Project Management - CPM/PERT

Network 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

Objectives Minimization of total time Minimization of total cost Mimimize production delays

Applications Construction of bridges Assembly line scheduling Installation of complex equipment Inventory planning and Control Construction of residential complex

Components 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

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

Critical 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)

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.

Float 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

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

PERT 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)

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

Q For the following PERT diagram : D A 3 4 H 2 B E I K 2 4 5 3 F 7 2 C 2 4 5 3 F 7 2 C G 4 J 6 3 2 9 4 7 8 1 6 5

Compute earliest event time and latest event time Critical path and total project duration

Forward pass method E1 =0 E2= E1 + t1,2 = 0 + 3 E6 = max ( E4 + t4,6 ; E5 + t5,6 ) = max ( 9, 6 ) = 9

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

Backward Pass Method L9= E9 = 15 L8 = L9 – t8,9 = 15 -3 =12 L4 = min( L6- t4,6; L7-t4,7) min( 9-7 ; 7 - 4) = 2

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

For the following PERT diagram : D A 3 4 H 2 B E I K 2 4 5 3 F 7 2 C 2 4 5 3 F 7 2 C G 4 J 6 3 7 10 2 3 6 9 15 15 4 2 2 7 6 7 8 11 12 1 0 0 6 9 9 5 2 5

Q 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 :

Activity Description Time(wk) A1-2 B 2-3 C 2-4 D 2-5 E 3-6 F 4-7 G 5-8 H 6-7 I 6-8 J 6-9 Organize sales office Hire salesmen Train salesmen Select advertising agency Plan advertising campaign Conduct advertising campaign Design package Set up packaging facilities Package initial stocks Order stock from manufacturer 2 4 14 6 3

K7-10 L 8-11 M 9-12 N 10-12 O 11-12 P 12-13 Select distributors Sell to distributors Ship stocks Check inventory Get customer feedback Re assess strategy 3 5 2

What is the critical path Calculate total and free float

6 3 2 3 2 2 4 14 4 3 3 2 3 3 3 5 4 4 20 7 23 26 10 26 29 12 31 31 13 34 34 1 0 0 6 20 20 9 24 28 2 2 2 3 6 6 5 4 20 8 23 23 11 28 28

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

Time Estimates Acti Dur Earliest Start Fin Latest Start Fin Float Total Free A 2 0 2 0 2 0 0 B 4 6 C 2 4 18 20 D 2 20 E 14 6 20 6 20 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

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 4 4 3 3 0 0

Q : A project schedule has the following characteristics : Activity Time(days) 1-2 4 5-6 1-3 1 5-7 8 2-4 6-8 3-4 7-8 2 3-5 6 8-10 5 4-9 9-10 7

Construct a network diagram Compute the total float Find the critical path and total project duration

4 1 5 7 1 1 8 2 9 10 15 2 4 9 10 22 22 4 5 10 1 0 0 7 15 15 8 17 3 1 1 5 7 7 6 11 16

Critical path is 1-3-5-7-8-10 and the project duration is 22 days Total Float = ( Lj-tij ) - Ei

Activit Duratio Earliest start Earliest finish Latest start Latest finish Total float 1-2 4 5 9 1-3 1 2-4 10 3-4 2 11 7 3-5 6 5-7 8 15 5-6 12 16

7-8 2 15 17 6-8 1 11 12 16 5 4-9 10 7 9 -10 22 8-10

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

Q 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.

activity Most optimistic time ( in days ) Most likely time m Most pessimistic time b 1-2 4 8 12 2-3 1 7 2-4 16 3-5 3 5 4-5 4-6 6 9 5-7

5-8 4 6 8 7-9 12 8-9 2 5 9-10 10 16 6-10

Calculate expected time te = a + 4m + b 6 Variance σ² = b-a ²

Activity a m b te σ² 1-2 4 8 12 1.78 2-3 1 7 2-4 16 3-5 3 5 0.44 4-5 4-6 6 9

5-7 3 6 9 1 5-8 4 8 0.44 7-9 12 1.78 8-9 2 5 9-10 10 16 6-10

Earliest Start Time E(μ1)= 0 E(μ2) = 0+8 = 8 E (μ3) = 8 + 4 = 12 E(μ5) = Max(12+5,20+0) = 20 E(μ6) = 20+6 = 26

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

Latest Start Time E(L10) = 44 E(L9) = 44-10 = 34 E(L8) = 34 – 5 = 29 E(L5) = Min (26-6, 29 -6) = 20

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

3 12 15 7 26 26 9 34 34 1 0 0 5 20 20 2 8 8 8 28 29 4 20 20 10 44 44 6 26 38

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

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 %

Q (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

Activity Preceding activity Activity duration ( days ) A - 4 B 7 C 6 D A,B 5 E F C,D,E G

