independent and Poisson. But are exiting passengers 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 passenger 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 not formed at turnstiles and the problem becomes a series of single channel queues.
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This is an M/M/1 system with = 12 per hour and = 15 per hour.
14-32.
a.
Wq = 0.27 hours
b.
Lq = 3.2
c.
L=4
d. W = 0.33 hours Pn3 = (12/15)3+1 = 0.4096
e.
This is an M/M/2 system with = 12 per hour and = 15 per hour.
14-33.
a.
Wq = 0.013 hours
b.
Lq = 0.152
c.
L = 0.952
14-35. a. 9 A.M.–3 P.M.; hour/doctor
W q
Lq
5
6 patients/hour;
5 patients/
6 1.20 5
minutes 0.0833 hour. W q
0.0833 implies
0.0833 or Lq 0.0833 or Lq 0.50
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
b.
W q 0.0409 minute
c.
W 0.2909 minutes
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 department 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
2, Lq
W
P0
Lq
0.0833 or Lq 0.0833 or
c. 8 P.M.–midnight; hour/doctor
12 patients/hour;
5 patients/
W Lq
0.0833 or Lq 0.0833 or Lq
1.00. m 4 doctors are needed. 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
1
1
⎡ M 1 1 ⎛ ⎞ n ⎤ 1 ⎛ ⎞ M M ⎢∑ ⎥ ⎠ ⎥⎦ M ! ⎜⎝ ⎟ ⎠ M ⎢⎣ n 0 n ! ⎜⎝ ⎟
1
⎡1 7 ⎞ ⎢ (1) 1 ⎛ 1 ⎜⎝ 5 ⎟ ⎠ ⎢⎣ 1
1
⎤ 1 ⎛ 7 ⎞ 2 ⎥ ⎜ ⎟ ⎥⎦ 2 ⎝ 5 ⎠
⎡ 2(5) ⎤ ⎢ 5(2) 7 ⎥ ⎣ ⎦
0.18
Therefore,
12 2.4 5
a.
P0
In this case
P0
Lq 0.03333. This means m 2 doctors.
Want W q 0.0833 hour or
( M 1)!( M )
2
0.0833 hour implies that
( / )
where
4 0.80 5
14-36.
d.
M
b. 3 P.M.–8 P.M.; 4 patients/hour; 5 patients/hour/doctor
3/4 0.75
SOLUTION TO NEW ENGLAND FOUNDRY CASE
Thus m 3 channels or doctors are needed (with m 0.6748; with m 3, Lq 0.0904).
W q
c.
d. P(n 1) 0.3409, P(n 2) 0.1278, P(n 3) 0.0479.
Want W q to be that
3 per minute and 4 per minute
M/M/1
a. Lq 0.1227
a.
b.
This is an M/M/2 system with 3 per minute and 4 per minute. Solving with QM for Windows we obtain the following:
P0 0.0377 3.8% (from formula) L 4.528 W 0.377 hour 22.6 minutes W q 0.127 hour 7.6 minutes Lq 1.5282 (from formula) 0.75 75% with m 5 barbers drops to 60%
W 1 minute
14-38.
12/hour; 4/hour/barber; M 4 channels
g.
P(n 3) 0.106
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS a. b. c. d. e. f. g.
W q 0.5 minute
P(n 2) 0.141
d. W = 0.079 hours
14-34.
f. 14-37.
225
Q UEUING THEORY MO D E L S
AND
5(7 / 5)2 1(10 7)2
(0.18) 1 / 5
0.396 hour, or 23 minutes and 45 seconds
Added to the travel times involved (6 minutes total for maintenance 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 singlechannel, single-line operations. Bob will serve the maintenance
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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
W
1
1 64
1 hour, or 30 minutes 2
The reduced travel time is equal to 2 minutes, making the total trip time equal to 32 minutes. This is an increase in time of 2 minutes and 15 seconds for the maintenance personnel. Pete can now service 7 people per hour ( 7). Three people arrive from the molding department every hour ( 3). The time in Pete’s department is
W
1 7 3
1 hour, or 15 minutes 4
The travel time is equal to 2 minutes, making the total trip time equal to 17 minutes. This is a decrease in time of 8 minutes and 45 seconds per trip for the molding personnel. 2. To evaluate systemwide savings, the times must be monetized. For the maintenance personnel who are paid $9.50 per hour, the 2Z\v minutes 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 factors should be considered before a final decision is made. For example, 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 noneconomic, 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.
SOLUTION TO WINTER PARK HOTEL CASE 1. Which of the two plans appears to be better? The current system 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 waiting and checking in is given by
W s
1
1 0.5 hour, or 30 minutes 20 18
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 average of 2 minutes ( 30 per hour) their average time in the system 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
average time in the system gives W 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
INTERNET CASE STUDY 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 service times, even though a UPC reader is going to be installed. This means that our waiting times are an upper bound for the new, better system (the M/M/s model). 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 customers per hour. Notice that this is not the same as taking 20 percent of the rate of 30 and 80 percent 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 minimum 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 minutes. Trying one more server leads to a waiting time of .64 minutes. Now we separate the express and regular. Assume that all express customers go into the express (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 between 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 slower than the regular lanes!
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