Name (print, please) _______________________________________________ ID ___________________________
Operations Management II 73-431 Fall 2001
Odette School of Business
University of Windsor
Midterm Exam Solution
Wednesday, October 24, 1:00 – 2:20 pm Odette B04
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a one-sided formula sheet.
Time available: 1 hour 20 min
Instructions:
 This exam has 11 pages including this cover page.
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/12
2
/6
3
/6
4
/6
5
/6
6
/9
7
/5
Total:
/50
Name:_________________________________________________
ID:_________________________
Question 1: (12 points)
1.1 Job shop is suitable for
a. make to stock production environment
b. products that are highly customized
c. high volume production system
d. standard products
e. None of the above
1.2 In a flow shop with m machines
a. every job must be processed through the m machines in the same order, and each job
is processed exactly once on each machine
b. not all jobs are assumed to require exactly m operations
c. some jobs may require multiple operations on some of the m machines
d. none of the above holds
1.3 The following rule minimizes total completion time on a single machine
a. Critical ratio
b. EDD
c. SPT
d. FCFS
e. LCFS
f. None of the above
1.4 What is makespan?
a. Maximum processing time
b. Summation of processing times of all the jobs (answer b is rue for a simple single machine
case, false in general)
c. Summation of completion times of all the jobs
d. Completion time of the last job processed
e. None of the above
1.5 If all the jobs are ready for processing at time zero, then
a. the scheduling problem is a dynamic scheduling problem
b. the scheduling problem is a stochastic scheduling problem
c. flow time of a job is the same as its completion time
d. none of the above
1.6 Which of the following is minimized by the EDD rule in the context of a single machine static
stochastic scheduling problem?
a. Maximum lateness
b. Maximum expected lateness
c. Maximum probability that a job is late
d. None of the above
1.7 SPT scheduling can significantly reduce the size of the queue in the dynamic scheduling problem
a. True
b. False
2
Name:_________________________________________________
ID:_________________________
1.8 If all the jobs are equally important (i.e., the weights are all 1), the c rule is exactly the same as
the SPT rule
a. True
b. False
Questions 1.9 and 1.10 Consider a warehouse at location (0,0) and two customers A and B at
locations (3,0) and (0,4) respectively
1.9 Which of the following number is nearest to the distance between A and B?
a. 0
b. 1
c. 2
d. 3
e. 4
f. 5
g. 6
h. 7
d  A, B 
 yB  yA 2  xB  xA 2

4  02  0  32
5
1.10 Which of the following number is nearest to the savings obtained when a vehicle visits both A
and B?
a. 0
b. 1
c. 2
S  A, B  d W , A  d W , B  d  A, B  3  4  5  2
d. 3
e. 4
f. 5
g. 6
h. 7
1.11
a.
b.
c.
d.
What is cycle time?
The summation of processing times of work elements of a product
The workload assigned to a worker
The time interval between two completed products
None of the above
1.12 Consider a project that requires completion of task A. Suppose that the pessimistic, normal,
and optimistic completion times of A are estimated to be 3, 5, and 9 days respectively. Then, the
variance of the completion time of A is ____________ day2.
a. 0
 2  b  a 2 / 36  9  32 / 36  1
b. 1
c. 2
d. 3
e. 4
f. 6
g. 17
3
Name:_________________________________________________
ID:_________________________
Question 2: (6 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:
E
8
7
B
6 A
D
9
C
12
a. (1 point) Consider a solution with station 1: {A,D}, station 2: {B,C,E}. Is the solution feasible? If so,
find the cycle time given by the solution.
Answer:
Not feasible, D cannot be assigned to Station 1 because B and C must be done before D.
b. (1 point) Consider a solution with station 1: {A,C}, station 2: {B,D,E}. Is the solution feasible? If so,
find the cycle time given by the solution.
Answer:
Yes, the solution satisfied all the precedence constraints. Station 1 workload=6+12=18 minutes
and Station 2 workload=8+9+7=24 minutes. The cycle time =max{18,24}=24 minutes
c. (2 points) If the cycle time is 15 minutes, then what is the theoretical minimum number of
workstations?
Answer:
Theoretical minimum number of workstation is obtained by dividing sum of all the activity times by
the required cycle time and rounding up. Hence, the theoretical minimum number of workstations
 6  8  12  9  7   42 

