Lesson Objective

1
Lesson Objective Understand what critical path
analysis is Be able to prioritise events and
create a precedence table Begin to use the
precedence table to create a network diagram
2
Consider the following problem You must toast
three pieces of bread on both sides. The grill
can take two pieces at a time. It takes 30
seconds to toast a side of bread. How quickly can
you toast the bread?
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A solution
A more efficient solution
0 seconds
30 seconds
60 seconds
90 seconds
120 seconds
4
In this problem time can be saved by planning. If
you have lots of complicated things to achieve in
a set sequence planning to find which things can
be run con-currently can save time. The study of
this type of problem is called critical path
analysis and the critical path is the sequence of
events that must run to schedule in order to
complete the task as efficiently as possible.
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A simple Critical path problem Jane, Sue and
Meena share a flat. Jane has an interview at 9.00
but she has woken up at 8.10. To get to the
interview she must Shower 3 mins Dry
her hair 8 mins Fetch the car 7
mins Iron her clothes 12 mins
Dress and make-up 10 mins Drive to the
interview 20 mins If she does all of these things
on her own will she make the interview? What if
she uses her friends to help her with some of the
tasks?
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To solve the problem algorithmically we should
first draw up a Precedence table. This is a table
that shows how each activity is dependent upon
the others
Activity Immediately preceding activities duration
A Shower - 3
B Dry hair A 8
C Fetch car - 7
D Iron clothes - 12
E Dress and make-up B,D 10
F Drive to interview C,E 20
7
We can then draw a Network diagram to illustrate
how each acticty is dependent upon preceding
activities
2
B(8)
A(3)
D(12)
1
5
3
E(10)
F(20)
C(7)
4
8
Event Times
Each event node needs two boxes, to mark in the
event times.

2
B(8)
A(3)


D(12)
1
5
3
E(10)

F(20)
C(7)
4

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Earliest Event Times
The EET for an event is the earliest time the
event can start taking into account the
activities that must complete before it can do
so.

3
2
B(8)
A(3)

