time cost-trade-off

Spring 2008, King Saud University

Time-Cost-Trade-offDr. Khalid Al-Gahtani

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Reasons to Reduce Project Durations:

1. To realize incentive pay2. To fit within the contractually required time (influences

Bid price).3. To avoid paying Liquated Damages/Damage for delay

causes (or avoid damaging the company’s relationship).

4. To complete a project early & move on to another project (Although the Job-specify profit may reduce, company profile may increase).

5. To complete a project when weather conditions make it less expensive (Avoid temporary Heating, avoid completing site work during raining season).



2

Methods to reduce durations:1. Have the existing crew work overtime. This increase

the labor costs due to increase pay rate and decrease productivity.

2. Bring in additional workers to enlarge crew size. This increases labor costs due to overcrowding and poor learning curve.

3. Use better/more advanced equipment. This will usually increase costs due to rental and transport fees. If labor costs (per unit) are reduced, this could reduce costs.

4. Add subcontracted labor to the activity. This almost always increases the cost of an activity unless the subcontracted labor is for more efficient.



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Project Costs Definitions



4

Practices to Estimate the Project cost

• Some company assigns Overhead office as % of direct cost.

• Most companies don’t consider profit as a cost of the job. Instead, analysis are completed by using Direct costs + Indirect costs + Company overhead vs. Budgeted (estimated costs).

• Similar to each activity, the project as a whole has an ideal (Lease expensive) Duration.



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Accelerating the Critical and Noncritical path



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• Determine Normal project duration, and cost.

• Identify Normal duration Critical Path.• For large network, using CRITICALITY

THEORM to eliminate the noncritical paths that do not need to be crashed.

• Compute the cost slop :cost slope = crash cost - normal cost

normal duration- crash duration

Network Compression Algorithm



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• shortening the CRITICAL ACTIVITIES beginning with the activity having the lowest cost-slope

• Determine the compression limit (Nil)

- Crash Limit, or Nil = Min - Free Float of any of the non critical activities in the parallel paths competing for critical path.*




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• Organize the data as in the following table:•

• Update the project network

Cycle #

Activity to Shorten

Can Be Shortened Nil Days

Shortened Cost per

Day Cost for Cycle

Total Cost

Project Duration

0 1 2 3




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• When a new Critical path is formed:– Shorten the combination of activity which

Falls on Both Critical Paths, OR– Shorten one activity from each of the critical

paths. Use the combined cost of shortening both activities when determining if it is cost effective to shorten the project.

• At each shortening cycle, compute the new project duration and project cost




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• Continue until no further shortening is possible • Tabulate and Plot the Indirect project Cost on

the same time-cost graph

Cycle #

Project Duration

Direct Cost

Indirect Cost

Total Cost

0 1 2 3




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• Add direct and indirect cost to find the project cost at each duration.

• Use the total project cost-time curve to find the optimum time.




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Example 1: The durations and direct costs for each activity in the network of a small construction contract under both normal and crash conditions are given in below Table. Establish the least cost for expediting the contract. Determine the optimum duration of the contract assuming the indirect cost amounts to SR 125/week.

Normal Crash Activity Preceded by Duration

(weeks) Cost (SR)

Duration (weeks)

Cost (SR)

