How do CPM and PERT differ in their time estimates?

CPM uses one deterministic estimate and PERT uses three

CPM uses three estimates and PERT uses one

Both use one estimate but different units

Which technique does time-cost trade-off (crashing) and which computes probability of completion using Z?

CPM does crashing; PERT computes probability

PERT does crashing; CPM computes probability

x has SD 0.67 (variance 0.44) and y has SD 0.50 (variance 0.25). Which is more certain?

y, because its SD and variance are smaller

Neither, because the t_e values differ

2066 Bhadra (old course) · 8 marks

CPM vs PERT; te and variance for x, y

Short answer

CPM uses deterministic single time, while PERT uses three probabilistic times. te = (t_o+4t_m+t_p)/6, σ = (t_p–t_o)/6. For x: te = 6.00, σ = 0.67, variance = 0.44; for y: te = 4.83, σ = 0.50, variance = 0.25. y is more certain because of smaller SD and variance.

What are the differences between CPM and PERT? Find the te (time estimates) for x and y where to, tm, tp are 4, 6, 8 for x and 3, 5, 6 for y. Also find which is more certain by using S.D. and variance.

Difference between CPM and PERT

Basis	CPM	PERT
Time estimate	One (deterministic)	Three: $t_o, t_m, t_p$ (probabilistic)
Nature	Activity oriented	Event oriented
Distribution	Not required	Beta distribution assumed
Suitable for	Repetitive, known work such as construction	New, research-type, uncertain work
Cost	Time-cost trade-off (crashing) is done	Mainly time; cost is not considered
Probability of completion	Not computed	Computed using $Z$ value
Developed by	Du Pont (1957)	US Navy (1958)

Calculation

$t_e = \dfrac{t_o + 4t_m + t_p}{6}$ , $\sigma = \dfrac{t_p - t_o}{6}$ , variance $= \sigma^2$ .

For x ( $4, 6, 8$ ):

\begin{aligned} t_e &= \frac{4 + 24 + 8}{6} = 6.00 \\ \sigma &= \frac{8 - 4}{6} = 0.67, \quad \text{Var} = 0.44 \end{aligned}

For y ( $3, 5, 6$ ):

\begin{aligned} t_e &= \frac{3 + 20 + 6}{6} = 4.83 \\ \sigma &= \frac{6 - 3}{6} = 0.50, \quad \text{Var} = 0.25 \end{aligned}

	$t_e$	$\sigma$	Variance
x	6.00	0.67	0.44
y	4.83	0.50	0.25

Answer: $t_e$ (x) = 6.00 and $t_e$ (y) = 4.83. Activity y has the smaller SD (0.50) and variance (0.25), so y is more certain.

✓ Answer checked by Claude Opus 5.5 (high effort)

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