1 Project Management Chapter 17 Lecture 5. 2 Project Management How is it different? Limited time frame Narrow focus, specific objectives Why.

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Project ManagementChapter 17

Lecture5

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Project ManagementProject Management How is it different?

Limited time frame Narrow focus, specific objectives

Why is it used? Special needs Pressures for new or improves products or services

Definition of a project Unique, one-time sequence of activities designed

to accomplish a specific set of objectives in a limited time frame

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Project ManagementProject Management What are the Key Metrics

Time Cost Performance objectives

What are the Key Success Factors? Top-down commitment Having a capable project manager Having time to plan Careful tracking and control Good communications

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Project ManagementProject Management

What are the tools? Work breakdown structure Network diagram Gantt charts

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Project ManagerProject Manager

Responsible for:

Work QualityHuman Resources TimeCommunications Costs

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Deciding which projects to implement

Selecting a project manager

Selecting a project team

Planning and designing the project

Managing and controlling project resources

Deciding if and when a project should be terminated

Key DecisionsKey Decisions

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Temptation to understate costs

Withhold information

Misleading status reports

Falsifying records

Compromising workers’ safety

Approving substandard work

http://www.pmi.org/

Ethical IssuesEthical Issues

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PERT and CPMPERT and CPM

PERT: Program Evaluation and Review TechniqueCPM: Critical Path Method

Graphically displays project activities Estimates how long the project will take Indicates most critical activities Show where delays will not affect project PERT and CPM have been used to plan, schedule, and control

a wide variety of projects: R&D of new products and processes Construction of buildings and highways Maintenance of large and complex equipment Design and installation of new systems

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PERT/CPMPERT/CPM

PERT/CPM used to plan the scheduling of individual

activities that make up a project. Projects may have as many as several

thousand activities. Complicating factor in carrying out the

activities some activities depend on the completion of

other activities before they can be started.

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

answer questions such as: What is the total time to complete the project? What are the scheduled start and finish dates for each

specific activity? Which activities are critical?

must be completed exactly as scheduled to keep the project on schedule?

How long can non-critical activities be delayed before they cause an increase in the project completion

time?

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Planning and SchedulingPlanning and Scheduling

Locate new facilities

Interview staff

Hire and train staff

Select and order furniture

Remodel and install phones

Furniture setup

Move in/startup

Activity 0 2 4 6 8 10 12 14 16 18 20

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Project NetworkProject Network

Project network constructed to model the precedence of the

activities. Nodes represent activities Arcs represent precedence relationships of the

activities Critical path for the network

a path consisting of activities with zero slack

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Project Network – An ExampleProject Network – An Example

A

B

C

E

F

Locatefacilities

Orderfurniture

Furnituresetup

Interview

RemodelMove in

D

Hire andtrain

GS

8 weeks

6 weeks

3 weeks

4 weeks9 weeks

11 weeks

1 week

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Management Scientist SolutionManagement Scientist Solution

Path Length (weeks)

Slack

A-B-F-G A-E-G C-D-G

18 20 14

2 0 6

Critical PathCritical Path

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Three-time estimate approach the time to complete an activity assumed to

follow a Beta distribution An activity’s mean completion time is:

t = (a + 4m + b)/6 a = the optimistic completion time estimate b = the pessimistic completion time estimate m = the most likely completion time estimate

An activity’s An activity’s completion time variancecompletion time variance is is 22 = (( = ((bb--aa)/6))/6)22

Uncertain Activity TimesUncertain Activity Times

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Uncertain Activity TimesUncertain Activity Times

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

The overall project completion time is assumed to have a normal distribution with mean equal to the sum of the means of

activities along the critical path, and variance equal to the sum of the variances of

activities along the critical path.

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ActivityImmediate

PredecessorOptimisticTime (a)

Most LikelyTime (m)

PessimisticTime (b)

A -- 4 6 8

B -- 1 4.5 5

C A 3 3 3

D A 4 5 6

E A 0.5 1 1.5

F B,C 3 4 5

G B,C 1 1.5 5

H E,F 5 6 7

I E,F 2 5 8

J D,H 2.5 2.75 4.5

K G,I 3 5 7

ExampleExample

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Management Scientist SolutionManagement Scientist Solution

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Network activities ES: early start EF: early finish LS: late start LF: late finish

Used to determine Expected project duration Slack time Critical path

Key TerminologyKey Terminology

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The Network Diagram (cont’d)The Network Diagram (cont’d) Path

Sequence of activities that leads from the starting node to the finishing node

AON path: S-1-2-6-7 Critical path

The longest path; determines expected project duration Critical activities

Activities on the critical path Slack

Allowable slippage for path; the difference the length of path and the length of critical path

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Advantages of PERTAdvantages of PERT

Forces managers to organize

Provides graphic display of activities

Identifies

Critical activities

Slack activities1

2

3

4

5 6

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Limitations of PERTLimitations of PERT

