Last Name _______________________ First Name ________________________ ID ___________________________
Operations Management II 73-431 Summer 2002
Odette School of Business
University of Windsor
Midterm Exam Solution
Wednesday, July 3, 1:00 – 2:20 pm
Last Name A-V Odette OB 212
Last Name W-Z Odette OB 321
Instructor : Mohammed Fazle Baki
Aids Permitted : Calculator, straightedge, and a one-sided formula sheet.
Time available : 1 hour 20 min
Instructions:

This exam has 13 pages.

Please be sure to put your name and student ID number on each page.

Show your work.
Grading:
Question
1
Marks:
/10
2 /10
3
4
5
6
7
/11
/6
/6
/10
/12
Total: /65

Last Name _______________________ First Name ________________________ ID ___________________________
Question 1: (10 points)
1.1 Continuous production is used when a. volume is low and customization is low b. volume is low and customization is high
c. volume is high and customization is low d. volume is high and customization is high
1.2 A job shop is a. useful when the production system contains only one machine b. useful when every job visits the machines in the same order
c. labor intensive d. capital intensive
1.3 Static scheduling means a. some jobs have non-zero arrival times
b. arrival time is zero for every job c. processing times are not known/fixed d. processing times are known and fixed
1.4 Total completion time means a. the time when the last job is completed b. sum of all the processing times
c. sum of all the completion times d. maximum of all the completion times
1.5 Minimize makespan if the concern is the cost of
a. utilities b. space c. material d. transportation
1.6 Consider a static stochastic scheduling problem with 2 identical parallel machines. Suppose that the processing times are exponentially distributed. Which of the following order minimizes the expected value of makespan? a. Shortest Expected Processing Time (SEPT)
b. Longest Expected Processing Time (LEPT) c. First Come First Served (FCFS) d. Earliest Expected Due Date (EEDD)
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1.7 Consider a static deterministic scheduling problem with 2 machines M1 and M2. Suppose that every job is processed first on M1, and then on M2. Which of the following rules minimizes makespan? a. Johnson’s rule b. Sequencing the jobs in the descending order of a j
 b j
, where a j is the rate processing of job j on M1 and b j is the rate processing of job j on M2 c. any of the above d. none of the above
1.8 A supply chain represents a. all suppliers and their suppliers b. all manufacturers and their manufacturers c. some suppliers and some manufactures
d. all the stages at which value is added in producing and delivery a product or service
1.9 The nearest neighbor rule a. produces a production schedule that minimizes total completion time b. produces a project schedule that minimizes project completion time
c. produces a delivery sequence that may need some improvement d. produces a delivery sequence that’s optimal
1.10 A project a. always has a single critical path b. always has a single critical activity c. may have multiple critical paths each with a different length
d. may have multiple critical paths each with the same length
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Question 2: (10 points)
Three jobs must be processed on a single machine that starts at 8:30 am. The processing times and due dates are given below:
Job
J1
J2
Processing Time (Hours)
3
3
Due Date
1:30 pm
12:30 pm
4 6:30 pm a. (7 points) Assuming that the jobs are processed in the sequence J1, J2, J3, compute makespan, total completion time, maximum lateness, and average tardiness.
Job
J3
Start Time
(Hours)
Processing
Time
Completion
Time
Due Date
(Hours)
Lateness
(Hours)
Tardiness
(Hours)
(Hours) (Hours)
(1 point)
(1 point) (1 point)
J1
J2
0
3
3
3
3
6
5
4
-2
2
0
2
J3
Maximum
6 4 10
10
10 0
2
0
2
Total 19
Makespan = Maximum completion time = Max (3, 6, 10) = 10 (1 point)
Total completion time = 3+6+10 = 19 (1 point)
Maximum lateness = Max (-2, 2, 0) = 2 (1 point)
Average tardiness = (0+2+0)/3 = 0.6667 (1 point)
0 2 b. (3 points) State the sequence that minimizes maximum lateness. What is the minimum maximum lateness?
The Earliest Due Date (EDD) sequence minimizes maximum lateness. (1 point)
Job
(1 point)
Start Time Processing
Time
Completion
Time
Due Date Lateness
J2
J1
J3
0
3
6
3
3
4
Maximum lateness = Max (-1,1,0) = 1 (1 point)
3
6
10
4
5
10
-1
1
0
4

