In: Finance
Given the following information, you are to determine the following
Task |
Predecessor |
Duration |
A |
None |
5 |
B |
A |
8 |
C |
A |
15 |
D |
A |
10 |
E |
A |
10 |
F |
B |
10 |
G |
C, D |
12 |
H |
E, |
15 |
I |
G, H |
7 |
J |
F,G,I |
10 |
K |
J |
12 |
Network Diagram -
Earliest Start (ES) = the earliest time when the activity can begin
Earliest Finish (EF) = ES + Activity Duration
Latest Finish (LF) = the latest time by whenthe activity can be finished without changing the project completion duration
Latest Start (LS) = LF - Activity Duration
Slack = LS-ES = LF-EF
Computing using the above gives us the below -
Earliest | Earliest | Latest | Latest | ||||
Activity | Predecessor | Duration | Start Time | Finish Time | Start Time | Finish Time | Slack |
A | - | 5 | 0 | 5 | 0 | 5 | 0 |
B | A | 8 | 5 | 13 | 21 | 29 | 16 |
C | A | 15 | 5 | 20 | 5 | 20 | 0 |
D | A | 10 | 5 | 15 | 10 | 20 | 5 |
E | A | 10 | 5 | 15 | 7 | 17 | 2 |
F | B | 10 | 13 | 23 | 29 | 39 | 16 |
G | C, D | 12 | 20 | 32 | 20 | 32 | 0 |
H | E | 15 | 15 | 30 | 17 | 32 | 2 |
I | G, H | 7 | 32 | 39 | 32 | 39 | 0 |
J | F, G, I | 10 | 39 | 49 | 39 | 49 | 0 |
K | J | 12 | 49 | 61 | 49 | 61 | 0 |
Possible Network Paths and their Duration are -
ABFJK --> 45
ACGJK --> 54
ACGIJK --> 61
ADGJK --> 49
ADGIJK --> 56
AEHIJK --> 59
Hence, Project duration is the duration of the longest path = 61
Critical path is the path with longest duration = ACGIJK