ISE 514 GEZA BOTTLIK
Fall 2005
Final 12/08/05 Solution 12/11/05
Problem No. 1
Consider the following problem that not only has earliness and tardiness penalties, but unlike our
examples, the penalties are different for each job and the due dates are also specific to the jobs.
Devise a heuristic of your own, explain your rationale, apply it to the problem and discuss the
effectiveness of your approach. (25 points)
J
1
2
3
4
5
6
7
8
p
4
2
5
4
3
2
3
2
d
6
9
7
8
7
15
7
12
Ep
1
3
2
3
1
2
3
2
Tp
3
5
4
2
2
2
2
5
The main thing we have to go on is to simulate the V shape penalty curve. Since we do not have a
common due date, one option is to use an average due date. The other is to work from both ends
alternately placing a job last and first, appropriately increasing the starting time and decreasing the
finishing time. I elected to do the latter. Calculating penalties if placed last at 25:
57
80
72
34
36
20
36
65
12
20
32
55
0
12
24
35
X
6
X
20
X
X
Place 6 last
Calculating penalties if placed first at 0:
2
21
4
12
4
X
Place 1 first
Calculating penalties if placed last at 23:
X
70
64
30
32
X
Place 4 last
Calculating penalties if placed first at 4:
X
9
8
X
0
X
Place 5 first
Calculating penalties if placed last at 19:
X
50
48
X
X
X
Place 7 last
Calculating penalties if placed first at 7:
X
0
20
X
X
X
Place 2 first
Calculating penalties if placed last at 16:
X
X
36
X
X
X
Place 8 last
Calculating penalties if placed first at 9:
X
X
28
X
X
X
Place 3 first
For a sequence of (1,5,2,3,8,7,4,6) and a total penalty of 124. Not knowing the optimum and not
having an easy way to calculate a lower bound it is hard to say how good this schedule is, but I
would feel it was pretty good based on a penalty of 170 of the original sequence. Using an
average due date of 9 and the procedure for equal penalties, using the early/tardy penalty as a
tie breaker, I got (1,4,6,2,8,7,4,3) and a penalty of 159. Later, when I had the time, I did an
exhaustive search and found a min of 108 and a max of 267, so both heuristics are reasonably
good. I did not expect you to comment on this, but of course a pair wise exchange following the
heuristic will probably improve the result quickly.
Page 1
ISE 514 GEZA BOTTLIK
Fall 2005
Final 12/08/05 Solution 12/11/05
Problem No. 2
A production line consists of four sequential machines with large intermediate buffers. The production
manager was interested in minimizing the average inventory of jobs passing through the line. The
processing times on the four machines are as follows (times are in minutes):
Machine
Mean processing time
Std. Deviation
Distribution
M1
3
0.2
Unknown
M2
2.3
0.05
Normal
M3
2.1
0.12
Exponential
M4
2.05
0.1
Exponential
What is the expected total time through this line (cycle time)? (4points)
Adding up the processing times, we have 9.45
What is your estimate for the maximum likely time to get through this line? (6 points)
We have to add about 3 standard deviations to the average:
9.45  3 * 0.2 2  0.052  0.12 2  0.12  10.23
If you had to guarantee a minimum hourly throughput for this line, what would that be? (5 points)
The throughput depends on the slowest machine at its slowest 3+3(.2) = 3.6minutes
60min/3.6min = 16.67 units
Would rearranging the sequence of the machines improve the objective? Why or why not? (5 points)
Since the buffers are large, the machines are effectively decoupled, so the sequence does not
matter for throughput, but does matter for inventory in the buffers. We want the slowest
machine first, so that no inventories will buildup. They are already in such a sequence.
Problem No. 3
The production manager in problem two wanted to increase throughput and decided to add a second
machine of the same type in parallel with one of the machines. Which of the machines would he/she
duplicate? (3 points) How will that change the answers to question 2? (7 points)
Add a second machine of type M1 because it is the slowest
The simplest assumption is to consider the machine twice as fast, so the mean time decreases by 1.5
minutes to 8.95 minutes
The standard deviation is unlikely to change so we add 0.78 as before to get 8.73
Another machine is now slowest 60/(2.3+3*0.05) = 24.48
We would now place M1s last, since they are the fastest, so no inventory will build up
Page 2
ISE 514 GEZA BOTTLIK
Fall 2005
Final 12/08/05 Solution 12/11/05
Problem No. 4
You have a machine that produces 8 different parts. Some of these parts have the same set-up
procedure (parts A, C, and E). This setup requires a half an hour. Parts B, F and G also have a
common set-up, requiring 1 ½ hours. Finally, the remaining parts have a yet different set-up and that
takes 1 hour.