   15   2.8  3
15

  
d. (1 point) Using eye inspection, provide a line balancing solution with cycle time at most 15
minutes and minimum (not necessarily theoretical minimum) number of workstations.
Answer:
Station 1:{A,B}, Station 2:{C}, Station 3:{D}, Station 4:{E}
e. (1 point) According to the ranked positional weight technique, what is the positional weight of B?
Answer:
Positional weight of B is its time plus the activity times of all its successors = 8+9+7=24 minutes.
4
Name:_________________________________________________
ID:_________________________
Question 3: (6 points)
The following four jobs must be processed through a three-machine flow shop.
Machine
Job
A
B
C
1
4
2
7
2
3
5
8
3
7
4
6
4
2
3
9
a. (4 points) Find the optimal sequence of the jobs in order to minimize the makespan.
Answer:
Equivalent two-machine problem
Job
Machine 1 Machine 2
1
6
9
2
8
13
3
11
10
4
5
12
Iteration
1
2
3
4
Minimum
Processing
Time
5
6
8
10
Final sequence:
Job
Machine
Corresponding to Corresponding to
Minimum
Minimum
Processing
Processing
Time
Time
4
1
1
1
2
1
3
2
Assign
Job 4 to the 1st position
Job 1 to the 2nd position
Job 2 to the 3rd position
Job 3 to the 4th position
4,1,2,3
b. (2 points) What is the makespan of the resulting schedule?
Answer:
As computed below, the last job processed, Job 2 is completed at time 35.
Hence, makespan = 35.
Job
4
1
2
3
Start
0
2
6
9
Machine A
Process End
2
2
4
6
3
9
7
16
Start
2
6
9
16
Machine B
Process End
3
5
2
8
5
14
4
20
5
Start
5
14
21
29
Machine C
Process End
9
14
7
21
8
29
6
35
Name:_________________________________________________
ID:_________________________
Question 4: (6 points)
Irving Bonner, an independent computer programming consultant, has contracted to complete four
computer programming jobs.
Job
Time required (days)
Due date (days)
1
6
10
2
10
17
3
3
18
4
8
26
a. (5 points) Find a sequence in which he should be performing the jobs in order to minimize the
number of tardy jobs.
Answer:
Step 1: Arrange the jobs according to the EDD rule
Processing Time Flow Time Due Date Tardiness
Job
(days)
(days)
(days)
(days)
1
6
6
10
0
2
10
16
17
0
3
3
19
18
1
Tardy
4
8
27
26
1
Tardy
First job tardy is Job 3
Among all the jobs processed before Job 3, Job 2 has the longest processing time
Eliminate Job 2
Job
1
3
4
Step 2: Eliminate Job 2 and arrange the remaining according to the EDD rule
Processing Time Flow Time Due Date Tardiness
(days)
(days)
(days)
(days)
6
6
10
0
3
9
18
0
8
17
26
0
Job
1
3
4
2
Step 3: Append the tardy jobs in the end in any order
Processing Time Flow Time Due Date Tardiness
(days)
(days)
(days)
(days)
6
6
10
0
3
9
18
0
8
17
26
0
10
27
17
10
Only Job 2 is tardy
The optimal sequence is 1-3-4-2.
b. (1 point) What is the minimum number of tardy jobs?
Answer:
As it is shown in Part a, only Job 2 is tardy in the optimal sequence 1-3-4-2. Hence, minimum
number of tardy jobs is 1.
6
Name:_________________________________________________
ID:_________________________
Question 5: (6 points)
Consider once again Irving Bonner’s problem stated in Question 4. The time required and due date
data are re-written below:
Job
Time required (days)
Due date (days)
1
6
10
2
10
17
3
3
18
4
8
26
Assume that some jobs must be completed in a certain sequence because they involve program
modules that will be linked. Precedence restrictions:
14
32
a. (5 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.
Answer:
Iteration 1:
Job
Candidate?
1
No
2
Yes
3
No
4
Yes
Due date
Completion time if scheduled
Tardiness if scheduled
17
6+10+3+8=27
27-17=10
26
6+10+3+8=27
27-26=1* (minimum)
Decision in iteration 1: Schedule Job 4 in the 4th position
Job 4
Iteration 2:
Job
Candidate?
Due date
Completion time if scheduled
Tardiness if scheduled
1
Yes
10
6+10+3=19
19-10=9
2
Yes
17
6+10+3=19
19-17=2* (minimum)
3
No
Decision in iteration 1: Schedule Job 2 in the 3 rd position
Job 2
Job 4
(Continued…)
7
Name:_________________________________________________
ID:_________________________
Iteration 3:
Job
Candidate?
Due date
Completion time if scheduled
Tardiness if scheduled
1
Yes
10
6+3=9
9-10 = -10*
(minimum)
3
Yes
18
9-18= -90*
(minimum)
Decision in iteration 1: Schedule any of Jobs 1 or 3 in the 2nd position and the other in the 1st
position.
Job 1
Job 3
Job 2
Job 4
or
Job 3
Job 1
Job 2
Job 4
Hence, two sequences are optimal, 1-3-2-4 or 3-1-2-4
b. (1 point) What is the maximum tardiness of the sequence you found in Part a?
Answer:
Maximum tardiness is the maximum of all the minimum values obtained in Iterations 1, 2 and 3.
Hence, the maximum tardiness is 2 days. The maximum tardiness can also be computed as
follows:
Job
Time
required
Completion
time
Due date
Lateness
Tardiness
(days)
(days)
(days)
(days)
(days)
1
6
6
10
6-10=-4
max(-4,0)=0
3
3
9
18
9-18=-9
max(-9,0)=0
2
10
19
17
19-17=2
max(2,0)=2
4
8
27
26
27-26=1
max(1,0)=1
8
Remark
maximum
Name:_________________________________________________
ID:_________________________
Question 6: (9 points)
A project consisting of five activities satisfies the following precedence constraints:
Activity
Time (Weeks)
Immediate Predecessors
A
5
-
B
7
-
C
3
A,B
D
6
A
a. (5 points) Construct a network for the project (Use activity on arc methods; you should need only
one pseudo activity)
Answer:
0
0
5
5
5 11
5 11
D
2
A
5
1
Dummy
5 5
8 8
6
C
7
8
B
0
1
7
8
7
3
3
Node 1 = starting node
Node 4 = finish node
4
10
11
ES EF
LS LF
(Continued…)
9
Name:_________________________________________________
ID:_________________________
b. (2 points) Identify the critical path and its length.
Answer:
The critical path is the longest path of the network. There are only three paths from the start to
end.
Path
Length in weeks
A,D
5+6=11* (maximum)
A,C
5+3=8
B,C
7+3=10
Since the path A,D is the longest of all 3 paths, the critical path A,D and its length is 11 weeks.
Note: Another way to find the critical path is that A,D is the only path from start to end for which
the following holds:

