project management - cpm/pert

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AMITY GLOBALBUSINESS SCHOOL Bangalore

Project Management - CPM/PERT

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AMITY GLOBALBUSINESS SCHOOL BangaloreNetwork Analysis

For planning, scheduling, monitoring and coordinating large and complex projects comprising a number of activities

- Defining the entire job to be done

- Formulating the logical sequence

- Controlling the progress

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AMITY GLOBALBUSINESS SCHOOL BangaloreObjectives

• Minimization of total time

• Minimization of total cost

• Mimimize production delays

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AMITY GLOBALBUSINESS SCHOOL BangaloreApplications

• Construction of bridges

• Assembly line scheduling

• Installation of complex equipment

• Inventory planning and Control

• Construction of residential complex

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AMITY GLOBALBUSINESS SCHOOL BangaloreComponents

• Activity : represented by an arrow, definite starting time, and a point where it ends

- Predecessor Activity : An activity which must be completed before one or more other activities start

- Successor Activity : An activity which started immediately after one or more activities are completed

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- Dummy Activity : any activity which does not consume time or resource ( -----)

• Event : represents the start of an activity and the end of an activity

activity i j

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AMITY GLOBALBUSINESS SCHOOL BangaloreCritical Path Method

• Ei = earliest start time for activity (i,j)

• Li = latest start time for activity (i,j)

• Ej = earliest finish time for activity (i,j)

• Lj = latest finish time for activity (i,j)

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AMITY GLOBALBUSINESS SCHOOL Bangalore Critical Path

• The longest continuous chain of activities through the network starting from the first to last event. ( )

• All activities lying in this critical path are called critical activities as any delay in their execution will lead to a delay in the completion of the entire project.

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AMITY GLOBALBUSINESS SCHOOL BangaloreFloat

• Total float :

The amount of time by which an activity can be delayed without delay in the project completion date

• Free Float :

The portion of the total float within which an activity can be manipulated without affecting floats of the subsequent activities

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• Independent Float :

That portion of the total float within which an activity can be delayed for start without affecting floats of the preceding activities.

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AMITY GLOBALBUSINESS SCHOOL BangalorePERT

programme evaluation and review technique

• Completion time is considered to be unknown

• Probability of activity completion time is estimated

– optimistic time (to or a)

- pessimistic time (tp or b )

- normal time / most likely time (tm or m)

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• Expected time te / μ = 1/6 ( to + 4 tm + tp)• Std deviation σ = 1/6 (tp – to)• Variance σ² = ( 1/6 (tp – to))²• Prob of completing the project by

scheduled time (Ts) is given by : Prob ( Z ≤ Ts – μ ) σ

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AMITY GLOBALBUSINESS SCHOOL BangaloreQ

For the following PERT diagram : D A 3 4 H 2 B E I K 2 4 5 3 F 7 2 C G 4 J 6

1

2 3

4 7 8 9

5 6

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• Compute earliest event time and latest event time

• Critical path and total project duration

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AMITY GLOBALBUSINESS SCHOOL BangaloreForward pass method

• E1 =0

• E2= E1 + t1,2 = 0 + 3

• E3 = E2 + t2,3 = 3 + 4 = 7

• E4 = E1 + t1,4 = 0 + 2 = 2

• E5 = E1 + t1,5 = 0 + 2 = 2

• E6 = max ( E4 + t4,6 ; E5 + t5,6 )

= max ( 9, 6 ) = 9

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• E7 = E4 + t4,7 = 2 + 4 = 6

• E8 = max ( E3 + t3,8 ; E7 + t7,8 )

= max ( 9, 11) = 11

• E9 = max ( E8 + t8,9 ; E6 + t6,9 )

= max ( 14, 15) = 15

Lower left portion

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AMITY GLOBALBUSINESS SCHOOL Bangalore Backward Pass

Method• L9= E9 = 15

• L8 = L9 – t8,9 = 15 -3 =12

• L7 = L8 – t7,8 = 12-5 =7

• L6 = L9 – t 6,9 = 15 – 6 = 9

• L 5 = L6 – t5,5 = 9-4 = 5

• L4 = min( L6- t4,6; L7-t4,7)

min( 9-7 ; 7 - 4) = 2

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• L3 = L8 – t 3,8 = 12 -2 = 10

