Post on 03-Apr-2018
7/28/2019 CM P Time-Cost
1/13
Time-CostTrade-Off
13/01/2002 Emad Elbeltagi
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
7/28/2019 CM P Time-Cost
2/13
13/01/2002 Emad Elbeltagi
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 flow
13/01/2002 Emad Elbeltagi
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
7/28/2019 CM P Time-Cost
3/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Activity Time-Cost Relationship
Decreasing activity duration will increase its cost
Normal duration
&
Normal cost
Crash duration
&
Crash cost
Time
Cost
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Activity Time-Cost Relationship
Crash duration
&
Crash cost
Time
Cost
Normal duration
&
Normal cost
7/28/2019 CM P Time-Cost
4/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Activity Time-Cost Relationship
For simplicity, linear relationship is adopted
Normal duration
&
Normal cost
Crash duration
&
Crash cost
Time
Cost
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Activity Time-Cost Relationship (Example)
Consider the following options
1 scaffold set, 2 labors, 2 carpenter, 1 foreman
2 scaffold set, 3 labors, 2 carpenter, 1 foreman
2 scaffold set, 3 labors, 3 carpenter, 1 foreman
5
6
7
166
204
230
Crew formationCrew size
(men)
Estimated daily
production
(square meter)
Consider the following rates: Labor LE 96/day;carpenter LE 128/day;foreman LE144/day andscaffolding LE60/day
7/28/2019 CM P Time-Cost
5/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
51 x (1x60 + 2x96 + 2x128 + 1x144) = 33252
42 x (2x60 + 3x96 + 2x128 + 1x144) = 33936
37 x (2x60 + 3x96 + 3x128 + 1x144) = 34632
50.6 (use 51)
41.2 (use 42)
36.5 (use 37)
5
6
7
Cost (LE)Duration (days)Crew
size
Activity Time-Cost Relationship (Example)
8400 Sq meters of scaffolds
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Activity Time-Cost Relationship (Example)
33000
33200
33400
33600
33800
34000
34200
34400
34600
34800
30 35 40 45 50 55
Duration(days)
Cost(LE)
1
2
3
7/28/2019 CM P Time-Cost
6/13
13/01/2002 Emad Elbeltagi
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 as project duration decreases
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Project Time-Cost Relationship
Project duration
Project cost
7/28/2019 CM P Time-Cost
7/13
13/01/2002 Emad Elbeltagi
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 will be applied
13/01/2002 Emad Elbeltagi
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
7/28/2019 CM P Time-Cost
8/13
13/01/2002 Emad Elbeltagi
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 itscrash 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)
13/01/2002 Emad Elbeltagi
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.
7/28/2019 CM P Time-Cost
9/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Example Application
Consider the following project
Indirect cost LE 125/day
Crash to 49 days
7200
5300
4600
5000
1050
3300
6300
2580
3150
10
6
12
23
4
4
15
11
10
7000
5000
4000
5000
1000
3000
6000
2500
3000
12
8
15
23
5
5
20
13
12
-
A
A
B
B
C
E, C
F
D, G, H
A
B
C
D
E
F
G
H
I
Cost (LE)Duration (day)Cost (LE)Duration (day)
CrashNormalPreceded byActivity
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
D (23)
24 47
20 43
I (12)
47 59
47 59
H (13)
34 47
32 45
F (5)
29 34
27 32
B (8)
14 22
12 20
A (12)
0 12
0 12
C (15)
12 27
12 27
E (5)
22 27
20 25
G (20)
27 47
27 47
2@40
2@100 1@502@150
3@200
5@60
1@300
2@75
Example Application
Total direct cost LE 36500
7/28/2019 CM P Time-Cost
10/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-OffExample Application
Activity G has the least cost slope 60 LE/day
Crash G by 2 days, cost increase by 2 x 60
D (23)
22 45
20 43
I (12)
45 57
45 57
H (13)
32 45
32 45
F (5)
27 32
27 32
B (8)
14 22
12 20
A (12)
0 12
0 12
C (15)
12 27
12 27
E (5)
22 27
20 25
G (18)
27 45
27 45
2@40
2@100 1@502@150
3@200
3@60
1@300
2@75
13/01/2002 Emad Elbeltagi
Time-Cost Trade-OffExample Application
Activity I has the least cost slope 75 LE/day
Crash I by 2 days, cost increase by 2 x 75
D (23)
22 45
20 43
I (10)
45 55
45 55
H (13)
32 45
32 45
F (5)
27 32
27 32
B (8)
14 22
12 20
A (12)
0 12
0 12
C (15)
12 27
12 27
E (5)
22 27
20 25
G (18)
27 45
27 45
2@40
2@100 1@502@150
3@200
3@60
1@300
7/28/2019 CM P Time-Cost
11/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-OffExample Application
Activity A has the least cost slope 100 LE/day
Crash A by 2 days, cost increase by 2 x 100
D (23)
20 43
18 41
I (10)
43 53
43 53
H (13)
30 43
30 43
F (5)
25 30
25 30
B (8)
12 20
10 18
A (10)
0 10
0 10
C (15)
10 25
10 25
E (5)
20 25
18 23
G (18)
25 43
25 43
2@40
1@502@150
3@200
3@60
1@300
13/01/2002 Emad Elbeltagi
Time-Cost Trade-OffExample Application
Activity H & G has the least cost slope 100 LE/day
Crash H & G by 2 days, cost increase by 2 x 100
D (23)
18 41
18 41
I (10)
41 51
41 51
H (11)
30 41
30 41
F (5)
25 30
25 30
B (8)
10 18
10 18
A (10)
0 10
0 10
C (15)
10 25
10 25
E (5)
20 25
18 23
G (16)
25 41
25 41
1@502@150
3@200
1@60
1@300
7/28/2019 CM P Time-Cost
12/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-OffExample Application
Activity C & B has the least cost slope 350 LE/day
Crash H & G by 2 days, cost increase by 2 x 350
D (23)
16 39
16 39
I (10)
39 49
39 49
H (11)
28 39
28 39
F (5)
23 28
23 28
B (6)
10 16
10 16
A (10)
0 10
0 10
C (13)
10 3
10 23
E (5)
18 23
16 21
G (16)
23 39
23 39
1@50
1@200
1@60
1@300
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Example Application
43.9956.12537.8749
43.5456.37537.1751
43.5956.62536.9753
43.6456.87536.7755
43.7457.12536.6257
43.8757.37536.5059
Total cost x 1000 LEIndirect cost x 1000 LEDirect cost X 1000 LEDuratio
n
7/28/2019 CM P Time-Cost
13/13
13/01/2002 Emad Elbeltagi
Time-Cost Trade-Off
Example Application
0
10
20
30
40
50
48 50 52 54 56 58 60
Time (days)
LEx
1000