The processing times and due times for the six parts are given below. You are not only interested in the
total time required to do all the jobs, but also in minimizing the maximum lateness. These two are
additive.
Find a sequence for this objective and the associated penalty, describe your method and discuss the
potential for its optimality. (5, 10 and 5 points)
Using MST was devised for minimizing Tbar and Cmax. Since we have Lmax here, we need to adjust the
procedure. I would start with EDD and the by exchanging adjacent jibs try to reduce the number of
setups. Then Lmax may also be reduced because of the reduced Cmax. The initial penalty is 40
By exchanging A and H, B and E, and D and G, we eliminate 4 set ups and reduce the penalty to 34
Processing time
set up
Set up
Total
Compl.
L
Job
(hrs)
Due time type
Time
0.5
2.5
2.5
0
A
2
4
1
1
3
5.5
0
H
2
7
3
0.5
2.5
8
0
C
2
8
1
1.5
4.5
12.5
3.5
B
3
9
2
0.5
4.5
17
4
E
4
13
1
1.5
4.5
21.5
7.5
F
3
14
2
1
3
24.5
8.5
D
2
16
3
1.5
4.5
29
11
G
3
18
2
Set
Set
Total
Compl.
L
up
up
Processing time
set up
Time
Job
(hrs)
Due time type
1
1
3
3
-4
H
2
7
3
0.5
0.5
2.5
5.5
1.5
A
2
4
1
0.5
0
2
7.5
-0.5
C
2
8
1
0.5
0
4
11.5
-1.5
E
4
13
1
1.5
1.5
4.5
16
7
B
3
9
2
1.5
0
3
19
5
F
3
14
2
1.5
0
3
22
4
G
3
18
2
1
1
3
25
9
D
2
16
3
Even better, group by type, then with EDD within groups and get 24 + 9 = 33. which also happens to
be the optimum. It can be shown to be the optimum since any change will either increase Cmax
Page 3
ISE 514 GEZA BOTTLIK
Fall 2005
Final 12/08/05 Solution 12/11/05
or Lmax.
Page 4
ISE 514 GEZA BOTTLIK
Fall 2005
Final 12/08/05 Solution 12/11/05
Problem No. 5
This is a one machine problem. The set up time for each job depends on the prior job. Find a good
schedule to minimize Cmax and explain why you chose the method. Calculate Cmax. Discuss the
potential for optimality (10 points)
Set up if preceded by Job
Job
1
2
3
4
5
6
1
2
10
20
5
17
10(6)
2
3
20
3
19
13
18(5)
3
9
9
17
10
6(3)
4
17
4
11
4
2 (1)
5
17
10
4
11
4 (2)
6
11
13
14
13
3(4)
Since only a good schedule is asked for, Baker’s heuristic is adequate – shortest set up first. Dynamic
programming, the best way to solve the problem, takes too long for an exam. If we start with job 1, pick
No 4 (2), pick 5 (4), then 3 (6), followed by 6 (3) and then we are forced to pick 2 (18), and return to 1
(10). Our total setup is 43. Note that we have picked a set up from each intersecting row and column. It
is very unlikely to be optimal because of the large variation of the coefficients of the matrix (ill –
conditioning). The optimum is 28, the worst 92, the initial sequence 50.
Problem No. 6
The individual tasks of a project are shown below. What is the probability of completing the project
before time 12? Explain the reasons for your method (15 points)
Given that we are given three values for each processing time, I will choose PERT. There are only 8
tasks, so we can find the CP by inspection. Using a mean of (best +4*most likely+ worst)/6 and a
variance of ((Worst – Best)/6)^2: (The standard deviation of the critical path is obtained by the square
root of the sum of the variances of the individual tasks). The z values are found by subtracting the
mean from 12 and dividing by the standard deviation
ACGH: 10.33, 0.67, z = 0.5
BFGH: 14.67. 0.85, z = 3.14
ACD: 9.67, 0.82, z = - 2.86
BE: 6.17, 0.6, z = - 9.7
Activity
Predessor
Most
Likely
Best
Worst
Mean
Variance
A
B
C
D
E
F
G
H
A
C
B
B
C, F
G
2
4
3
4
2
3
5
2
2
3
2
3
1
3
4
1
4
6
4
7
3
6
6
3
2.33
4.17
3.00
4.33
2.00
3.50
5.00
2.00
0.11
0.25
0.11
0.44
0.11
0.25
0.11
0.11
The largest z is greater than 3, hence practically 0 chance of achieving 12.
Page 5