Network Planning With Pert

Network planning with PERT

Network analysis: methodology PERT CPM

project evaluation and review technique critical path method

• enumeration of necessary activities, with indication of immediately preceding and necessarily finished activities • activities placed in a logical network by means of presentation with arrows. Each arrow presents an activity and is preceded and followed by an event

Network analysis • Analysing and planning of the subsequent activities, necessary to complete the project • Basis of cost calculation and timing of project • Evaluation of possible measures to reduce the duration of the project • Basis of monitoring and evaluation of the project

Definitions • A network is a logical presentation of the activities of a project. It shows the relations that exist between activities • An activity is a basic task necessary to complete the project. The activity needs time and resources • An event is a situation that defines the beginning or the end of a (number of) activit(y)(ies)

0

0

7

24

7

24

B 2

A

1

Start

7

5

110 17

D 24

K

26

E H

3

23 0

0

24

G

0

C

50

L

M 4 50

33

0

Critical path sequence TL Dummy Activity = TE min= slack (TLkf of normal max((TEkp activities time –TA) activity = LF for + TA) –which with EF duration TE = TL0

31 6

15 14

F

4 38

46 8

0

J 7

7 57

57 0

Finish

Activity Schedule – ES earliest starting time • ES = TE start event of the activity

– EF earliest finishing time • EF = ES + TA

– LF latest finishing time • LF = TL end event of the activity

– LS latest starting time • LS = LF – TA

ES (TE)

LS (LF-TA)

EF (ES+TA)

LF (TL)

Activity slack (LF-EF)

A

0

0

7

7

0

B

7

13

18

24

6

C

24

26

55

57

2

D

7

7

24

24

0

E

24

24

24

24

0

F

24

24

50

50

0

G

0

1

23

24

1

H

24

35

39

50

11

J

50

50

57

57

0

K

0

5

33

46

13

L

24

32

38

46

8

M

38

46

42

50

8

Float • • • •

Free float = ES of the next activity – EF of the activity Total float = LF – ES -TA interfering float = total float – free float independent float = ES of the next activity – LF of the activity

Crash activities

4

4

10

15

6 B

0

D

4

0

2

17

4

7 2

C

4

8

23

23

6

6 A

17

E

10

10

F

G

The optimal crash sequence therefore is; A - B @ £500 per unit time. F - G @ £1000 per unit time. B - E @ £2000 per unit time. E - F @ £3000 per unit time

Crash A-B .Crash A -B by 2 weeks. • Project completion reduces by 2 weeks. • Project cost increases by £1000. • Revised project cost = £19,500 + £1000 = £20,500 • Revised project completion = 23 - 2 = 21 weeks. • Check the critical path 2

2

8

13

6 B

0

D

2

0

2

15

4

7 2

C

4

6

21

21

6

6 A

15

E

8

8

F

G

Crash F-G Crash F -G by 1 week. • Project completion reduces by 1 weeks. • Project cost increases by £1000. • Revised project cost = £20,500 + £1000 = £21,500 • Revised project completion = 21 -1 = 20 weeks. • Check the critical path 2

2

8

13

6 B

0

D

2

0

2

15

4

7 2

C

4

6

20

20

5

6 A

15

E

8

8

F

G

Crahs B-E Crash B - E by 1 week. • Project completion reduces by 1 weeks. • Project cost increases by £2000. • Revised project cost = £21,500 + £2000 = £23,500 • Revised project completion = 20 -1 = 19 weeks • Check the critical path 2

2

8

12

6 B

0

D

2

0

2

14

4

7 2

C

4

5

19

19

5

5 A

14

E

7

7

F

G

Crash E-F Crash E - F by 1 week. • Project completion reduces by 1 weeks. • Project cost increases by £3000. • Revised project cost = £23,500 + £3000 = £26,500 • Revised project completion = 19 -1 = 18 weeks • Final critical path 2

2

8

11

6 B

0

D

2

0

2

13

4

6 2

C

4

5

18

18

5

5 A

13

E

7

7

F

G

Final crash curve Cost

18 weeks @ £26500

£26.5K

19 weeks @ £23500

Crash E - F

20 weeks £23.5K

@ £21500 Crash E - F 21 weeks

£21.5K

@ £20500

Crash E - F

£20.5K Crash E - F £19.5K Original condition 18

19

20

21

23

Time

Typical crash curve Cost

Cost-ineffective crash. Maximum effective crash

Minimum cost Origin

Time

Typical extended crash curve Cost

Non-critical crashing Maximum crash Optimum time-cost point

Best cost

Fixed costs

Best time

CPM with Uncertainty • Steps leading to network are same as before • Added complexity of having time estimates – Optimistic Time =a – Most Likely Time=m – Pessimistic time=p

• Expected time calculated as ET=(a+4m+b)/6

• Variance calculated as 2 σ

= ( b - a) 2 / 36

CPM with uncertainty • Given any path we can calculate – Expected completion time of path • Just add the expected times (ET)

– Variance of the path • Just add the variances

• Critical Path Determined Exactly as before – Only difference, we will now use expected times.

Completion Probability Assuming (which is the case) that the completion time follows a normal distribution with mean as ET and 2 σ variance as , we can calculate the following: What is the probability that the project is completed in T time units.

Completion Probability Take the critical path( determined as before, but now with expected times). We know ET for this critical path. We also know its std. dev (σ ) . We will always assume that the time of completion is Normally distributed. Hence: If X=Time of completion.

T − ET P( X ≤ T ) = P( Z ≤ ) σ