Cost Crashing 14

8/10/2019 Cost Crashing 14

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2007 Pearson Education

ImmediateActivity Description Predecessor(s) Responsibility

A Select administrative and medical staff. Johnson

B Select site and do site survey.

TaylorC Select equipment. A AdamsD Prepare final construction plans and layout. B TaylorE Bring utilities to the site. B BurtonF Interview applicants and fill positions in A Johnson

nursing, support staff, maintenance,and security.

G Purchase and take delivery of equipment. C AdamsH Construct the hospital. D TaylorI Develop an information system. A SimmonsJ Install the equipment. E,G,H AdamsK Train nurses and support staff. F,I,J Johnson

St. Adolfs HospitalExamp le 3.1


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St. Adolfs HospitalDiagramm ing the Network

FinishStart

A

B

C

D

E

F

G

H

I

J

K

A B

C AD BE BF AG CH DI A

J E,G,HK F,I,J

Immediate

Predecessor


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St. Adolfs Hospital

FinishStart

A

B

C

D

E

F

G

H

I

J

K

Path Time (wks)

A-I-K 33

A-F-K 28A-C-G-J-K 67B-D-H-J-K 69B-E-J-K 43

Paths are the sequence ofactivities between a

projects start and finish.


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St. Adolfs Hospital

FinishStart

A

B

C

D

E

F

G

H

I

J

KPath Time (wks)

A-I-K 33A-F-K 28A-C-G-J-K 67B-D-H-J-K 69B-E-J-K 43

Project ExpectedTime is 69 wks.

The critical pathis thelongest path!


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Earliest Start Time(ES) is the latest earliest finishtime of the immediately preceding activities.

Earliest Finish Time(EF) is an activitys earlieststart time plus its estimated duration.

Latest Start Time(LS) is the latest finish timeminus the activitys estimated duration.

Latest Finish Time(LF) is the earliest latest starttime of the activities that immediately follow.

For simplicity, all projects start at time zero.

St. Adolfs HospitalDevelop ing the Schedu le


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What AON Nodes look l ike

LatestFinish

LatestStart

Activity

ActivityDuration

Slack

The earliest you can completean activity -- determined byadding the activity time to theearliest start time.

The latest you can finish anactivity without delaying theproject completion date. It is thesame as the Latest Start time ofthe next activity. If there are twoor more subsequent activities,this time is the same as the

earliest of those Latest Starttimes.

Determined by the earliest finish

time of the precedent activity. Ifthere are two or more precedentactivities, this time is the same asprecedent activity with the latestEarliest Finish time.

This is the LatestFinish time minus

the activity time.

Slack is the difference, if any,

between the earliest start and lateststart times (or the earliest finish andlatest finish times).

S = LSES or S = LFEF


EarliestStart

Earli

estFinish


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Earl iest Start and Earl iest Finish Times

K

6

C

10

G

35

J

4

H

40

B

9

D

10

E

24

I

15

FinishStart

A

12

F

10


0

Earliest start time

12

Earliest finish time

0 9

9 33

9 19 19 59

22 5712 22

59 63

12 27

12 22 63 69

Example 3.2


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Earl iest Start and Earl iest Finish Times


Crit ical Path

The Crit ical Path

takes 69 weeks

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

Example 3.2


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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

LatestStart and LatestFinish Times


48 63

53 63

59 63

24 59

19 59

35 59

14 24

9 19

2 14

0 9

Latestfinishtime

63 69

Lateststarttime

Example 3.2


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2007 Pearson Education 2007 Pearson Education

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

Example 3.2


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Project Schedule

A Gantt Chartis a project schedule, usually createdby the project manager using computer software,that superimposes project activities, with their

precedence relationships and estimated durationtimes, on a time line.Activity slack is useful because it highlights activities that

need close attention.

Free slackis the amount of time an activitys

earliest finish time can be delayed without delayingthe earliest start time of any activity that immediatelyfollows.Activities on the critical path have zero slack and cannot be

delayed without delaying the project completion.


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Act iv i ty Slack Analys is

Node Duration ES LS Slack

A 12 0 2 2B 9 0 0 0C 10 12 14 2D 10 9 9 0

E 24 9 35 26F 10 12 53 41G 35 22 24 2H 40 19 19 0I 15 12 48 36

J 4 59 59 0

K 6 63 63 0

K

6

C

10

G

35

J4

H40

B9

D10

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

Example 3.3


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Analyzing Cos t-Time

Trade-Offs

There are always cost-time trade-offs inproject management.