Important activities may be omitted

Precedence relationships may not be correct

Estimates may include a fudge factor

May focus solely on critical path

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George Dantzig – 1914 -2005 Concerned with optimal allocation of limited

resources such as Materials Budgets Labor Machine time

among competitive activities under a set of constraints

Linear ProgrammingLinear Programming

George Dantzig – 1914 -2005

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Product Mix Example (from session 1)Product Mix Example (from session 1)

Type 1 Type 2

Profit per unit $60 $50

Assembly time per unit

4 hrs 10 hrs

Inspection time per unit

2 hrs 1 hr

Storage space per unit

3 cubic ft 3 cubic ft

Resource Amount available

Assembly time 100 hours

Inspection time 22 hours

Storage space 39 cubic feet

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Maximize 60X1 + 50X2

Subject to

4X1 + 10X2 <= 100

2X1 + 1X2 <= 22

3X1 + 3X2 <= 39

X1, X2 >= 0

Linear Programming ExampleLinear Programming ExampleVariables

Objective function

Constraints

What is a Linear Program?

• A LP is an optimization model that has

• continuous variables

• a single linear objective function, and

• (almost always) several constraints (linear equalities or inequalities)

Non-negativity Constraints

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Decision variables unknowns, which is what model seeks to determine for example, amounts of either inputs or outputs

Objective Function goal, determines value of best (optimum) solution among all feasible (satisfy

constraints) values of the variables either maximization or minimization

Constraints restrictions, which limit variables of the model limitations that restrict the available alternatives

Parameters: numerical values (for example, RHS of constraints)

Feasible solution: is one particular set of values of the decision variables that satisfies the constraints Feasible solution space: the set of all feasible solutions

Optimal solution: is a feasible solution that maximizes or minimizes the objective function

There could be multiple optimal solutions

Linear Programming ModelLinear Programming Model

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Another Example of LP: Diet Another Example of LP: Diet ProblemProblem

Energy requirement : 2000 kcal Protein requirement : 55 g Calcium requirement : 800 mgFood Energy (kcal) Protein(g) Calcium(mg) Price per

serving($)

Oatmeal 110 4 2 3

Chicken 205 32 12 24

Eggs 160 13 54 13

Milk 160 8 285 9

Pie 420 4 22 24

Pork 260 14 80 13

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Example of LP : Diet ProblemExample of LP : Diet Problem

oatmeal: at most 4 servings/day chicken: at most 3 servings/day eggs: at most 2 servings/day milk: at most 8 servings/day pie: at most 2 servings/day pork: at most 2 servings/day

Design an optimal diet plan which minimizes the cost per

day

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Step 1: define decision variablesStep 1: define decision variables

x1 = # of oatmeal servings x2 = # of chicken servings x3 = # of eggs servings x4 = # of milk servings x5 = # of pie servings x6 = # of pork servings

Step 2: formulate objective function• In this case, minimize total cost

minimize z = 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

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Step 3: ConstraintsStep 3: Constraints

Meet energy requirement110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000 Meet protein requirement4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55 Meet calcium requirement2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800 Restriction on number of servings0x14, 0x23, 0x32, 0x48, 0x52, 0x62

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So, how does a LP look like?So, how does a LP look like?

minimize 3x1 + 24x2 + 13x3 + 9x4 + 24x5 + 13x6

subject to110x1 + 205x2 + 160x3 + 160x4 + 420x5 + 260x6 2000

4x1 + 32x2 + 13x3 + 8x4 + 4x5 + 14x6 55

2x1 + 12x2 + 54x3 + 285x4 + 22x5 + 80x6 800

0x14

0x23

0x32

0x48

0x52

0x62

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Optimal Solution – Diet ProblemOptimal Solution – Diet ProblemUsing LINDO 6.1Using LINDO 6.1

Cost of diet = $96.50 per day

Food # of servings

Oatmeal 4

Chicken 0

Eggs 0

Milk 6.5

Pie 0

Pork 2

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Optimal Solution – Diet ProblemOptimal Solution – Diet ProblemUsing Management ScientistUsing Management Scientist

Cost of diet = $96.50 per day

Food # of servings

Oatmeal 4

Chicken 0

Eggs 0

Milk 6.5

Pie 0

Pork 2

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Guidelines for Model FormulationGuidelines for Model Formulation

Understand the problem thoroughly. Describe the objective. Describe each constraint. Define the decision variables. Write the objective in terms of the decision

variables. Write the constraints in terms of the decision

variables Do not forget non-negativity constraints

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Product Mix ProblemProduct Mix Problem• Floataway Tours has $420,000 that can be used to

purchase new rental boats for hire during the summer. • The boats can be purchased from two different

manufacturers.• Floataway Tours would like to purchase at least 50 boats.• They would also like to purchase the same number from

Sleekboat as from Racer to maintain goodwill. • At the same time, Floataway Tours wishes to have a total

seating capacity of at least 200.

