The document discusses the relationship between time and cost for activities and projects. It explains that reducing the duration of an activity typically increases its cost in a linear relationship. For projects, reducing the overall duration lowers indirect costs but raises direct costs due to crashing activities. An example demonstrates calculating optimal project duration that results in lowest total cost by shortening critical path activities based on their cost slopes.

1. Time-Cost
Trade-Off
Time-Cost Trade-Off
Duration and cost of activity depends on the amount an type
of resources used
Assuming more workers will normally reduce time and will
change its cost
This relationship between activity time and cost called
time-cost Trade off
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2. Time-Cost Trade-Off
Why do we need to reduce project time?
Finish the project in a predefined deadline date
Recover early delays to avoid liquidated damages
Free key resources early for other projects
Avoid adverse weather conditions that might affect productivity
Receive an early completion-bonus
Improve project cash flowscheduling techniques are duration
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Time-Cost Trade-Off
What is the way to reduce activity duration?
Working extended hours (Over time)
Offering incentive payments to increase the productivity
Using additional resources
Using materials with faster installation methods
Using alternate construction methods or sequence
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3. Time-Cost Trade-Off
Activity Time-Cost Relationship
Decreasing activity duration will increase its cost
Cost
Crash duration
&
Crash cost
Normal duration
&
Normal cost
Time
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Time-Cost Trade-Off
Activity Time-Cost Relationship
Cost Crash duration
&
Crash cost
Normal duration
&
Normal cost
Time
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4. Time-Cost Trade-Off
Activity Time-Cost Relationship
For simplicity, linear relationship is adopted
Cost Crash duration
&
Crash cost
Normal duration
&
Normal cost
Time
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Time-Cost Trade-Off
Activity Time-Cost Relationship (Example)
Consider the following options
Estimated daily
Crew size
production Crew formation
(men)
(square meter)
166 5 1 scaffold set, 2 labors, 2 carpenter, 1 foreman
204 6 2 scaffold set, 3 labors, 2 carpenter, 1 foreman
230 7 2 scaffold set, 3 labors, 3 carpenter, 1 foreman
Consider the following rates: Labor LE 96/day;
carpenter LE 128/day;
foreman LE144/day and
scaffolding LE60/day
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5. Time-Cost Trade-Off
Activity Time-Cost Relationship (Example)
8400 Sq meters of scaffolds
Crew
Duration (days) Cost (LE)
size
5 50.6 (use 51) 51 x (1x60 + 2x96 + 2x128 + 1x144) = 33252
6 41.2 (use 42) 42 x (2x60 + 3x96 + 2x128 + 1x144) = 33936
7 36.5 (use 37) 37 x (2x60 + 3x96 + 3x128 + 1x144) = 34632
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Time-Cost Trade-Off
Activity Time-Cost Relationship (Example)
34800
3
34600
34400
34200
Cost (LE)
34000 2
33800
33600
33400 1
33200
33000
30 35 40 45 50 55
Duration (days)
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6. Time-Cost Trade-Off
Project Time-Cost Relationship
Project direct cost = summation of direct cost of individual
activities
Indirect cost comprises site and head office overheads
Project indirect cost = daily indirect cost x project duration
Direct cost increases as project duration decreases
Indirect cost decreases s project duration decreases
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Time-Cost Trade-Off
Project Time-Cost Relationship
Project cost
Project duration
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7. Time-Cost Trade-Off
Shortening Project Duration (Optimum Duration)
It might be necessary to shorten project duration to meet
specific deadline
Or to determine the optimum project duration that is
correspond to the least project total cost
Many approaches can be used
Heuristic Cost-Slope method wii be applied
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Time-Cost Trade-Off
Procedure for Shortening Project Duration
Draw the project network.
Perform CPM calculations and identify the critical path, use
normal durations and costs for all activities.
Compute the cost slope for each activity from the following
equation: cost slope = crash cost – normal cost /
normal duration – crash duration
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8. Time-Cost Trade-Off
Procedure for Shortening Project Duration
Start by shortening the activity duration on the critical path
which has the least cost slope and not been shortened to its
crash duration.
Reduce the duration of the critical activities with least cost slope
until its crash duration is reached or until the critical path
changes.
When multiple critical paths are involved, the activity(ies) to be
shorten is determined by comparing the cost slope of the
activity which lies on all critical paths (if any)
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Time-Cost Trade-Off
Procedure for Shortening Project Duration
Having shortened a critical path, you should adjust activities
timings, and floats.
