Last Name ________________________ First Name ________________________ ID __________________________
Operations Management II 73-431 Summer 2002
Odette School of Business
University of Windsor
Final Exam Solution
Tuesday, August 16, 12:00 – 3:00 pm ED Gym Area 1-3
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and 3 one-sided formula sheet.
Time available: 3 hours
Instructions:
 This exam has 26 pages including this cover page and 6 pages of table
 Please be sure to put your name and student ID number on each page.
 Show your work.
Marking Scheme:
Question
1
3
5
7
9
11
13
15
Score
/15
/4
/8
/4
/5
/3
/5
/2
Question
2
4
6
8
10
12
14
Total
Score
/6
/8
/14
/10
/6
/5
/5
/100
Name:_________________________________________________
ID:_________________________
Question 1: (15 points)
1.1 For which of the following should we use an X control chart to monitor process quality?
a. Windshield cracked or not
b. Flash lights work or not
c. Number of flaws in a paint job
d. Weight errors in cans of soup
e. All of the above
f. None of the above
1.2 For which of the following should we use an R control chart to monitor process quality?
a. The volume of scrap parts
b. Lengths of boards cut in a mil
c. The ounce of beer poured in a can
d. Errors in the length of pencils produced
e. All of the above
f. None of the above
1.3 You want to develop a three-sigma X control chart with subgroup size 4. You know that the
standard deviation of the sample means is 4 and the average of the sample means is 24. Which
of the following is the resulting UCL
a. 40
b. 36
c. 30
d. 24
e. 12
f. 4
1.4 Consider the problem of economic design of X control chart. If k is increased keeping
n unchanged
a. Searching cost increases
b. Searching cost decreases
c. Cost of operating in out-of-control state increases
d. Cost of operating in out-of-control state decreases
e. a and d
f. b and c
1.5 Which of the following is sampling cost?
a. Personnel cost for scrap or repair
b. Personnel cost for the fixing the problem
c. Cost of item
d. Cost of warranty claims
e. All of the above
f. None of the above
2
Name:_________________________________________________
ID:_________________________
1.6 Consider a single sampling plan with sample size n and acceptance number c . The producer’s
risk increases if
a. only n increases
b. only c increases
c. both n and c increase proportionately
d. a and c
e. all of the above
f. none of the above
1.7 What is AOQ?
a. Proportion of good items in the outgoing lot
b. Proportion of defective items in the outgoing lot
c. The maximum proportion of good items in the outgoing lot
d. The maximum proportion of defective items in the outgoing lot
e. Proportion of items inspected
f. Proportion of items replaced by the good items
1.8 Reliability increases if
a. the components are in series and the number of components increases
b. the components are in series and the number of components decreases
c. the components are in parallel and the number of components increases
d. the components are in parallel and the number of components decreases
e. a and d
f. b and c
1.9 The average life is called
a. Mean Time Between Failure (MTBF) if the system is repairable
b. Mean Time Between Failure (MTBF) if the system is non-repairable
c. Mean Time To Failure if the system is repairable
d. Mean Time To Failure if the system is non-repairable
e. a and d
f. b and c
1.10 Failure rate
a. decreases during the normal operating period
b. is the inverse of average life
c. is constant during the wear-out period
d. increases during the infant-mortality period
e. all of the above
f. none of the above
1.11 Process layout is used if
a. product is standardized
b. volume is high
c. efficiency is more important than flexibility
d. demand is fluctuating
e. all of the above
f. none of the above
3
Name:_________________________________________________
ID:_________________________
1.12
a.
b.
c.
d.
e.
f.
Cellular layouts
group machines into cells that resemble small assembly lines based on part families
identify families of parts with similar flow paths
may provide poorly balanced cells
all of the above
a and b
none of the above
1.13
a.
b.
c.
d.
e.
f.
In a job shop
every job visits all the machines in the same order
special purpose equipment is used
scheduling is part of line balancing
aisles are narrow
all of the above
none of the above
1.14
a.
b.
c.
d.
g.
h.
A scheduling problem is called stochastic if
jobs are processed on two or more machines in series
jobs are processed on two or more machines in parallel
all the jobs are not available for processing at time zero
processing times are not known at time zero
a and d
b and c
1.15
a.
b.
c.
d.
e.
f.
A critical path is the
longest path
one along which ES = LS for every activity
one along which EF = LF for every activity
one along which slack time = 0 for every activity
all of the above
none of the above
Question 2: (6 points)
An amplifier has a constant failure rate of 4% per 500 hour.
a. (2 points) What is the probability that the amplifier will survive 2000 hours?
Plife  2000   1     1  0.04  0.8493
t
4
b. (2 points) What is the average life?

