cm p time-cost

7/28/2019 CM P Time-Cost

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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


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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


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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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


Time-Cost Trade-Off


Crash duration

&

Crash cost

Time

Cost

Normal duration

&

Normal cost


4/13


Time-Cost Trade-Off


For simplicity, linear relationship is adopted

Normal duration

&

Normal cost

Crash duration

&

Crash cost

Time

Cost


Time-Cost Trade-Off

Activity Time-Cost Relationship (Example)

Consider the following options

1 scaffold set, 2 labors, 2 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


5/13


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


8400 Sq meters of scaffolds


Time-Cost Trade-Off


33000

33200

33400

33600

33800

34000

34200

34400

34600

34800

30 35 40 45 50 55

Duration(days)

Cost(LE)

1

2

3


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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


Time-Cost Trade-Off

Project Time-Cost Relationship

Project duration

Project cost


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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


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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Time-Cost Trade-Off


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)


Time-Cost Trade-Off


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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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


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


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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



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


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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



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


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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


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


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Time-Cost Trade-Off

Example Application

0

10

20

30

40

50

48 50 52 54 56 58 60

Time (days)

LEx

1000

cm p time-cost

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