• L2= L3- t2,3 = 10 -4 = 6

• L1= min ( L2-t1,2 ; L4- t1,4; L5 – t1,5)

= min ( 6-3; 2-2; 5-2 ) = 0

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For the following PERT diagram : D A 3 4 H 2 B E I K 2 4 5 3 F 7 2 C G 4 J 6

1

0 0

23 6

37 10

42 2

76 7

811 12

915 15

52 5

69 9

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An established company has decided to add a new product to its line. It will buy the product from a mfg concern, package it, and sell it to a number of distributors selected on a geographical basis. Market Research has indicated the volume expected and the size of sales force reqd. the steps in the following table are to be planned :

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Activity Description Time(wk)A1-2 B 2-3C 2-4D 2-5E 3-6F 4-7G 5-8H 6-7I 6-8J 6-9

Organize sales officeHire salesmenTrain salesmenSelect advertising agencyPlan advertising campaignConduct advertising campaignDesign packageSet up packaging facilitiesPackage initial stocksOrder stock from manufacturer

24221463334

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K7-10L 8-11M 9-12N 10-12

O 11-12P 12-13

Select distributorsSell to distributorsShip stocksCheck inventoryGet customer feedbackRe assess strategy

353233

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What is the critical path

Calculate total and free float

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6 3

2 3 2 2 4 14 4 3 3

2 3 3

3 5

10 0

2

2 2

36 6

620 20

924 28

1231 31

1334 34

44 20

7

23 26

10

26 29

54 20

823 23

11

28 28

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• Total float :

The amount of time by which an activity can be delayed without delay in the project completion date

= (Lj – tij) - Ei• Free Float :

The portion of the total float within which an activity can be manipulated without affecting floats of the subsequent activities = (Ej – Ei) - tij

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AMITY GLOBALBUSINESS SCHOOL Bangalore Time EstimatesActi Dur

Earliest

Start Fin

Latest

Start Fin

Float

Total Free

A 2 0 2

0 2 0 0

B 4 2 6 2 6 0 0

C 2 2 4 18 20 16 0

D 2 2 4 18 20 16 0

E 14 6 20 6 20 0 0

F 6 4 10 20 26 16 13=23-4-6

G 3 4 7 20 23 16 16 =23-4-3

H 3 20 23 23 26 3 0

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Activity Duration

Earliest

Start Finish

Latest

Start Finish

Float

I 3

J 4

K 3

L 5

M 3

N 2

O 3

P 3

20 23

20 24

23 26

23 28

24 27

26 28

28 31

31 34

20 23

24 28

26 29

23 28

28 31

29 31

28 31

31 34

0 0

4 0

3 0

0 0

4 4

3 3

0 0

0 0

28


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AMITY GLOBALBUSINESS SCHOOL BangaloreQ : A project schedule has the

following characteristics :Activity Time(days) Activity Time(days)

1-2 4 5-6 4

1-3 1 5-7 8

2-4 1 6-8 1

3-4 1 7-8 2

3-5 6 8-10 5

4-9 5 9-10 7

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• Construct a network diagram

• Compute the total float

• Find the critical path and total project duration

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4 1 5 7

1 1 8 2

1

0 0

2

4 9

31 1

4

5 10

57 7

715 15

910 15

611 16

8

17 17

1022 22

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• Critical path is 1-3-5-7-8-10 and the project duration is 22 days

• Total Float = ( Lj-tij ) - Ei

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Activit Duratio Earliest start

Earliest finish

Latest start

Latest finish

Total float

1-2 4 0 4 5 9 5

1-3 1 0 1 0 1 0

2-4 1 4 5 9 10 5

3-4 1 1 2 11 10 7

3-5 6 1 7 1 7 0

5-7 8 7 15 7 15 0

5-6 4 7 11 12 16 5

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7-8 2 15 17 15 17 0

6-8 1 11 12 16 17 5

4-9 5 5 10 12 17 7

9 -10 7 10 17 15 22 5

8-10 5 17 22 17 22 0

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steps in developing PERT network• Draw a network diagram

• Time estimates are obtd

- m , a , b

• Calculate te

• Determine the critical path

• Calculate σ²

• Calculate Z

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Various activities of a small project and other relevant information have been shown in the adjoining table. Using the given info the resulting network is shown in the following figure. Map out the critical path and find the prob of completing the project within 48 days.

