Lesson Objective Understand what critical path analysis is Be able to prioritise events and create a...

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

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?

A solution A more efficient solution

30 seconds

0 seconds

60 seconds

90 seconds

120 seconds

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.

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?

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

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:

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

2

We can then draw a Network diagram to illustrate how each acticty is dependent upon preceding activities:

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

2

Earliest event timeEET

Latest event timeLET

Each event node needs two boxes, to mark in the event times.

Event Times:

Earliest Event Times:

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

2

0

3

12

22

42

To find EETs, work forwards through the network from the start node to the finish node.

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.

Latest Event Times:

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

23

0

12

22

42 42

22

12

4

0

To find LETs, work backwards through the network from the finish node to the start node.

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.

Critical Activities:

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

23 4

0 0

12 12

22 22

42 42

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

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.

A(3)

D(12)

B(8)

E(10)

C(7) F(20)

1 5

4

3

23 4

0 0

12 12

22 22

42 42

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.

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.

Total

float

Independent

float

Interfering

float

A 1 0 1

B 1 0 1

C 15 15 0

Hint:Total 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.

We can put the results of the floats into a table:

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

Example from the book wit h an easy diagram and no dummies.

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

Activity Depends on

A Prepare foundations −

B Have foundations passed by inspector A

C Obtain bricks −

D Erect walls B, C

E Construct roof D

F Install plumbing D

G Install wiring E

H Plaster walls F, G

I Decorate H

J Landscape garden E

Warm Up –

Draw a Network diagram for this precedence table

Activity networks

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

The table below shows the tasks involved in a project, with their durations and immediate predecessors.

Draw an activity network for this table.

Activity networks

Activities A, B and C do not depend on any other activity, so they all begin at node 1.

A(3)

B(4)1

C(6)

First draw a start node labelled 1.

Activity networks

Activity D depends on A, so add event node 2 at the end of A.

A(3)

B(4)1

C(6)

D(5)

Now add activity D.

2

Activity networks

Activities E and F both depend on B, so add event node 3 at the end of B.

A(3)

B(4)1

C(6)

Now add activities E and F.

E(1)

D(5)

F(6)

3

2

C(6)

B(4) E(1)

F(6)

3

Activity networks

Activity G depends on C, D and E, so all these three events need to end at the same node.

A(3)

This is easiest if you redraw the network so that C is between A and B.

D(5)2

B(4)

C(6)

E(1)

F(6)

31

Activity networks

A(3)

D(5)2

C(6)

B(4) E(1)

F(6)

3

Now add node 4, with C, D and E leading into it.

Now add activity G.

1 4

D(5)

G(7)

Activity networks

A(3)

2

C(6)

B(4) E(1)

F(6)

3

A finish node is now needed. Any activities not leading into a node must end at the finish node.

1 4

D(5)

G(7)

5

F(6)

G(7)

Activity networks

A(3)

2

C(6)

B(4) E(1)

3

1 4

D(5)

5

F(6)

G(7)

Try drawing a network for this table.

What problems arise?

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.

3

2

A

B

D

C

C B

D A,B

dummy

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

Official Practice Paper B

Official Practice Paper A

D

4

Lesson Objective

Be able to use a network from a precedence table to find the shortest completion time and the critical paths and events

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.

Activity networks – Example 1

A(3)

2

C(6)

B(4) E(1)

3

Event 1 occurs at time zero.

1 4

D(5)

5

F(6)

G(7)0


A(3)

2

C(6)

B(4) E(1)

3

Event 2 cannot occur until A is finished. The earliest time for this is 3.

1 4

D(5)

5

F(6)

G(7)0

3


A(3)

2

C(6)

B(4) E(1)

3

Event 3 cannot occur until B is finished. The earliest time for this is 4.

1 4

D(5)

5

F(6)

G(7)0

3

4


A(3)

2

C(6)

B(4) E(1)

3

Event 4 cannot occur until C, D and E are all finished.

1 4

D(5)

5

F(6)

G(7)0

3

4

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.

So the earliest time for event 4 is 8.

8


A(3)

2

C(6)

B(4) E(1)

3

Event 5 cannot occur until F and G are both finished.

1 4

D(5)

5

F(6)

G(7)0

3

4

The earliest F can finish is 4 + 6 = 10. The earliest G can finish is 8 + 7 = 15.

So the earliest time for event 5 is 15.

8

15


A(3)

2

C(6)

B(4) E(1)

3

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.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15


A(3)

2

C(6)

B(4) E(1)

3

Event 5 must occur by time 15, or the project will not finish in the minimum possible time.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15 15


A(3)

2

C(6)

B(4) E(1)

3

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.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15 15

8


A(3)

2

C(6)

B(4) E(1)

3

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.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15 15

8

7


A(3)

2

C(6)

B(4) E(1)

3

The only activity leading from event 2 is D, which must start by time 3. So event 2 must occur by time 3.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15 15

8

7

3


A(3)

2

C(6)

B(4) E(1)

3

Finally, event 1 must occur by time zero.

1 4

D(5)

5

F(6)

G(7)0

3

4

8

15 15

8

7

3

0


A(3)

C(6)

B(4) E(1)

3

D(5)

F(6)

G(7)0

3

4

8

15 15

8

7

3

0

The completed network shows that the project can be completed in 15 hours.

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 critical activities are A, D and G.

For analysis of the float in this example, see the Notes and Examples.

1

2

4

5


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

The table below shows the tasks involved in a project, with their durations and immediate predecessors.

Draw an activity network and use it to find the critical activities and the minimum duration of the project.


Begin with a start node, labelled 1.

Activities A, B and C have no preceding activities, so can all begin at the start node.

A(2)

B(3)1

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.

A(2)

B(3)1

D(6)

2

3

C(5)


Activities E and F both depend on activity C.

A(2)

B(3)1

D(6)

2

3

C(5)

4

E(8)

F(2)

E(8)


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.

A(2)

B(3)1

D(6)

2

3G(4)

5

C(5)

4

E(8)

F(2)


Finally, activities F and G must finish at the end node.

A(2)

B(3)1

D(6)

2

3G(4)

5 6

E(8)

F(2)

C(5)

4

F(2)


The next step is to find the earliest event times (EETs).

A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

Event 1 occurs at time zero.


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

The earliest that event 2 can occur is after A has finished, at time 2.

0

2


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

Event 3 cannot occur until both A and B have finished, so the earliest time at which event 3 can occur is 3.

3


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

Event 4 cannot occur until C has finished, so the earliest time at which event 5 can occur is 5.

5


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5The 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.

13


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

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.

17


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17

The next step is to find the latest event times (LETs), starting from the finish node and working backwards.


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17

Event 6 must not occur later than time 17, or the project will be delayed .

17

The latest G can start is at time 13, so the latest time for event 5 is 13.


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13

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


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13

5The latest that D can start is at time 7, so the latest possible time for event 3 is 7.

7


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13

5

7

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.

7


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13

5

7

7

The latest that event 1 can start is at time 0.

0


A(2)

B(3)1

D(6)

2

3G(4)

C(5) E(8)

4

5 6

F(2)

0

2

3

5

13

17 17

13

5

7

7

0

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.


A(2)

B(3)

D(6)

2

3G(4)

C(5) E(8)

F(2)

0

2

3

5

13

17 17

13

5

7

7

0

The critical activities are C, E and G.

The project can be completed in 17 days.

For analysis of the float in this example, see Example 2 in the Notes and Examples.

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