SME
Construction Engineering
CPM scheduling
Use this for “A (4 days) and B (3 days) start a job; C (7 days) follows A, D (3 days) follows A and B, E (2 days) follows C and D. How long is the job, and which activities can slip?”
Table
Activity
d
(days)
ES
(days)
EF
(days)
LS
(days)
LF
(days)
TF
(days)
FF
(days)
A (start)
4
0
4
0
4
0
0
B (start)
3
0
3
5
8
5
1
C (after A)
7
4
11
4
11
0
0
D (after A, B)
3
4
7
8
11
4
4
E (after C, D)
2
11
13
11
13
0
0
This is an example. Type your own values to start a new problem.
dA
Duration of A
example · days
dB
Duration of B
example · days
dC
Duration of C
example · days
dD
Duration of D
example · days
dE
Duration of E
example · days
ESD
Early start of D
worked out (not typed) · days
ESE
Early start of E
worked out (not typed) · days
T
Project duration
calculated · days
TFB
Total float of B
worked out (not typed) · days
TFD
Total float of D
worked out (not typed) · days
Formulas
ESD = max(dA, dB)
4 = max(4, 3)
ESE = max(dA + dC, ESD + dD)
11 = max(4 + 7, 4 + 3)
T
=
ESE
+
dE
13 = 11 + 2
TFD = ESE − dD − ESD
4 = 11 − 3 − 4
TFB = ESE − dD − dB
5 = 11 − 3 − 3
Assumptions
•
Activity on node: A and B start; C follows A; D follows A and B; E follows C and D.
•
Each link is finish-to-start with no lag; each duration is fixed and known.
•
Total float is how long an activity can slip without delaying the project.
•
Activities with zero total float form the critical path.
Step-by-step
WE KNOW
dA = 4 days, dB = 3 days, dC = 7 days, dD = 3 days, dE = 2 days
FIND
early start of D (ESD), early start of E (ESE), project duration (T), total float of D (TFD), total float of B (TFB)
Step 1 · Find early start of D (ESD)
Use ESD = max(dA, dB)
Forward pass: an activity starts when the last one before it ends, so D waits for the later of A and B.
ESD = max(4, 3)
ESD = 4 days
Step 2 · Find early start of E (ESE)
Use ESE = max(dA + dC, ESD + dD)
E waits for both C and D: C ends at dA + dC, D at ESD + dD, and E starts at the later.
ESE = max(4 + 7, 4 + 3)
ESE = 11 days
Step 3 · Find project duration (T)
Use T = ESE + dE
E ends the project, so the project lasts until E finishes. The longest path, A–C–E, is critical: its activities have zero total float.
T
=
11
+
2
T
=
13
days
Step 4 · Find total float of D (TFD)
Use TFD = ESE − dD − ESD
Backward pass: D must finish by the late start of E, which is ESE (E is critical). So LSD = ESE − dD, and TF = LS − ES.
TFD = 11 − 3 − 4
TFD = 4 days
Step 5 · Find total float of B (TFB)
Use TFB = ESE − dD − dB
Only D follows B, so B must finish by LSD = ESE − dD. B starts at 0, so its float is that late finish less dB.
TFB = 11 − 3 − 3
TFB = 5 days
Check
Put the numbers back in. Both sides should match.
4 = max(4, 3) ✓
11 = max(4 + 7, 4 + 3) ✓
13 = 11 + 2 ✓
4 = 11 − 3 − 4 ✓
5 = 11 − 3 − 3 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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