1
1

 25 in 500 hours = 12,500 hours
 0.04
4
Name:_________________________________________________
ID:_________________________
c. (2 points) What failure rate is required to have a probability of survival of 95% at 2000 hour?
1   4  0.95
or, 1    0.951 / 4
or,   1 - 0.951 / 4  0.0127
Hence, the failure rate is 0.0127 1.27% per 500 hour
Question 3: (4 points)
Two components A and B are used to make a subassembly in a circuit board. The reliabilities of A
and B are 0.60 and 0.75 respectively.
a. (2 points) Find the reliability of the subassembly, if the components are in parallel.
R p  1  1  0.601  0.75  0.90
b. (2 points) Find the reliability of the subassembly, if the components are in series.
Rs  0.60  0.75  0.45
Question 4: (8 points)
The next-door neighbours are currently installing a new cistern that will provide water for their
household. The different components for the system and their reliability are described in the diagram
below. Determine the reliability of the system.
0.65
0.80
0.75
0.50
0.95
0.75
0.70
0.85
0.85
0.75
0.60
0.70
Consider the following subsystems:
3
0.80
7
0.65
1
4
0.75
0.75
0.95
0.50
6
5
2
0.85
0.85
0.75
0.70
0.60
0.70
5
Name:_________________________________________________
ID:_________________________
R1  0.75  0.75  0.5625, R2  0.85  0.85  0.7225
R3  1  1  0.651  R1 1  R2   1  1  0.651  0.56251  0.7225  0.9575
R4  0.95  0.751  0.95  0.9875
R5  1  1  0.701  0.70  0.91, R6  0.60  R5  0.60  0.91  0.546
R7  1  1  0.501  R6   1  1  0.501  0.546  0.773
Rsy stem  0.80  R3  R4  R7  0.80  0.9575  0.9875  0.773  0.5847
Question 5: (8 points)
In one of the acceptance sampling plans, 15 items were to be tested for 100 hours with replacement
and with an acceptance number of 2. Plot an Operating Characteristic curve showing probability of
acceptance as function of average life. Consider average life,  = 250, 500, 1000, 2500, and 5000
hours. For each average life compute and plot the probability of acceptance.
T  15100  1,500 hours, n  15, c  2