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activity Most optimistic time ( in days )

Most likely time m

Most pessimistic

time b

1-2 4 8 12

2-3 1 4 7

2-4 8 12 16

3-5 3 5 7

4-5 0 0 0

4-6 3 6 9

5-7 3 6 9

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5-8 4 6 8

7-9 4 8 12

8-9 2 5 8

9-10 4 10 16

6-10 4 6 8

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Calculate expected time te = a + 4m + b

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Variance σ² = b-a ²

6

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Activity a m b te σ²

1-2 4 8 12 8 1.78

2-3 1 4 7 4 1

2-4 8 12 16 12 1.78

3-5 3 5 7 5 0.44

4-5 0 0 0 0 0

4-6 3 6 9 6 1

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5-7 3 6 9 6 1

5-8 4 6 8 6 0.44

7-9 4 8 12 8 1.78

8-9 2 5 8 5 1

9-10 4 10 16 10 4

6-10 4 6 8 6 0.44

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AMITY GLOBALBUSINESS SCHOOL BangaloreEarliest

Start Time• E(μ1)= 0

• E(μ2) = 0+8 = 8

• E (μ3) = 8 + 4 = 12

• E(μ4)= 12 + 8 = 20

• E(μ5) = Max(12+5,20+0) = 20

• E(μ6) = 20+6 = 26

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• E (μ7) = 20 + 6 = 26

• E(μ8)= 20 + 6 = 26

• E(μ9) = Max(26+8, 26+5) = 34

• E(μ10) =Max(34 + 10, 26 + 6 ) = 44

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AMITY GLOBALBUSINESS SCHOOL BangaloreLatest Start

Time• E(L10) = 44

• E(L9) = 44-10 = 34

• E(L8) = 34 – 5 = 29

• E(L7) = 34 – 8 = 26

• E(L6) = 44-10 = 34

• E(L5) = Min (26-6, 29 -6) = 20

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• E(L4) = Min(20 – 0, 36 – 6 ) = 20

• E(L3) = 20 – 5 = 15

• E(L2) = Min(15-4, 20 -12) = 8

• E(L1) = 8 – 8 = 0

Draw network diagram and find critical path

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10 0

28 8

312 15

420 20

520 20

726 26

934 34

828 29

626 38

1044 44

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• Critical path is

1---2---4---5---7----9---10

• Expected time in completing the project

8 + 12 + 0 + 6 + 8 + 10 = 44 days

• Project variance

1.78 + 1.78 + 0 +1 + 1.78 + 4 = 10.34

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• To find out the prob of completing the project, within , 48 days

Z = X - X¯ = 48 – 44 = 1.24

σ 3.216

value from normal tables = 0.8925

Hence prob of completing project within 48 days = 89 %

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AMITY GLOBALBUSINESS SCHOOL BangaloreQ (Dummy Variable )

A small project consists of seven activities for which the relevant data are given :

• Draw the network and find the project completion time

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Activity Preceding activity

Activity duration ( days )

A - 4

B - 7

C - 6

D A,B 5

E A,B 7

F C,D,E 6

G C,D,E 5

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A D1 G D3

B E F

C D D2

10 0

24 7

37 7

46 14

514 14

614 20

720 20

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CRITICAL PATH IS 1-3-5-7

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• Q has been commissioned by the food manufacturer Z to carry out market research in a new product development project, prior to a test market launch. The table below lists required activities and their duration in weeks

• Draw the new product development network. State and explain the critical path and its duration

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Prepare a table of the earliest start and finish times, the latest start and finish times and the total and free float

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Activity Immediate predecessor

duration

A - 6

B - 4

C A 20

D B 12

E B 11

F C 5

G C,D,E 4

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H E 30

I G 8

J G 10

K F,I 7

L H,J,K 6

M H,J,K 4

N L 10

P M,N 3

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Use only 2 dummy variables

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• F P C A R I K M N

B D G J L S E H

10 0

26 6

34 4

426 26

838 38

626 26

515 15

730 30 9

45 4510

51 51

1161 61

12

64 64

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Activity

duration

ES EF LS LF Tot F Free

A 6 0 6 0 6 0 0

B 4 0 4 0 4 0 0

C 20 6 26 6 26 0 0

D 12 4 16 14 26 10 10

E 11 4 15 4 15 0 0

F 5 26 31 33 38 7 7

G 4 26 30 26 30 0 0

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H 30 15 45 15 45 0 0

I 8 30 38 30 38 0 0

J 10 30 40 35 45 5 5

K 7 38 45 38 45 0 0

L 6 45 51 45 51 0 0

M 4 45 49 57 61 12 12

N 10 51 61 51 61 0 0

P 3 61 64 61 64 0 0

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• Floats are useful to management as they indicate how much an activity can be delayed without delaying the entire project. Resources can be redeployed from an activity with slack to be used elsewhere

