A brief summary of Critical Path (or network) Analysis 

Critical Path Analysis is a way of planning complex projects. •It allows one to find those parts of the project that cannot be delayed (the critical activities), and those where ‘slack’ time exists. •It also forces managers to think in a logical manner, to plan activities in a logical sequence, and shows which other activities a part of the project is dependant upon.

Diagrams are made up of three parts….

 

•Nodes These are shown as circles and represent the start or finish of a particular activity.

2.Arcs These are lines and they represent the activity being undertaken. (The length of the arc does represent the length of the activity). They should have an arrow showing the direction of the project.

Dummies Shown as dotted lines. These are used to link up dependent activities. For example two that start at the same node and finish at the same node as each other. Or where only some activities that leave a node are dependent on all the activities that enter the node. They may be thought of as activities with a duration of 0.

Earliest start times (EST) (shown in the top right hand corner of a node). This shows the earliest the next activity(ies) can start. Add the duration of the task to the previous EST. If there is a choice of EST´s at a node take the highest. (IT GOES IN THE TOP OF THE CIRCLE, SO TAKE THE TOP NUMBER.)

Latest Finishing Time (LFT) (shown in the bottom left hand corner of the node) They show the latest an activity can finish before the whole project over runs. Take the duration of activity away from the LFT of the next activity. When faced with a choice, choose the lower LFT. (IT GOES IN THE BOTTOM OF THE CIRCLE, SO TAKE THE BOTTOM NUMBER)

Critical Path Those activities that can not over run without effecting the total length of the project, are those where the EST = LFT.

Total Float LFT of the activity- the duration- EST of the activity. This shows how much ´slack´ there is on a particular route of the network. If the total float is 0 then an activity lies on the critical path.Free Float EST of the next activity – Duration – EST of this activity. This shows the ´slack´ on an individual activity before it delays the start of the next activity.

There are several key points to remember when facing a network question...

1. Don’t be afraid of drawing the network twice. Lines are NOT allowed to cross, and you do not know where they will go until the diagram is done.

2. Draw it firstly freehand, and then again, neatly. The second one will take only moments.

3. ALL lines must be activities YOU CAN NOT ADD LINES TO MAKE ACTIVITIES AND NODES JOIN.

4. They are easy to score highly on if you know what you are doing.

There are two types of question.

One where they give you the network, and the more common, harder one, where you have to construct the newtork as well.

The question will have a table like the one below. This is where you will get the information from

Activity Must be prededed by

Duration (days)

A - 4

B - 5

C A 6

D B 7

E B 14

F C,D 6

G E 12

H E,F 14

I G 10

J H,I 4

K J 6

Activity Must be prededed by

Duration (days)

A - 4

B - 5

C A 6

D B 7

E B 14

F C,D 6

G E 12

H E,F 14

I G 10

J H,I 4

K J 6

We know that there can only ever be one starting point, so as A and B are not preceded by any thing, then they must both come out of the first node.

1

A

B

Activity Must be prededed by

Duration (days)

C A 6

D B 7

E B 14

F C,D 6

G E 12

H E,F 14

I G 10

J H,I 4

K J 6

Activity C is preceded by A. Like so...

1

A

B

2C

Activity Must be prededed by

Duration (days)

D B 7

E B 14

F C,D 6

G E 12

H E,F 14

I G 10

J H,I 4

K J 6

In the same way as before, activities D and E are preceded by B. This time however (if look down to activity F), we see that both C and D need to be finished before it can start. So these two need to join up.

A

B

C

D

E

1

2

3

4 F

Activity Must be prededed by

Duration (days)

G E 12

H E,F 14

I G 10

J H,I 4

K J 6Fairly straight forward! Activity G follows on from E.

A

B

C

D

E

1

2

3

4 F

6 G

Activity Must be prededed by

Duration (days)

H E,F 14

I G 10

J H,I 4

K J 6

Slightly more tricky this time. Both E and F need to be finished before H can start. We can not just move our lines so that E and F join up. If we did that, then it would men that G also needed both E and F finished. It only needs E, so we must use a dummy from E to F

A

B

C

D

E

1

2

3

4 F

6 G

7 H

Activity Must be prededed by

Duration (days)

I G 10

J H,I 4

K J 6

I follows on from G (nice and easy), but if we look ahead to J we see that both H and I need to be finished before J can start, so they need to join up.

A

B

C

D

E

1

2

3

4 F

6 G

7 H

8

9 J

I

Activity Must be prededed by

Duration (days)

K J 6

To finish the network, J follows K.

A

B

C

D

E

1

2

3

4F

6G

7 H

8

9

I

JK

1011

Activity Must be prededed by

Duration (days)

A - 4

B - 5

C A 6

D B 7

E B 14

F C,D 6

G E 12

H E,F 14

I G 10

J H,I 4

K J 6

The next stage is to add the durations to from the table to the network.