250

1

0.004
  T
 1  PNumber of defective  c  1  3 |   T 
 1 p
where p is obtained from Table A - 3 for x'  3, m    T
1500(0.004)
=6
500
0.002
1500(0.002)
=3
1000
0.001
1500(0.001)
=1.5
2500
0.0004
1500(0.0004)
=0.60
5000
0.0002
Pa  PNumber of defective  c  2 |   T 
1500(0.0002)
=0.30
Pa  PNumber of defective  c  2 |   6 
 1  PNumber of defective  c  1  3 |   6 
 1  0.9380  0.0620
where 0.9380 is obtained from Table A - 3 for x'  3, m  6
Pa  PNumber of defective  c  2 |   3
 1  PNumber of defective  c  1  3 |   3
 1  0.5768  0.4232
where 0.5768 is obtained from Table A - 3 for x'  3, m  3
Pa  PNumber of defective  c  2 |   1.5
 1  PNumber of defective  c  1  3 |   1.5
 1  0.1912  0.8088
where 0.1912 is obtained from Table A - 3 for x'  3, m  1.5
Pa  PNumber of defective  c  2 |   0.60 
 1  PNumber of defective  c  1  3 |   0.60 
 1  0.0231  0.9769
where 0.0231 is obtained from Table A - 3 for x'  3, m  0.60
Pa  PNumber of defective  c  2 |   0.30 
 1  PNumber of defective  c  1  3 |   0.30 
 1  0.0036  0.9964
where 0.5768 is obtained from Table A - 3 for x'  3, m  0.30
6
Name:_________________________________________________
ID:_________________________
Probability of acceptance,
Pa
Operating Characteristic Curve, Question #5
1.2
1
0.8
0.6
0.4
0.2
0
0
2000
4000
6000
Average life
Question 6: (14 points)
A film processing service monitors the quality of the developing process with light-sensitive
equipment. The accuracy measure is a number with a target value of zero. Suppose that an X chart
with subgroup of size four is used to monitor the process and the control limits are UCL = 1.10 and
LCL = -1.10. Assume that the process mean is zero and the process standard deviation is 0.80.
a. (2 points) What is the Type I error probability,  for this control chart?


1.1  0 

  2 P X  1.1 |   0,   0.8, n  4  2 P z 
  2 Pz  2.75  20.50  0.4970  0.006
0.8 / 4 

b. (3 points) Suppose that the process mean shifts to 1.00. What is the probability that the shift is
detected on the first subgroup after the shift occurs?
1.1  0 
  1.1  0
  P 1.1  x  1.1 |   1,   0.8, n  4  P
z

0.80 / 4 
 0.80 / 4
 P 5.25  z  0.25  P 5.25  z  0  P0  z  0.25  0.5  0.0987  0.5987
Pshift is detected   1    1  0.5987  0.4013
c. (2 points) Suppose that there is a probability of 0.05 that the process shifts from an in-control
state to an out-of-control state in any period. What is the expected number of periods per cycle
that the process remains in control? Assuming each search costs $10, compute the expected
searching cost per cycle.
1   1  0.05

 19

0.05
Search cost per cycle  a 2 1  E T   101  0.006  19  $11.14
E T  
7
Name:_________________________________________________
ID:_________________________
d. (2 points) Assume that when the process shifts out of control, the process mean shifts to 1.00.
What is the expected number of periods per cycle that the process remains in an out-of-control
state until a detection is made? If the cost of operating in an out-of-control state is $100 per
period, what is the expected cost of operating in out-of-control state per cycle?
E S  
1
1

 2.4919 periods
1   1  0.5987
Cost of operating in out-of-control state = a3 E S   1002.4919  $249.19
e. (3 points) What is the expected number of periods per cycle? If the cost of sampling is $10 per
unit, what is expected cost of sampling per cycle? What is the expected total cost of sampling,
searching, and operating in out-of-control state per period?
E C   E T   E S   19  2.4919  21.4919 periods
Sampling cost per cycle  a a nE c   10  4  21.4919  $859.68
Total cost per period 
11.14  249.19  859.68 1120.01

 $52.11 per period
21.4919
21.4919
f. (2 points) Find the UCL and LCL based on three-sigma control limits.
UCLX    3
LCLX    3

n

n
 03
 03
0.80
4
0.80
4
 1.2
 1.2
Question 7: (4 points)
An R chart is used to monitor the variation in the weights of packages of chocolate chip cookies
produced by a large national producer of baked goods. An analyst has collected a baseline of 200
observations to construct the chart. Suppose the computed value of R is 5.068.
a. (2 point) If subgroups of size six are to be used, compute the value of three-sigma limits for the
R chart.
n6
UCLR  d 4 R  2.00  5.068  10.136
LCLR  d 3 R  0  5.068  0
b. (2 points) If an X chart based on three-sigma limits is used, what is the difference between UCL
and the LCL? (You can compute UCL X  LCL X without knowing X )
UCLX  X  3
LCLX  X  3

n

n

  
 

