Network Diagram -
Hence, (d) is the correct answer
Network paths and their duration are -
ABCDH --> 3+4+4+3+4 = 18
ABCEH --> 3+4+4+7+4 = 22
ABCFGH --> 3+4+4+7+4+4 = 26
Hence, critical path is ABCFGH and project duration = 26 days
(c)
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 -
Activity | Precedes | Activity Time | ES | EF | LS | LF | Slack |
A | - | 3 | 0 | 3 | 0 | 3 | 0 |
B | A | 4 | 3 | 7 | 3 | 7 | 0 |
C | B | 4 | 7 | 11 | 7 | 11 | 0 |
D | C | 3 | 11 | 14 | 19 | 22 | 8 |
E | C | 7 | 11 | 18 | 15 | 22 | 4 |
F | C | 7 | 11 | 18 | 11 | 18 | 0 |
G | F | 4 | 18 | 22 | 18 | 22 | 0 |
H | D, E, G | 4 | 22 | 26 | 22 | 26 | 0 |
Hence, slack for D = 8 and slack for E = 4 days
Slack for all the other activities = 0 days
Slack for activities on critical path is always 0
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