ops work 2
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School
Liberty University *
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Course
361
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
Pages
2
Uploaded by BrigadierWillpower11048
1. Consider the following project with activities A through I.
Activity
Immediate Predecessor
Duration (weeks)
A
–
6
B
–
1
C
B
1
D
C
6
E
C
3
F
E
1
G
E
4
H
A,F
3
I
D,H,G
2
a. Using the activity-on-arc method, draw the precedence diagram of the process.
b. Using the critical path method learned in class, determine the earliest start, earliest finish, latest start
and latest finish time of each activity. Finally, list the slack of each activity.
After labeling the precedence diagram and performing the critical path method, we obtain the following:
The slack is given in the chart below:
Activity
Slack
A
0
B
0
C
0
D
1
E
0
F
0
G
0
H
0
I
0
c. What are the critical activities and the critical path(s) of the project?
The critical activities
for the project are A, B, C, E, F, G, H and I. The critical paths for the project are then: A-H-I, B-C-E-G-I,
and B-C-E-F-H-I
c. If activity A will take 6 weeks instead of 3 weeks to complete, what is the new project completion time?
Since the activity A has a slack of 3, the project completion time will not change.
d. Suppose now that we are given the following crashing information. For each week that we are able to
reduce the project completion time, we are paid an extra $1000. Using the method learned in class, crash
the project for as many weeks as it is profitable to do so. Clearly show which activities you will crash for
each week, the cost of crashing for each week, and all of the critical paths at the end of each week.
We begin week 1 of crashing with the 3 critical paths: A-H-I, B-C-E-G-I, and B-C-E-F-H-I
B, C, and F can not be crashed. Therefore The only crashable activities are activity A&E which costs
$300 and is less than the $1000 per week extra payment. We therefore crash Activity A&E by 1 week and
bring the project completion time down to 10 weeks. We obtain the following after running the critical path
method.
Activity
Immediate
Predecessor
Normal Duration
(weeks)
Crash Duration
(weeks)
Normal Cost
Crash Cost
Crash cost
per week
A
–
6
2
$400
$600
$200
B
–
1
-
$200
-
-
C
B
1
-
$100
-
-
D
C
6
4
$400
$1200
$400
E
C
3
1
$600
$800
$100
F
E
1
-
$300
-
-
G
E
4
3
$800
$850
$50
H
A,F
3
2
$100
$500
$400
I
D,H,G
2
1
$1400
$2700
$1300
We are now in week 2 of crashing and the critical paths are: A-H-I, B-C-E-G-I, B-C-D-I, B-C-E-F-H-I
Activity B,C,F cannot be crashed, so the crashable options are:
We could crash activity I, which costs $1300
We could crash activity A,D,E, which costs $700
We could crash activity A,D,G,H, which costs $1050
We could crash activity D,E,H, which costs $900
We could crash activity D,G,H, which costs $850
We therefore crash activity A&E&D by 1 week and bring the project completion time down to 9 weeks. We
could obtain the following after running critical path method:
We are now in week 3 of crashing and the critical paths are: A-H-I, B-C-E-G-I, B-C-D-I, B-C-E-F-H-I
Activity B,C,E,F cannot be crashed, so the crashable options are:
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