In: Operations Management
Paulson is in charge of a project at the local community center.
The center needs to remodel one of the rooms in time for the start
of a new program. Delays in the project mean that the center must
rent other space at a nearby church at additional cost (and Paulson
gets a roasting).
Time and cost data for your project are contained in Table below.
Paulson’s interest is in minimizing the cost of the project to the
community center.
TABLE
DATA FOR THE COMMUNITY CENTER PROJECT
Activity
Normal Time (days)
Normal Cost($)
Crash Time (days)
Crash Cost ($)
Immediate Predecessor(s)
START
0
0
0
0
-
A
10
50
8
150
START
B
4
40
2
200
START
C
7
70
6
160
B
D
2
20
1
50
A, C
E
3
30
3
30
A, C
F
8
80
5
290
B
G
5
50
4
180
D
H
6
60
3
180
E, F
FINISH
0
0
0
0
G, H
a. Using the normal times for each activity, what is the earliest date Paulson can complete the project?
b. Suppose the variable overhead costs are $50 per day for your project. Also, suppose that the center must pay $40 per day for a temporary room on day 15 or beyond. Find the minimum-cost project schedule (best chance that Paulson is not hauled by Church authorities). What is the minimum cost? (3+7 = 10 Marks)
(a)
Project paths | Duration | |||
A-D-G | 17 | |||
A-E-H | 19 | |||
B-C-D-G | 18 | |||
B-C-E-H | 20 | |||
B-F-H | 18 |
The critical path under normal time estimate is B-C-E-H and the completion time is 20 days.
(b)
Activity | Normal Time (days) | Normal Cost($) | Crash Time (days) | Crash Cost ($) |
Slope ($/ day) |
START | 0 | 0 | 0 | 0 | - |
A | 10 | 50 | 8 | 150 | 50 |
B | 4 | 40 | 2 | 200 | 80 |
C | 7 | 70 | 6 | 160 | 90 |
D | 2 | 20 | 1 | 50 | 30 |
E | 3 | 30 | 3 | 30 | - |
F | 8 | 80 | 5 | 290 | 70 |
G | 5 | 50 | 4 | 180 | 130 |
H | 6 | 60 | 3 | 180 | 40 |
FINISH | 0 | 0 | 0 | 0 | - |
Totals | 400 |
Note that the total cost increases in step-5. Hence, Step-4 is the optimal point i.e. the minimum-cost schedule.
So, the minimum-cost schedule is 16 days with a total cost = $1,470