You can completing a project early by hiring moreworkers or running extra shifts.

There are often penalties if projects extendbeyond some specific date, and a bonus may beprovided for early completion.

Crashinga project means expediting someactivities to reduce overall project completiontime and total project costs.


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Project Costs

The total project costsare the sum of direct costs,indirect costs, and penalty costs.

Direct costsinclude labor, materials, and any othercosts directly related to project activities.

Indirect costsinclude administration, depreciation,financial, and other variable overhead costs that can

be avoided by reducing total project time.

The shorter the duration of the project, the lowerthe indirect costs will be.


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Cost to Crash

To assess the benefit of crashing certain activities,either from a cost or a schedule perspective, theproject manager needs to know the following timesand costs.

Normal time(NT) is the time necessary to completeand activity under normal conditions.

Normal cost(NC) is the activity cost associatedwith the normal time.

Crash time(CT) is the shortest possible time tocomplete an activity.

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


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Cost to Crash per Per iod

The Cost to Crash per Period =

CC NC

NT CT

CrashCost NormalCost

NormalTime CrashTime


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Linear cost assumption

8000

7000

6000

5000

4000

3000

0

Directcost(do

llars)

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

Time (weeks)

Crash cost (CC)

Normalcost (NC)

(Crash time) (Normal time)

Estimated costs fora 2-week reduction,from 10 weeks to8 weeks

5200

St. Adolfs HospitalCost-Time Relat ionsh ips in Cost Analysis


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The objective of cost analysisis todetermine the project schedule thatminimizes total project costs.

A minimum-cost scheduleis determinedby starting with the normal time scheduleand crashing activities along the critical path

in such a way that the costs of crashing donot exceed the savings in indirect andpenalty costs.

St. Adolfs HospitalMinim izing Costs


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Use these steps to determine the minimum costschedule:

1. Determine the projects critical path(s).

2. Find the activity or activities on the critical path(s)with the lowest cost of crashing per week.

3. Reduce the time for this activity untila. It cannot be further reduced or

b. Until another path becomes critical, or

c. The increase in direct costs exceeds the savings that resultfrom shortening the project (which lowers indirect costs).

4. Repeat this procedure until the increase in directcosts is larger than the savings generated by

shortening the project.

St. Adolfs HospitalMinimum Cost Schedule


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Direct Cost and Time Data fo r the

St. Adolfs Hospital Project

A 12 $ 12,000 11 $ 13,000 1 $ 1,000

B 9 50,000 7 64,000 2 7,000C 10 4,000 5 7,000 5 600D 10 16,000 8 20,000 2 2,000E 24 120,000 14 200,000 10 8,000F 10 10,000 6 16,000 4 1,500G 35 500,000 25 530,000 10 3,000H 40 1,200,000 35 1,260,000 5 12,000I 15 40,000 10 52,500 5 2,500J 4 10,000 1 13,000 3 1,000K 6 30,000 5 34,000 1 4,000

Totals $1,992,000 $2,209,500

Maximum Cost ofNormal Normal Crash Crash Time Crashing per

Time Cost Time Cost Reduction WeekActivity (NT) (NC) (CT) (CC) (wk) (CC-NC)

Shortenfirst


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St. Adolfs HospitalFinding the min imum cost sch edule: Stage 1

Step 1: The critical path is: B-D-H-J-K.

Step 2: The cheapest activity to crash is J at $1000.

Step 3: Crash activity J by its limit of three weeksbecause the critical path remains unchanged.

The new project length becomes 66 weeks.

The project completion time is 69 weeks. The direct costs for that schedule are $1,992,000. The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week. Total cost is $2,624,000 for 69 weeks

($1,992,000 + 69($8000) + (6965)($20,000)

Example 3.4


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St. Adolfs HospitalFinding the min imum co st schedule: Stage 1

The project completion time is 69 weeks. The direct costs for that schedule are $1,992,000. The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week. Total cost is $2,624,000 for 69 weeks

Crashing by 3 weeks saves $81,000 for a new total cost of$2,543,000.

