Time-Cost Optimisation

ByDr. Mahdi Damghani

Objective

To determine a reduced or optimum length of contract and still save money

Introduction

The use of CPM is not limited to calculation of project duration

Reducing the duration of projects leads to cost consideration

Reducing the time of project may lead to increase in cost due to use of more resources

There must be a balanced point wherein the project finishes at minimum cost and optimum duration

Direct costs

As the project is conceived, cost is incurred on the purchase of Materials Plant Equipment Manpower

Relationship of direct costs with time is of rime importance

Direct costs

There is an optimum duration for a project for which the direct cost is a minimum

During this time the resources are optimally used

This duration is called Normal Duration

Crash point is a point beyond which the project duration can not be decreased irrespective of the increase in direct costs

Direct costs

Cost

Time Crash time Normal time

Crash Point

Normal PointNormal cost

Crash cost

Indirect costs

Includes Overheads Administrative expenses Depreciation Loss of profit Loss of revenue Penalty

The loss in profit, penalty comes under a category called outage loss

Indirect costs

Cost

Time Crash time Normal time

Supervisory and overheads

Outage loss

Total cost

Direct cost + Indirect costCost

Time Crash time Normal time

Indirect cost

Direct cost

Total cost

Optimum time

Optimum cost

Some terms

Normal time (tn) Standard time that an estimator normally allow for an activity

Crash time (tc) Minimum possible time in which an activity can be

completed, by employing extra resources Normal cost(Cn) Direct cost required to complete the activity in the normal

time duration Crash cost(Cc) Direct cost corresponding to the completion of the activity

within crash time Cost slope (Cs) Slope of the line(s) used to approximate the direct cost curve

Cost slope

t

c

tt

CCC

cn

ncs

Increase in cost

Decrease in time

Note:If the cost curve is approximated by one single line it will have only one cost slope whereas if it is approximated by more than one line it would adopt more cost slopes and more calculations, respectively.

Using network analysis

The project time is controlled by the activities lying on the critical path

To minimize the cost we crash the critical activities.

Particularly, critical activity with minimum cost slope must be crashed first

At each and every time check whether the critical path is unique

If not, crash the activity one by one

Example 1

Solution 1

Critical path is 1-2-3-5-6 Length of critical path

=6+14+12+8=40 Calculate the cost slop for each

activity and rank them in descending order

Activity

Cost slope

Ranking

1-2 20 3

2-3 14 2

3-5 12 1

1-4 20 4

4-5 120 6

5-6 100 5

Solution 1

Instruction Extra direct cost

New project duration

New direct cost

New indirect cost

New project cost

Normal overall duration

_ 40 1030 0 1030

3-5 (5) 5x12=60 40-5=35 1090 0 1090

2-3 (5) 5x14=70 35-5=30 1160 0 1160

1-2 (2) 2x20=40 30-2=28

1-2 (1) 1x20=20 30-1=29 1180 0 1180

1-4 (1) 1x20=20 29-1=28 1200 0 1200

Activity

Cost slope

Ranking

1-2 20 3

2-3 14 2

3-5 12 1

1-4 20 4

4-5 120 6

5-6 100 5

By doing this the critical path changes to 1-4-5-6 so I reduce 1-2 by 1 day

Solution 1 Instruction Extra

direct cost

New project duration

New direct cost

New indirect cost

New project cost

Normal overall duration

_ 40 1030 0 1030

3-5 (5) 5x12=60 40-5=35 1090 0 1090

2-3 (5) 5x14=70 35-5=30 1160 0 1160

1-2 (2) 2x20=40 30-2=28

1-2 (1) 1x20=20 30-1=29 1180 0 1180

1-4 (1) 1x20=20 29-1=28 1200 0 1200

By doing this the critical path changes to 1-4-5-6 so I reduce 1-2 by 1 day

Now we have two parallel critical paths with duration 29 days. (i.e. 1-2-3-5-6 & 1-4-5-6)The following are the activities yet to be crashed. They are 1 - 4, 4 - 5 and 5 - 6. Of the 3 activities 1 - 4 is with minimum cost slope.

Solution 1

Therefore, a 28-day project schedule costs 1200 pounds

Example 2 The following data shows the

duration and costs of each activity of a project. The indirect cost of the project is £3000 per week. Determine the optimum duration of project and the corresponding minimum cost.

Activity Normal Duratio

n (week)

Normal Cost (£)

Crash Duration (week)

Crash Cost (£)

1-2 6 7000 3 14500

1-3 8 4000 5 8500

2-3 4 6000 1 9000

2-4 5 8000 3 15000

3-4 5 5000 3 11000

Solution 2

1 4

2

3

N6C3

N8C5

N4C1

N5C3

N5C3

Solution 2

Critical path is 1-2-3-4 = 15 weeks

Notes:1. The critical path must remain critical whilst

crashing2. When you have more than one critical path,

crash the common activities to the same amount to maintain the first condition

Solution 2

Activity ΔC (£) Δt (weeks)

Cost slope (£/week)

Rank

1-2 14500-7000=7500

6-3=3 2500 3

1-3 8500-4000=4500

8-5=3 1500 2

2-3 9000-6000=3000

4-1=3 1000 1

2-4 15000-8000=7000

5-3=2 3500 5

3-4 11000-5000=6000

5-3=2 3000 4

Solution 2

Instruction Extra direct cost

New project duration

New direct cost

New indirect cost

New project cost

Normal overall duration

_ 15 30000 15x3000= 45000

45000+30000=75000

2-3 (2) 2x1000=2000

15-2=13 32000 13x3000 71000

2-3 (1) & 1-3 (1)

1x1000+1x1500=2500

13-1=12 34500 12x3000 70500

3-4 (1) 1x3000=3000

12-1=11 37500 11x3000 70500

1-2 (2) & 1-3 (2)

2x1500+2x2500=8000

11-2=9 45500 9x3000 72500

3-4 (1) & 2-4 (1)

3000+3500=6500

9-1=8 52000 8x3000 76000

Activity Cost slope (£/week)

Rank

1-2 2500 3

1-3 1500 2

2-3 1000 1

2-4 3500 5

3-4 3000 4

Solution 2

From the table on the last slide the minimum cost is 70500 for duration of 11 weeks

Example 3

Activity Normal Duratio

n (week)

Normal Cost (£)

Crash Duration (week)

Crash Cost (£)

10-20 2 1000 2 1000

10-30 7 500 3 900

20-30 6 300 3 420

20-40 5 200 4 250

30-40 0 0 0 0

30-50 9 600 4 900

40-60 11 600 6 1000

50-60 6 700 3 910

Determine minimum cost and optimum duration

Indirect cost is £80 per day

Solution 3

Optimum project duration is 18 days Corresponding cost=£4320