• Slack --- events

• Head slack = Lj - Ej

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• Tail Slack = Li – Ei

• Total float = Lj – Ei - tij

• Free float = total float – head slack

• Independent float = free float – tail slack

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Activity Preceding activity

to tm tp

A Nil 4 7 16

B Nil 1 5 15

C A 6 12 30

D A 2 5 8

E C 5 11 17

F D 3 6 15

G B 3 9 27

H E , F 1 4 7

I G,H 4 19 28

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• Draw the PERT network diagram• Identify the critical path• Find the expected duration and variance for each

activity. What is expected project length ?• What is the prob that the project is completed

between 35 and 40 weeks ?

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activity to tm tp te σ²

Time estimates(wks)

A 1-2 4 7 16 8 4

B 1-3 1 5 15 6 49 / 4

C 2-4 6 12 30 14 16

D 2-5 2 5 8 5 1

E 4-6 5 11 17 11 4

F 5-6 3 6 15 7 4

G 3-7 3 9 27 11 16

H 6-8 1 4 7 4 1

I 7-8 4 19 28 18 16

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10 0

28 8

36 8

422 22

633 33

513 26

717 19

8

37 3755 55

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• Critical path

1---2 ---4 –6---7----8---9

• Expected project length

8 + 14 + 11 + 4 = 55 weeks

• Variance

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• Prob that the project will be completed between 35 and 40 weeks

prob { 35 - 55 ≤ Z = Ts – Te ≤ 40 – 55 }

σ σ e σ

= p ( - ≤ Z ≤ )

= P ( - ≤ Z ≤ 0 ) + P ( 0 ≤ Z ≤ 0.6 )

=

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AMITY GLOBALBUSINESS SCHOOL BangaloreCrashing

The process of reducing an activity time by putting extra effort is called crashing the activity. It is the process of shortening a project and is usually achieved by adding extra resources to an activity.

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• Direct costs are the costs directly associated with each activity such as machine costs, labor costs and so on.

• Indirect costs are the costs due to management services, rentals, insurance etc.

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• Duration of any activity cannot be reduced indefinitely

• Crash time represents minimum activity duration time that is possible.

• Activity cost corresponding to the crash time is called the crash cost.

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AMITY GLOBALBUSINESS SCHOOL BangaloreActivity crashing

Act

ivit

y co

st

Activity time

Crashing activity

Crash time

Crash cost

Normal Activity

Normal time

Normal cost

Slope = crash cost per unit time

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• Crashing costs increase as project duration decreases

• Indirect costs increase as project duration increases

• Reduce project length as long as crashing costs are less than indirect costs

Time-Cost Relationship

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Time-Cost Tradeoff

cost

time

Direct cost

Indirect cost

Total project costMin total cost = optimal project time

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Cost slope =

Crash cost – normal cost

Normal time – crash time

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The following table gives the activities in a construction project and other relevant information

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Activity Predecessor

Time Direct Costs

Normal Crash Normal Crash

A - 4 3 60 90

B - 6 4 150 250

C - 2 1 38 60

D A 5 3 150 250

E C 2 2 100 100

F A 7 5 115 175

G D,B,E 4 2 100 240

Total 713

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Indirect costs vary as follows :

Days 15 14 13 12 11 10 9 8 7 6

Cost-Rs 600 500 400 250 175 100 75 50 35 25

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• Draw an arrow diagram of the network

• Determine the project duration which will result in minimum total cost.

• Activity F & G end at the same time.

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3

24 4

513 13

• 2 4

C 2 E 2

B 4 G

6 9 9 4

A 4 D 5 F 7

10 0

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The critical path is 1-2-4-5.

Critical activities are 1-2, 2-4 and 4-5

Project duration is 13 days

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Activity A B C D E F G

1-2 1-4 1-3 2-4 3-4 2-5 4-5

Cost slope

30 50 22 50 - 30 70

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Among the critical activities, choose activity with least cost slope

i.e A (1-2)

Crash activity A by 4-3 = 1 day.

