project crashing

6-1

Project Crashing

Crashing means shortening duration of activities Because some activities were delayed Or client is willing to pay more for earlier completion

Crashing changes the schedule for remaining activities It has impact on schedules for all the subcontractorsOften introduces unanticipated problemsThe faster an activity is completed, the more it costsThere is always a lower bound on task duration

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6-2

Linear Time / Cost Tradeoff

Time

Cost

Crash point

Normal point

Normal time =Crash time =

Normal cost =

Crash cost =

tjNtj

c

Cjc

CjN

Slope (bj) = increase in cost from reducing task duration by one time unit

Time Normal - TimeCrash

Cost Normal -Cost Crash Slope

04/21/23 Ardavan Asef-Vaziri -4

6-3

Crashing Algorithm

Assume each task can be crashed one day at a time (simplifying assumption, but not necessary)

Crash Only critical activities. Crashing other activities can only increase cost without changing project duration

To decrease project duration by one day, the critical path or paths must decrease by one day.

1. Find the critical path or paths

2. If there is no other critical activity which could still be reduced, and shorten the critical path. Stop.

3. Crash the cheapest critical activity (or combination of activities) to shorten the critical path (or paths) by one day.

4. Go to 1


6-4

Project Crashing Example


Activity Predecesor Time(N) Time(C) Cost(N) Cost(C) C/T Lefta - 6 5 60 90 30 1b - 7 4 50 150 33.3 3c a 6 4 100 160 30 2d a 7 7 30 30 - -e b 5 4 70 85 15 1f c 9 7 40 120 40 2g d,e 7 4 50 230 60 3

6-504/21/23 Ardavan Asef-Vaziri -5

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b

c

d

e

g

f

21 Days Network and Its Total Cost

Normal Cost = 60+50+100+30+70+40+50 = 400

6-6

From 21 to 20 Days

a-c-f are on the critical path a and c are the least-cost choice. We crash a because it

affects two paths

04/21/23 Ardavan Asef-Vaziri 6

Activity Predecesor Time(N) Time(C) Cost(N) Cost(C) C/T Lefta - 6 5 60 90 30 1b - 7 4 50 150 33.3 3c a 6 4 100 160 30 2d a 7 7 30 30 - -e b 5 4 70 85 15 1f c 9 7 40 120 40 2g d,e 7 4 50 230 60 3


0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b e

c

d g

f

21 to 20 Days Network

Normal Cost = 400 + a(30) = 430

6-8

20 Days Network


1Activity Predecesor Time(N) Time(C) Cost(N) Cost(C) C/T Left

a - 6 5 60 90 30 1b - 7 4 50 150 33.3 3c a 6 4 100 160 30 2d a 7 7 30 30 - -e b 5 4 70 85 15 1f c 9 7 40 120 40 2g d,e 7 4 50 230 60 3

Activities a-c-f are still on the critical path a cannot be crashed any more c is the least-cost choice. Lower c’s normal time by one day.


0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b e

c

d g

f

20 to 19 Days Network


0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b e

c

d g

f

19 Days Network

Normal Cost = 430 +c(30) =460

6-11

At 19 Days All are Critical


1

1

Activity Predecesor Time(N) Time(C) Cost(N) Cost(C) C/T Lefta - 6 5 60 90 30 1b - 7 4 50 150 33.3 3c a 6 4 100 160 30.0 2d a 7 7 30 30 - -e b 5 4 70 85 15 1f c 9 7 40 120 40 2g d,e 7 4 50 230 60 3

All activities on CP, do not think of e-b because of d-a Activities a-c-f, and d-g are on the critical path a and d cannot be crashed any more g and one of c-f and c30, f40 g60 c and g


0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b e

c

d g

f

19 To 18 Days Network


0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b e

c

d g

f

18 Days Network

Normal Cost = 460 + c(30) + g(60) =550

6-14

At 18 Days All are Critical


1

2


f40, g60

1


17 Days Network

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b

c

d

e

g

f


From 17 Day To 16

1

2

2


1

6-17

Crashing- 17,16 Days

We can shorten the project to 16 days by crashing g and f still another day.

The cost of the project is $400+ 30(a) + 30(c2) + 60(g3)+40(f2) = $750

Activities a,c,f, and g have been crashed to their limits. No further crashing will help so b,d, and e remain at their

normal times and costs.

04/21/23 Ardavan Asef-Vaziri 17


16 Days Network

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

a

b

c

d

e

g

f


Project Crashing

400

450

500

550

600

650

700

750

15 16 17 18 19 20 21

Project Duration

Pro

ject

Co

st

Example 1

Trade-off: Cost-Time

6-20

Assignment 1: Crash the Following Network. Prepare Cost-time Curve


6-21

Prepare Cost-Time Curve



Activity Predecesor T(N) Time(C) Cost(N) Cost(C) C/T Lefta - 6 5 60 90 30 1b - 7 4 50 150 33.3 3[c] a 6 4 100 160 60 2d a 7 7 30 30 - -e b 5 4 70 85 15 1f c 9 7 40 120 40 2g d,e 7 4 50 230 60 3

Assignment 2

The same example as we solved.Activity c either 0 or 2 but not 1. Solve it in the easiest way. Prepare Time-Cost Trade-off Curve

project crashing

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