Time, Rate, and Productivity 1

Summer 2002 1

Project Costs and Crashing

Management of OperationsBrad C. Meyer

Summer 2002 2

“Crashing”

When we say that an activity will take acertain number of days or weeks, whatwe really mean is: normally, this activitytakes this many days or weeks. Wecould make it take less time – but itwould cost more money. To spendmore money so as to get somethingdone more quickly is called “crashing”the activity.

Summer 2002 3

Crash time and crash cost

NT = normal time to complete an activityNC = normal cost to complete an activityCT = crash time to complete an activity,

that is, the shortest possible time itcould be completed in.

CC = crash cost - the cost to completethe activity if it is performed in it’sshortest possible time.

Time, Rate, and Productivity 2

Summer 2002 4

Parameters for crashing

maximum time reduction for an activity= NT – CT

cost to crash per period CC – NC = ----------- NT - CT

Summer 2002 5

By the way

the cost to crash per period assumesthat the relationship between addingmore money to the activity andreducing the time is linear. Spend halfof the money, and get half the timereduction, spend _ of the money andget _ time reduction. This is not alwaystrue in practice, but works alright for arough planning technique.

Summer 2002 6

Computing crash data

given:activitiesnormal timenormal costcrash timecrash cost

compute:maximum time reductioncost to crash per period

Time, Rate, and Productivity 3

Summer 2002 7

Example

9100670009E

190005100008D

200002150004C

5500240003B

6000430007A

CCCTNCNTAct.

Summer 2002 8

Example

9-6 = 3

8-5 = 3

4-2 = 2

3-2 = 1

7-4 = 3

Max Red

9100670009E

190005100008D

200002150004C

5500240003B

6000430007A

CCCTNCNTAct.

Summer 2002 9

Example

For activity A: (6000-3000)/(7-4) = 1000

1000

Cost to crashper period

3

3

2

1

3

MaxRed

9100670009E

190005100008D

200002150004C

5500240003B

6000430007A

CCCTNCNTAct.

Time, Rate, and Productivity 4

Summer 2002 10

Example

700

3000

2500

1500

1000

Cost to crashper period

3

3

2

1

3

MaxRed

9100670009E

190005100008D

200002150004C

5500240003B

6000430007A

CCCTNCNTAct.

Summer 2002 11

Finding the minimum cost schedule

To shorten a project, crash only activities thatare critical.Crash from least expensive to mostexpensive.Each activity can be crashed until

it reaches it’s maximum time reductionit causes another path to also become criticalit is more expensive to crash than not to crash

Continue until no more activities should becrashed.

Summer 2002 12

Example continued

Same problem as earlier. Times shown are normaltimes. There are three paths in this network:

ABD 18ACD 19ACE 20

A7

B3

C4

D8

E9

Time, Rate, and Productivity 5

Summer 2002 13

This project, under normal conditions takes 20 days.Suppose each day the project runs incurs an indirectproject cost of $1400 (overhead). What activitiesshould be crashed if any?

ABD 18ACD 19ACE 20 *Start by looking at activities on the critical path:

A, C, and E. E is least expensive to crash.

Summer 2002 14

E has maximum time reduction of 3, but if it is crashedby 1, then ACD also becomes a critical path. Also,we save $1400 per day the project is shortened andwould spend $700 per day to crash E, so it isprofitable to crash E.

crash E by 1

ABD 18 18ACD 19 19 *ACE 20 * 19 *

Summer 2002 15

Now there are two critical paths. To finish the projectearlier, we would need to shorten both paths.Crashing A or C does this, since those two activitiesare on both paths. Alternately, we could crash bothD and E together.

crash E by 1

ABD 18 18ACD 19 19 *ACE 20 * 19 *

Time, Rate, and Productivity 6

Summer 2002 16

Cost to crash A = 1000Cost to crash C = 2500Cost to crash both D and E = 3700Best is to crash A. How much should it be crashed? crash E crash A by 1 by ?

ABD 18 18ACD 19 19 *ACE 20 * 19 *

Summer 2002 17

Maximum reduction for A is 3When A is crashed, all paths are affected, so no new

paths will become critical.Cost savings remain at $1400 vs $1000 cost to crash. crash E crash A by 1 by 3

ABD 18 18 15ACD 19 19 * 16 *ACE 20 * 19 * 16 *

Summer 2002 18

Stopping condition

To continue, we could crash C or both Dand E. But in each case, the cost wouldbe greater than the $1400 savings perday. So, we stop at this point.

We can compute the cost to perform theproject in 16 days.

Time, Rate, and Productivity 7

Summer 2002 19

Figuring Total Project Cost

Sum of normal costs = 39,000Indirect costs = 16 dys * 1400 = 22400Crashing cost

E by 1 = 700A by 3 = 3000

Total cost = $65100

Summer 2002 20

Minimum time schedule atminimum cost

Sometimes it is necessary to complete aproject in as short a time as possible.To find the shortest time possible, crashall activities completely and then findthe times for all paths.The longest path is, of course, criticaland tells us how long the project musttake.

Summer 2002 21

Minimum time schedule continued

Activities on non-critical paths may notneed to be fully crashed in order for theproject to be finished in the shortestpossible time. These can be“uncrashed” one at a time, startingfrom the most expensive, until nothingis crashed that does not need to be.

Time, Rate, and Productivity 8

Summer 2002 22

Example

Same problem as earlier. Times shown are crash times.Same three paths, but now with shorter times.

ABD 11ACD 11ACE 12*

A4

B2

C2

D5

E6

Summer 2002 23

Minimum time schedule continued

Critical activities are A,C, and E. B and D are not criticaland could be relaxed. B costs $1500 to crash. Dcosts $3000 to crash. To save some money, but stillcomplete in 12 days, uncrash D by 1.

uncrash D by 1

ABD 11 12*ACD 11 12*ACE 12* 12*

Summer 2002 24

Minimum time schedule continued

Uncrashing anything else would not allowcompletion in the 12 days, so we are finished.

uncrash D by 1ABD 11 12*ACD 11 12*ACE 12* 12*

Time, Rate, and Productivity 9

Summer 2002 25

Look over the solved problems and examples inthe text for more practice.