1 1 slide © 2005 thomson/south-western q 5 – 13 x 1 = the probability that station a will take...

1 1 Slide

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© 2005 Thomson/South-Western© 2005 Thomson/South-Western

Q 5 – 13

x1 = the probability that Station A will take Sitcom Rerun

x2 = the probability that Station A will take News Program

x3 = the probability that Station A will take Home Improvement

x4 = the value of the game for Station A

y1 = the probability that Station B will select Red

y2 = the probability that Station B will take News Program

y3 = the probability that Station B will take Home Improvement

y4 = the value of the game for Station B

2 2 Slide

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Q 5 – 13 cont’d

x*1 = 0, x*

2 = 0.6, x*3 = 0.4, x*

4 = 6.4

Max x4

s.t.10 x1 + 8 x2 + 4 x3 - 1 x4 ≧ 0

- 5 x1 + 7 x2 + 8 x3 - 1 x4 ≧ 03 x1 + 6 x2 + 7 x3 - 1 x4 ≧ 0

x1 + x2 + x3 = 1x1, x2, x3 ≧ 0

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Q 5 – 13 cont’d

y*1 = 0.2, y*

2 = 0, y*3 = 0.8, y*

4 = 6.4

Min y4

s.t.10 y1 - 5 y2 + 3 y3 - 1 y4 ≦ 08 y1 + 7 y2 + 6 y3 - 1 y4 ≦ 04 y1 + 8 y2 + 7 y3 - 1 y4 ≦ 0

y1 + y2 + y3 = 1y1, y2, y3 ≧ 0

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Q 5 – 13 cont’d

y*1 = 0.2, y*

2 = 0, y*3 = 0.8, y*

4 = 6.4

The optimal strategies for Station A are to take News Program with probability 0.6, and Home Improvement with probability 0.4. For Station B, the optimal strategies are to take Sitcom Rerun with probability 0.2, and Home Improvement with probability 0.8.The value of game is 6.4

x*1 = 0, x*

2 = 0.6, x*3 = 0.4, x*

4 = 6.4

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Q 5 – 15

x1 = the probability that Player A will select Red

x2 = the probability that Player A will select White

x3 = the probability that Player A will select Blue

x4 = the value of the game for Player A

y1 = the probability that Player B will select Red

y2 = the probability that Player B will select White

y3 = the probability that Player B will select Blue

y4 = the value of the game for Player B

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Q 5 – 15 cont’d

x*1 = 0.7, x*

2 = 0.3, x*3 = 0, x*

4 = 0.5

Max x4

s.t.0 x1 + 5 x2 + 2 x3 - 1 x4 ≧ 0

- 1 x1 + 4 x2 + 3 x3 - 1 x4 ≧ 02 x1 - 3 x2 - 4 x3 - 1 x4 ≧ 0

x1 + x2 + x3 = 1x1, x2, x3 ≧ 0

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OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables)

Variable Value Cost

1 Red 0.7000 0.0000

2 White 0.3000 0.0000

4 Blue 0.5000 1.0000

Slack Variables

5 CONSTR 1 1.0000 0.0000

Objective Function Value = 0.5

Constraint Type RHS Slack Dual price

1 CONSTR 1 >= 0.0000 1.000 0.0000

2 CONSTR 2 >= 0.0000 0.000 -0.5000

3 CONSTR 3 >= 0.0000 0.000 -0.5000

4 CONSTR 4 = 1.0000 0.000 0.5000

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Q 5 – 15 cont’d

y*1 = 0, y*

2 = 0.5, y*3 = 0.5, y*

4 = 0.5

Min y4

s.t.0 y1 - 1 y2 + 2 y3 - 1 y4 ≦ 05 y1 + 4 y2 - 3 y3 - 1 y4 ≦ 02 y1 + 3 y2 - 4 y3 - 1 y4 ≦ 0

y1 + y2 + y3 = 1y1, y2, y3 ≧ 0

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Q 5 – 15 cont’d

y*1 = 0, y*

2 = 0.5, y*3 = 0.5, y*

4 = 0.5

The optimal strategies for Player A are to select

Red with probability 0.7, and White with

probability 0.3. For Player B, the optimal

strategies are to select White with probability

0.5, and Blue with probability 0.5.

