Historical productivity data exist but the WBS is only partially decomposed. Which estimating technique and range is most appropriate?

Parametric estimate with wide range and documented assumptions

Three-point estimate with narrow optimistic-pessimistic spread

Bottom-up estimate using only the current partial WBS

Parametric estimate using a single historical data point

The WBS is incomplete for several deliverables yet bottom-up estimating is required. Which sequence enables a reliable estimate?

Decompose then estimate bottom-up

Create flat task list then estimate

Use analogous methods on every deliverable

Estimate bottom-up on high-level deliverables

Deliverables differ in repeatability with metrics available for some. Which approach per deliverable minimizes risk?

Parametric for repeatable, analogous for novel

Analogous estimates for all work packages

Bottom-up on every deliverable regardless

Parametric for novel regulated items

A work package estimate includes development effort of 120 hours. Historical data shows acceptance testing and criteria validation typically require 25% additional effort. What is the total estimate?

145 hours rounded up to the next sprint boundary

120 hours plus a 10% contingency reserve

130 hours using parametric scaling from prior projects

An activity is estimated at optimistic 6 days, most likely 9 days and pessimistic 24 days. Using the beta (PERT) estimate and its standard deviation, within which range should the activity finish about two times in three?

8 to 14 days, the expected value plus or minus one standard deviation

6 to 24 days, the full span between the optimistic and pessimistic estimates

6 to 12 days, the most likely value plus or minus one standard deviation

5 to 17 days, the expected value plus or minus two standard deviations

An activity is estimated at optimistic 10 days, most likely 16 days, and pessimistic 34 days. What is the approximate standard deviation of the PERT estimate, and what does it suggest?

4 days; the wide range signals lower confidence, so the activity needs closer monitoring or a buffer

8 days; dividing the range by three shows how far the actual duration may stray from the expected value

16 days; the variance of the estimate is the spread the team should add on top of the expected duration

2 days; the difference between the expected duration of 18 days and the most likely value of 16 days

Based on the provided network path estimates, identify the path posing the greatest schedule risk and determine its expected duration and standard deviation.

Path 3: 18.0 days expected duration; 5.0 days standard deviation

Path 4: 17.0 days expected duration; 5.44 days standard deviation

Path 2: 18.0 days expected duration; 1.33 days standard deviation

Path 3: 18.0 days expected duration; 25.0 days standard deviation

2.2.5.2

Concept · under Schedule management

PERT, three-point and the other estimating techniques on the PMP®

The PERT formula, or beta three-point estimate, weights the most likely value four times: (O + 4M + P) / 6. The triangular version weights all three points equally: (O + M + P) / 3. Both turn an uncertain activity into an expected duration with a visible spread, and three points are not always the right tool: an analogous, parametric or bottom-up estimate can fit the data better.

Exam: PMP
Domain: Process
Domain weight: 41%
Packages: 9
Updated: Oct 13, 2026

01The beta estimate

Beta (PERT) expected value

E=(O + 4M + P) divided by 6

O: optimistic: the duration if things go well
M: most likely: the duration you would bet on
P: pessimistic: the duration if the known risks land

Worked example

O = 5 days
M = 8 days
P = 17 days

Weight the most likely value: 4 × 8 = 32.
Add the three terms: 5 + 32 + 17 = 54.
Divide by 6: 54 / 6 = 9.

E = 9.0 days

Reading itOne day above the most likely value, because the pessimistic tail (9 days past M) is longer than the optimistic one (3 days).

02Beta or triangular: which comes out higher

The same formulas on two skewed ranges

Estimates (O, M, P)	Beta	Triangular	Higher
5, 8, 17 days: long pessimistic tail	9.0	10.0	Triangular
1, 10, 13 days: long optimistic tail	9.0	8.0	Beta

Beta always sits closer to M than triangular does, so which result is larger depends on which way the range stretches. Neither formula is always the lower one. When the stem does not name a distribution, PERT usually means beta.