Project duration reduces by 1 day

= 13-1 = 12days

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Total project cost = total direct cost + increased direct cost due to crashing +

indirect cost for 12 days

= 713 + 1 * 30(cost slope) + 250

= 993

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AMITY GLOBALBUSINESS SCHOOL BangaloreNew network

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23 3

512 12

C 2 E 2

1 6 4 4

0 0 B 8 G

A 3 D 5 F 7

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The critical path is 1-2-4-5 Total time = 12 days;

The critical activities are 1-2, 2-4, 4-5

The first one A 1-2 has already been crashed

From the other two choose the one with the smallest cost slope and crash it.

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Activity D(2-4) is crashed by 5-3=2 days.

90


3

23 3

510 10

2 2

C 2 E 2

B 4 G

6 6 6 4

A 3 D 3 F 7

10 0

91


There are 3 critical paths

1-2-5

1-4-5

1-2-4-5

Each with 10 days

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Total cost = 743 +2*50 +100 = Rs. 943

since there are more than one parallel critical path, crashing must be performed.

As A & D are crashed, crash activities F & G by 2 days each to reduce the project duration by 2 days.

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3

23 3

58 8

2 2

C 2 E 2

B 4 G

6 6 6 2

A 3 D 3 F 5

10 0

94


There are 3 critical paths1-4-51-2-51-2-4-5Each with 8 days

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• Total cost = 843 +(2*30 +2*70) +50

= Rs. 1093

Since all the activities in the critical path 1-2-5 are crashed , further crashing is not possible.

Decision : minimum total cost is Rs. 943 and the optimum duration is 10 days.

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Q

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The following table gives data on normal time, and cost and crash time and cost for a project.

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AMITY GLOBALBUSINESS SCHOOL BangaloreActivity Normal Crash

Time(weeks) Cost (INR) Time(weeks) Cost (INR)

1-2 3 300 2 400

2-3 3 30 3 30

2-4 7 420 5 580

2-5 9 720 7 810

3-5 5 250 4 300

4-5 0 0 0 0

5-6 6 320 4 410

6-7 4 400 3 470

6-8 13 780 10 900

7-8 10 1000 9 1200

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Indirect cost is Rs. 50 per week.

1. Draw the network for the project and identify the critical path

2. What are the normal project duration and associated cost

3. Crash the relevant activities systematically and determine the optimal project completion time and cost.

100


10 0

2

3 3

4

10 12

36 7

512 12

618 18

722 22

832 32

101


The critical path is

1-2-5-6-7-8

Project duration is 32 weeks and associated cost is as follows :

= direct normal cost +indirect cost for 32 weeks

= 4220 + 50*32 = Rs. 5820

102

AMITY GLOBALBUSINESS SCHOOL BangaloreCrash cost slope for critical

activities are 1-2 400-300 = 100

3-2

2-5 810 – 720 = 45

9-7

5-6 45

6-7 70

7-8 200

103


The minimum crash cost per week is for activity 2-5 & 5-6.

Hence, crash activity 2-5 by 2 weeks.

But, the activity should b reduced only by 1 week – so that another critical path be raised

104


• 81

0 0

2

3 3

4

10 11

36 6

511 11

617 17

721 21

831 31

105


Network is as shown in previous slide.

New project time is 31 weeks

The critical paths are

1-2-5-6-7-8

1-2-3-5-6-7-8

106


The new total cost

= total direct normal cost + increased direct cost due to crashing + indirect cost for 31 weeks.

= (4220 + 1*45) + 31 * 50 = Rs. 5815

107

AMITY GLOBALBUSINESS SCHOOL BangaloreWrt new network

Crashed Activity Slope

1-2 100

2-5 Crashed

2-3 No need(not in first path)

3-5 50

5-6 45

6-7 70

7-8 200

108


Crash 5-6 by 2 weeks.

109


• 8 41

0 0

2

3 3

4

10 11

36 6

511 11

615 15

719 19

829 29

110


The new total cost

= total direct normal cost + increased direct cost due to crashing of 5-6 + indirect cost for 29 weeks.

= (4220 + 1*45 + 2 *45) + 29 * 50 = Rs. 5805

111


If we go in for further crashing the project cost wil;l go up (Rs.5825), hence stop here.

project management - cpm/pert

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