1 project management chapter 17 lecture 5. 2 project management how is it different? limited time...
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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
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