03How wide the uncertainty is

Standard deviation of a beta estimate

σ=(P − O) divided by 6

P − O: the full range between the pessimistic and optimistic estimates

Worked example

P = 17 days
O = 5 days

Range: 17 − 5 = 12.
Divide by 6: 12 / 6 = 2.

σ = 2.0 days

Reading itσ is in the same unit as the estimate, here days. It describes how far apart the optimistic and pessimistic values sit, and nothing about which of them is more likely.

04Paths, padding and where the numbers go

Adding up a path

Expected durations along a path simply add. Standard deviations do not: square each one to get the variance, add the variances, then take the square root. Two activities with σ of 3 and 4 days give the path a σ of 5 days, not 7.

Padding is a hidden range

Why does a padded estimate count against you? One number with 20% quietly added on the way up is a reserve nobody can see or manage. A three-point estimate puts the same worry where the sponsor can read it, as a pessimistic value and a spread.

Where the estimates feed in

Expected durations go into the forward and backward pass on the critical path. The weighting idea, probability times outcome, returns in expected monetary value for risks. Estimating as a whole belongs to schedule management.

05Four techniques and what each one needs

Estimating techniques by the data they depend on and where they break

Technique	What it needs	Where it breaks
Analogous	One comparable past project, plus expert judgment to adjust for differences	When the comparison is only superficially similar: same product name, different technology or team
Parametric	A unit rate from history, such as hours per interface or cost per square meter	When the work does not scale in a straight line, or the rate rests on too few past jobs
Bottom-up	Scope decomposed into work packages, each estimated by the people who will do it	When the breakdown is unfinished; the total can only be as complete as the list beneath it
Three-point	Optimistic, most likely and pessimistic values from people who know the work	When the three numbers are guesses around a target date; a narrow range on risky work is false comfort

06Which estimate, which number?

Two kinds of item here: ones you calculate, and ones where you judge which technique the available data supports.