• Formulate this problem as a linear program

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Maximum Expected Daily

Boat Builder Cost Seating Profit

Speedhawk Sleekboat $6000 3 $ 70

Silverbird Sleekboat $7000 5 $ 80

Catman Racer $5000 2 $ 50

Classy Racer $9000 6 $110

Product Mix ProblemProduct Mix Problem

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Define the decision variables

x1 = number of Speedhawks ordered

x2 = number of Silverbirds ordered

x3 = number of Catmans ordered

x4 = number of Classys ordered Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units)

Max: 70x1 + 80x2 + 50x3 + 110x4

Product Mix ProblemProduct Mix Problem

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Define the constraints(1) Spend no more than $420,000:

6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000 (2) Purchase at least 50 boats: x1 + x2 + x3 + x4 > 50 (3) Number of boats from Sleekboat equals number

of boats from Racer: x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0

(4) Capacity at least 200: 3x1 + 5x2 + 2x3 + 6x4 > 200

Nonnegativity of variables: xj > 0, for j = 1,2,3,4

Product Mix ProblemProduct Mix Problem

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Max 70x1 + 80x2 + 50x3 + 110x4

s.t.

6000x1 + 7000x2 + 5000x3 + 9000x4 < 420,000

x1 + x2 + x3 + x4 > 50

x1 + x2 - x3 - x4 = 0

3x1 + 5x2 + 2x3 + 6x4 > 200

x1, x2, x3, x4 > 0

Product Mix Problem - Complete FormulationProduct Mix Problem - Complete Formulation

Daily profit = $5040

Boat # purchased

Speedhawk 28

Silverbird 0

Catman 0

Classy 28

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Marketing Application: Media SelectionMarketing Application: Media Selection

Advertising budget for first month = $30000 At least 10 TV commercials must be used At least 50000 customers must be reached Spend no more than $18000 on TV adverts Determine optimal media selection plan

Advertising Media # of potential customers reached

Cost ($) per advertisement

Max times available per month

Exposure Quality Units

Day TV 1000 1500 15 65

Evening TV 2000 3000 10 90

Daily newspaper 1500 400 25 40

Sunday newspaper 2500 1000 4 60

Radio 300 100 30 20

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Media Selection FormulationMedia Selection Formulation Step 1: Define decision variables

DTV = # of day time TV adverts ETV = # of evening TV adverts DN = # of daily newspaper adverts SN = # of Sunday newspaper adverts R = # of radio adverts

Step 2: Write the objective in terms of the decision variables Maximize 65DTV+90ETV+40DN+60SN+20R

Step 3: Write the constraints in terms of the decision variables

DTV <= 15

ETV <= 10

DN <= 25

SN <= 4

R <= 30

1500DTV + 3000ETV + 400DN + 1000SN + 100R <= 30000

DTV + ETV >= 10

1500DTV + 3000ETV <= 18000

1000DTV + 2000ETV + 1500DN + 2500SN + 300R >= 50000

BudgetBudget

Customers Customers reachedreached

TV TV ConstraintConstraint

ss

Availability of Availability of MediaMedia

DTV, ETV, DN, SN, R >= 0DTV, ETV, DN, SN, R >= 0

Exposure = 2370 units

Variable Value

DTV 10

ETV 0

DN 25

SN 2

R 30

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Applications of LPApplications of LP

Product mix planning Distribution networks Truck routing Staff scheduling Financial portfolios Capacity planning Media selection: marketing

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Possible Outcomes of a LPPossible Outcomes of a LP

A LP is either Infeasible – there exists no solution which satisfies

all constraints and optimizes the objective function or, Unbounded – increase/decrease objective

function as much as you like without violating any constraint

or, Has an Optimal Solution Optimal values of decision variables Optimal objective function value

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Infeasible LP – An ExampleInfeasible LP – An Example minimize

4x11+7x12+7x13+x14+12x21+3x22+8x23+8x24+8x31+10x32+16x33+5x34

Subject to x11+x12+x13+x14=100 x21+x22+x23+x24=200 x31+x32+x33+x34=150

x11+x21+x31=80 x12+x22+x32=90 x13+x23+x33=120 x14+x24+x34=170

xij>=0, i=1,2,3; j=1,2,3,4

Total demand exceeds total supply

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Unbounded LP – An ExampleUnbounded LP – An Example

maximize 2x1 + x2

subject to

-x1 + x2 1

x1 - 2x2 2

x1 , x2 0

x2 can be increased indefinitely without violating any constraint

=> Objective function value can be increased indefinitely

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Multiple Optima – An ExampleMultiple Optima – An Example

maximize x1 + 0.5 x2

subject to

2x1 + x2 4

x1 + 2x2 3

x1 , x2 0• x1= 2, x2= 0, objective function = 2

• x1= 5/3, x2= 2/3, objective function = 2