Critical path analysis

A-level Further Maths (9FM0) · Decision 1

ORIGINAL

Forward pass takes the largest, backward pass takes the smallest — then float picks out the critical activities.

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Question

A project consists of the activities below, with durations in days.

A

(3) and

B

(5) have no predecessors.

C

(4) and

D

(5) depend on

A

E

(2) depends on

B

and

C

F

(4) depends on

D

and

E

G

(3) depends on

E

. Find the minimum completion time, the critical path and the total float of each non-critical activity.

Every step worked, with the reasoning.

1
Earliest starts: $A = 0$ , $B = 0$ , $C = 0 + 3 = 3$ , $D = 3$ , $E = \max(0 + 5,\ 3 + 4) = 7$
Forward pass: an activity starts only when all of its predecessors have finished, so take the largest finish time.
2
$F = \max(3 + 5,\ 7 + 2) = 9$ , $G = 7 + 2 = 9$ ; minimum completion time $= \max(9 + 4,\ 9 + 3) = 13$ days
Continue the forward pass to the end of the project.
3
Latest finishes: $F = 13$ , $G = 13$ , $D = 13 - 4 = 9$ , $E = \min(13 - 4,\ 13 - 3) = 9$
Backward pass: an activity must finish by the latest start of every successor, so take the smallest.
4
$C = 9 - 2 = 7$ , $B = 9 - 2 = 7$ , $A = \min(7 - 4,\ 9 - 5) = 3$
$A$ has two successors, $C$ (latest start 3) and $D$ (latest start 4).
5
Total float $=$ latest finish $-$ duration $-$ earliest start: $B = 7 - 5 - 0 = 2$ , $D = 9 - 5 - 3 = 1$ , $G = 13 - 3 - 9 = 1$ ; $A$ , $C$ , $E$ , $F$ each have float $0$ .
Check the zero floats: $A = 3 - 3 - 0$ , $C = 7 - 4 - 3$ , $E = 9 - 2 - 7$ , $F = 13 - 4 - 9$ .
6
Critical path: $A\,C\,E\,F$ , of length $3 + 4 + 2 + 4 = 13$ days
The zero-float activities form an unbroken chain from start to finish whose length equals the completion time.

Answer: 13 days; critical path $ACEF$ ; total floats $B = 2$ , $D = 1$ , $G = 1$