R
5.068
Hence, UCLX  LCLX   X  3
   X  3
  6
6
6
 4.8990
n 
n
n
d2 n
2.534 6

Question 8: (10 points)
8
Name:_________________________________________________
ID:_________________________
Consider a single sampling plan with n  20 , c  2 , AQL=0.01, and LTPD=0.2.
a. (2 points) Compute producer’s risk  . Use Poisson probability table.
  Producer’s risk = PRejection | p  AQL  0.01
= Pc  1 or more defective | n  20 items are drawn, p  0.01
= P3 or more defective |   np  20  0.01  0.20
= 0.0011
(Table A-3 gives Poisson cumulative probabilities for x' or more defectives. The value 0.0011 is
obtained from x'  3, m    np  0.20 )
b. (3 points) Compute consumer’s risk  . Use Normal probability table.
  Consumer’s risk = PAcceptance | p  LTPD  0.2
= Pc or fewer defective | n  20 items are drawn, p  0.20

= P 2 or fewer defective |   np  20  0.20  4,   20  0.20  0.80  1.7889
2.50  


|   4,   1.7889  (continuity correction)
= P z 




2.50  4 

 P z 
  P z  0.84   P   z  0  P 0.84  z  0  0.5  P0  z  0.84
1.7889 

 0.50 - 0.2995 (From Table A-1) = 0.2005
The following questions refer to the sampling plan stated above: n  20 , c  2 .
c. (4 points) Compute the probability of acceptance, Pa at proportion of defective, p  0.3 . Use
Binomial probability formula.
Pa
= P2 or fewer defective | n  20 items are drawn, p  0.30
= P0 defective | n  20 , p  0.30 + P1 defective | n  20 , p  0.30 + P2 defectives | n  20 , p  0.30
 20 
0
20 0  20 
1
201  20 
2
20 2
=  0.30 1  0.30 +  0.30 1  0.30 +  0.30 1  0.30
0
1
2
20!
0.300 1  0.30200 + 20! 0.301 1  0.30201 + 20! 0.302 1  0.30202
=
0!20  0!
1!20  1!
2!20  2!
20!
0.300 0.7020 + 20! 0.301 0.7019 + 20! 0.302 0.7018
=
0!20!
1!19!
2!18!
1
19
0
20
2
18
= 10.30 0.70 + 200.30 0.70 + 1900.30 0.70
=0.000798+0.0006839+0.027845873
= 0.0355
d. (1 point) Use the result of part c. Compute AOQ at p  0.3 if the lot size, N is 500.
AOQ 
pN  nPa 0.30500  200.0355

 0.01022
N
500
9
Name:_________________________________________________
ID:_________________________
Question 9: (5 points)
A large national producer of cookies and baked goods uses a c chart to monitor the number of
chocolate chips in its chocolate chip cookies. The company would like to have an average of six
chips per cookie. One cookie is sampled each hour. The results of the last 12 hours were:
Hour
Number of chips per
cookie
Hour
Number of chips per
1
7
7
3
2
4
8
6
3
3
9
3
4
3
10
2
5
5
11
4
6
4
12
4
a. (2 points) Using the above data, set up a c -chart with z  3.
c
7  4  3  3  5  4  3  6  3  2  4  4 48

