3-Points Estimating

last edited by: Brian Anderson on Jul 15, 2024 12:59 PM

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Contents
1 Application
2 Example
3 Other Estimating Techniques
4 References
5 Tools

3-Points is a technique used by professionals in estimating. This technique use the three figures that are produced initially for every distribution that is required, based on prior experience or best-guesses:

The first is a most likely (M)/best guess (BG) which is the average amount of work the task might take if the team member performed it 100 times. The second estimate is the pessimistic (P) estimate which is the amount of work the task might take if the negative factors they identified do occur. The third estimate is the optimistic (O) estimate which is the amount of work the task might take if the positive risks they identified do occur.

Application

Two popular formula:

1. Triangular distribution:

Triangular Distribution: E = (o + m + p ) / 3

where E is Estimate; o = optimistic estimate; p = pessimistic estimate; m = most likely estimate

2. Beta (or PERT):

Beta Distribution (PERT): E = (o + 4m + p ) / 6

The beta distribution is a weighted average in which more weight is given to the most likely estimate. This alteration to the formula and placing more weight on the most likely estimate is made to increase the accuracy of the estimate by making it follow the Normal Distribution shape. Hence, in most of the cases, the Beta (PERT) distribution has been proven to be more accurate than the 3-Point triangular estimation.

By using beta distribution you can determine the level of certainty of this prediction The variance is obtained by the difference between the pessimistic and the optimistic forecast divided by six squared. The variance is the square of the Standard deviation.

Example

For Activity A:

o = 4 hours , m = 8 hours , p = 16 hours

Triangular Distribution:

E = (4 + 8 + 16 ) / 3

E = 28 / 3

E = 9.3 hours

Beta Distribution (PERT):

E = (4 + 4(8) + 16) / 6

E = 52 / 6

E = 8.7 hours

The Standard Deviation is a measure of variability from the mean and is defined as (p - o)/6 so in the example above

S.D = (16 - 4 )/6 = 12/6 = 2 hours

This means that using the normal distribution function where there is a 68% of the probability of one SD range from the mean It means that 68% chance of the schedule having a duration of 8.7+/-2 which is 6.7 to 10.7 hours