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Transcript of PROJECT NETWORKING
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NETWORKING PROJECTACTIVITIES
By
Prof. (Dr.) Indrasen Singh
Professor In-Charge, NICMAR
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Crashing and Updating
o Direct cost of an activity
o The direct cost of an activity depends upon the amount ofresources employed in the execution of the activity.
o This is the cost of materials consumed, machines, equipment and
labour employed to perform the activity.
o In CPM projects, the direct cost of the activity is generally afunction of its duration.
o By increasing the resources that is the activity cost, its duration
can be decreased.
o For Example, the white washing of a building can be done in 3days by a team of 4 workers.
o If the number of men is decreased to 3, it takes 4 days.
o Thus each time 12 man-days at a cost of say Rs. 360 (Rs. 30 per
man-day) are required.
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o the number of men is further decreased to say 2, they may take
say 7 days i.e. 14 man-days at a cost of Rs. 420.
o On the other hand if the resources that is the men are increased to
say 6, the work may takes 2.5 days or 15 man-days at a cost of
Rs. 450.
o The variation of activity cost with activity duration is illustrated
in figure 1.
o The variation of cost with time is rarely linear.
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2 2.5 3 4 5 6 7
Activity Duration
Figure 1
420
450
Cost
360
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Straight Line Approximation
o The cost-time curve of the activity is rarely available. Generally,
the normal and crash durations and costs of the activities are
estimated.
o Even if the cost-time curve of an activity is known, it is very
cumbersome to analyse it for cost calculations.
o For the convenience of computations, the cost-time curve,
depending upon its curvature is approximated by straight lines.
o If the curvature is less, a single straight line approximation can
be made, as in figure 2 and if the curvature is more, multi-
straight lines (segments) approximation can be adopted, as in
figure 3.
o The slope of the straight line (or a segment of line) gives the
increase in cost per unit time for expediting the activity. This is
called cost slope (CS).
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CT NT
Figure 2
CC
NC
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CT2 CT1 NT
Figure 3
CC2
CC1
NC
DuaratinCrashdurationNoraml
CostNormalCostCrashCSSlopeCost
,
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o Indirect cost of the project.
o The indirect cost of the project can be subdivided into tow parts;
fixed indirect cost and variable indirect cost.
o The fixed indirect cost is due to the general and administrative
expenses, license fee, consultants fee, insurance cost, and taxes
etc., and does not depend upon the progress of the project.
o The variable indirect cost depends upon the project duration, but
can not be attributed to individual activities. It consists of
overhead expenses, supervision, interest on capital and
depreciation, etc.
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Crashing the Network
o A network can be crashed or contracted by identifying the least
cost slope, critical activity and the float available on the paths
parallel to the critical path.
o The smallest amount of float on the parallel paths, determines the
duration by which the selected activity can be crashed without
making it non-critical.
o This process is continued, until either all the activities become
critical or all activities with cost slopes less than the indirect cost
are crashed.
o In the former case, that is when all activities become critical, the
project duration can be reduced by crashing more than one
activities in parallel.
o The combination of activities, which together give the least cost
slope less than the indirect cost, are selected for crashing.
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o They are crashed as far as they can without making any activity
non-critical.
o The process is continued, until the total cost of the project does
not start increasing.
o The project duration corresponding to the minimum cost is the
optimum duration.
o The crashing process can be continued if required, until the
minimum possible duration is reached.
o Example:
o Let us consider the following time and cost data for a project
network.
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Activity Normal
Duration (Days
Cost (Rs.) Crash Duration
(Days)
Cost (Rs.)
1-2
1-3
2-4
2-5
3-5
4-7
5-6
6-7
6-8
7-98-9
15
7
9
0
5
6
8
12
10
1010
10,000
7,000
16,000
0
6,000
8,000
7,000
8,000
9,500
8,50015,000
11
5
6
0
2
3
5
8
7
76
12,400
8,400
20,500
0
9,000
10,400
8,350
9,400
10,850
10,00018,000
o The indirect cost of the project is Rs. 1,000 per day, and it is
required to determine the optimum project duration.
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o The project network corresponding to the normal activity times is
shown in figure 4.
o The normal activity times, along-with crash activity times in
parenthesis are written along the activity arrows. The cost slope
of each activity is computed in Table 1 and the same is shown in
the network below the corresponding activity arrow.