42

12
D(12)
1
5
3
E(10)

0
F(20)
C(7)
4

22
To find EETs, work forwards through the network
from the start node to the finish node.
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Latest Event Times
The LET for an event is the latest time you can
leave the event so that you can still complete
the remaining activities without over running.
3
4
2
B(8)
A(3)
42
42
12
12
D(12)
1
5
3
E(10)
0
0
F(20)
C(7)
4
22
22
To find LETs, work backwards through the network
from the finish node to the start node.
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Critical Activities
Critical activities are activities that cannot
run late. For critical activities Latest finish
Earliest start length of activity In this
case D, E and F
3 4
2
B(8)
A(3)
42 42
12 12
D(12)
1
5
3
E(10)
0 0
F(20)
C(7)
4
22 22
The green arrows mark the critical activities,
which form the critical path. The critical
path(s) must form a continuous route from the
start node to the finish node.
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Non-Critical Activities
Activities that are not critical have some
flexibility over when they start/finish. The
spare time they have to work in is called the
float.
3 4
2
B(8)
A(3)
42 42
12 12
D(12)
1
5
3
E(10)
0 0
F(20)
C(7)
4
22 22
In some cases using the float to delay the start
time of an activity will not affect the other
activities (in the case of C above). The float of
C is said to be independent. But in other cases
it will (as in the case of A and B). We say that
the floats of A and B are interfering.
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We can put the results of the floats into a table
Total float Independent float Interfering float
A 1 0 1
B 1 0 1
C 15 15 0
HintTotal float is maximum possible float so
take outside nos. Independent float is minimum
possible float so take inside nos. Interfering
float is the difference between the two floats.
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Have a go at this Draw an activities network for
this situation and find the critical path.
Task Duration (hours) Immediate predecessors
A 4 -
B 4 -
C 6 -
D 4 A
E 2 B
F 8 D,E
G 3 C
H 4 G,F
I 3 B
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Example from the book wit h an easy diagram and
no dummies.
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Lesson Objective Be able to use the precedence
table to create a network diagram Understand the
importance and use of dummies when creating a
network diagram from a precedence table
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Warm Up Draw a Network diagram for this
precedence table
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Activity networks
The table below shows the tasks involved in a
project, with their durations and immediate
predecessors.
Task Duration (hours) Immediate predecessors
A 3 -
B 4 -
C 6 -
D 5 A
E 1 B
F 6 B
G 7 C, D, E
Draw an activity network for this table.
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Activity networks
A(3)
B(4)
C(6)
First draw a start node labelled 1.
Activities A, B and C do not depend on any other
activity, so they all begin at node 1.
21
Activity networks
D(5)
A(3)
1
B(4)
C(6)
Activity D depends on A, so add event node 2 at
the end of A.
Now add activity D.
22
Activity networks
D(5)
2
A(3)
E(1)
1
B(4)
F(6)
C(6)
Activities E and F both depend on B, so add event
node 3 at the end of B.
Now add activities E and F.
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Activity networks
D(5)
2
A(3)
1
Activity G depends on C, D and E, so all these
three events need to end at the same node.
This is easiest if you redraw the network so that
C is between A and B.
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Activity networks
D(5)
2
D(5)
A(3)
G(7)
1
Now add node 4, with C, D and E leading into it.
Now add activity G.
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Activity networks
2
D(5)
A(3)
G(7)
1
C(6)
G(7)
E(1)
B(4)
F(6)
F(6)
A finish node is now needed. Any activities not
leading into a node must end at the finish node.
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Activity networks
2
D(5)
A(3)
1
C(6)
G(7)
E(1)
B(4)
F(6)
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Try drawing a network for this table. What
problems arise?
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Using dummy activities Case 1 Activity C and D
both have dependency on the same letter, but not
exactly the same letters. We need to use a dummy.
C B
D A,B
A
D
3
dummy
2
B
C
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Using dummy activities Case 2 Activity B and C
both share the same start and finish node. This
will cause problems when we use our EETs and LETs
so we cannot let this happen. We need to use a
dummy.
A -
B A
C A
D BC
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Official Practice Paper B
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Official Practice Paper A
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Lesson Objective Be able to use a network from a
precedence table to find the shortest completion
time and the critical paths and events
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Reminder The critical path is the path through
the Network that takes the minimum amount of time
(there may be more than one) but manages to
complete all the activities. Critical Activities
are those that cannot be delayed (there is no
float) if the task is to complete on time. The
float of an activity is how long it can be
delayed without slowing down the overall
completion time. There are three types of
float Total float is maximum possible float so
take outside nos. Independent float is minimum
possible float so take inside
nos. Interfering/dependent float is the
difference between the two floats.
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Activity networks Example 1
2
D(5)
A(3)
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
Event 1 occurs at time zero.
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Activity networks Example 1
3
2
D(5)
A(3)
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
Event 2 cannot occur until A is finished. The
earliest time for this is 3.
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Activity networks Example 1
3
2
D(5)
A(3)
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
4
Event 3 cannot occur until B is finished. The
earliest time for this is 4.
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Activity networks Example 1
3
2
D(5)
A(3)
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
4
Event 4 cannot occur until C, D and E are all
finished.
So the earliest time for event 4 is 8.
The earliest C can finish is 6. The earliest D
can finish is 3 5 8. The earliest E can
finish is 4 1 5.
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Activity networks Example 1
3
2
D(5)
A(3)
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
4
Event 5 cannot occur until F and G are both
finished.
So the earliest time for event 5 is 15.
The earliest F can finish is 4 6 10. The
earliest G can finish is 8 7 15.
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Activity networks Example 1
3
2
D(5)
A(3)
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
4
The next step is to find the late event times
(LETs), working backwards through the network. A
LET is the latest time that an event can occur