The value of game is 0.5

x*1 = 0.7, x*

2 = 0.3, x*3 = 0, x*

4 = 0.5

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Additional Question

x*1 = 0.0526, x*

2 = 0.7368 x*

3 = 0.2105, x*4 = 2.3684

Max x4

s.t.5 x1 + 2 x2 + 3 x3 - 1 x4 ≧ 00 x1 + 4 x2 + 2 x3 - 1 x4 ≧ 03 x1 + 3 x2 + 0 x3 - 1 x4 ≧ 01 x1 + 2 x2 + 4 x3 - 1 x4 ≧ 0

x1 + x2 + x3 = 1x1, x2, x3 ≧ 0

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OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables) Variable Value Cost1 VAR 1 0.0526 0.00002 VAR 2 0.7368 0.00003 VAR 3 0.2105 0.00004 VAR 4 2.3684 1.0000

Objective Function Value = 2.368421

Constraint Type RHS Slack Shadow price

1 CONSTR 1 >= 0.0000 0.000 -0.15792 CONSTR 2 >= 0.0000 1.000 0.00003 CONSTR 3 >= 0.0000 0.000 -0.36844 CONSTR 4 >= 0.0000 0.000 -0.47375 CONSTR 5 = 1.0000 0.000 2.3684

Because of complementally slackness condition, y*

1 = 0.1579, y*2 = 0, y*

3 = 0.3684, y*

4 = 0.4737, y*5 = 2.3684

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Chapter 9Chapter 9Project Scheduling: Project Scheduling:

PERT/CPMPERT/CPM Project Scheduling with Known Project Scheduling with Known

Activity TimesActivity Times

Project Scheduling with Uncertain Project Scheduling with Uncertain Activity TimesActivity Times

Considering Time-Cost Trade-OffsConsidering Time-Cost Trade-Offs

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PERT/CPMPERT/CPM PERTPERT

• Program Evaluation and Review TechniqueProgram Evaluation and Review Technique•Developed by U.S. Navy for Polaris missile projectDeveloped by U.S. Navy for Polaris missile project•Developed to handle uncertain activity timesDeveloped to handle uncertain activity times

CPMCPM•Critical Path MethodCritical Path Method•Developed by Du Pont & Remington RandDeveloped by Du Pont & Remington Rand•Developed for industrial projects for which Developed for industrial projects for which

activity times generally were knownactivity times generally were known Today’s project management software packages Today’s project management software packages

have combined the best features of both have combined the best features of both approaches.approaches.

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PERT/CPMPERT/CPM

PERT and CPM have been used to PERT and CPM have been used to plan, schedule, and control a wide plan, schedule, and control a wide variety of projects:variety of projects:

•R&D of new products and processesR&D of new products and processes

•Construction of buildings and Construction of buildings and highwayshighways

•Maintenance of large and complex Maintenance of large and complex equipmentequipment

•Design and installation of new Design and installation of new systemssystems

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PERT/CPMPERT/CPM

PERT/CPM is used to plan the PERT/CPM is used to plan the scheduling of individual scheduling of individual activitiesactivities that make up a project.that make up a project.

Projects may have as many as Projects may have as many as several thousand activities.several thousand activities.

A complicating factor in carrying A complicating factor in carrying out the activities is that some out the activities is that some activities depend on the activities depend on the completion of other activities completion of other activities before they can be started.before they can be started.

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PERT/CPMPERT/CPM Project managers rely on PERT/CPM to Project managers rely on PERT/CPM to

help them answer questions such as:help them answer questions such as:

•What is the What is the total timetotal time to complete the to complete the project?project?