Earliest Start, ES = Latest Start, LS and

Earliest Finish, EF = Latest Finish, LF
c. (2 points) Compute the earliest and latest starting and finishing time for activity C.
Answer:
Earliest start, ES = 7
Earliest finish, EF = 10
Latest start, LS = 8
Latest finish, LF = 11
10
Name:_________________________________________________
ID:_________________________
Question 7: (5 points)
Joe has now been released from his government job. Based on his excellent performance, he was
able to land a job as a production scheduler in a brand-new custom refinishing auto service shop
located near the border. The sequence is customizing first, followed by repainting. The processing
times are independent exponentially distributed random variables with the mean times as stated
below.
Car
Customizing Time (Hours)
Painting (Hours)
1
4.0
2.0
2
1.1
1.0
3
3.0
1.2
Find a schedule that minimizes expected makespan.
Answer:
Car, j
1
2
3
Customizing
time
(Hours)
Aj
4.0
1.1
3.0
Customizing
rate
(Units/Hour)
1
aj 
Aj
1/4.0 = 0.25
1/1.1 = 0.91
1/3.0 = 0.33
Painting
time
(Hours)
Bj
2.0
1.0
1.2
Painting
rate
(Units/Hour)
1
bj 
Bj
1/2.0 = 0.50
1/1.0 = 1.00
1/1.20 = 0.83
Now, arrange the jobs (cars) from largest a j  b j values to smallest a j  b j values.
Since -0.09 > -0.25 > -0.50, the optimal sequence is 2-1-3.
11
a j  bj
-0.25
-0.09
-0.50