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The PERT use of statistical techniques is confined to a formula that gives a value which can be used as an estimate in a schedule based upon best and worst estimates.

The above gives an example of how a duration can be calculated using PERT based upon a three point analysis. The three point analysis applying to risk and uncertainty is discussed more fully in the 'The Complete Risk Management package'.

The formula is based upon giving three estimates for the duration of a task.

The first is the best (shortest) estimate assuming all goes well, ‘**O**’ above.
The second (middle) estimate is what you realistically expect to happen, ‘**R**’ above.
The third (longest) estimate is what you might think if everything went awry, ‘**P**’ above.

The duration will be (O + 4R + P) / 6 For example,

O = 3 days R = 5 days P = 7 days

Then the duration will be (3 + 4x5 + 7) / 6 = 5 days

The reason for the estimated time being the same as the realistic time is due to the optimistic and pessimistic times being equally 2 days either side of the realistic estimate of 5 days. If say, O = 2 days, R = 6 days and P = 7 days the estimated time would be:

(2 + 4x6 + 7) / 6 = 5.5 days

Whilst this method provides an ‘average’ it does not display any of the advantages of a full risk analysis.

The PERT **standard deviation** is given as:

(P – O) / 6 = (7 – 3) / 6 = 4 / 6 = 0.67

Knowing the standard deviation of each task (of the critical path) you can use this information to calculate the standard deviation for the entire project duration. In a similar manner, knowing the individual standard deviations of a series of non critical tasks can provide valuable information

Let us say we have 5 tasks on the critical path. Their durations and standard deviation data are:

Duration | PERT | Standard | ||||
---|---|---|---|---|---|---|

Task | Best | Realistic | Worst | Duration | Deviation | Variance |

1 | 3 | 6 | 11 | 6.3 | 1.3 | 1.8 |

2 | 6 | 10 | 16 | 10.3 | 1.7 | 2.8 |

3 | 2 | 4 | 7 | 4.2 | 0.8 | 0.7 |

4 | 4 | 9 | 12 | 8.7 | 1.3 | 1.8 |

5 | 5 | 11 | 14 | 10.5 | 1.5 | 2.3 |

Total | 20 | 40 | 60 | 40 | 9.3 |

The variance is the square of the standard deviation.

Total duration = 40.0 Total variance = 9.3 (A) (the Excel spreadsheet rounds this accurately)

Average variance = 1.9 (B) [(A) / 5] The square root of this = 1.4 (C) (average standard deviation)

This can be summarised as:

To find the standard deviation of the total project we look at the critical path, tasks 1 to 5. Calculate the variance as the square of the individual task standard deviations. Find the total of the individual variances = A Find their average over 5 tasks = B Find the square root of this = C This is the standard deviation of the overall project with critical path as tasks 1 to 5.

So for this simple 5 task project.

The overall duration = 40.0 days As there is a standard deviation = 1.4 days

There is:

Probability Duration range 68% 38.6 to 41.4 days 95% 37.2 to 42.8 days 99.7% 35.8 to 44.2 daysRemember

68.27% of all measurements fall within one standard deviation of the mean. 95.45% of all measurements fall within two standard deviations of the mean. 99.73% of all measurements fall within three standard deviations of the mean.

Strictly, the probabilities above refer to repeating the project enough times to collect many values of the total duration. Had we repeated the project 100 times we would get 100 values of the total duration. In this example we are saying that on 68% of the occasion that we carry out the task the result will lie between 38.6 and 41.4 and outside of this range on the other 32% of occasions.

When we carry out a project we in effect carry it out usually **once only** so we have to be careful in saying there is a 68% probability of a
total project duration (for example) lying between two values. This would be the case if we were able to carry out the same project many times.
In effect when someone estimates a task duration value they are doing this on the basis of experience by mentally simulating what is likely to happen if they repeated it many times.

What does this mean in practice? If you only carried out this particular project once then there is always the chance that absolutely everything will go according to plan and you may complete in your minimum duration time = 20 days or it could be a complete disaster and fail completely.

Statistics is all about looking at data over a greater population. Thus, if you were able to carry out the same project 100 times then there is a probability that on 68 occasions you would complete in the range 38.6 to 41.4 days.

If you were able to carry out the same project 1000 times then there is a probability that on 683 (taking 68.27%) occasions you would complete in the range 38.6 to 41.4 days. Also, if you were able to carry out the same project 1000 times then there is a probability that on 997 (taking 99.73%) occasions you would complete in the range 35.8 to 44.2 days.

So, there is an element of judgement in doing the project only once but basing estimates on the experience of individuals.

All things being equal, for the total duration, you can say that there is a probability that it will lie in the following ranges:

68% 38.6 to 41.4 days (32% it will lie outside)

95% 37.2 to 42.8 days (5% it will lie outside)

99.7% 35.8 to 44.2 days (0.3% it will lie outside)

Let us look at the following example:

Duration | PERT | Standard | ||||
---|---|---|---|---|---|---|

Task | Best | Realistic | Worst | Duration | Deviation | Variance |

1 | 4 | 13 | 27 | 13.8 | 3.8 | 14.7 |

2 | 3 | 9 | 19 | 9.7 | 2.7 | 7.1 |

3 | 8 | 17 | 24 | 16.7 | 2.7 | 7.1 |

4 | 2 | 6 | 16 | 7.0 | 2.3 | 5.4 |

5 | 10 | 15 | 25 | 15.8 | 2.5 | 6.3 |

Total | 27 | 60 | 111 | 63.0 | 40.6 |

The variance is the square of the standard deviation.

Total duration = 63.0 Total variance = 40.6 (A)

Average variance = 8.1 (B) [(A) / 5] The square root of this = 2.9 (C)

This can be summarised as:

To find the standard deviation of the total project we look at the critical path, tasks 1 to 5. Calculate the variance as the square of the individual task standard deviations. Find the total of the individual variances = A Find their average over 5 tasks = B Find the square root of this = C This is the standard deviation of the overall project with critical path as tasks 1 to 5.

So for this simple 5 task (non critical) project.

The overall duration = 63.0 days As there is a standard deviation = 2.9 days

There is:

Probability Duration range 68% 60.1 to 65.9 days 95% 57.2 to 68.8 days 99.7% 54.3 to 71.7 daysThe example above shows that this project has a very high probability (99.7%) of completing within 8.7 days of the projected 63.0 days. It also shows a reasonable probability (68%) of completion within the narrower (one standard deviation) of 2.9 days. This information will have an impact on how the Project Manager views managing the tasks.

If the above series of tasks has a total float = 2 days we can consider the above information. As the series of tasks has a 68% probability of completing within 2.9 days it has a 100 – 68 = 32% probability of completing outside of this range. As the total float is only 2 days the Project Manager may consider there is a high risk of these tasks becoming critical. Hence, the Project Manager would focus more attention on this series of tasks.

Risk management is covered in more detail in 'The Complete Risk Management package'

Next we can look at the use of the Cumulative Probability curve.

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