4
12
12
UCL c  c  3 c  4  3 4  10, LCL c  c  3 c  4  3 4  2  0 (negative)
b. (2 points) A purchaser of a bag of chocolate chip cookies discovers a cookie that has no chips in
it and charges the company with fraudulent advertising. If the expected number of chips per
cookie is four, what is the probability that a cookie would have no chips? Use Poisson probability
table.
Px  0 |   4  1  Px  1 |   4  1  0.9817 (from Table A - 3 for x'  1, m  4)
 0.0183
c. (1 point) Suppose that the company produces 300,000 cookies per year. How many cookies
baked each year would have no chips?
E X   np  300,000  0.0183  5,490 cookies
Question 10: (6 points)
Frank Brown, an independent TV repairman, is considering purchasing a home in Ames, Iowa, that
he will use as a base of operations for his repair business. Frank’s primary sources of business are 5
industrial accounts located throughout the Ames area. He has overlaid a grid on a map of the city
and determined the following locations for these clients as well as the expected number of calls per
month he receives:
Client
Grid Location
Expected Calls per Month
1
(6,4)
6
2
(8,3)
4
3
(5,9)
5
4
(4,5)
3
5
(7,8)
2
Find the optimal location of his house, assuming the goal is to
10
Name:_________________________________________________
ID:_________________________
a. (3 points) minimize weighted squared of the Euclidean distances.
Client, i
ai
bi
wi
wi ai
wi bi
1
6
4
6
36
24
2
8
3
4
32
12
3
5
9
5
25
45
4
4
5
3
12
15
5
7
8
2
14
16
w
i
 w a = 119/20 = 5.95
w
 w b = 112/20 = 5.60

w
X* 
i
w a
 20
i
i
 119
w b
i
i
 112
i
i
Y*
i i
i
Optimal location = (5.95, 5.60)
b. (3 points) minimize the maximum rectilinear distance to any client.
Client
ai
bi
ai  bi
 ai  bi
1
6
4
10
-2
2
8
3
11
-5
3
5
9
14
4
4
4
5
9
1
5
7
8
15
1
Min
c1  9
c3  -5
Max
c2  15
c4  4
c5  max c 2  c1 , c 4  c3   max 15  9,4   5  max 6,9  9
x1  c1 c3  / 2  9   5 / 2  7.0
y1  c1  c3  c5  / 2  9  5  9 / 2  6.5
x 2  c 2 c 4  / 2  15  4 / 2  5.5
y 2  c 2  c 4  c5  / 2  15  4  9 / 2  5.0
Any point along (7.0, 6.5) and (5.5, 5.0) is optimal.
11
Name:_________________________________________________
ID:_________________________
Question 11: (3 points)
Activities of a project and their immediate predecessors are shown below:
Activity
Time (Weeks)
Immediate Predecessors
A
5
---
B
7
A
C
3
A
D
2
B,C
E
6
C
F
5
B,E
Construct a network for the project. Use activity on arc methods.
D
4
6
B
1
A
F
2
C
5
3
E
Node 1 = starting node
Node 6 = finish node
Question 12: (5 points)
Following network shows PERT time estimates (a,m,b)
in weeks:
4
(2,5,9)
C
a. (3 points) Determine a critical path based on most
likely times. What is the mean and standard
deviation of the length of the critical path?
There are only 4 paths from the start node 1 to the
finish node 7. These paths ad their lengths are listed
below:
Path
ACEG
ABEG
ABDG
ABFH
Length
6+5+7
+3=21
6+12+7
+3=28
6+12+1
1+3=32
6+12+5
+6=29
The longest path, ABDG is the critical path.
12
(5,7,9)
E
1
A
(4,6,8)
2
5
P1
(1,3,7)
G
D
(10,11,12)
B
(9,12,14)
3
F
(3,5,7)
6
7
H
(4,6,9)
Name:_________________________________________________
Critical activity

ID:_________________________
a  4m  b
6
2 
4  46   8
6
6
A
B
b  a 2
8  42
14  92
10  411  12
 11
6
12  102
36
 0.6944
 0.1111
36
1  43  7
 3.33
6
G
 0.44
36
9  412   14
 11.83
6
D
36
7  12
6
  6  11.83  11  3.33  32.16
 1.0
 2  0.44  0.69  0.11  1.0  2.25
   2  2.25  1.5
b. (2 points) Assuming that the critical path is the one identified in a, find the probability that the
project will be completed in 35 days.
35  32.16 