Work packages 9

WP 01Historical productivity data exist but the WBS is only partially decomposed. Which estimating technique and range is most appropriate?
Debrief answer to open the notes
1. ACorrect: historical productivity data supports a parametric estimate, and a wide, documented range reflects the incomplete WBS.
2. BA narrow three-point spread implies more certainty than a partially decomposed WBS provides.
3. CA bottom-up estimate on a partial WBS misses work that has not yet been decomposed.
4. DA single data point gives a weak statistical basis for a parametric model.
More notes below ↓
WP 02The WBS is incomplete for several deliverables yet bottom-up estimating is required. Which sequence enables a reliable estimate?
Debrief answer to open the notes
1. AA flat task list lacks the deliverable hierarchy that makes bottom-up estimates complete.
2. BAnalogous estimating is top-down, which contradicts the bottom-up requirement.
3. CEstimating bottom-up on high-level deliverables is a contradiction, since there is nothing below them to sum.
4. DCorrect: bottom-up estimates sum work-package estimates, so the WBS must be decomposed first.
More notes below ↓
WP 03Deliverables differ in repeatability with metrics available for some. Which approach per deliverable minimizes risk?
Debrief answer to open the notes
1. AAnalogous estimates everywhere waste the metrics available for repeatable work.
2. BCorrect: repeatable work with metrics suits parametric estimating, while novel work without data suits analogous comparison.
3. CBottom-up everywhere is costly and may be impossible where detail is lacking.
4. DNovel items lack the historical data that parametric estimating requires.
More notes below ↓
WP 04A work package estimate includes development effort of 120 hours. Historical data shows acceptance testing and criteria validation typically require 25% additional effort. What is the total estimate?
Debrief answer to open the notes
1. A145 hours adds only 25 hours, about 21%, and sprint rounding is a scheduling choice, not part of the estimate.
2. BCorrect: parametric scaling applies the historical ratio, 120 × 1.25 = 150 hours.
3. CA 10% contingency is not the historically observed 25% acceptance effort.
4. D130 hours reflects about 8% added effort, not the 25% the data shows.
More notes below ↓
WP 05A project manager is planning a technology implementation project. The detailed requirements for one component are still uncertain. Which approach would be most appropriate for estimating this component?
Debrief answer to open the notes
1. AAnalogous estimating gives a quick top-down figure, but it does not express the range of uncertainty in this component.
2. BCorrect: three-point estimating uses optimistic, most likely, and pessimistic values, for example (O + 4M + P) / 6, to reflect the uncertainty.
3. CBottom-up estimating requires detailed, decomposed requirements that this component does not yet have.
4. DParametric estimating needs reliable historical rates, which uncertain requirements rarely provide.
More notes below ↓
WP 06A project manager asks a subject matter expert for a duration estimate for a data migration activity. The expert gives an optimistic estimate of 9 days, a most likely estimate of 12 days, and a pessimistic estimate of 21 days. Using a beta (PERT) distribution, what is the expected duration?
Debrief answer to open the notes
1. AThis is the most likely estimate alone; a three-point estimate also accounts for the optimistic and pessimistic values.
2. BCorrect: (9 + 4 × 12 + 21) / 6 = 78 / 6 = 13 days.
3. CThis is the triangular mean, (9 + 12 + 21) / 3, which gives the most likely value no extra weight.
4. DThis is the midpoint of the optimistic and pessimistic values, (9 + 21) / 2, which ignores the most likely estimate.
More notes below ↓
WP 07An activity is estimated at optimistic 6 days, most likely 9 days and pessimistic 24 days. Using the beta (PERT) estimate and its standard deviation, within which range should the activity finish about two times in three?
Debrief answer to open the notes
1. AThe full range from optimistic to pessimistic covers nearly every outcome, not the roughly two-in-three band.
2. BOne standard deviation goes around the expected value of 11 days, not around the most likely value of 9.
3. CCorrect: E = (6 + 36 + 24) / 6 = 11 days and σ = (24 − 6) / 6 = 3 days, so about two times in three the finish falls within 11 ± 3, between 8 and 14 days.
4. DPlus or minus two standard deviations covers about 95% of outcomes; the two-in-three band is one σ either side of the mean.
More notes below ↓
WP 08An activity is estimated at optimistic 10 days, most likely 16 days, and pessimistic 34 days. What is the approximate standard deviation of the PERT estimate, and what does it suggest?
Debrief answer to open the notes
1. ACorrect: σ ≈ (P − O) / 6 = (34 − 10) / 6 = 4 days; a wider range means more uncertainty around the 18-day expected duration.
2. BDividing the range by three, (34 − 10) / 3, is not the PERT standard deviation; the divisor is six.
3. C16 is the variance (σ²), not the standard deviation, and it is in squared days, so it cannot be added to a duration.
4. DThe gap between the mean and the most likely value shows skew, not the standard deviation of the estimate.
More notes below ↓
WP 09Based on the provided network path estimates, identify the path posing the greatest schedule risk and determine its expected duration and standard deviation.
Exhibit for WP 09
Network Path Optimistic (O) Most Likely (M) Pessimistic (P)
Path 1 12 15 24
Path 2 14 18 22
Path 3 9 15 39
Path 4 12 16 26
Debrief answer to open the notes
1. ACorrect: Path 3 TE = (9 + 4x15 + 39) / 6 = 18.0; SD = (39 - 9) / 6 = 5.0, the widest spread of the four paths.
2. BPath 4 TE is (12 + 64 + 26) / 6 = 17.0, but its SD is (26 - 12) / 6 = 2.33, not 5.44.
3. CPath 2 has the same 18.0-day mean but an SD of only 1.33, so it carries the least schedule risk.
4. D25.0 is Path 3's variance (5.0 squared), not its standard deviation.
More notes below ↓

Network Path	Optimistic (O)	Most Likely (M)	Pessimistic (P)
Path 1	12	15	24
Path 2	14	18	22
Path 3	9	15	39
Path 4	12	16	26

Open a row to answer. Keys: 1–4 or A–D answer, N opens the next row.

A single number for a risky activity is still a range. It just hides the uncertainty from whoever reads it.