Last Name _______________________ First Name ________________________ ID ___________________________
Question 3: (11 points)
A production facility assembles inexpensive telephones on a production line. The assembly requires
5 tasks with precedence relationships and activity times in minutes are shown below:
A
10
B
17
D
6
C
8
E
8 a. (5 points) For each task compute positional weight and rank
Task Positional Weight
A
(2.5 points)
10+17+8+6+8=49
Rank
(2.5 points)
1
B
C
D
E
17+8+6+8=39
8+8=16
6+8=14
8
2
3
4
5 b. (3 points) According to the ranked positional weight method what is the assembly line balancing solution if the cycle time is 20 minutes?
Step 1: Assign A, the task with Rank 1 to Station 1.
Station 1: {A} 10
Step 2: B, the task with Rank 2 does not fit in Station 1. So, assign B to Station 2.
Station 1: {A} 10 , Station 2: {B} 17
Step 3: C, the task with Rank 3 cannot be assigned before Station 2 because it must be done after B. C does not fit in Station 2. So, assign C to Station 3.
Station 1: {A} 10, Station 2: {B} 17 , Station 3: {C} 8
Step 4: D, the task with Rank 4 cannot be assigned before Station 2 because it must be done after B. D does not fit in Station 2. D fits in Station 3. So, assign D to Station 3.
Station 1: {A} 10 , Station 2: {B} 17 , Station 3: {C,D} 14
Step 5: E, the task with Rank 5 cannot be assigned before Station 3 because it must be done after C and D. E does not fit in Station 3. So, assign E to Station 4.
Station 1: {A} 10 , Station 2: {B} 17 , Station 3: {C,D} 14 Station 4: {E} 8
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Last Name _______________________ First Name ________________________ ID ___________________________ c. (2 points) If the cycle time is 20 minutes, what is the theoretical minimum number of station?
Theoretical minimum number of stations



10

167

20
8

6

8






40
20



2 .
45


3 d. (1 point) Does there exist a line balancing solution that contains the theoretical minimum number of stations (assume 20 minute cycle time)? If so, find the solution.
No, there is no solution that contains 3 stations with cycle time 20 minutes.
Question 4: (6 points)
A project has the following precedence relationships and activity times:
Activity Activity Time (weeks) Immediate Predecessor(s)
A
B
4
10
-
-
C
D
7
9
A
B a. (3 points) Draw the network diagram. Use activity on arc method. You need no dummy arc.
2
A
C
1 4
B
3 D b. (3 points) Suppose that the project has another activity, E which has an activity time of 8 weeks and the immediate predecessors of which are B and C. Draw the network diagram that contains all five activities A, B, C, D and E. Use activity on arc method. You need only one dummy arc.
2
Marking scheme (both parts a and b ):
A
C
4
0.5 point is taken off for violation of each network construction rule and precedence constraint.
1
B
3
E
5
D
6

Last Name _______________________ First Name ________________________ ID ___________________________
Question 5: (6 points)
An activity-on-arc project network consisting of four activities is shown below.
B
3
D
A C
1 2 4
Node 1 = starting node
Node 4 = finish node
Suppose that the activity times are uncertain. Minimum, most likely and maximum times are shown in days below. Compute the mean and variance of each activity time. What is the probability that the project will be completed in under 21 days?
A B C D
Min ( a )
Most Likely ( m )
Max ( b )
6
9
6

4
10

9

10
5
7
5

4
9

7

9
2
5
2

4
8

5

8
6
6
6

4

8
9

10 Mean,
  a

4 m
 b
(1 point)
6
Variance,
 2 
 b

36 a

2
(1 point)

10

8 .
67
6

36
6

2

0 .
4444

9

7

36
5

2
6

0 .
4444

5

8

36
2
6

2

1

8

6 .
33
6

36
6

2

0 .
1111
Critical path: A-B-D
 
8 .
67

7 .
00

6 .
33

22 (1 point)
 2 
0 .
4444

0 .
4444

0 .
1111

1
   2 
1

1 (1 point)
Probability(completion time

21 |
 
22 ,
 
1 )

P z


0 .
50

21
 
P

0

 z


P

1

 z

21
0 .
50


1
22

0 .
3413
P
 z
 
1


P



0 .
1587
  z

0
 

1
 z

0

(2 points)
7

Last Name _______________________ First Name ________________________ ID ___________________________
Question 6: (10 points)
Irving Bonner, an independent computer programming consultant, has contracted to complete four computer programming jobs.
Job
1
2
Time required (days)
8
9
Due date (days)
12
18
3
4
5
7
22
29
Assume that some jobs must be completed in a certain sequence because they involve program modules that will be linked. Precedence restrictions:
1