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 36,500



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2@100

0 12 12

0 0 12 A

1@300

27 5 32

29 2 34 F

3@200

12 15 27

12 0 27 C

2@150

12 8 20

14 2 22 B

0

20 23 43

24 4 47 D

1@50

20 5 25

22 2 27 E

5@60

27 20 47

27 0 47 G

2@75

47 12 59

47 0 59 I

2@40

32 13 45

34 2 47 H ES D EF

LS F LF Activity

Crash limit

4

2

Cycle #

Activity to Shorten


Shortened Cost per

Week Cost for Cycle

Total Cost

Project Duration

0 36,500 59 1 G 5 2 2 60 120 36,620 57



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2@100

0 12 12

0 0 12 A

1@300

27 5 32

27 0 32 F

3@200

12 15 27

12 0 27 C

2@150

12 8 20

14 2 22 B

0

20 23 43

22 2 45 D

1@50

20 5 25

22 2 27 E

3@60

27 18 45

27 0 47 G

0

45 12 57

45 0 55 I

2@40

32 13 45

32 0 45 H ES D EF

LS F LF Activity

Crash limit

Cycle #

Activity to Shorten


Shortened Cost per

Week Cost for Cycle

Total Cost

Project Duration

0 36,500 59 1 G 5 2 2 60 120 36,620 57 2 I 2 2 2 70 140 36,760 55



16

2@100

0 12 12

0 0 12 A

1@300

27 5 32

27 0 32 F

3@200

12 15 27

12 0 27 C

2@150

12 8 20

14 2 22 B

0

20 23 43

22 2 45 D

1@50

20 5 25

22 2 27 E

3@60

27 18 45

27 0 45 G

0

45 10 55

45 0 55 I

2@40

32 13 45

32 0 45 H ES D EF

LS F LF Activity

Crash limit

0

Cycle #

Activity to Shorten


Shortened Cost per

Week Cost for Cycle

Total Cost

Project Duration

0 36,500 59 1 G 5 2 2 60 120 36,620 57 2 I 2 2 70 140 36,760 55 3 A 2 2 100 200 36,960 53



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0

0 10 10

0 0 10 A

1@300

25 5 30

25 0 30 F

3@200

10 15 25

10 0 25 C

2@150

10 8 18

12 2 20 B

0

18 23 41

20 2 43 D

1@50

20 5 25

22 2 27 E

3@60

25 18 43

25 0 43 G

0

43 10 53

43 0 53 I

2@40

30 13 43

30 0 43 H ES D EF

LS F LF Activity

Crash limit

2

Cycle #

Activity to Shorten


Shortened Cost per

Week Cost for Cycle

Total Cost

Project Duration

0 36,500 59 1 G 5 2 2 60 120 36,620 57 2 I 2 2 70 140 36,760 55 3 A 2 2 100 200 36,960 53 4 H, G 2 2 2 60+40 200 37,160 51



18

0

0 10 10

0 0 10 A

1@300

25 5 30

25 0 30 F

3@200

10 15 25

10 0 25 C

2@150

10 8 18

10 0 18 B

0

18 23 41

18 0 41 D

1@50

18 5 23

20 2 25 E

1@60

25 16 41

25 0 41 G

0

41 10 51

41 0 51 I

0

30 11 41

30 0 41 H ES D EF

LS F LF Activity

Crash limit

0

Cycle #

Activity to Shorten


Shortened Cost per

Week Cost for Cycle

Total Cost

Project Duration

0 36,500 59 1 G 5 2 2 60 120 36,620 57 2 I 2 2 70 140 36,760 55 3 A 2 2 100 200 36,960 53 4 G, H 2 2 2 60+40 200 37,160 51 5 B, C 2 2 150+200 700 37,860 49



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0

0 10 10

0 0 10 A

1@300

23 5 28

23 0 28 F

1@200

10 13 23

10 0 23 C

0

10 6 16

10 0 16 B

0

16 23 39

16 0 39 D

1@50

16 5 21

18 2 23 E

1@60

23 16 39

23 0 39 G

0

39 10 49

39 0 49 I

0

28 11 39

28 0 39 H ES D EF

LS F LF Activity

Crash limit

Cycle #

Project Duration

Direct Cost

Indirect Cost

Total Cost

0 59 36500 7375 43875 1 57 36620 7125 43745 2 55 36760 6875 43635 3 53 36960 6625 43585 4 51 37160 6375 43535 5 49 37860 6125 43985



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Project Optimal Duration

0

5000

10000

15000

20000

25000

30000

35000

40000

45000

50000

48 49 50 51 52 53 54 55 56 57 58 59 60

Project Duration (Week)

Cos

t (SR

)

DirectCostIndirect costTotalCost



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Class Work #2

Data on small maintenance project is given as below:

Depends Normal Crash Activity on Time Cost Time Cost

A 6 days $700 4 days $800 B 4 days 400 4 days 400 C 5 days 650 4 days 700 D A 8 days 625 5 days 700 E B 10 days 200 7 days 350 F B 7 days 500 5 days 700 G C 3 days 600 3 days 600 H D, E 6 days 300 5 days 400 I F, G 7 days 350 4 days 425

On completion, the project will give a return of $110/day. Using time-cost trade-off method, how much would you like to compress the project for maximizing the return? Show all calculations.

time cost-trade-off

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