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172 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Supplement

C

Part 2 Designing and Managing Processes WAITING LINE MODELS

PROBLEMS 1. Paula Caplin. Little’s Law.

a. = 120 jobs/day W = 4 days Current work-in-process = L = W = (120 jobs/day)(4 days) = 480 jobs.

b. L must be reduced to 240 jobs. Therefore, either the average number of repairs, λ, or the

time in the system, W, must be cut in half (or some combination). Paula has little or no control over the number of repairs, but has several options for reducing the time in the system. First, she can identify the bottleneck in the total repair process and apply the theory of constraints to utilize the bottleneck to its maximum performance. Second, she can do a process analysis and improve the work methods at the bottleneck as well as all other processes feeding the bottleneck to improve overall throughput in the repair process. Finally, if all else fails, she can add capacity until the goal has been met.

2. Banco Mexicali. Little’s Law.

= 20 customers/hour L = 4 customers L = W , or W = L/ W = (4 customers)/(20 customers/hour) = 0.20 hour, or 12 minutes.

3. Hasty Burgers. Single-server model, 20

a. Find resulting in L = 4.

20420

4 80 204 100

25

L

The required service rate is 25 customers per hour. b. Find the probability that more than four customers are in the system. This is one minus the probability of four or fewer customers in the system.

First we calculate average utilization of the drive-in window.

Waiting Lines SUPPLEMENT C 173

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

20 0.825

The probability that more than four customers are in line and being served is: 0 1 2 3 41P P P P P P

where

0 1 2

3 4

0 1 2 3 4

1

1 1 1 1

1 1

1 1

nnP

P

P

when 0.8

1 0.2 1 0.8 0.64 0.512 0.4096

0.3277

P

P

Consequently, there is about a 33 percent chance of more than four customers in the system.

c. Find the average time in line. 1

10.825 20

qW W

0.16qW hours or 9.6 minutes Ten minutes borders on being unbearable, particularly in the atmosphere of exhaust fumes. Keep in mind that this is an average, and some people must wait longer.

4. Precision Machine Shop. Single-server model. With the junior attendant, the average number of idle machinists, L

8 410 8

L

Average hourly idle machinist cost = $20(L) = $20(4) = $80 With the senior attendant, average number of idle machinists, L

8 116 8

L

Average hourly cost of idle machinists drops to $20(L) = $20(l) = $20 Adding the attendant pay gives a total cost of $85 per hour ($80 + $5) for the junior attendant and

$32 per hour ($20 + $12) for the senior attendant. The best choice is the senior attendant.

174 PART 2 Managing Processes


5. KRAN radio. Single-server model.

60 min hr 6 calls hr10 min call

60 min hr 2.4 calls hr25 min call

2.4 0.46

A caller will not receive a busy signal when there are zero, one, or two callers in the system. Therefore, the probability of receiving a busy signal is one minus the probability of two or fewer callers in the system.

22

11

00

1

1 0.4 0.4 0.096

1 0.4 0.4 0.240

1 0.4 0.4 0.6000.936

nnP

P

P

P

1 0.936 0.064 Jake’s callers will get busy signals 6.4 percent of the time. 6. Local Bank

Service rate 60 3 min. per customer = 20 customers/hour. a. Average utilization, 50 3 20 0.8333 s .

b.

11

0

132

0

1

1! ! 1

50 20 50 20 1! 3! 1 0.8333

1 2.5 3.125 15.6220.0449

n ss

on

n

n

Pn s

n

b.

3

20.0449 50 20 0.8333

3.50633! 1 0.8333! 1

so

qP

Ls

c. 3.5063 0.070150

qq

LW

hours, or 4.2 minutes

d. 1 50 0.0701 1 20 6.005qL W W or 6 customers



7. Fantastic Styling Salon a. A common queue, two-server model.

8 , 5 , s = 2

8 0.8

2 2 5

Customer’s average waiting time in queue = q

qL

W

2! 1

so

q

PL

s

11

0

10 1 2

12

1

1! ! 1

1 0! 1! 2! 1

8 5 11 8 52 1 0.80

1 1.6 6.41 9 0.1111

n s

on

Pn s

2

20.1111 8 5 0.80 0.2275 2.844

0.082! 1 0.80qL

2.844 0.368qW hours or 21.6 minutes

b. Two separate, single-server models. The results for Jenny and Jill are identical.

