Analyzing Cost-Time Trade-Offs There are always cost-time trade-offs in project management. – You can completing a project early by hiring more workers

Analyzing Cost-Time Trade-Offs

• There are always cost-time trade-offs in project management.– You can completing a project early by hiring more

workers or running extra shifts.– There are often penalties if projects extend

beyond some specific date, and a bonus may be provided for early completion.

• Crashing a project means expediting some activities to reduce overall project completion time and total project costs.

Cost to Crash

• To assess the benefit of crashing certain activities, either from a cost or a schedule perspective, the project manager needs to know the following times and costs.

• Normal time (NT) is the time necessary to complete and activity under normal conditions.

• Normal cost (NC) is the activity cost associated with the normal time.

• Crash time (CT) is the shortest possible time to complete an activity.

• Crash cost (CC) is the activity cost associated with the crash time.

Cost to Crash per Period

The Cost to Crash per Time Period =

CC − NC

NT − CT

Crash Cost − Normal Cost

Normal Time − Crash Time

Linear cost assumption

8000 —

7000 —

6000 —

5000 —

4000 —

3000 —

0 —

Dire

ct c

ost (

dolla

rs)

| | | | | |5 6 7 8 9 10 11

Time (weeks)

Crash cost (CC)

Normal cost (NC)

(Crash time) (Normal time)

Estimated costs for a 2-week reduction, from 10 weeks to 8 weeks

5200

Cost-Time Relationships in Cost Analysis

• The objective of cost analysis is to determine the project schedule that minimizes total project costs.

• A minimum-cost schedule is determined by starting with the normal time schedule and crashing activities along the critical path in such a way that the costs of crashing do not exceed the savings in indirect and penalty costs.

Minimizing CostsMinimizing Costs

• Use these steps to determine the minimum cost schedule:

1. Determine the project’s critical path(s).2. Find the activity or activities on the critical path(s) with

the lowest cost of crashing per time unit.3. Reduce the time for this activity until…

a. It cannot be further reduced orb. Until another path becomes critical, orc. The increase in direct costs exceeds the savings that result from

shortening the project (which lowers indirect costs & penalties).

4. Repeat this procedure until the increase in direct costs is larger than the savings generated by shortening the project.

Minimum Cost ScheduleMinimum Cost Schedule

Regular & Crash Times & Cost to Crash

Activity Regular Time Crash TimeTotal Costto Crash

Cost Per Unit of Crash Time

$ A* 2 1 $1,000

$ B* 4 2 $ 500

$ C* 2 2 C

$ D* 1 1 C

$ E 3 1 $ 200

$ F* 2 2 C

$ G* 1 1 C

$ H* 12 8 $ 800

$ I* 8 5 $1,200

$ J 6 6 C

$ K 5 3 $ 600

$ L*TOTAL

* Activities on critical path

3 3 C$4,300

START A2

B4

C2

D1

F2

G1

H12

J6

I8

K5

L3

END

E3

Network for a Software Purchasing Project using Normal Times

IF ALL ACTIVITIES ARE CRASHED

STARTA1

B2

C2

D1

F2

G1

H8

J6

I5

K3

L3

END

E1

(1,5,4)

(8,8,0)(6,6,0)(5,5,0)(0,0,0) (1,1,0) (3,3,0)

(26)

(23,23,0)

(17,18,1)

(17,17,0)

(9,20,11)

(9,9,0)

*What is the minimum amount of time to complete the project?*What is the additional cost involved in achieving this reduction in time?

START A2

B4

C2

D1

F2

G1

H12

J6

I8

K5

L3

END

E3

What is the minimum cost to achieve minimum time schedule for project



$ A 2 1 $1,000 $1,000

$ B 4 2 $ 500 $ 250

$ C 2 2 C C

$ D 1 1 C C

$ E 3 1 $ 200 $ 100

$ F 2 2 C C

$ G 1 1 C C

$ H 12 8 $ 800 $ 200

$ I 8 5 $1,200 $ 400

$ J 6 6 C C

$ K 5 3 $ 600 $ 300

$ L 3 3 C$4,300

C

STARTA2

B4

C2

D1

F2

G1

H12

J6

I8

K5

L3

END

E3

What is the minimum cost to achieve minimum time schedule for project



$ A 2 1 $1,000 $1,000

$ B 4 2 $ 500 $ 250

$ C 2 2 C C

1 1 C C

$ E 3 1 $ 200 $ 100

$ F 2 2 C C

$ G 1 1 C C

$ H 12 8 $ 800 $ 200

$ I 8 5 $1,200 $ 400

$ J 6 6 C C

$ K 5 3 $ 600 $ 300

$ L 3 3 C$4,300

C

STARTA2

B4

C2

D1

F2

G1

H12

J6

I8

K5

L3

END

E3

What is the minimum cost to reduce project time to 30 months



$ A 2 1 $1,000 $1,000

$ B 4 2 $ 500 $ 250

$ C 2 2 C C

1 1 C C

$ E 3 1 $ 200 $ 100

$ F 2 2 C C

$ G 1 1 C C

$ H 12 8 $ 800 $ 200

$ I 8 5 $1,200 $ 400

$ J 6 6 C C

$ K 5 3 $ 600 $ 300

$ L 3 3 C$4,300

C

STARTA2

B4

C2

D1

F2

G1

H12

J6

I8

K5

L3

END

E3

What is the optimal schedule if I incur a penalty of $280 for each month above the absolute minimum project time of 26 months?



