Program Evaluation and Review Technique

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PERT network chart for a seven-month project with five milestones (10 through 50) and six activities (A through F).

The Program Evaluation and Review Technique commonly abbreviated PERT is a model for project management invented by Booz Allen Hamilton, Inc. under contract to the United States Department of Defense's US Navy Special Projects Office in 1958 as part of the Polaris mobile submarine-launched ballistic missile project. This project was a direct response to the Sputnik crisis.

PERT is basically a method to analyze the tasks involved in completing a given project, especially the time needed to complete each task, and identifying the minimum time needed to complete the total project.

PERT was developed in the 1950’s, primarily to simplify the planning and scheduling of large and complex projects. It was able to incorporate uncertainty by making it possible to schedule a project not knowing precisely the details and durations of all the activities. It is more of an event-oriented technique rather than start- and completion-oriented, and is used more in R&D-type projects where not cost but time is a major factor.

This project model was the first of its kind, a revival for scientific management, founded in Fordism and Taylorism. Though every company now has its own "project model" of some kind, they all resemble PERT in some respect. Only DuPont corporation's critical path method was invented at roughly the same time as PERT.

The most famous part of PERT is the "PERT Networks", charts of timelines that interconnect. PERT is intended for very large-scale, one-time, complex, non-routine projects.

1 PERT terminology and conventions
- 1.1 Conventions
- 1.2 Terminology
2 Implementing PERT
3 See also
4 External links

[edit] PERT terminology and conventions

[edit] Conventions

A PERT chart is a tool that facilitates decision making; The first draft of a PERT chart will number its events sequentially in 10s (10, 20, 30, etc.) to allow the later insertion of additional events.
Two consecutive events in a PERT chart are linked by activities, which are conventionally represented as arrows in the diagram above.
The events are presented in a logical sequence and no activity can commence until its immediately preceding event is completed.
The planner decides which milestones should be PERT events and also decides their “proper” sequence.
A PERT chart may have multiple pages with many sub-tasks.
PERT Charts are one of the tools used in the Earned Value Management Technique EVMT, used by many corporations today to track Earned Value.

[edit] Terminology

A PERT event: is a point that marks the start or completion of one (or more) tasks. It consumes no time, and uses no resources. It marks the completion of one (or more) tasks, and is not “reached” until all of the activities leading to that event have been completed.
A predecessor event: an event (or events) that immediately precedes some other event without any other events intervening. It may be the consequence of more than one activity.
A successor event: an event (or events) that immediately follows some other event without any other events intervening. It may be the consequence of more than one activity.
A PERT activity: is the actual performance of a task. It consumes time, it requires resources (such as labour, materials, space, machinery), and it can be understood as representing the time, effort, and resources required to move from one event to another. A PERT activity cannot be completed until the event preceding it has occurred.
Optimistic time (O): the minimum possible time required to accomplish a task, assuming everything proceeds better than is normally expected
Pessimistic time (P): the maximum possible time required to accomplish a task, assuming everything goes wrong (but excluding major catastrophes).
Most likely time (M): the best estimate of the time required to accomplish a task, assuming everything proceeds as normal.
Expected time (T_E): the best estimate of the time required to accomplish a task, assuming everything proceeds as normal (the implication being that the expected time is the average time the task would require if the task were repeated on a number of occasions over an extended period of time).

T_E = (O + 4M + P) ÷ 6

Critical Path: the longest possible continuous pathway taken from the initial event to the terminal event. It determines the total calendar time required for the project; and, therefore, any time delays along the critical path will delay the reaching of the terminal event by at least the same amount.
Lead time: the time by which a predecessor event must be completed in order to allow sufficient time for the activities that must elapse before a specific PERT event is reached to be completed.
Lag time: the earliest time by which a successor event can follow a specific PERT event.
Slack: the slack of an event is a measure of the excess time and resources available in achieving this event. Positive slack(+) would indicate ahead of schedule; negative slack would indicate behind schedule; and zero slack would indicate on schedule.

[edit] Implementing PERT

The first step to scheduling the project is to determine the tasks that the project requires and the order in which they must be completed. The order may be easy to record for some tasks (i.e. When building a house, the land must be graded before the foundation can be laid) while difficult for others (There are two areas that need to be graded, but there are only enough bulldozers to do one). Additionally, the time estimates usually reflect the normal, non-rushed time. Many times, the time required to execute the task can be reduced for an additional cost or a reduction in the quality.

