A directed graph can be used to represent an activity that consists of many smaller activities that all play a crucial role in forming the final product.

An activity is described as critical (**critical activity**), if it is delayed it will also delay the entire project.

Critical activities form part of the **critical path**. The critical path is the **LONGEST** path is a directed graph. An example of a critical path graph is shown below.

*Essential Further Mathematics 4ed 2012 (Figure 24.6)*

**NOTE:** Even though the critical path is the longest path, it represents the minimal time taken to complete the entire project.

In order to find the critical path there are two new terminologies that you must understand before we can continue.

This represents the earliest time an activity can start. When finding ESTs you are adding.

This represents the latest time an activity can be left for before it delays the entire project. You can find the LSTs by subtracting.

This is the time that represents that an activity can start later than its EST. It can be found by subtracting the LST from EST for each activity. **NOTE:** Activities on the critical path should have a slack time of 0.

The following diagrams have been obtained from** VCAA Further Mathematics 2013 Exam 2 (p.34)** and have been adapted to fit the purpose needed.

The diagram above shows the ESTs for each activity. The process to find the ESTs is as follows:

- Draw boxes and split them into two. Place them
**in front**of each arrow as shown in red. - The ESTs for the activities at the start is always 0.
- To find the EST for the other activities
**add**the EST from the previous activity with the number on the edges of each activity. For example the EST of activity F is 5, because the EST of A is 0 and the weighting for activity A is 5 (5 + 0 = 5). - Continue that process for the other activities until you get to the finish.
- If there are one or more activities feeding into another activity such as for activity J whose predecessors are activities B, D and H, then choose the
**largest values**out of those three activities. For example, for B will give J an EST of 7, D will give an EST of 6 and H will give an EST of 15. Therefore, we would choose activity H to form the EST of J.

The above diagram shows the EST and LST (**Right box**) for each activity. The process to find the LSTs is as follows:

- The EST for the finishing time is 37. Therefore the LST will also be 37.
- Using 37, work your way
**backwards**to fill in the boxes on the right for each activity by**subtraction**. For example, for activity M its LST will be 37 – 8 = 29. - Repeat Step 2 until you end up back at the start point.
- If you encounter something like activity J and G feeding into (look from backwards i.e. right to left), activities B, D and H then out of the two you must pick the number that will give the
**smallest**LST for each activity. E.g. the LST for J is 17 and G is 16, the smallest is obviously 16 therefore we pick activity G to find the LSTs for B, D and H.

By finding the ESTs and LSTs for each activity it makes finding the critical path easier. All you have to do is find the boxes that contain the same number. For this example the critical path is: **A, F, I, M.**

As the title suggests crashing involves reducing the overall completion time by completing some of the activities more quickly.

It is often a trial and error process but it should be noted:

- Always reduce the duration of activities that are on the CRITiCAL PATH first.
- Reducing an activity that is not on the CRITICAL PATH may not reduce the project completion time, because the critical path as we know is the longest path.
- The process of crashing can change the critical path. Do not reduce too much as that path may no longer be a critical. A new critical path may be created.
- It is possible to have two critical paths that give you the same completion time after crashing.

This is an artificial activity that has a time duration of 0. It is added to a network diagram to ensure that all predecessor activities are properly accounted for.

When performing a critical path analysis and you have to construct your own network from a table you need to ensure that:

- An activity can only be represented by one edge.
- Two vertices can only be connected by ONE edge.

*Essential Further Mathematics 4ed 2012 (Example 8 p.654)*

In the above example, a network has been constructed using the information contained in the table. A dummy activity (D1) was needed because the predecessor of activities C and D is A and B. The dummy activity makes it easier for us to calculate the ESTs and LSTs correctly.

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