$ A 2 1 $1,000 $1,000

$ B 4 2 $ 500 $ 250

$ C 2 2 C C

1 1 C C

$ E 3 1 $ 200 $ 100

$ F 2 2 C C

$ G 1 1 C C

$ H 12 8 $ 800 $ 200

$ I 8 5 $1,200 $ 400

$ J 6 6 C C

$ K 5 3 $ 600 $ 300

$ L 3 3 C$4,300

C

© 2007 Pearson Education

Calculating total CostCalculating total Cost Adding Adding Direct, Penalty & Overhead CostsDirect, Penalty & Overhead Costs

The overhead (indirect) costs are $200 per dayThere is a penalty of $100/day for completing project in more than 12 days

Network Diagram


Total Cost ProblemTotal Cost Problem

Assessing Risks

• Risk is a measure of the probability and consequence of not reaching a defined project goal.

• A major responsibility of the project manager at the start of a project is to develop a risk-management plan.

• A Risk-Management Plan identifies the key risks to a project’s success and prescribes ways to circumvent them.

Probabilistic Probabilistic Time EstimatesTime Estimates

MeanMeanmmaa bb TimeTime

Pro

babi

lity

Pro

babi

lity

Beta Distribution

PessimisticOptimistic

TimeTime

Pro

babi

lity

Pro

babi

lity

Normal Normal DistributionDistribution

MeanMeanaa bbmm

33 33

Area under curve Area under curve between a and b is between a and b is 99.74%99.74%

Probabilistic Probabilistic Time EstimatesTime Estimates

te = a + 4m + b

6

Mean

22 = = ( )

bb – – aa

66

22

VarianceVariance

Probabilistic Time EstimatesProbabilistic Time Estimates

Calculating Means and Variances for a Beta Distribution

Where:

a is the Optimistic estimate

b is the pessimistic estimate

m is the most likely estimate

Optimistic Likely PessimisticActivity (a) (m) (b)

Time Estimates (wk)

A 11 12 13

B 7 8 15

C 5 10 15

D 8 9 16

E 14 25 30

F 6 9 18

G 25 36 41

H 35 40 45

I 10 13 28

J 1 2 15

K 5 6 7

St. John’s HospitalSt. John’s HospitalProbabilistic Time EstimatesProbabilistic Time Estimates

A F

I

C G Finish

D

E

HB J

K

Start

Activity BMost

Optimistic Likely Pessimistic(a) (m) (b)7 8 15

A F

I

C G Finish

D

E

HB J

K

Start


te = = 9 weeks7 + 4(8) + 15

6

2 = = 1.78( )15 - 7

6

2

Calculating Means and Variances

Optimistic Likely Pessimistic Expected VarianceActivity (a) (m) (b) Time (te ) (2 )

Time Estimates (wk) Activity Statistics

A 11 12 13 12 0.11B 7 8 15 9 1.78C 5 10 15 10 2.78D 8 9 16 10 1.78E 14 25 30 24 7.11F 6 9 18 10 4.00G 25 36 41 35 7.11H 35 40 45 40 2.78I 10 13 28 15 9.00J 1 2 15 4 5.44K 5 6 7 6 0.11



K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 63 690 12

48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

63 69

Earliest start time Earliest finish time

Latest start time Latest finish time

2 = (variances of activities) z = T – TE

2

2 = 1.78 + 1.78 + 2.78 + 5.44 + 0.11 = 11.89

z =72 – 69

11.89

What is the Probability of finishing the project in 72 weeks?

Given that: Critical Path = B - D - H - J - K

T = 72 Weeks TE = 69 Weeks

St. John’s HospitalSt. John’s HospitalAnalyzing ProbabilitiesAnalyzing Probabilities

From Normal Distribution appendix Pz = .8078 .81

= .87

.00 .01 .02 .03 .04 .05 .06 .07 .08 .09

.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

NORMAL DISTRIBUTION TABLE

z0 Ğ

Project duration (weeks)Project duration (weeks)

6969 7272

Normal distribution: Normal distribution: Mean = 69 weeks; Mean = 69 weeks; = 3.45 weeks = 3.45 weeks

Probability of Probability of exceeding 72 exceeding 72 weeks is 0.1922 weeks is 0.1922

St. John’s HospitalSt. John’s HospitalProbability of Completing Project On TimeProbability of Completing Project On Time

Probability of Probability of meeting the meeting the schedule is schedule is 0.80780.8078

Length of Length of critical pathcritical path

Optimistic Likely Pessimistic Expected VarianceActivity (a) (m) (b) Time (te ) (2 )

Time Estimates (Months) Activity Statistics

A 1 2 3B 3 4 6C 2 2 2D 1 1 1E 1 3 5F 1 2 3G 1 1 1H 8 12 14I 7 8 10J 5 6 7K 4 5 6L 2 3 6

Software Purchasing ProjectSoftware Purchasing ProjectProbabilistic Time EstimatesProbabilistic Time Estimates

START A2

B 4.17

C2

D1

F2

G1

H11.67

J6

I8.17

K5

L3.33

END

E3

Network for a Software Purchasing Project using the Mean TimesCritical Path: A—B—C—D—F—G—H—I—L

Expected Time to complete the project: 2+4.17+2+1+2+1+11.67+8.17+3.33 = 35.33

What is the probability in finishing in less than 37 months?

What is the probability in finishing in less than 31 months?

Documents

Analyzing Cost-Time Trade-Offs There are always cost-time trade-offs in project management. – You can completing a project early by hiring more workers