•What are the What are the scheduled start and scheduled start and finish datesfinish dates for each specific activity? for each specific activity?

•Which activities are Which activities are criticalcritical and must and must be completed exactly as scheduled to be completed exactly as scheduled to keep the project on schedule?keep the project on schedule?

•How long can How long can noncritical activitiesnoncritical activities be be delayed before they cause an increase delayed before they cause an increase in the project completion time?in the project completion time?

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Project NetworkProject Network

A A project networkproject network can be can be constructed to model the constructed to model the precedence of the activities. precedence of the activities.

The The nodesnodes of the network of the network represent the activities. represent the activities.

The The arcsarcs of the network reflect the of the network reflect the precedence relationships of the precedence relationships of the activities. activities.

A A critical pathcritical path for the network is a for the network is a path consisting of activities with path consisting of activities with zero slack.zero slack.

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Example: Frank’s Fine FloatsExample: Frank’s Fine Floats

Frank’s Fine Floats is in the business of Frank’s Fine Floats is in the business of building elaborate parade floats. Frank and building elaborate parade floats. Frank and his crew have a new float to build and want to his crew have a new float to build and want to use PERT/CPM to help them manage the use PERT/CPM to help them manage the projectproject . .

The table on the next slide shows the The table on the next slide shows the activities that comprise the project. Each activities that comprise the project. Each activity’s estimated completion time (in days) activity’s estimated completion time (in days) and immediate predecessors are listed as well.and immediate predecessors are listed as well.

Frank wants to know the total time to Frank wants to know the total time to complete the project, which activities are complete the project, which activities are critical, and the earliest and latest start and critical, and the earliest and latest start and finish dates for each activity.finish dates for each activity.

19 19 Slide

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Immediate CompletionImmediate Completion

ActivityActivity DescriptionDescription PredecessorsPredecessors Time (days)Time (days)

A Initial Paperwork A Initial Paperwork --- --- 3 3

B Build Body B Build Body A A 3 3

C Build Frame C Build Frame A A 2 2

D Finish Body D Finish Body B B 3 3

E Finish Frame E Finish Frame C C 7 7

F Final Paperwork F Final Paperwork B,C B,C 3 3

G Mount Body to Frame D,EG Mount Body to Frame D,E 6 6

H Install Skirt on Frame CH Install Skirt on Frame C 2 2

20 20 Slide

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StartStart FinishFinish

BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

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Earliest Start and Finish TimesEarliest Start and Finish Times

Step 1:Step 1: Make a forward pass through the Make a forward pass through the network as follows: For each activity network as follows: For each activity i i beginning at the Start nodebeginning at the Start node, , compute:compute:

•Earliest Start TimeEarliest Start Time = the maximum of = the maximum of the earliest finish times of all activities the earliest finish times of all activities immediately preceding activity immediately preceding activity ii. (This is . (This is 0 for an activity with no predecessors.)0 for an activity with no predecessors.)

•Earliest Finish TimeEarliest Finish Time = (Earliest Start = (Earliest Start Time) + (Time to complete activity Time) + (Time to complete activity i i ).).

The project completion time is the The project completion time is the maximum of the Earliest Finish Times at maximum of the Earliest Finish Times at the Finish node.the Finish node.

22 22 Slide

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Earliest Start and Finish TimesEarliest Start and Finish Times


BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

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Latest Start and Finish TimesLatest Start and Finish Times

Step 2:Step 2: Make a backwards pass Make a backwards pass through the network as follows: Move through the network as follows: Move sequentially backwards from the Finish sequentially backwards from the Finish node to the Start node. At a given node to the Start node. At a given node, node, jj, consider all activities ending at , consider all activities ending at nodenode j j. For each of these activities, . For each of these activities, ii, , compute:compute:

•Latest Finish TimeLatest Finish Time = the minimum of = the minimum of the latest start times beginning at the latest start times beginning at node node jj. (For node . (For node NN, this is the , this is the project completion time.)project completion time.)

