Activity |
Predecessor |
Duration (weeks) |
A |
None |
2 |
B |
A |
4 |
C |
A |
3 |
D |
A |
2 |
E |
B |
3 |
F |
C |
6 |
G |
C, D |
5 |
H |
E, F |
6 |
I |
G |
5 |
J |
H, I |
5 |
Activities B and H can be shortened by a minimum of 2 time units. Which activity would you shorten to reduce the project duration of 2 weeks. Why? (be specific)
ACTIVITY |
DURATION |
ES |
EF |
LS |
LF |
SLACK |
A |
2 |
0 |
2 |
0 |
2 |
0 |
B |
4 |
2 |
6 |
4 |
8 |
2 |
C |
3 |
2 |
5 |
2 |
5 |
0 |
D |
2 |
2 |
4 |
5 |
7 |
3 |
E |
3 |
6 |
9 |
8 |
11 |
2 |
F |
6 |
5 |
11 |
5 |
11 |
0 |
G |
5 |
5 |
10 |
7 |
12 |
2 |
H |
6 |
11 |
17 |
11 |
17 |
0 |
I |
5 |
10 |
15 |
12 |
17 |
2 |
J |
5 |
17 |
22 |
17 |
22 |
0 |
FORWARD PASS
1. Go from left to right in a network.
2. ES = maximum of the EF value from all the direct predecessors.
3. EF = ES + Duration.
BACKWARD PASS
1. Go from right to left in a network.
2. LF = minimum of the LS value from all the direct predecessors.
3. LS = LF - Duration.
CRITICAL PATH
1. Total slack of all activities should be 0.
2. The total duration of all the activities should be the highest.
3. Slack = LF - EF or LS - ES
CRITICAL PATH & DURATION
A + C + F + H + J = 22
SINCE B IS NOT A CRITICAL ACTIVITY, WE WILL SHORTEN ACTIVITY H BY 2 PERIODS TO GET THE NECESSARY CRASHING, BECAUSE B HAS A SLACK OF 2 AND WOULD INCUR ADDITIONAL COST WITHOUT EVEN AFFECTING THE CRITICAL PATH. a CRITICAL ACTIVITY, ON THE OTHER HAND, WOULD ENABLE US TO SHORTEN THE ENTIRE PROJECT BY 2 PERIODS.
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