The cost increase due to activity shortening is calculated as the
cost slope multiplied by the time of time units shortened.
Continue until no further shortening is possible, and then the
crash point is reached.
The results may be represented graphically by plotting project
completion time against cumulative cost increase.
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9. Time-Cost Trade-Off
Example Application
Consider the following project
Indirect cost LE 125/day
Crash to 49 days
Normal Crash
Activity Preceded by
Duration (day) Cost (LE) Duration (day) Cost (LE)
A - 12 7000 10 7200
B A 8 5000 6 5300
C A 15 4000 12 4600
D B 23 5000 23 5000
E B 5 1000 4 1050
F C 5 3000 4 3300
G E, C 20 6000 15 6300
H F 13 2500 11 2580
I D, G, H 12 3000 10 3150
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Time-Cost Trade-Off
Example Application
Total direct cost LE 36500
20 43
D (23)
24 47
0 12 12 20 20 25 27 47 47 59
A (12) B (8) E (5) G (20) I (12)
0 12 14 22 22 27 27 47 47 59
2@100 2@150 1@50 5@60 2@75
12 27 27 32 32 45
C (15) F (5) H (13)
12 27 29 34 34 47
3@200 1@300 2@40
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10. Time-Cost Trade-Off
Example Application
Activity G has the least cost slope 60 LE/day
Crash G by 2 days, cost increase by 2 x 60
20 43
D (23)
22 45
0 12 12 20 20 25 27 45 45 57
A (12) B (8) E (5) G (18) I (12)
0 12 14 22 22 27 27 45 45 57
2@100 2@150 1@50 3@60 2@75
12 27 27 32 32 45
C (15) F (5) H (13)
12 27 27 32 32 45
3@200 1@300 2@40
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Time-Cost Trade-Off
Example Application
Activity I has the least cost slope 75 LE/day
Crash I by 2 days, cost increase by 2 x 75
20 43
D (23)
22 45
0 12 12 20 20 25 27 45 45 55
A (12) B (8) E (5) G (18) I (10)
0 12 14 22 22 27 27 45 45 55
2@100 2@150 1@50 3@60
12 27 27 32 32 45
C (15) F (5) H (13)
12 27 27 32 32 45
3@200 1@300 2@40
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11. Time-Cost Trade-Off
Example Application
Activity A has the least cost slope 100 LE/day
Crash A by 2 days, cost increase by 2 x 100
18 41
D (23)
20 43
0 10 10 18 18 23 25 43 43 53
A (10) B (8) E (5) G (18) I (10)
0 10 12 20 20 25 25 43 43 53
2@150 1@50 3@60
10 25 25 30 30 43
C (15) F (5) H (13)
10 25 25 30 30 43
3@200 1@300 2@40
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Time-Cost Trade-Off
Example Application
Activity H & G has the least cost slope 100 LE/day
Crash H & G by 2 days, cost increase by 2 x 100
18 41
D (23)
18 41
0 10 10 18 18 23 25 41 41 51
A (10) B (8) E (5) G (16) I (10)
0 10 10 18 20 25 25 41 41 51
2@150 1@50 1@60
10 25 25 30 30 41
C (15) F (5) H (11)
10 25 25 30 30 41
3@200 1@300
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12. Time-Cost Trade-Off
Example Application
Activity C & B has the least cost slope 350 LE/day
Crash H & G by 2 days, cost increase by 2 x 350
16 39
D (23)
16 39
0 10 10 16 16 21 23 39 39 49
A (10) B (6) E (5) G (16) I (10)
0 10 10 16 18 23 23 39 39 49
1@50 1@60
10 23 23 28 28 39
C (13) F (5) H (11)
10 3 23 28 28 39
1@200 1@300
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Time-Cost Trade-Off
Example Application
Duratio
Direct cost X 1000 LE Indirect cost x 1000 LE Total cost x 1000 LE
n
59 36.50 7.375 43.875
57 36.62 7.125 43.745
55 36.77 6.875 43.645
53 36.97 6.625 43.595
51 37.17 6.375 43.545
49 37.87 6.125 43.995
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13. Time-Cost Trade-Off
Example Application
50
40
LE x 1000
30
20
10
0
48 50 52 54 56 58 60
Time (days)
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