Savings is 3 weeks of indirect costs (3 * $8000 = $24,000)

plus 3 weeks of penalties (3 * $20,000 = $60,000)less the cost of crashing (3 * $1,000 = $3,000)

Example 3.4


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Step 1: The critical path is still B-D-H-J-K.

Step 2: The cheapest activity to crash per week is now D at $2,000 a

week.Step 3: Crash D by 2 weeks.

The first week of reduction saves $28,000 by eliminating both thepenalty and indirect costs (but $2,000 goes toward crashing costs.)

The second week of reduction had no penalty, so it saves onlythe indirect costs of $8,000.

Total cost is now $2,511,00($2,543,00 - $28,000 - $8,000 + $4,000)


The indirect costs are $8000 per week. Penalty costs after week 65 are $20,000 per week.

Example 3.4


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Start

A12

B9

D8

E24

F10

FinishC10

G35

H40

I15

J1

K6

Shor tening D and J h ave

created a second cr i t ical

path , A-C-G-J-K. Both

crit ic al paths are 64

weeks.

Bo th must now be

sho rtened to real ize any

savings in indirect costs.

Example 3.4


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The alternatives are to crash one of the followingcombination of activities: A-B, A-H, C-B, C-H, G-B, G-H,

orCrash activity K which is on both critical paths.

(J and D have already been crashed.)

The cheapest alternative is to crash activity K.

It can only be crashed by one week at a cost of $4,000

The net savings are $8,000 $4,000 = $4,000

Total project cost now becomes$2,507,000

A $ 1,000

B 7,000C 600D 2,000E 8,000F 1,500G 3,000H 12,000

I 2,500J 1,000K 4,000

The indirect costs are $8000 per week.

Example 3.4


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A $ 1,000

B 7,000C 600D 2,000E 8,000F 1,500G 3,000H 12,000

I 2,500J 1,000K 4,000

Start

A12

B9

D8

E24

F10

FinishC10

G35

H40

I15

J1

K5 63 wks

The critical paths remain the samebut are now both 63 weeks.

Example 3.4


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B and C are the only remaining activities that canbe crashed simultaneously without exceeding the

potential savings of $8000 per week in indirectcosts.

Crash activities B and C by two weeks (the limit foractivity B)

Net savings are 2($8,000) 2($7,600) = $800 Total project costs are now $2,506,200

A $ 1,000

B 7,000C 600D 2,000E 8,000F 1,500G 3,000H 12,000

I 2,500J 1,000K 4,000

St. Adolfs HospitalFinding the min imum cost sch edule: Stage 4

The indirect costs are $8000 per week.

Example 3.4


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St. Adolfs HospitalSummary

The minimum cost schedule is 61 weeks. Activities J, D, K, B,and C were crashed for a total savings of $117,800

Example 3.4


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App l icat ion 3.3


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App l icat ion 3.3


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App l icat ion 3.3


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App l icat ion 3.3


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App l icat ion 3.3


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App l icat ion 3.3


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Assess ing Risks

Riskis a measure of the probability andconsequence of not reaching a definedproject goal.

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

A Risk-Management Planidentifies the keyrisks to a projects success and prescribesways to circumvent them.


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Categories o f

Project Risk

Strategic Fit: Projects should have a purpose that supportsthe strategic goals of the firm.

1. Service/Product Attributes: If the project involves newservice or product, several risks can arise.

Market riskcomes from competitors. Technological riskcan arise from advances made once the

project has started, rendering obsolete the technology chosen forservice or product.

Legal riskfrom liability suits or other legal action.

2. Project Team Capability: Involves risks from the projectteam itself such as poor selections and inexperience.

3. Operations Risk: Information accuracy, communications, andproject timing.


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Stat ist ical Analys is

The Statistical Analysis approach requires that

activity times be stated in terms of three reasonable

time estimates for each activity.

1. Optimistic Time (a)is the shortest time in which a activity

can be completed if all goes exceptionally well.