3
4

2 a. (8 points) Using Lawler’s algorithm, find the sequence in which he should be performing the jobs in order to minimize the maximum tardiness subject to the precedence constraints.
Iteration 1:
Job Candidate?
(1 point)
1
2
3
Yes
Yes
Due date Completion time if scheduled
(1 point)
18
22
8+9+5+7=29
8+9+5+7=29
Tardiness if scheduled
(1 point)
Max(0,29-18)=11
1
2
4
Yes
Yes
Decision in iteration 2:
Max(0,29-22)=7 * (Min)
(1 point)
4
Decision in iteration 1: (1 point)
Job 3
Iteration 2: (1 point)
Job Candidate? Due date Completion time if scheduled Tardiness if scheduled
12
18
8+9+7=24
8+9+7=24
Max(0,24-12)=12
Max(0,24-18)=6 * (Min)
Job 2 Job 3
(Continued…)
8

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Iteration 3: (1 point)
Job Candidate? Due date Completion time if scheduled Tardiness if scheduled
1 Yes 12 8+7=15 Max(0,15-12)=3
29 8+7=15 Max(0,15-29)=0 * 4 Yes
Decision in iteration 3:
Job 4 Job 2 Job 3
Final Sequence: (1 point)
Job 1 Job 4 Job 2 Job 3 b. (2 points) What is the maximum tardiness of the sequence you found in Part a ?
Job
1
4
2
3
Start time
(days)
0
8
15
24
Time required
(days)
8
7
9
5
Maximum tardiness = Max(0,0,6,7) = 7
Completion time
(days)
8
15
24
29
Due date
(days)
12
29
18
22
Lateness
(days)
-4
-14
6
7
Tardiness
(days)
0
0
6
7
9

Last Name _______________________ First Name ________________________ ID ___________________________
Question 7: (12 points)
All-Weather Oil and Gas Company has two delivery vehicles and is planning delivery routes to four natural gas customers. The customer locations and gas requirements (in gallons) are given below:
Customer
1
2
Location
(0,4)
(0,12)
Requirements (gallons)
550
400
3
4
(5,12)
(5,0)
650
250
Assume that the depot is located at the origin of the grid and that the delivery trucks have capacity
1,200 gallons. Also assume that the cost of travel between any two locations is the straight-line
(Euclidean) distance between them. Find the route schedule obtained from the savings matrix method and the nearest neighbor method. a. (5 points) Compute the distance matrix (Each cell 0.5 point

10 cells = 5 points)
Distance Matrix
Depot
Customer 1
Customer 2
Customer 3
Customer 4
Distance (1,3) =
Depot Customer 1
4

4

12
 
0

5

2 
   
2 
Customer 2
12
8
89

9 .
43
Customer 3
13
9.43
5
Customer 4
5
6.40
13
12
Distance (1,4) =

4

0
 
0

5

2     
2 
41

6 .
40 b. (3 points) Compute the savings matrix (Each cell 0.5 point

6 cells = 3 points)
Customer 1
Savings Matrix
Customer 2 Customer 3
Customer 1
Customer 2
4+12-8=8
Customer 4
4+13-9.43=7.57 =4+5-6.40=2.60
12+13-5=20 12+5-13=4
13+5-12=6 Customer 3
Customer 4
Saving(1,3)=Dist(D,1)+Dist(D,3)-Dist(1,3)=4+13-9.43=7.57
Saving(1,4)=Dist(D,1)+Dist(D,4)-Dist(1,4)=4+5-6.40=2.60 c. (2 points) Rank the savings
10

Last Name _______________________ First Name ________________________ ID ___________________________
1. (2,3)
2. (1,2)
3. (1,3)
4. (3,4)
5. (2,4)
6. (1,4) d. (2 points) Find the delivery routes.
Since (2,3) has Rank 1, assign Customers 2 and 3 to Vehicle 1.
Vehicle 1  (2,3). Capacity used = 400+650 = 1050. Capacity unused = 1200-1050=150.
Demand of each of Customers 1 and 3 is more than 150.
Hence, assign Customers 1 and 4 to Vehicle 2.
Vehicle 2  (1,4). Capacity used = 550+250 = 800. Capacity unused = 1200-800=400.
Using nearest neighbor rule, the sequences are as follows:
Vehicle 1 
Depot  2  3  Depot
Vehicle 2

Depot  1  4  Depot
11