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.
