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Elements of Crashing a Network

Every activity may have two types of completion times-normal time and crash time. Accordingly costs are also two types i.e., normal cost and crash cost. Obviously, the crash cost is higher than the normal cost and the normal time is higher than the crash time.

 

Crashing of a network implies that crashing of activities. During crashing direct cost increases and there is a trade-off between direct cost and indirect cost. So the project can be crashed till the total cost is economical. The following procedures are carried out :

 

(a) Calculate the critical path (CP) with normal times of the activities.

(b) Calculate the slope as given below of each activity.

 

Slope   =Crashing cost - Normal cost / Normal time - Crash time

 

(c) Identify the critical activity with lowest slope.

 

(d) Compress that activity within crash limit. Compression time can also be calculated by

taking min. (crash limit, free float limit).

 

If there are more than one critical path then select a common critical activity with least

slope . If there is no such activity then select the critical activity with least slope from   each critical path and compress them simultaneously within the crash limit.

 

(e) Continue crashing until it is not possible to crash any more.

 

(f) Calculate the total cost (TC) after each crashing as follows:

TC = Previous TC + Increase in direct cost - Decrease in indirect cost..

 

If the current TC is greater than the previous TC then the crashing is uneconomical and stop. Suggest the previous solution as optimal crashing solution.

 

Example 5. A project consists of six activities with the following times and costs estimates

 

Activity           Normal            Normal                        Crash                           Crash

time (weeks)     cost ($)                       time (weeks)               cost ($)

1-2                   9                      4 0 0                                       7                     900

1-3                   5                      5 0 0                                        3                      800

1-4                   10                    4 5 0                                        6                      1000

2-5                   8                      600                                          6                     1000

3-5                   7                     1000                                        5                     1300

4-5                   9                     900                                          6                      1200

 

If the indirect cost per week is $ 1 2 0, find the optimal crashed project completion time.

 

Solution. The slope calculations and the crash limit are given in the following table :

 

Activity                       Slope                 Crash  limit

(weeks)

1-2                               250                              2

1-3                               150                               2

1-4                               1 37.5                         4

2-5                               200                               2

3-5                               150                             2

4-5                               100                              3

 

Iteration 1

The CP calculations are shown in Fig. 8.5.

 

 

CP : 1-4-5

 

Normal project duration = 19 weeks

Total direct (i.e., normal) cost = $ 3850

Indirect cost = $ (19 x 120) = $ 2280

Total Cost (TC) = $ 3850 +  2280 = $ 6 1 30

 

The slopes and crash limits of critical activities are summarised below

 

Critical                        Slope                   Crash limit

Activity                                        (weeks)

1-4                   137.5                           4

4--5                  100*                            3

 

Since 100 is the minimum slope, crash the activity 4-5 by I week i.e., from 9 weeks to. 8 weeks.

Iteration 2

The CP calculations are shown in Fig. 8.6.

CP : 1-4-5

 

Under the crashing, the project duration reduces to 1 8 weeks.

 

New TC = $ (6130 + 100 - 120) = $ 6 1 1 0

 

Since the new TC is less than the previous TC, the present crashing is economical and proceed for further crashing.

 

The slopes and crash limits of critical activities are summarised below :

 

Critical                        Slope               Crash limit

Activity                                               (weeks)

1-4                               1 37.5                  4

4-5                               100*                     2

 

Crash the activity 4-5 by 1 week i.e., from 8 weeks to 7 weeks.

 

Iteration 3

The CP calculations are shown in Fig. 8.7.

We obtain two CPS : 1--4-5 and 1-2-5.

 

New TC = $ (6 110 + 100 - 120) = $ 6090.

 

Since the new TC is less than the previous TC, the present crashing is economical and proceed for further crashing. The slopes and crash limits of critical activities are summarised below :

 

Critical                        Slope                           Crash limit

activity                                                               (weeks)

1-4                               137.5                                       4

4-5                                 100                                        1

 

Since there is no common critical activity, let us crash 4-5 by 1 week and 2-5 by 1 week.

 

Iteration 4

The CP calculations are shown in Fig. 8.8.

New TC = $ 6090 + $ 100 + $ 200 - $ 120 = $ 6270.

Since the new TC is greater than previous TC, stop the iteration.

The previous iteration solution is the best for implementation.

Therefore, the final crashed project completion time is 17 weeks and the CPS are 1-2-5 and 1-4-5

 

PROBLEMS

 

1 .A project consists of seven activities with the following times and costs estimates

 

Activity           Normal           Normal           Crash              Crash

time (weeks)    cost ($)           time (weeks)    cost $)

1-2                    12                     500                  8                      900

1-3                  6                      600                 5                     700

1-4                   8                      700                 5                     850

2-5                  11                    500                 10                   820

3-5                  7                      1000               5                       1200

4-6                   6                      900                 4                      1000

5-6                  10                    1200               8                     1450

 

If the indirect cost per week is $ 150, find the optimal crashed project completion time.

 

2. Consider the data of a project as shown in the following table :

 

Activity           Normal                      Normal                        Crash               Crash

time (weeks)                cost ($)                        time (weeks)    cost ($)

1-2                   9                                    500                             8                      600

1-3                   7                                    800                               6                      1100

1-4                   8                                     900                              6                      1200

2-5                   6                                    850                              5                       1200

3-4                   10                                  1200                            8                      14400

4-5                   4                                   700                               3                      870

5-6                   5                                     1000                            4                      1200

 

If the indirect cost per week is Rs. 160, find the optimal crashed project completion time.

 

3. The table below provides the costs and times for a seven activity project :

 

Activity                       Time estimates (weeks)                      Direct cost estimates ($ .000)

(i, J)                             Normal            Crash                           Normal           Crash

(1 , 2)                            2                     1                                     10                   15

(1,3)                             8                      5                                  15                     21

(2,4)                             4                      3                                    20                    24

(3,4)                             1                        1                                 7                      7

(3,5)                             2                        1                                  8                     15

(4,6)                             5                      3                                    10                    16

(5,6)                             6                      2                                   12                     36

(i) Draw the project network corresponding to normal time.

(ii) Determine the critical path and the normal duration and cost of the project.

(iii) Crash the activities so that the project completion time reduces to 11 weeks irrespective of   The costs.

 

ANSWERS

 

1 . Total crashed cost = $ 10100 with crashed project completion time = 27 weeks.

2 . Total crashed cost = $ 9990 with crashed project completion time = 24 weeks.

3. (ii) CP : 1-3-5-6. Normal project duration 15 weeks and normal cost = $ 82000.

(iii) All paths becomes CP.