Early/Late Times

E S_{j} = max (E F_{i}) E F = E S + D L S = L F - D T F = L S - E S

Forward pass: ES of an activity is the max EF of predecessors; EF = ES + duration. Backward pass: LF is the min LS of successors; LS = LF − duration. Total float TF = LS − ES. Critical activities have TF = 0.

$E S, E F$: Early start / finish
$L S, L F$: Late start / finish
$D$: Duration
$T F$: Total float

Step-by-step solved example

Activity D=5, ES=8. This activity’s LF = 16. TF?

Forward pass ES/EF, backward pass LS/LF.

1. EF
EF = 8+5 = 13.
2. LS
LS = 16−5 = 11.
3. TF
TF = LS−ES = 11−8 = 3 (also 16−13=3).
$T F = 3$

Answer: TF = 3

10 practice questions

0/10 correct

1.Critical path activities have TF =

2.EF =

3.LS =

4.Two predecessors finish at 4 and 7. Successor ES =

5.Free float is

6.A dummy has D =

7.If TF = 5 and you delay the activity 7 days the project

8.Backward pass starts from

9.LS < ES would mean

10.Two successors with LS = 12 and 15. Predecessor LF =