SME
Construction Engineering · Problem type
PERT: chance of finishing by a date
Use this for “The critical path is expected to take 42 days, and its activity variances add to 9 days². What is the chance the job is done in 46 days?”
Chart
X is normal with Tₑ = 42 days and σ = 3 days.
P(X ≤ 46) = 0.9088.
Tₛ = 46 is 1.3333 standard deviations above the mean: z = 1.3333.
This is an example. Type your own values to start a new problem.
σ (days)
=
Σσ² (days²)
z
=
Tₛ (days)
−
Tₑ (days)
σ (days)
P
Probability of finishing by Tₛ
calculated
Formulas
σ
=
√Σσ²
3
=
√
9
z
=
Tₛ − Tₑ
σ
1.3333
=
46 − 42
3
P
=
1
−
Q(
z
)
0.9088 = 1 − Q(1.3333)
Assumptions
•
Each activity has tₑ = (a + 4m + b) ÷ 6 and σ = (b − a) ÷ 6 from three time estimates.
•
Tₑ and Σσ² add up along the critical path; the path’s total is about normal.
•
Other near-critical paths are ignored, so P comes out a little high.
Step-by-step
WE KNOW
Tₑ = 42 days, Σσ² = 9 days², Tₛ = 46 days
FIND
standard deviation of the path (σ), standard normal score (z), probability of finishing by Tₛ (P)
Step 1 · Find standard deviation of the path (σ)
Use σ = √Σσ²
Variances add along the path, standard deviations do not: take the root of the sum.
σ
=
√
9
σ
=
3
days
Step 2 · Find standard normal score (z)
Use z = (Tₛ − Tₑ) ÷ σ
Count how many standard deviations the scheduled date sits past the expected duration.
z
=
46 − 42
3
z
=
4
3
z
=
1.3333
Step 3 · Find probability of finishing by Tₛ (P)
Use P = 1 − Q(z)
The
chance
of
finishing
by
Tₛ
is
the
area
of
the
normal
curve
left
of
z
;
Q(
z
)
is
the
area
right
of
it.
P
=
1
−
Q(1.3333)
P
=
1
−
0.09122
P
=
0.9088
Check
Put the numbers back in. Both sides should match.
3
=
√
9
✓
1.3333
=
46 − 42
3
✓
0.9088 = 1 − Q(1.3333) ✓
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