•Latest Start TimeLatest Start Time = (Latest Finish = (Latest Finish Time) - (Time to complete activity Time) - (Time to complete activity i i ).).

24 24 Slide

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Latest Start and Finish TimesLatest Start and Finish Times


BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

6 96 9 9 9 1212

0 30 3

3 53 5

12 12 1818

15 15 1818

16 16 1818

5 125 12

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Determining the Critical PathDetermining the Critical Path

Step 3:Step 3: Calculate the slack Calculate the slack time for each activity by: time for each activity by:

SlackSlack = (Latest Start) - = (Latest Start) - (Earliest Start), or (Earliest Start), or

= (Latest Finish) - = (Latest Finish) - (Earliest Finish).(Earliest Finish).

26 26 Slide

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Activity Slack TimeActivity Slack Time

ActivityActivity ESES EFEF LSLS LFLF SlackSlack A 0 3 0 3 0 A 0 3 0 3 0

(critical)(critical) B 3 6 6 9 3B 3 6 6 9 3 C 3 5 3 5 0 C 3 5 3 5 0

(critical)(critical) D 6 9 9 12 3D 6 9 9 12 3 E 5 12 5 12 0 E 5 12 5 12 0

(critical)(critical) F 6 9 15 18 9F 6 9 15 18 9 G 12 18 12 18 0 G 12 18 12 18 0

(critical)(critical) H 5 7 16 18 11H 5 7 16 18 11

27 27 Slide

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• A A critical pathcritical path is a path of activities, from the is a path of activities, from the Start node to the Finish node, with 0 slack Start node to the Finish node, with 0 slack times.times.

• Critical Path: A – C – E – GCritical Path: A – C – E – G

• The The project completion timeproject completion time equals the equals the maximum of the activities’ earliest finish maximum of the activities’ earliest finish times.times.

• Project Completion Time: 18 daysProject Completion Time: 18 days


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Critical PathCritical Path


BB33

DD33

AA33

CC22

GG66

FF33

HH22

EE77

0 30 3

3 63 6 6 96 9

3 53 5

12 12 1818

6 96 9

5 75 7

5 125 12

6 96 9 9 9 1212

0 30 3

3 53 5

12 12 1818

15 15 1818

16 16 1818

5 125 12

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In the In the three-time estimate approachthree-time estimate approach, the , the time to complete an activity is assumed to time to complete an activity is assumed to follow a Beta distribution. follow a Beta distribution.

An activity’s An activity’s mean completion timemean completion time is: is:

tt = ( = (aa + 4 + 4mm + + bb)/6)/6

• aa = the = the optimisticoptimistic completion time completion time estimateestimate

• bb = the = the pessimisticpessimistic completion time completion time estimateestimate

• mm = the = the most likelymost likely completion time completion time estimateestimate

Uncertain Activity TimesUncertain Activity Times

30 30 Slide

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An activity’s An activity’s completion time variancecompletion time variance is: is:

22 = (( = ((bb--aa)/6))/6)22

• aa = the = the optimisticoptimistic completion time completion time estimateestimate

• bb = the = the pessimisticpessimistic completion time completion time estimateestimate

• mm = the = the most likelymost likely completion time completion time estimateestimate


31 31 Slide

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In the three-time estimate approach, In the three-time estimate approach, the critical path is determined as if the the critical path is determined as if the mean times for the activities were mean times for the activities were fixed times. fixed times.

The The overall project completion timeoverall project completion time is is assumed to have a normal distribution assumed to have a normal distribution with mean equal to the sum of the with mean equal to the sum of the means along the critical path and means along the critical path and variance equal to the sum of the variance equal to the sum of the variances along the critical path.variances along the critical path.