In the following example there are seven tasks, labeled a through g. Some tasks can be done concurrently (a & b) while others cannot be done until their predecessor task is complete (c cannot begin until a is complete). Additionally, each task has three time estimates: the optimistic time estimate (a), the most likely or normal time estimate (m), and the pessimistic time estimate (b). The expected time (T_E) is computed using the formula (a + 4m + b)/6.

Activity	Predecessor	Opt. a	Norm. m	Pess. b	T_E (a + 4m + b)/6
a	--	2	4	6	4.00
b	--	3	5	9	5.33
c	a	4	5	7	5.17
d	a	4	6	10	6.33
e	b, c	4	5	7	5.17
f	d	3	4	8	4.50
g	e	3	5	8	5.17

Note: All times listed are in work days (Mon - Fri, 8 _A.M. to 5 _P.M. with a one hour lunch break).

Once this step is complete, one can draw a Gantt chart or a network diagram (project management).

A Gantt chart created using Microsoft Project (MSP). Note (1) the critical path is in red, (2) the slack is the black lines connected to non-critical activities, (3) when using MSP, you must use the task ID when labeling predecessor activities, and (4) since Saturday and Sunday are not work days (as described above) some bars on the Gantt chart are longer if they cut through a weekend.

A Gantt chart created using OmniPlan. Note (1) the critical path is highlighted, (2) the slack is not specifically indicated on task 5 (d), though it can b observed on tasks 3 and 7 (b and f), (3) when using OmniPlan, you may use the GUI to easily link dependencies, or you may enter them by reference to task ID, and (4) since weekends are indicated by a thin vertical line, and take up no additional space on the work calendar, bars on the Gantt chart are not longer or shorter when they do or don't carry over a weekend.

A network diagram can be created by hand or by using software such as Microsoft Project. There are two types of network diagrams, activity on arrow (AOA) and activity on node (AON). Activity on node diagrams are generally easier to create and interpret. To create an AON diagram, it is recommended (but not necessary) to start with a node named start. This "activity" has a duration of zero (0). Then you draw each activity that does not have a predecessor activity (a and b in this example) and connect them with an arrow. Next, since both c and d list a as a predecessor activity, their nodes are drawn with arrows coming from a. Activity e is listed with b and c as predecessor activities, so node e is drawn with arrows coming from both b and c, signifing that e cannot begin until both b and c have been completed. Activity f has d as a predecessor activity, so an arrow is drawn connecting the activities. Likewise, and arrow is drawn from e to g. Since there are no activities that come after f or g, it is recommended (but again not necessary) to connect them to a node labeled finish.

A network diagram created using Microsoft Project (MSP). Note the critical path is in red.

A node like this one (from Microsoft Visio) can be used to display the activity name, duration, ES, EF, LS, LF, and slack.

By itself, the network diagram pictured above does not give much more information than a Gantt chart; however, it can be expanded to display more information. The most common information shown is:

The activity name
The normal duration time
The early start time (ES)
The early finish time (EF)
The late start time (LS)
The late finish time (LF)
The slack

In order to determine this information it is assumed that the activities and normal duration times are given. The first step is to determine the ES and EF. The ES is defined as the maximum EF of all predecessor activities, unless the activity in question is the first activity, which the ES is zero (0). The EF is the ES plus the task duration (EF = ES + duration).

The ES for start is zero since it is the first activiy. Since the duration is zero, the EF is also zero. This EF is used as the ES for a and b.
The ES for a is zero. The duration (4 work days) is added to the ES to get an EF of four. This EF is used as the ES for c and d.
The ES for b is zero. The duration (5.33 work days) is added to the ES to get an EF of 5.33.
The ES for c is four. The duration (5.17 work days) is added to the ES to get an EF of 9.17.
The ES for d is four. The duration (6.33 work days) is added to the ES to get an EF of 10.33. This EF is used as the ES for f.
The ES for e is the greatest EF of its predecessor activities (b and c). Since b has an EF of 5.33 and c has an EF of 9.17, the ES of e is 9.17. The duration (5.17 work days) is added to the ES to get an EF of 14.34. This EF is used as the ES for g.
The ES for f is 10.33. The duration (4.5 work days) is added to the ES to get an EF of 14.83.
The ES for g is 14.34. The duration (5.17 work days) is added to the ES to get an EF of 19.51.
The ES for finish is the greatest EF of its predecessor activities (f and g). Since f has an EF of 14.83 and g has an EF of 19.51, the ES of finish is 19.51. Finish is a milestone (and therefore has a duration of zero), so the EF is also 19.51.

