Critical Path Analysis

1
Critical Path Analysis
2
Precedence Tables
Complex problems can be broken down into tasks
called ACTIVITIES

• .
• Making an instant black coffee can be broken down
  into the following activities

You only need to put the step before as previous
steps are implied in this
3
Draw a precedence table showing how to start a car
4
Answers
5
Activity Networks

• We now turn the precedence table into a graph

D, E F depend only on B.
H depends on E
The start is called the SOURCE NODE
The number in brackets is the DURATION of the
activity
Tail Event for A
E(8)
2
4
Head Event for A
B(3)
I depends on F, G H
F(6)
A(1)
H(2)
D(2)
0
1
C(6)
Activity A can start at any time
3
5
6
G(5)
I(1)
Activities B C depend on A so can only start
when A is complete (node 1)
G depends on C D
The end is called the SINK NODE
6
Dummies

• If two events in different chains are
  independent but rely on a common event then the
  link between them is a logical link and not an
  activity in the process. This is represented by
  a dummy activity.

A
C
1
Dummy
B
D
1
This is because D does not depend on A and B,
only on B
7
Finding the critical path

• Events are the numbers in circles
• Activities are the letters

Step 2 Finding the latest time an event can
start without changing the length of the critical
path
Step 1 Finding the earliest time an event can
start
There are 3 routes to event 5. Event 2 add
F(6)10 Event 3 add G(5)12 Event 4 add
H(2)14 Therefore the earliest event 5 can start
is 14
Three routes backwards so find the latest can
start. Event 4 E(8) 4 Event 5 F(6)
8 Event 3 D(2) 7 So the latest event 2 can
start is 4 or it will lengthen the critical path
Event 2 only depends on B. So as soon as B is
complete. Node 1 add B(3) gives earliest start
time 4
Event 3 requires C D to be complete so the
earliest it can start is at the longest of these
two routes. Event 2 has a start time of 4 add
D(2) gives a start time of 6 Alternatively event
1 has a start time of 1 add C(6) gives a start
time of 7. So although B and D are finished
they have to wait until C is finished. Therefore
the earliest start time for event 3 is 7
Event 4 only depends on E. Event 2 add E(8) gives
earliest start time 12
Only one route so latest start is 12
4
4
12
12
As the only way to event 6 is through event 5
then the latest event 5 can start is 14
The earliest the project can be completed is
15. The CRITICAL PATH is 15
Only one route to event 6, so event 5 add I(1) 15
The latest that event 6 can start is obviously 15
1
1
0
0
Two routes Event 2 B(3) 1 Event 3 C(6)
3 So the latest start time is 1
Node 1 can start as soon as A is finished. A(1)
so the earliest start time is 1
Node 0 can start immediately so at time 0
Obviously 0
Only one route backwards so event 5 G(5) 9
7
9
14
14
15
15
8
Notation

• Let e stand for the earliest start time and the
  subscript stand for the event
• Let l stand for the latest start time and the
  subscript stand for the event
• e1 3 means the earliest event 1 can start is 3.
• Let i stand for any event

9
Finding the critical path

• Events are the numbers in circles
• Activities are the letters

Step 2 Finding the latest time an event can
start without changing the length of the critical
path
Step 1 Finding the earliest time an event can
start