P x  35 |   32.16,   1.5  P z 
  P z  1.89 
1. 5


 P   z  0   P0  z  1.89   0.50  0.4706  0.9706
Question 13: (5 points)
Shop starts at 8:30 am. Following are the information about two jobs.
Job
Time when the job
arrives
Time when the job is
completed
Time when the job is
due
A
8:30 am
12:30 pm
1:30 pm
B
11:30 am
4:30 pm
2:30 pm
Find
a. (1 point) completion time of each job.
Set t  0 at 8:30 a.m.
C A  12 : 30  8 : 30  4 hours
C B  4 : 30  8 : 30  8 hours
b. (1 point) flow time of each job.
FA  C A  rA  4  0  4 hours
FB  C B  rB  8  3  5 hours
c. (1 point) tardiness of each job.
13
Name:_________________________________________________
ID:_________________________
TA  max 0, C A  d A   max 0,4  5  0 hours
TB  max 0, C B  d B   max 0,8  6  2 hours
d. (1 point) total completion time.
C  C
A
 CB  4  8  12 hours
e. (1 point) makespan.
Makespan = max C A , CB   max 4,8  8 hours
Question 14: (5 points)
Jobs processed through Percy’s machine shop pass through milling, grinding and turning in the
stated order. The hours required for each of these operations is as follows:
Job
Milling
Grinding
Turning
A
7
2
2
B
2
1
7
C
4
1
1
D
5
2
6
E
3
1
8
a. (2 points) Determine a sequence to minimize makespan.
First, consider the following fictitious 2-machine problem:
Job
Machine 1’
Machine 2’
A
7+2=9
2+2=4
B
2+1=3
1+7=8
C
4+1=5
1+1=2
D
5+2=7
2+6=8
E
3+1=4
1+8=9
Iteration 1: Min{processing times} = 2 for Job C on Machine 2’. Assign Job C to Position 5 (near
end). Eliminate Job C.
C
Position 1
Position 2
Position 3
Position 4
Position 5
Iteration 2: Min{processing times, except Job C} = 3 for Job B on Machine 1’. Assign Job B to
Position 1 (near beginning). Eliminate Job B.
B
C
Position 1
Position 2
Position 3
Position 4
Position 5
Iteration 3: Min{processing times, except Jobs B,C} = 4 for Job A on Machine 2’. Assign Job A to
Position 4 (near end). Eliminate Job A.
B
C
Position 1
Position 2
Position 3
Position 4
Position 5
Iteration 4: Min{processing times, except Jobs A,B,C} = 4 for Job E on Machine 1’. Assign Job E
to Position 2 (near beginning). Eliminate Job E.
B
E
A
C
Position 1
Position 2
Position 3
Position 4
Position 5
Iteration 5: Assign Job D to the only remaining position 3.
B
E
D
A
Position 1
Position 2
Position 3
Position 4
14
C
Position 5
Name:_________________________________________________
ID:_________________________
Final sequence: B, E, D, A, C.
b. (2 points) Draw a Gantt chart illustrating your solution.
Milling
B
E
B
Grinding
D
E
Turning
A
A
D
B
2
4
C
C
E
6
8
10 12
Time
A
D
14
16
18
20
22
24
C
26
28
30
c. (1 point) What is the optimal makespan? 27.
Question 15: (2 points)
A production facility assembles inexpensive telephones on a production line. The assembly requires
7 tasks with precedence relationships and activity times in minutes are shown below:
5
E
5
A
3
G
7
C
6
B
F
8
D
12
a. (1 point) Consider a solution with station 1: {A,B}, station 2: {C,D}, station 3: {E,F,G}. Is the
solution feasible? If so, find the cycle time given by the solution.
Yes. Cycle time = max 5  6,7  12,5  8  3  max 11,19,16  19 min
b. (1 point) Using eye inspection, provide a line balancing solution with cycle time at most 23
minutes and minimum number of workstations.
Station 1: {A, B, C, E}, time = 5+6+7+5 = 23 min
Station 2: {D, F, G}, time = 12+8+3 = 23 min
Cycle time = max{23, 23} = 23 min as required.
15