0.5 8 4 , 5 , 4 0.805

Waiting in line is given by 1qW W

0.8 0.805 4qW

hours or 48 minutes

c. On average, a customer’s waiting will be more than twice that of the multiple server

design. The reason is that a server will serve any customer as they enter the system in the multiple-server system, thereby reducing the wait time on average.



8. Moore, Akin, and Payne (dental clinic). Multiple-server model.

3s , 5 , 2 , 5 0.8333

3 2

s

a. Probability of no patients, 0 P

1 11 2

0 50 0 6

10 1 2 3

56

1

5 2 5 21 1! ! 1 ! 3! 1

5 2 5 2 5 2 5 2 10! 1! 2! 3! 1

1 2.5 3.125 2.604 6

1 0.044946.625 15.625

n s n ss

n nP

n s n

b. The probability of 6 or more customers in the clinic is: 0 0.04494P (from part a), s = 3

for n s

0

152

1

252

2

!

0.04494 0.112351!

0.04494 0.140442!

n

nP Pn

P

P

for n s

0

352

3 0

452

4 1

552

5 2

!

0.04494 0.117033!3

0.04494 0.097533!3

0.04494 0.081273!3

n

n n sP Ps s

P

P

P

0 1 2 3 4 51

1 0.04494 0.11235 0.14044 0.11703 0.09753 0.08127 0.40644

P P P P P P

c. The average number of patients waiting in the lobby,

35 50 2 6

2 2 1566

0.04494 0.585163.5109

! 1 3! 1

s

q

PL

s

d. The average time spent in the clinic,

W = 1 1 3.5109 11.2022

5 2q

q

Lw

hours



9. Benny’s Arcade Because there are only six machines, we must use the finite source model.

a. To calculate the Jimmy’s utilization, we need to compute the probability that he will have no machines to repair.

1/50 0.02 machines per hour 1/15 0.0667 machines per hour

16 6! 0.026 ! 0.0667

n

on o

Pn

113.92 0.0718

Jimmy’s utilization 1 0.0718 0.9282 b. Average number of machines out of service.

0.06676 1 0.0718 6 3.095 2.905

0.02 L machines

c. Average time a machine is out of service. 2.905 6 2.905 0.02 46.93 W hours

10. Solomon, Smith and Samson

a. Single-server model, average utilization rate. 8 0.8

10

or 80% utilization

b. The probability of four or fewer documents in the system is 0.6723 as shown following. Therefore, the probability of more than four documents in the system is 1 – 0.6723 = 0.3277.

44

33

22

11

00

1

1 0.8 0.8 0.0819

1 0.8 0.8 0.1024

1 0.8 0.8 0.1280

1 0.8 0.8 0.1600

1 0.8 0.8 0.20000.6723

nnP

P

P

P

P

P

c. The average number of pages of documents waiting to be typed, 8 8 3.2 pages

10 10 8qL L



ADVANCED PROBLEMS

11. Quarry

a. Current System: Single-server model 9 hour ; 10 hour

Average waiting line in the system 1 1 110 9

W hour or 60 minutes

b. First Alternative: Improved single-server model 6 hour ; 15 hour

Average time in the system 1 1 0.111115 6 9

W hour or 6.67 minutes

c. Second Alternative: Two-server model

2

11

0

12

1

2

9 hour ; 10 hour; 22 9 2 10 9 20 0.45

1 1

! 1

1! ! 1

0.90 11 0.902! 1 0.45

1 0.90 0.73640.3793

0.3793 9 10 0.45

2! 1

q q

sq o

n ss

on

q

s

W W L

L P s

Pn s

L

2 0.22850.45

0.2285 9 1 100.1254 hours or 7.52 minutes

W

The second alternative results in an 87 percent reduction in waiting time relative to the current system, however, the first alternative dominates the second alternative in time and cost. Nonetheless, to make a final determination between the current system and alternative 1, the cost of the waiting time of the trucks must be considered.



