In: Operations Management
Consider the following data for a project to install a new server at the Northland Pines High School.
Activity |
Activity Time (days) |
Immediate Predecessor(s) |
A |
2 |
— |
B |
4 |
A |
C |
5 |
A |
D |
2 |
B |
E |
1 |
B |
F |
8 |
B, C |
G |
3 |
D, E |
H |
5 |
F |
I |
4 |
F |
J |
7 |
G, H, I |
(a)
(b)
The critical path of a project is composed of activities with zero total slack. The critical path(s) is/ are the longest path(s) in a project network. The slacks for each activity can be computed by carrying out forward pass and backward pass in the project network.
Forward Pass:
ES of the starting activities = 0
ES of all other activities = Max. (EF of their immediate
predecessors)
EF of an activity = Its ES + Its duration
Backward Pass:
LF of ending activities = Max. (All the EFs)
LF of all other activities = Min. (LS of their immediate
successors)
LS of an activity = Its LF - Its duration
Total slack (or, float) = LF - EF for each activity.
Activity | Durations | ES | EF | LS | LF | Slack |
A | 2 | 0 | 2 | 0 | 2 | 0 |
B | 4 | 2 | 6 | 3 | 7 | 1 |
C | 5 | 2 | 7 | 2 | 7 | 0 |
D | 2 | 6 | 8 | 15 | 17 | 9 |
E | 1 | 6 | 7 | 16 | 17 | 10 |
F | 8 | 7 | 15 | 7 | 15 | 0 |
G | 3 | 8 | 11 | 17 | 20 | 9 |
H | 5 | 15 | 20 | 15 | 20 | 0 |
I | 4 | 15 | 19 | 16 | 20 | 1 |
J | 7 | 20 | 27 | 20 | 27 | 0 |
So, the critical path is A-C-F-H-J.with a duration of 27 days.
(c)
As we have already computed in part (b), the slack of G, H, and I are 9, 0, and 1 day respectively.