2. Most Likely Time (m)is the probable time for an activity.

3. Pessimistic Time (b)is the longest time required.

The expected time for an activity thus becomes

te=a+ 4m+ b

6


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Probabi l is t ic

Time Estimates

Mean

ma b Time

Probability

Beta

Distr ibut ion

PessimisticOptimistic


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Time

Probability

Normal

Distr ibut ion

Meana bm

3 3

Area under curvebetween aand bis 99.74%

Probabi l is t ic

Time Estimates


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A F

I

C G Finish

D

E

HB J

K

Start

te=a+ 4m+ b

6

Mean

2=( )ba62

Variance

St. Adolfs HospitalProbabi l ist ic Time Estim ates

Example 3.5

Calculating Means and Variances


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Activity B

Most

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

A F

I

C G Finish

D

E

HB J

K

Start

St. Adolfs HospitalProbabi l ist ic Time Estim ates

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

6

2= = 1.78( )15 - 762

Example 3.5

Calculating Means and Variances


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

2 )

Time Estimates (wk) Activity Statistics

A 11 12 13 12 0.11

B 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 4.00

G 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

St. Adolfs HospitalProbabi l ist ic Time Estim atesExample 3.5


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App l icat ion 3.4


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2= (variances of activities) z=T

TE2

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

z=72 69

11.89

Probabilities

Critical Path = B - D - H - J - K

T= 72 days TE= 69 days

St. Adolfs HospitalAnalyzing Probabi l i t ies

From Normal Distribution appendixPz= .8078 .81

Example 3.6


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Project duration (weeks)

69 72

Normal distribution:Mean = 69 weeks;

= 3.45 weeks

Probability ofexceeding 72weeks is 0.1922

St. Adolfs HospitalProbabi li ty of Completing Project On Time

Example 3.6

Probabilityof meetingthe scheduleis 0.8078

Length ofcritical path


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2= (variances of activities) z=T

TE2

2= 0.11 + 2.78 + 7.11 + 5.44 + 0.11 = 15.55

z= = 1.2772 67

15.55

Probabilities

Critical Path = A - C - G - J - K

T= 72 days TE= 67 days

From Normal Distribution appendix

Pz= .8980 .90

St. Adolfs HospitalProbabi li ty o f Completing Project On Time

Example 3.6


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App l icat ion 3.5


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App l icat ion 3.5


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Resource-Related Prob lems

Excessive Activity Duration Estimates:Many time estimates come with a built-incushion that management may not realize.

Latest Date Mentality: The tendency foremployees to procrastinate until the lastmoment before starting.

Failure to Deliver Early, even if the workis completed before the latest finish date.


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Path Mergersoccur when two or moreactivity paths combine at a particular node.Both paths must be completed up to thispoint, which will eliminate any built-up slack.

Multitaskingis the performance of multipleproject activities at the same time. Work on

some activities is delayed for other work. Loss of Focusby a manager can happen if

the critical path changes frequently.

Resource-Related Prob lems


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The Cri tical Chain Approach

A Critical Chainis the sequence of dependentevents that prevents a project from completing in ashorter interval and recognizes resource as well asactivity dependencies.

Time Estimates: The most likely time (m) is used to build thecritical chain project plan. The difference between it and thepessimistic time (b m) is used to develop the time buffers.

Buffers: Once the critical chain and all paths feeding it areidentified, time buffers can be added to protect the chain.

Using Latest Start Scheduleshas the advantage ofdelaying project cash outlays.

Project Control: Managers, using the critical chainapproach, must control the behavioral aspects of theirprojects.


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A F

I

C G Finish

D

E

HB J

K

Start

Activity Duration ES LSSlack

C 10 16 142G 35 26 242

J 4 61 59

2K 6 65 632D 10 10 91H 40 20 191E 24 10 3525I 15 16 4832

F 10 16 5337

SLACK CALCULATIONS

AFTER ACTIVITIES A AND BHAVE BEEN COMPLETED

Mon itor ing Project

Resources


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Project L ife Cyc le

Start Finish

Resourcerequ

irements

Time

Definitionand

organization

Planning Execution Close out

Solved Problem 1


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Solved Problem 1


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Solved Problem 1


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Solved Prob lem 2

What is the probabi l i ty of

com plet ing the pro ject in

23 weeks?

S l d P bl 2


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Solved Problem 2


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Solved Prob lem 2

Finish

Start

A

4.0

0.04.0

4.08.0

D

12.0

4.08.0

16.020.0

E

6.5

9.09.0

15.515.5

G

4.5

15.5

15.5

20.0

20.0

C

3.5

5.55.5

9.09.0

F

9.0

5.56.5

14.515.5

B

5.5

0.0

0.0

5.5

5.5

Solved Prob lem 2


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Solved Prob lem 2

Using the Normal Distribution appendix,we find that the probability of completing

Cost Crashing 14

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