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Example: ABC Associates Example: ABC Associates

Consider the following project:Consider the following project:

Immed. Optimistic Most Likely PessimisticImmed. Optimistic Most Likely Pessimistic

ActivityActivity Predec.Predec. Time (Hr.Time (Hr.) ) Time (Hr.)Time (Hr.) Time (Hr.)Time (Hr.) A A -- 4 -- 4 6 6 8 8 B B -- 1 -- 1 4.5 4.5

5 5 C C A A 3 3 3 3

3 3 D D A 4 5 A 4 5 6 6 E E A 0.5 1 A 0.5 1

1.51.5 F F B,C 3 4 5 B,C 3 4 5 G G B,C B,C 1 1.5 5 1 1.5 5 H H E,F E,F 5 6 7 5 6 7 I I E,F 2 5 8 E,F 2 5 8 J J D,H D,H 2.5 2.75 2.5 2.75

4.5 4.5 K K G,I 3 5 G,I 3 5

77

33 33 Slide

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Example: ABC AssociatesExample: ABC Associates


E

Start

A

H

D

F

J

I

K

Finish

B

C

G

E

Start

A

H

D

F

J

I

K

Finish

B

C

G

6666

4444

3333

5555

5555

2222

4444

11116666

3333

5555

34 34 Slide

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Activity Expected Times and VariancesActivity Expected Times and Variances

tt = ( = (aa + 4 + 4mm + + bb)/6 )/6 22 = (( = ((bb--aa)/6))/6)22

ActivityActivity Expected TimeExpected Time VarianceVariance A A 6 6 4/9 4/9

B B 4 4 4/9 4/9 C C 3 3 0 0 D D 5 5 1/9 1/9 E E 1 1 1/36 1/36 F F 4 4 1/9 1/9 G G 2 2 4/9 4/9 H H 6 6 1/9 1/9 I I 5 5 1 1 J J 3 3 1/9 1/9 K K 5 5 4/9 4/9

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Earliest/Latest Times and SlackEarliest/Latest Times and Slack

ActivityActivity ESES EF EF LSLS LFLF SlackSlack A A 0 6 0 6 0 * 0 6 0 6 0 *

B B 0 4 5 9 5 0 4 5 9 5 C 6 9 6 9 0 *C 6 9 6 9 0 * D D 6 11 15 20 9 6 11 15 20 9 E E 6 7 12 13 6 6 7 12 13 6 F F 9 13 9 13 0 * 9 13 9 13 0 * G 9 11 16 18 7G 9 11 16 18 7 H H 13 19 14 20 1 13 19 14 20 1 I I 13 18 13 18 0 * 13 18 13 18 0 * J J 19 22 20 23 1 19 22 20 23 1 K K 18 23 18 23 0 * 18 23 18 23 0 *

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• A A critical pathcritical path is a path of activities, from the is a path of activities, from the Start node to the Finish node, with 0 slack Start node to the Finish node, with 0 slack times.times.

• Critical Path: A – C – F – I – KCritical Path: A – C – F – I – K

• The The project completion timeproject completion time equals the equals the maximum of the activities’ earliest finish maximum of the activities’ earliest finish times.times.

• Project Completion Time: 23 hoursProject Completion Time: 23 hours


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Critical Path (A-C-F-I-K)Critical Path (A-C-F-I-K)

E

Start

A

H

D

F

J

I

K

Finish

B

C

G

E

Start

A

H

D

F

J

I

K

Finish

B

C

G

6666

4444

3333

5555

5555

2222

4444

11116666

3333

5555

0 60 60 60 6

9 139 139 139 13

13 1813 1813 1813 18

9 119 1116 1816 18

13 1913 1914 2014 20

19 2219 2220 2320 23

18 2318 2318 2318 23

6 76 712 1312 13

6 96 96 96 9

0 40 45 95 9

6 116 1115 2015 20

38 38 Slide

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Probability the project will be completed within 24 Probability the project will be completed within 24 hrshrs

