Consider a project to produce custom door moldings for GMC Acadia cross-over vehicles, described in the following table.
Activity Activity Time (weeks) Immediate
Predecessor(s)
START 0 -
A 2 START
B 6 START
C 1 B
D 2 A
E 8 A, B
F 8 D, E
G 2 C, E
FINISH 0 F, G
You are interested in completing your project as soon as possible. You have only one option. Suppose you could assign Employee A, currently assigned to Activity G, to help Employee B, currently assigned to Activity F. Each week that Employee A helps Employee B will result in Activity G increasing its time by one week and Activity F reducing its time by one week. How many weeks should Employee A work on Activity F?
Employee A should work. weeks on Activity F and the project's duration will be reduced t weeks.
(Enter
your responses as
integers.)
ACTIVITY |
DURATION |
ES |
EF |
LS |
LF |
SLACK |
CRITICAL |
A |
2 |
0 |
2 |
4 |
6 |
6 - 2 = 4 |
|
B |
6 |
0 |
6 |
0 |
6 |
6 - 6 = 0 |
YES |
C |
1 |
6 |
7 |
19 |
20 |
20 - 7 = 13 |
|
D |
2 |
2 |
4 |
12 |
14 |
14 - 4 = 10 |
|
E |
8 |
6 |
14 |
6 |
14 |
14 - 14 = 0 |
YES |
F |
8 |
14 |
22 |
14 |
22 |
22 - 22 = 0 |
YES |
G |
2 |
14 |
16 |
20 |
22 |
22 - 16 = 6 |
ES = MAX(EF OF ALL PREDECESSORS); 0 FOR FIRST ACTIVITY
EF = ES + DURATION
LF = MIN(LS OF ALL SUCCESSOR ACTIVITIES); MAX EF AS LF FOR LAST ACTIVITY
LS = LF - DURATION
SLACK = LF- EF
CRITICAL PATH = LONGEST PATH WITH 0 SLACK:
CRITICAL PATH = B---E---F
DURATION OF PROJECT = 22
2. A SHOULD WORK ON F FOR 3 PERIODS FOR THE PROJECT TO HAVE THE LOWEST POSSIBLE SCHEDULE
ACTIVITY |
DURATION |
ES |
EF |
LS |
LF |
SLACK |
CRITICAL |
A |
2 |
0 |
2 |
4 |
6 |
6 - 2 = 4 |
|
B |
6 |
0 |
6 |
0 |
6 |
6 - 6 = 0 |
YES |
C |
1 |
6 |
7 |
13 |
14 |
14 - 7 = 7 |
|
D |
2 |
2 |
4 |
12 |
14 |
14 - 4 = 10 |
|
E |
8 |
6 |
14 |
6 |
14 |
14 - 14 = 0 |
YES |
F |
5 |
14 |
19 |
14 |
19 |
19 - 19 = 0 |
YES |
G |
5 |
14 |
19 |
14 |
19 |
19 - 19 = 0 |
YES |
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