A D1 G D3 B E F C D D2 2 4 7 6 14 20 3 7 7 1 0 0 5 14 14 7 20 20 4 6 14

CRITICAL PATH IS 1-3-5-7

Q 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

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

Activity Immediate predecessor duration A - 6 B 4 C 20 D 12 E 11 F 5 G C,D,E

H E 30 I G 8 J 10 K F,I 7 L H,J,K 6 M 4 N P M,N 3

Use only 2 dummy variables

F P C A R I K M N B D G J L S E H 12 64 64 4 26 26 8 38 38 11 61 61 2 6 6 1 0 0 7 30 30 6 26 26 9 45 45 10 51 51 3 4 4 5 15 15

Activity duration ES EF LS LF Tot F Free A 6 B 4 C 20 26 D 12 16 14 10 E 11 15 F 5 31 33 38 7 G 30

H 30 15 45 I 8 38 J 10 40 35 5 K 7 L 6 51 M 4 49 57 61 12 N P 3 64

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

Tail Slack = Li – Ei Total float = Lj – Ei - tij Free float = total float – head slack Independent float = free float – tail slack

Q Activity Preceding activity to tm tp A Nil 4 7 16 B 1 5 15 C 6 12 30 8 E 11 17 F 3 G 9 27 H E , F I G,H 19 28

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 ?

activity to tm tp te σ² Time estimates(wks) A 1-2 4 7 16 8 B 1-3 1 5 15 6 49 / 4 C 2-4 12 30 14 D 2-5 2 E 4-6 11 17 F 5-6 3 G 3-7 9 27 H 6-8 I 7-8 19 28 18

4 22 22 6 33 33 2 8 8 5 13 26 1 0 0 8 37 37 55 55 3 6 8 7 17 19

Critical path 1---2 ---4 –6---7----8---9 Expected project length 8 + 14 + 11 + 4 = 55 weeks Variance

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 ) =

Crashing 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.

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.

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.

Activity crashing Crash cost Crashing activity Activity cost Activity time Crashing activity Crash time Crash cost Slope = crash cost per unit time Normal Activity Normal time Normal cost

Time-Cost Relationship 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 Tradeoff Min total cost = optimal project time Direct cost Total project cost Indirect cost

Cost slope = Crash cost – normal cost Normal time – crash time

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

Activity Predecessor Time Direct Costs Normal Crash A - 4 3 60 90 B 6 150 250 C 2 1 38 D 5 E 100 F 7 115 175 G D,B,E 240 Total 713

Indirect costs vary as follows : 15 14 13 12 11 10 9 8 7 6 Days 15 14 13 12 11 10 9 8 7 6 Cost-Rs 600 500 400 250 175 100 75 50 35 25

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.

3 2 4 C 2 E 2 B 4 G 6 9 9 4 A 4 D 5 F 7 5 13 13 1 0 0 2 4 4

The critical path is 1-2-4-5. Critical activities are 1-2, 2-4 and 4-5 Project duration is 13 days

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 - 70

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

Total project cost = total direct cost + increased direct cost due to crashing + indirect cost for 12 days = 713 + 1 * 30(cost slope) + 250 = 993

New network

C 2 E 2 1 6 4 4 0 0 B 8 G A 3 D 5 F 7 5 12 12 2 3 3

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.

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

3 2 2 C 2 E 2 B 4 G 6 6 6 4 A 3 D 3 F 7 5 10 10 1 0 0 2 3 3

There are 3 critical paths 1-2-5 1-4-5 1-2-4-5 Each with 10 days

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.

3 2 2 C 2 E 2 B 4 G 6 6 6 2 A 3 D 3 F 5 5 8 8 1 0 0 2 3 3

There are 3 critical paths 1-4-5 1-2-5 1-2-4-5 Each with 8 days

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.

Q

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

Activity Normal Crash 1-2 3 300 2 400 2-3 30 2-4 7 420 5 580 2-5 9 720 Time(weeks) Cost (INR) 1-2 3 300 2 400 2-3 30 2-4 7 420 5 580 2-5 9 720 810 3-5 250 4 4-5 5-6 6 320 410 6-7 470 6-8 13 780 10 900 7-8 1000 1200

Indirect cost is Rs. 50 per week. Draw the network for the project and identify the critical path What are the normal project duration and associated cost Crash the relevant activities systematically and determine the optimal project completion time and cost.

7 22 22 4 10 12 1 0 0 5 12 12 6 18 18 2 3 3 8 32 32 3 6 7

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

Crash 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

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

7 21 21 8 4 10 11 1 0 0 5 11 11 6 17 17 2 3 3 8 31 31 3 6 6

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

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

Wrt 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

Crash 5-6 by 2 weeks.

7 19 19 8 4 4 10 11 1 0 0 5 11 11 6 15 15 2 3 3 8 29 29 3 6 6

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

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