Project Scheduling - PERT/CPM | Finding Critical Path - YouTube

Welcome to this video on finding the critical path on a project network.

I’ll be working with this activity schedule for a project.

I’ll be constructing a project network, doing forward and backward passes, determining

the project completion time, calculating slack values, and finally, stating the critical

path.

I will be using this node convention here as you will find in Quantitative Methods for

Business by Anderson, Sweeny and Williams.

A here is the activity being described, and t represents the expected activity duration

or time.

ES is the earliest time the activity can start; EF is the earliest finish time; LS is the

latest start time, and LF is the latest finish time without extending the minimum completion

time of the project.

I usually like to start with a sketch to make it easier when drawing the full network.

Activities A and B have no predecessors so they can begin at start.

Activity C needs A to be completed before it can start.

D needs both A and B completed.

E needs D, F needs C and E, and G depends on E. Since F and G have no successors, they

go to Finish.

So here is the network with the activity nodes, displaying the letters and times.

So let’s do the forward pass.

A has no predecessor so its earliest start time will be 0 or right away.

Since it has 7 weeks to be completed, its earliest finish time will be 0 + 7 which gives

7.

B also has an earliest start time of 0, and with and activity time of 9, it will have

an earliest finish time of 9.

Now C needs A to be completed before it can start.

Since the Earliest Finish time for A is 7, then the Earliest Time C can start is 7.

And with an activity time of 12 weeks, C will have an Earliest Finish time of 12 + 7 which

gives 19.

D on the other hand needs A and B to finish before it can start.

Since the Earliest Finish time for A and B are 7 and 9 respectively, and D needs both

of them to finish in order to start, then the Earliest Time D can start is 9.

In order words, the highest of the Earliest Finish Times preceding an activity will be

the activity’s Earliest Start Time.

So D finishes at 8 + 9 which gives 17.

E here has only one predecessor D, and so can start at 17 and finish earliest

at 26.

F has predecessors, C and E. Since the higher Earliest finish time is 26, F can start earliest

at 26 and finish at 32.

G also can start earliest at 26 since it has only one predecessor, E. And G can finish

earliest at 26 + 5 which gives 31.

Note here that although G is the last letter, it does not have the highest Earliest Finish

Time because F has 32.

So we say that the project’s completion time is 32 weeks.

In essence, the project’s completion time is the highest of the Earliest Finish Times

at the Finish node.

Now let’s do the backward pass.

Since the project completion time is 32 weeks, the latest finish times for the activities

at the finish node, F and G, has to be 32.

That is, F and G cannot be completed in longer than 32 weeks.

Next we obtain the latest start times by subtracting the activity times from the latest finish

times.

For G, the latest start time will be 32 minus 5 to give 27.

For F, the Latest Start will be 32 minus 6 and that gives 26.

Now E has 2 successors, F and G.

Their latest start times are 26 and 27 respectively.

As a result, the latest time E has to finish has to be 26 in order for F to start.

In essence, when doing backward pass, the latest finish time of an activity must be

the minimum of the latest start times of its successors.

Thus the Latest start time for E will be 26 minus 9 which gives 17.

Now D has only one successor, E. So the latest finish time for D will be the latest start

time for E, 17.

And the LS will be 17 minus 8 which gives 9 for D.

Activity C also has one successor, F. Therefore, latest finish will be 26 for C, and latest

start will be 14.

A has two successors, C and D. The minimum of their latest starts is 9.

So the latest finish for A will be 9 and its latest start will be 2.

Activity B has one successor, D, with latest start of 9.

So the latest finish for B will be 9, and its latest start will be 0.

The backward pass is now complete.

Now slack for an activity is defined as how long the activity can be delayed without extending

or increasing the project completion time.

And it is calculated as LS – ES or LF – EF.

So the slack for A will be 2 minus 0 or 9 minus 7.

Which will be 2.

The slack for B will then be 0, for C it will be 7, for D 0, for E 0, for F 0, and for G

1.

Note, for example, that activity C can begin anytime between week 7 and 14 and it can finish

anytime between week 19 and 26.

Thus C can be delayed for up to 7 weeks and the project will still be completed in week

32.

Activities, B, D, E, and F on the other hand cannot be delayed at all without extending

the project’s completion time.

So, for example if D is delayed by 2 weeks, then the project completion time will be extended

by 2 weeks as well, from 32 to 34.

The activities with 0 slack are called critical activities.

And they form the critical path which is the longest path in the network.

So the critical path here is B-D-E-F.

And that’s it for this video.

Thanks for watching.