Barring any unforseen events, the project should take 19.51 work days to complete. The next step is to determine the late start (LS) and late finish (LF) of each activity. This will eventually show if there are activities that have slack. The LF is defined as the minimum LS of all successor activities, unless the activity is the last activity, for which the LF equals the EF. The LS is the LF minus the task duration (LS = LF - duration).

The LF for finish is equal to the EF (19.51 work days) since it is the last activity in the project. Since the duration is zero, the LS is also 19.51 work days. This will be used as the LF for f and g.
The LF for g is 19.51 work days. The duration (5.17 work days) is subtracted from the LF to get a LS of 14.34 work days. This will be used as the LF for e.
The LF for f is 19.51 work days. The duration (4.5 work days) is subtracted from the LF to get a LS of 15.01 work days. This will be used as the LF for d.
The LF for e is 14.34 work days. The duration (5.17 work days) is subtracted from the LF to get a LS of 9.17 work days. This will be used as the LF for b and c.
The LF for d is 15.01 work days. The duration (6.33 work days) is subtracted from the LF to get a LS of 8.68 work days.
The LF for c is 9.17 work days. The duration (5.17 work days) is subtracted from the LF to get a LS of 4 work days.
The LF for b is 9.17 work days. The duration (5.33 work days) is subtracted from the LF to get a LS of 3.84 work days.
The LF for a is the minimum LS of its successor activities. Since c has a LS of 4 work days and d has a LS of 8.68 work days, the LF for a is 4 work days. The duration (4 work days) is subtracted from the LF to get a LS of 0 work days.
The LF for start is the minimum LS of its successor activities. Since a has a LS of 0 work days and b has a LS of 3.84 work days, the LS is 0 work days.

The next step is to determine the critical path and if any activities have slack. The critical path is the path that takes the longest to complete. To determine the path times, add the task durations for all available paths. Activities that have slack can be delayed without changing the overall time of the project. Slack is computed in one of two ways, slack = LF - EF or slack = LS - ES. Activities that are on the critial path have a slack of zero (0).

The duration of path adf is 14.83 work days.
The duration of path aceg is 19.51 work days.
The duration of path beg is 15.67 work days.

The critical path is aceg and the critical time is 19.51 work days. It is important to note that there can be more than one critical path (in a project more complex than this example) or the critical path can change. For example, let's say that activities d and f take their pessimistic (b) times to complete instead of their expected (T_E) times. The critical path is now adf and the critical time is 22 work days. On the other hand, if activity c can be crashed to one work day, the path time for aceg is reduced to 15.34 work days, which is slightly less than the time of the new critical path, beg (15.67 work days).

Assuming these scenarios do not happen, the slack for each activity can now be determined.

Start and finish are milestones and by definition have no duration, therefore they can have no slack (0 work days).
The activities on the critical path by definition have a slack of zero; however, it is always a good idea to check the math anyway when drawing by hand.
- LF_a - EF_a = 4 - 4 = 0
- LF_c - EF_c = 9.17 - 9.17 = 0
- LF_e - EF_e = 14.34 - 14.34 = 0
- LF_g - EF_g = 19.51 - 19.51 = 0
Activity b has a LF of 9.17 and a EF of 5.33, so the slack is 3.84 work days.
Activity d has a LF of 15.01 and a EF of 10.33, so the slack is 4.68 work days.
Activity f has a LF of 19.51 and a EF of 14.83, so the slack is 3.84 work days.

Therefore, activity b can be delayed almost 4 work days without delaying the project. Likewise, activity d or activity f can be delayed 4.68 work days without delaying the project (alternatively, d and f can be delayed 2.34 work days each).