without delaying the project. The LET is found by
finding the latest time that each activity
leading out of the event can begin the LET is
the earliest of these.
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Activity networks Example 1
3
2
D(5)
A(3)
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
15
4
Event 5 must occur by time 15, or the project
will not finish in the minimum possible time.
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Activity networks Example 1
3
2
D(5)
A(3)
8
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
15
4
The only activity leading from event 4 is G,
which must start by time 8 if the project is not
to be delayed. So event 4 must occur by time 8.
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Activity networks Example 1
3
2
D(5)
A(3)
8
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
15
4
7
The activities leading from event 3 are E (which
must start by time 7) and F (which must start by
time 9). So event 3 must occur by time 7.
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Activity networks Example 1
3
3
2
D(5)
A(3)
8
8
1
C(6)
G(7)
0
E(1)
B(4)
F(6)
15
15
4
7
The only activity leading from event 2 is D,
which must start by time 3. So event 2 must occur
by time 3.
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Activity networks Example 1
3
3
2
D(5)
A(3)
8
8
1
C(6)
G(7)
0
0
E(1)
B(4)
F(6)
15
15
4
7
Finally, event 1 must occur by time zero.
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Activity networks Example 1
3
3
2
D(5)
A(3)
8
8
1
C(6)
G(7)
0
0
E(1)
B(4)
F(6)
15
15
4
7
The critical activities are the activities (i, j)
for which the LET for j the EET for i is equal
to the activity duration.
The completed network shows that the project can
be completed in 15 hours.
The critical activities are A,
D
and G.
For analysis of the float in this example, see
the Notes and Examples.
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Activity networks Example 2
The table below shows the tasks involved in a
project, with their durations and immediate
predecessors.
Task Duration (days) Immediate predecessors
A 2 -
B 3 -
C 5 -
D 6 A, B
E 8 C
F 2 C
G 4 D, E
Draw an activity network and use it to find the
critical activities and the minimum duration of
the project.
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Activity networks Example 2
A(2)
1
B(3)
C(5)
Begin with a start node, labelled 1.
Activities A, B and C have no preceding
activities, so can all begin at the start node.
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Activity networks Example 2
2
A(2)
D(6)
1
3
B(3)
C(5)
Activity D depends on both A and B. Since A and B
must not start and finish at the same node, a
dummy activity is needed to ensure unique
numbering.
The dummy activity has zero duration. Now
activity D can be drawn in, following on from
both A and B.
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Activity networks Example 2
2
A(2)
D(6)
1
3
B(3)
C(5)
E(8)
4
Activities E and F both depend on activity C.
F(2)
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Activity networks Example 2
2
A(2)
D(6)
G(4)
1
3
5
B(3)
C(5)
E(8)
E(8)
4
Since G depends on both D and E, these two
activities must both lead into the same node.
Activity G can now be drawn in.
F(2)
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Activity networks Example 2
2
A(2)
D(6)
G(4)
1
3
5
6
B(3)
C(5)
E(8)
F(2)
4
Finally, activities F and G must finish at the
end node.
F(2)
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Activity networks Example 2
2
A(2)
D(6)
G(4)
1
3
5
6
B(3)
C(5)
E(8)
F(2)
4
The next step is to find the earliest event times
(EETs).
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Activity networks Example 2
2
2
A(2)
D(6)
G(4)
1
3
5
6
B(3)
0
C(5)
E(8)
F(2)
4
Event 1 occurs at time zero.
The earliest that event 2 can occur is after A
has finished, at time 2.
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Activity networks Example 2
2
2
A(2)
D(6)
G(4)
1
3
5
6
B(3)
0
3
C(5)
E(8)
F(2)
4
Event 3 cannot occur until both A and B have
finished, so the earliest time at which event 3
can occur is 3.
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Activity networks Example 2
2
2
A(2)
D(6)
G(4)
1
3
5
6
B(3)
0
3
C(5)
E(8)
F(2)
4
Event 4 cannot occur until C has finished, so the
earliest time at which event 5 can occur is 5.
5
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Activity networks Example 2
2
2
A(2)
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
C(5)
E(8)
F(2)
4
5
The earliest that D can finish is at time 9, and
the earliest that E can finish is at time 13, so
the earliest that event 5 can occur is time 13.
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Activity networks Example 2
2
2
A(2)
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
C(5)
E(8)
F(2)
4
5
The earliest that F can finish is at time 7, and
the earliest that G can finish is at time 17, so
the earliest that event 5 can occur is time 17.
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Activity networks Example 2
2
2
A(2)
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
C(5)
E(8)
F(2)
4
5
The next step is to find the latest event times
(LETs), starting from the finish node and working
backwards.
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Activity networks Example 2
2
2
A(2)
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
C(5)
E(8)
F(2)
4
5
Event 6 must not occur later than time 17, or the
project will be delayed .
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Activity networks Example 2
2
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
C(5)
E(8)
F(2)
4
The latest G can start is at time 13, so the
latest time for event 5 is 13.
5
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Activity networks Example 2
2
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
C(5)
E(8)
F(2)
4
5
The latest that E can start is at time 5, and the
latest that F can start is at time 15, so the
latest possible time for event 4 is 5.
5
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Activity networks Example 2
2
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
7
C(5)
E(8)
F(2)
4
5
5
The latest that D can start is at time 7, so the
latest possible time for event 3 is 7.
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Activity networks Example 2
2
7
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
7
C(5)
E(8)
F(2)
4
5
5
The latest that the dummy activity (with zero
duration) can start is at time 7, so the latest
possible time for event 2 is 7.
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Activity networks Example 2
2
7
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
7
0
C(5)
E(8)
F(2)
4
5
5
The latest that event 1 can start is at time 0.
68
Activity networks Example 2
2
7
2
A(2)
13
13
D(6)
G(4)
1
3
5
6
B(3)
0
3
17
17
7
0
C(5)
E(8)
F(2)
4
5
5
The critical activities are activities for which
the float is zero i.e. the latest event time for
activity j the earliest event time for activity
i is equal to the activity duration.
69
Activity networks Example 2
2
7
2
A(2)
13
13
D(6)
G(4)
3
6
5
1
B(3)
0
3
17
17
7
0
C(5)
E(8)
F(2)
4
5
5
The critical activities are C,
For analysis of the float in this example, see
Example 2 in the Notes and Examples.
E
and G.
The project can be completed in 17 days.
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