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ActivityDuration
(Days) Direct Cost
T C
Cost
slope
C/T
Nor
mal
Cra
sh
Normal Crash
1-2
1-3
2-42-5
3-5
4-7
5-6
6-7
6-8
7-9
8-9
15
7
90
5
6
8
12
10
10
10
11
5
60
5
3
5
8
7
7
6
10,000
7,000
16,0000
6,000
8,000
7,000
8,000
9,500
8,500
15,000
12,400
8,400
20,5000
6,000
10,400
8,350
9,400
10,850
10,000
18,000
4
2
3-
-
3
3
4
3
3
4
2400
1400
4500-
-
2400
1350
1400
1350
1500
3000
600
700
1500-
-
800
450
350
450
500
750
Table: 1
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0 0 15 15 24 29
107
35 3545 45
15 15
23 23
33 35
1
9
8
765
4
3
215(11)
600
5(5)
750
9(6)
450
7(5)
350
10(7)
500
12(8)
1500
6(3)
800
8(5)
450
10(6)10(7)
Solution
Figure 4
700
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0 0 15 15 24 27
107
33 3343 43
15 15
23 23
33 33
1
9
8
765
4
3
215(11)
600
5(5)
750
9(6)
450
7(5)
350
10(7)
500
10(8)
1500
6(3)
800
8(5)
450
10(6)10(7)
700
Figure 5
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0 0 15 15 24 24
107
30 3040 40
15 15
20 20
30 30
1
9
8
765
4
3
215(11)
600
5(5)
750
9(6)
450
7(5)
350
10(7)
500
10(8)
1500
6(3)
800
5(5)
450
10(6)10(7)
700
Figure 6
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0 0 12 12 21 21
77
27 2737 37
12 12
17 17
27 27
1
9
8
765
4
3
212(11)
600
5(5)
750
9(6)
450
7(5)
350
10(7)
500
10(8)
1500
6(3)
800
5(5)
450
10(6)10(7)
700
Figure 7
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0 0 12 12 21 21
77
27 2734 34
12 12
17 17
24 24
1
9
8
765
4
3
212(11)
600
5(5)
750
9(6)
450
7(5)
350
7(7)
500
10(8)
1500
6(3)
800
5(5)
450
10(6)7(7)
700
Figure 8
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0 0 11 11 20 20
66
26 2633 33
11 11
16 16
23 23
1
9
8
765
4
3
211(11)
600
5(5)
750
9(6)
450
6(5)
350
7(7)
500
10(8)
1500
6(3)
800
5(5)
450
10(6)7(7)
700
Figure 9
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0 0 11 11 20 20
66
24 2431 31
11 11
16 16
23 23
1
9
8
765
4
3
211(11)
600
5(5)
750
9(6)
450
6(5)
350
7(7)
500
8(8)
1500
4(3)
800
5(5)
450
8(6)7(7)
700
Figure 10
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o The critical path of the normal network is 1-2-5-6-7-9 and the
normal project duration is 45 days.
o The sum of normal costs of the activities is Rs. 95,000.
Therefore, total project cost = 95,000 + 1000 x 45 = 1,40,000
o On the critical path the least cost activity is 6-7. On the
parallel path 6-8-9, there is a slack of 2 days while on path 2-
4-7 there is a slack of 5 days. Activity 6-7 is thus crashed by 2
days. The project duration becomes 43 days.
Total cost = 1,40,000 + 350 x 2 2 x 1,000 = 1,38,700.
The corresponding network is given in figure 5.
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o Activity 5-6 can be crashed by 3 days. The project duration
becomes 40 days.
Total Cost = 1,38,700 + 3 x 450 - 3 x 1,000 = 1,37,050.
The corresponding network is given in figure 6.
o Activity 1-2 can be crashed by 3 days project duration becomes
37 days, and
Total Cost = 1,37,050 + 3 x 600
3 x 1,000 = 1,35,850.The Corresponding network is shown in figure 7.
o Activity 6-8 and 7-9 combined together can be crashed by 3
days. With this crashing, the project duration becomes 34 days
and Total cost = 1,35,850 + 3 x 950
3 x 1,000 = 1,35,700.The corresponding network is given in figure 8. Thus, for the
given cost data, the optimum project duration is 34 days and the
optimum project cost is Rs. 1,35,700.
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Updating the Project
To illustrate the procedure of updating, let us consider the
network shown in figure 11.
00 108
1715
2424
88
1210
3636
2016
1818
74
52
3 9861
8
8 8
8
1016
6
10
57
Figure 11
12
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Suppose the progress of work is checked after 15 days, that is at
the end of 15 days, and it is observed that:
- Activities 1-2, 1-3, 1-4 and 2-5 are completed.
- Activity 2-6 is in progress and needs 2 days more.
- Activity 3-6 is in progress and needs 5 days more.
- Activity 4-7 is in progress and needs 1 day more and
- Activity 5-9 is in progress and needs 14 days more.- Also it is estimated that due to the non-availability of fast setting
cement, activity 7-8 will take 12 days while due to the arrival of
a new crane, activity 8-9 will now require only 10 days.
This information can be put into a tabular form, as given below:
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Activity Status Time Required
1-2
1-31-4
2-5
2-6
3-6
4-7
5-9
6-8
7-8
8-9
Completed
CompletedCompleted
Completed
In Progress
In Progress
In Progress
In Progress
Not Started
Not Started
Not Started
0
00
0
2
5
1
14
6
12
10
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Solution
o The method, which is comparatively more convenient, is to
bunch all the activities completed up to the date of review, into
one activity and represent this by one arrow called the elapsedtime arrow.
o All the activities in progress will burst from the end node of the
elapsed time arrow and their durations will now be the times
required for their completion, from the date of review. Theremaining activities will follow in their precedence order and will
carry their revised time estimates.
o The nodes in the new network will be numbered in a different
fashion.o Then following the forward and backward computations the
network can be analysed.
o Activity 10-20 in figure 12 represents the elapsed time of 15
days.
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o Activity 20-60, 20-65, 20-70 and 20-90 represent the activities 2-
6, 3-6, 4-7 and 5-9 of the previous network, which respectively
require 2,5,1 and 14 days.
o The critical path for the revised project network is 10-20-70-80-90, which gives the project duration of 38 days.
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o Thus the updated version of the project is of increased duration,
with some change in critical activities of the project.
00
20
3838
70
1515
65
2220
10
221760
2828
80 90Elapsed Time
15
14
1616
5 610
2