A

B

C

D

E

1

2

3

4 F

6 G

7 H

8

9

I

JK

10 11

4

5

55

6

7

14

6

12

14

10

4 6

Next add the Earliest Start Times (EST), this is the earliest start time from the activity before plus the duration of the activity. If we have two arcs joining then we take the highest number. This goes in the top right quarter of the node. (highest quarter, so the highest number.)

A

B

C

D

E

1

2

3

4F

6G

7 H

A

B

C

D

E

1

2

3

4F

6G

7

A

B

C

D

E

1

2

3

4F

6G

7 H

8

9

I

JK

1011

4

5

6

7

14

6

12

14

10

4 6

0

4

EST shows

0 +4=4

5

0+5=5

12

Either 4+6=10

Or 5+7=12Choose the biggest so it’s 12

19

5+14=19

Either 12+6= 18

Or 19+0=19 ( (from the dummy)

19

31

19+12=31

41

Either 19+14=33

Or 31+10=41

45

41+4=45

51

45+6=51

Next add the Latest Finish Times (LFT), this is the LFT time from the activity at the end minus the duration of the activity. If we have two arcs joining then we take the lowest number. This goes in the bottom right quarter of the node. (bottom quarter, so the lowest number.) This time you start at the end of the network.shows the time by which

the previous activity must be finished if the project is not to be delayed.

A

B

C

D

E

1

2

3

4F

6G

7H

A

B

C

D

E

1

2

3

4F

6G

A

B

C

D

E

1

2

3

4F

6G

H

8

9

I

JK

1011

4

5

6

7

14

6

12

14

10

4 6

0

4

5

12

19

19

31

41 45 51

5145

51-6=45

41

45-4=41

27

41-14=27

31

41-10=31

Either 27-0=27 from the dummy, or 31-12=19.

We choose the lowest, so 19

19

19

27-6=19

5

Either 19-7=12 (from activity D) or, 19-14=5 (from activity E)

13

19-6=13

0

Must always be 0!

It is usual for a question to then ask you to identify the critical path. These are the activities for which any delay will mean a pdelay to the whole project. They can be sotted as those activities that have the same EST and LFT at their begining and end.

A

B

C

D

E

1

2

3

4F

6G

7H

B

C

D

E

1

2

3

4F

6G

B

E

2

3

4

6G

8

9

I

J K10

11

4

5

6

7

14

6

12

14

10

4 6

0

4

5

12

19

19

31

41 45 51

514541

27

3119

19

5

13

0

The critical path is B E G I J K.

Questions may then ask you to calculate the float for each activity. The float is the amount of spare time.

There are two kinds of float.TOTAL FLOAT and FREE FLOAT.

Total float: The amount of spare time before the activity effects the length of the whole project.

LFT (at end of activity) – duration – EST (start of activity)

Critical path activities will have a float of 0.

Free float: The amount of spare time before the activity delays the start of the next activity.

EST (at the end) – duration – LFT (at the start)

Most activities will have a free float of 0.

A

B

C

D

E

1

2

3

4F

6G

7H

A

B

C

D

E

1

2

3

4F

6G

A

B

C

D

E

1

2

3

4F

6G

H

8

9

I

J K10

11

4

5

6

7

14

6

12

14

10

4 6

0

4

5

12

19

19

31

41 45 51

514541

27

3119

19

5

13

0

Activity Must be prededed by

Duration (days) Total float Free float

A - 4 13 – 4 – 0 = 7 4 – 4 – 0 = 0

B - 5 5 – 5 – 0 = 0 5 – 5 – 0 = 0

C A 6 19 – 6 – 4 = 9 12 – 6 –4 = 2

D B 7 19 - 7 - 5 = 7 12 – 7 – 5 = 0

E B 14 19 – 14 – 5 = 0 19 – 14 – 5 = 0

F C,D 6 27 – 6 – 12 = 9 19 – 6 – 12 = 1

G E 12 31 – 12 –19 = 0 31– 12 – 19 =0

H E,F 14 41 –14 – 19 = 8 41 – 14 –19 =8

I G 10 41 – 10 – 31 = 0 41 – 10 –31 =0

J H,I 4 45 – 4 –41 = 0 45 – 4 –41=0

K J 6 51 – 6 –45 =0 51 – 6 –45 =0

Finally, it is usual to ask you to outline the advantages and disadvantages to a firm of using network analysis.

Minimises time lost between tasks, so the project should run smoothly.

Planning is always good! It forces you to think through potential problems, and allows you to spot them before they occur.

It forces you to think logically what order activities should be in.

Money is saved by not having resources sat around doing nothing, as you have planned when you will require them.

May be good for organistions with cash flow problems.

Drawing the diagram does not guarantee times will be stuck to. It still requires managers to manage so that times are stuck to.

A network quickly becomes massively complex, so expensive computers are needed to plan it, putting off many firms from using the method.

The network is only as good as the data used to construct it. If the time estimates are wrong, then the network will be useless.

ADVANTAGES DISADVANTAGES