LESSON 2. TIME-COST OPTIM.IZATION

ALGORITHM

The process of shortening a project is called crashing and is usually achieved by adding extra resources to an activity. Project crashing involves following steps:

Step 1: Critical path. Find the normal critical path and identify the critical activities.

Step 2: Cost slope. Calculate the cost slope for the different activities by using the formula:

Module 14 Lesson 2 fig.1.1

Step 3: Ranking. Rank the activities in the ascending order of cost slope.

Step 4: Crashing. Crash the activities in the critical path as per the ranking, i.e., activity having lower cost slope would be crashed first to the maximum extent possible. Calculate the new direct cost by cumulative adding the cost of crashing to the normal cost.

Step 5: Parallel crashing. As the critical path duration is reduced by the crashing in Step 3, other paths also become critical, i.e., we get parallel critical paths. This means that project duration can be reduced duly by simultaneous crashing of activities in the parallel critical paths.

Step 6: Optimal duration. Crashing as per step 3 and step 4, optimal project duration is determined. It would be the time duration corresponding to which the total cost (i.e., direct cost plus indirect cost) is a minimum.

SAMPLE PROBLEMS

The following table gives the activities in a construction project and other, relevant information:

 

Activity

i -- j

Normal time

(days)

Crash time

(days)

Normal cost

(Rs.)

Crash cost

(Rs.)

1-2

20

17

600

720

1-3

25

25

200

200

2-3

10

8

300

440

2-4

12

6

400

700

3-4

5

2

300

420

4-5

10

5

300

600

4-6

5

3

600

900

5-7

10

5

500

800

6-7

8

3

400

700

 

(a) Draw the activity network of the project

(b) Find the total float and free float for each activity

(c) Using the above information crash the activity step by step until all paths are critical.

Solution: (a) using the normal time duration the network is given below:

(b) Considering the normal time of the project, the earliest times and latest times as well as the total floats and free floats in respect of the node points is obtained in the following table:

 

Activity

(i-j)

Normal duration

(days)

Earliest time

Latest time

Float

Start

Finish

Start

Finish

Total

Free

1-2

20

0

20

0

20

0

0

1-3

25

0

25

5

30

5

5

2-3

10

20

30

20

30

0

0

2-4

12

20

32

23

35

3

3

3-4

5

30

35

30

35

0

0

4-5

10

35

45

35

45

0

0

4-6

5

35

40

42

47

7

7

5-7

10

45

55

45

55

0

0

6-7

8

40

48

47

55

7

7

 

The critical path of project is

          1 → 2 → 3 → 4 → 5 → 7

and duration of the project is 55 days with total cost as Rs. 3600

(c)The cost slopes of the activities of the above network are computed as follows:

          Activity                                                                Cost slope

          1-2                                                              (720 – 600)/(20 – 17) = 40

          1-3                                                              (200 – 200)/(25 – 25) =0

          2-3                                                              (440 – 300)/(10 – 8) = 70

          2-4                                                              (700 – 400)/(12 - 6) = 50

          3-4                                                              (420 – 300)/(5 – 2) = 40

          4-5                                                              (600 – 300)/(10 – 5) = 60

          4-6                                                              (900 – 600)/(5 – 3) = 150

          5-7                                                              (800 – 500)/(10 – 5) = 60

          6-7                                                              (700 – 400)/(8 – 3) =60

 

Crashing of Activities

Since the activities lying on the critical path control the project duration, we crash the activities lying on the critical path.

First crashing. First of all we crash that activity of critical path which involves the minimum cost slope. Since the activities (1, 2) and (3, 4) give the minimum cost slope, we compress the duration of activity (3, 4) from 5 to 2 days with an additional cost Rs. 3 x 40, i.e., Rs. 120.

Thus the revised network is

Duration of project is now 52 days and total cost

                   = Rs. 3,600 + Rs. 40 x 3 = Rs. 3,720.

Second crashing. Now, Since there are two parallel critical paths, we choose the minimum cost slope of the activity which lies on any of the two critical paths. As the minimum cost slope is for the activity (1, 2) we compress this activity from 20 to 17 days with an additional cost of Rs. 40 x 3 i.e. Rs. 120.

Thus we have the following network:

Duration of project is now 49 days and total cost

          = Rs. 3,720 + Rs. 40 x 3 = Rs. 3,840.

Third crashing. Minimum cost slope now is for the activities (4, 5), (5, 7), and (6, 7). Therefore crashing the activities (4, 5) and (5, 7) by 5 days each and activity (6, 7) by 3 days at an extra cost of Rs. 60 per day, we have

Duration of project is now 39 days and total cost

          = Rs. 3,840 + Rs. 60 x (5 + 5+ 3) = Rs. 4,620.

Fourth crashing. Finally, crashing the activities (2, 3) and (2, 4) by two days each, we find that all the activities are lying on the critical path. Thus, we have

Duration of project is now 39 days and total cost

                   = Rs. 4,620 + Rs. 70 x 2 + Rs. 50 x 2 = Rs. 4,860.

Since all the activities are now on critical paths, the process of crashing is completed.

Last modified: Thursday, 27 March 2014, 9:24 AM