22 = = 22AA + + 22

CC + + 22FF + + 22

HH + + 22KK

= 4/9 + 0 + 1/9 + 1 + 4/9 = 4/9 + 0 + 1/9 + 1 + 4/9

= 2= 2

= 1.414= 1.414

zz = (24 - 23)/ = (24 - 23)/(24-23)/1.414 (24-23)/1.414 = .71= .71

From the Standard Normal Distribution From the Standard Normal Distribution table: table:

P(z P(z << .71) = .5 + .2612 = .7612 .71) = .5 + .2612 = .7612


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EarthMover is a manufacturer of road EarthMover is a manufacturer of road constructionconstruction

equipment including pavers, rollers, and graders. equipment including pavers, rollers, and graders. TheThe

company is faced with a newcompany is faced with a new

project, introducing a newproject, introducing a new

line of loaders. Managementline of loaders. Management

is concerned that the project mightis concerned that the project might

take longer than 26 weeks totake longer than 26 weeks to

complete without crashing somecomplete without crashing some

activities.activities.

Example: EarthMover, Inc.Example: EarthMover, Inc.

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Immediate Immediate CompletionCompletion

ActivityActivity DescriptionDescription PredecessorsPredecessors Time (wks)Time (wks) A Study Feasibility A Study Feasibility --- ---

6 6 B Purchase Building B Purchase Building A A 4 4 C Hire Project Leader C Hire Project Leader A A 3 3 D Select Advertising StaffD Select Advertising Staff B B 6 6 E Purchase Materials E Purchase Materials B B 3 3 F Hire Manufacturing Staff B,CF Hire Manufacturing Staff B,C 10 10 G Manufacture Prototype E,FG Manufacture Prototype E,F 2 2 H Produce First 50 Units GH Produce First 50 Units G 6 6 II Advertise Product D,G Advertise Product D,G 8 8


41 41 Slide

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PERT NetworkPERT Network


C

Start

D

E

I

A

Finish

H

G

B

F

C

Start

D

E

I

A

Finish

H

G

B

F

66664444

333310101010

3333

6666

2222 6666

8888

42 42 Slide

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Earliest/Latest TimesEarliest/Latest Times

ActivityActivity ESES EFEF LSLS LFLF SlackSlack A A 0 6 0 6 0 * 0 6 0 6 0 * B B 6 10 6 10 0 * 6 10 6 10 0 * C C 6 9 7 10 1 6 9 7 10 1 D 10 16 16 22 6D 10 16 16 22 6 E E 10 13 17 20 7 10 13 17 20 7 F F 10 20 10 20 0 * 10 20 10 20 0 * G G 20 22 20 22 0 * 20 22 20 22 0 * H H 22 28 24 30 2 22 28 24 30 2 I I 22 30 22 30 0 * 22 30 22 30 0 *


43 43 Slide

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Critical ActivitiesCritical Activities

C

Start

D

E

I

A

Finish

H

G

B

F

C

Start

D

E

I

A

Finish

H

G

B

F

66664444

333310101010

3333

6666

2222 6666

88880 60 60 60 6

10 2010 20 10 2010 20

20 2220 2220 2220 22

10 1610 1616 2216 22 22 3022 30

22 3022 30

22 2822 2824 3024 30

6 96 9 7 107 10

10 1310 1317 2017 20

6 106 10 6 106 10

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CrashingCrashing

The completion time for this project using The completion time for this project using normalnormal

times is 30 weeks. Which activities should be times is 30 weeks. Which activities should be crashed,crashed,

and by how many weeks, in order for the and by how many weeks, in order for the project to beproject to be

completed in 26 weeks?completed in 26 weeks?

45 45 Slide

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Crashing Activity TimesCrashing Activity Times

In the In the Critical Path Method (CPM)Critical Path Method (CPM) approach to approach to project scheduling, it is assumed that the project scheduling, it is assumed that the normal time to complete an activity, normal time to complete an activity, ttj j , which , which can be met at a normal cost, can be met at a normal cost, ccj j , can be crashed , can be crashed to a reduced time, to a reduced time, ttjj’, under maximum crashing ’, under maximum crashing for an increased cost, for an increased cost, ccjj’.’.

