1. Project Management - CPM/PERT

2. 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

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

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

5. 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

6. 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

7. 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)

8. 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.

9. 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

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

11. 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)

12. 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

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

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

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

16. E7 = E4 + t4,7 = = 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

17. 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

18. L3 = L8 – t 3,8 = = 10
L2= L3- t2,3 = = 6
L1= min ( L2-t1,2 ; L4- t1,4; L5 – t1,5)
= min ( 6-3; 2-2; 5-2 ) = 0

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

20. 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 :

21. 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

22. 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

23. What is the critical path
Calculate total and free float

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

25. 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

26. Time Estimates Acti Dur Earliest Start Fin Latest Start Fin Float
Total Free A B 6 C 4 D 20 E F
16 13=23-4-6 G
=23-4-3 H

27. Activity Duration
Earliest
Start Finish
Latest
Start Finish
Float I J K L M N O P

29. 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

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

31. 9
10 15 2
4 9 10
22 22 4
5 10 1
0 0 7
15 15 8 17 3 5
7 7 6
11 16

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

33. 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

34. 7-8 2 15 17
6-8 1 11 12 16 5
4-9 10 7
9 -10 22
8-10

36. 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

37. 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.

38. 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

39. 5-8 4 6 8
7-9 12
8-9 2 5
9-10 10 16
6-10

40. Calculate expected time te = a + 4m + b6
Variance σ² = b-a ²

41. 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

42. 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

43. 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

44. E (μ7) = = 26
E(μ8)= = 26
E(μ9) = Max(26+8, 26+5) = 34
E(μ10) =Max( , ) = 44

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

46. E(L4) = Min(20 – 0, 36 – 6 ) = 20
E(L3) = 20 – 5 = 15
E(L2) = Min(15-4, ) = 8
E(L1) = 8 – 8 = 0
Draw network diagram and find critical path

47. 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

48. Critical path is
Expected time in completing the project
= 44 days
Project variance
= 10.34

49. To find out the prob of completing the project, within , 48 days
Z = X - X¯ = 48 – 44 = 1.24 σ
value from normal tables =
Hence prob of completing project within 48 days = 89 %

50. 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

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

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

53. CRITICAL PATH IS

54. 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

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

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

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

58. Use only 2 dummy variables

59. F P C
A R I K M N
B D G J L S
E H 12
64 64 4
26 26 8 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

60. 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

61. 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

62. 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

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

64. Q Activity Preceding activity to tm tp A Nil 4 7 16 B 1 5 15 C 6 12 308 E 11 17 F 3 G 9 27 H
E , F I
G,H 19 28

65. 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 ?

66. 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

67. 4
22 22 6
33 33 2
8 8 5
13 26 1
0 0 8 3
6 8 7

68. Critical path –
Expected project length
= 55 weeks
Variance

69. Prob that the project will be completed between 35 and 40 weeks
prob { ≤ Z = Ts – Te ≤ 40 – 55 }
σ σ e σ
= p ( ≤ Z ≤ )
= P ( - ≤ Z ≤ 0 ) + P ( 0 ≤ Z ≤ 0.6 ) =

70. 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.

71. 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.

72. 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.

73. 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

74. 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

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

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

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

78. 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

79. 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

80. 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.

81. 3
2 4
C E 2
B G
A D F 7 5
13 13 1 2

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

83. 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

84. 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

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

86. New network

87. C E 2
B G
A D F 7 5 2

88. 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.

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

90. 3
2 2
C E 2
B G
A D F 7 5 1
0 0 2
3 3

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

92. Total cost = * = 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.

93. 3
2 2
C E 2
B G
A D F 5 5 1
0 0 2
3 3

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

95. Total cost = 843 +(2*30 +2*70) +50
= Rs. 1093
Since all the activities in the critical path are crashed , further crashing is not possible.
Decision : minimum total cost is Rs. 943 and the optimum duration is 10 days.

96. Q

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

98. 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

99. 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.

100. 7
22 22 4
10 12 1
0 0 5
12 12 6
18 18 2
3 3 8
32 32 3
6 7

101. The critical path is
Project duration is 32 weeks and associated cost is as follows :
= direct normal cost +indirect cost for 32 weeks
= *32 = Rs. 5820

102. Crash cost slope for critical activities are
1-2
= 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. 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

105. Network is as shown in previous slide.
New project time is 31 weeks
The critical paths are

106. The new total cost
= total direct normal cost + increased direct cost due to crashing + indirect cost for 31 weeks.
= ( *45) + 31 * 50 = Rs. 5815

107. 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

108. Crash 5-6 by 2 weeks.

109. 7
19 19 4
10 11 1
0 0 5
11 11 6
15 15 2
3 3 8
29 29 3
6 6

110. The new total cost
= total direct normal cost + increased direct cost due to crashing of indirect cost for 29 weeks.
= ( * *45) + 29 * 50 = Rs. 5805

111. 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.