12. Benton University, Finite Source Model a. To calculate the utilization in a finite source waiting line situation, we must first compute the probability that the maintenance person will have no customers.

1

00

11 2 3

4 5

!!

5! 0.4 5! 0.4 5! 0.415 1 ! 2.5 5 2 ! 2.5 5 3 ! 2.5

5! 0.4 5! 0.45 4 ! 2.5 5 5 ! 2.5

1 0.

nN

n

NPN n

1

0

8 0.512 0.246 0.079 0.0131 0.3774

2.65P

01 1 0.3774 0.6226 P b. Copy machines in repair system

01

2.55 1 0.37740.4

5 3.8911.109

L N P

c. Time spent in repair system

1

1.1095 1.109 0.4

0.712 days5.7 hours, assuming an 8-hour day

W L N L

13. Pinball Wizard. Multiple-server model. In this analysis we determine the expected total labor

and machine failure costs for the existing complement of three employees and then compare it to larger maintenance complements until costs begin to rise. Three maintenance people: s = 3, = 0. 333, = 0. 125, Average utilization

0.333 0.888

3 0.125

s



Probability of an empty system

1 130.333 0.3331 20.125 0.125

00 0

10 1 2 30.333 0.333 0.333 0.3330.125 0.125 0.125 0.125

1 1! ! 1 ! 3! 1 0.888

10! 1! 2! 3! 1 0.888

1

n s ns

n nP

n s n

1

0

2.664 3.548 3.151 8.929

1 0.02837.212 28.135

P

Average number of machines in waiting line

30.3330 0.1252 2

0.0283 0.888

! 1 3! 1 0.8880.4751 6.3100.0753

s

q

q

PL

s

L

The average time waiting in line 6.310 18.9490.333

qq

LW

hrs.

Average time in system 1

118.9490.125

26.949 hrs

qW W

Average number of machines in system 0.333 26.949

8.974 machines

L WL

The total expected hourly costs for the crew size of three employees is: Labor: 3 ($8 per hour) $ 24.00 Machine downtime: 8.974 ($10 per hour) 89.74 TOTAL $113.74

Four maintenance people: Average utilization

0.333 0.666

4 0.125

s




1 140.333 0.3331 30.125 0.125

00 0

10 1 2 3 4

1 1! ! 1 ! 4! 1 0.666

2.664 2.664 2.664 2.664 2.664 10! 1! 2! 3! 4! 1 0.666

1 2.664 3.54

n s ns

n n

Pn s n

1

0

8 3.151 2.099 2.994

10.0601

10.363 6.284P

Average number of machines in waiting line

40

2 20.0601 2.664 0.666

! 1 4! 1 0.6662.016 0.7532.677

s

q

q

PL

s

L


qq

LW

hrs.


12.2610.125

10.261

qW W


3.417

L WL

The total expected hourly costs for the crew size of four employees is: Labor: 4 ($8 per hour) $ 32.00 Machine downtime: 3.417 ($10 per hour) 34.17 TOTAL $ 66.17 Five maintenance people: Average utilization

0.333 0.5328

5 0.125

s




1 150.333 0.3331 40.125 0.125

00 0

10 1 2 3 4 5

1 1! ! 1 ! 5! 1 0.5328

2.664 2.664 2.664 2.664 2.664 2.664 10! 1! 2! 3! 4! 5! 0.4672

1 2

n s ns

n nP

n s n

1

0

.664 3.548 3.151 2.099 1.118 2.140

1 0.067312.462 2.393

P

A

verage number of machines in waiting line

40

2 20.0673 2.664 0.5328

! 1 5! 1 0.53284.811 0.184

26.193

s

q

q

PL

s

L


qq

LW


10.5530.125

8.553

qW W


2.848

L WL

The total expected hourly costs for the crew size of five employees is: Labor: 5 ($8 per hour) $ 40.00 Machine downtime: 2.848 ($10 per hour) 28.48 TOTAL $ 68.48

This total is higher than that for employing four maintenance people. Therefore, the manager of the Pinball Wizard should add only one more maintenance person.