Using CPM, activity Using CPM, activity jj's maximum time 's maximum time reduction, reduction, MMj j , may be calculated by: , may be calculated by: MMj j = = ttjj - - ttjj'. '. It is assumed that its cost per unit reduction, It is assumed that its cost per unit reduction, KKj j , is linear and can be calculated by: , is linear and can be calculated by: KKjj = ( = (ccjj' - ' - ccjj)/)/MMjj..

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Normal CrashNormal Crash ActivityActivity TimeTime CostCost TimeTime CostCostA) Study Feasibility A) Study Feasibility 6 $ 80,000 5 6 $ 80,000 5 $100,000$100,000B) Purchase Building B) Purchase Building 4 100,000 4 4 100,000 4 100,000100,000C) Hire Project Leader C) Hire Project Leader 3 50,000 2 3 50,000 2 100,000100,000D) Select Advertising Staff D) Select Advertising Staff 6 150,000 3 6 150,000 3 300,000300,000E) Purchase Materials E) Purchase Materials 3 180,000 2 3 180,000 2 250,000250,000F) Hire Manufacturing Staff F) Hire Manufacturing Staff 10 300,000 7 10 300,000 7 480,000480,000G) Manufacture Prototype G) Manufacture Prototype 2 100,000 2 2 100,000 2 100,000 100,000H) Produce First 50 Units H) Produce First 50 Units 6 450,000 5 6 450,000 5 800,000800,000 I) Advertising Product 8 350,000 4 I) Advertising Product 8 350,000 4 650,000650,000

Normal Costs and Crash CostsNormal Costs and Crash Costs

47 47 Slide

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Min 20Min 20YYAA + 50 + 50YYCC + 50 + 50YYDD + 70 + 70YYEE + 60 + 60YYFF + 350 + 350YYHH + + 7575YYII

s.t. s.t. YYAA << 1 1 XXAA >> 0 + (6 - 0 + (6 - YYII) ) XXGG >> XXFF + + (2 - (2 - YYGG) ) YYCC << 1 1 XXBB >> XXAA + (4 - + (4 - YYBB) ) XXHH >> XXGG + (6 - + (6 - YYHH) ) YYDD << 3 3 XXCC >> XXAA + (3 - + (3 - YYCC) ) XXII >> XXDD + (8 - + (8 - YYII) ) YYEE << 1 1 XXDD >> XXBB + (6 - + (6 - YYDD) ) XXII >> XXGG + (8 - + (8 - YYII)) YYFF << 3 3 XXEE >> XXBB + (3 - + (3 - YYEE)) XXHH << 26 26

YYHH << 1 1 XXFF >> XXBB + (10 - + (10 - YYFF) ) XXII << 26 26 YYII << 4 4 XXFF >> XXCC + (10 - + (10 - YYFF) ) XXGG >> XXEE + (2 - + (2 - YYGG) ) XXii, , YYjj >> 0 0 for all i for all i


Linear Program for Minimum-Cost CrashingLinear Program for Minimum-Cost Crashing

Let: Let: XXii = earliest finish time for activity = earliest finish time for activity ii YYii = the amount of time activity = the amount of time activity ii is crashed is crashed

48 48 Slide

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EMGT 501

Home Work 5Chapter 9 – Prob 3&4

Chapter 9 – Prob 10

Due Day: Oct 7

49 49 Slide

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RemindRemind Mid-Term Exam will be Mid-Term Exam will be

listed at my HP.listed at my HP.

The Exam will be a Take-The Exam will be a Take-Home Style for one week.Home Style for one week.

The Due Day will be Oct 20 The Due Day will be Oct 20 (Monday-Noon).(Monday-Noon).

1 1 slide © 2005 thomson/south-western q 5 – 13 x 1 = the probability that station a will take...

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