2-1 copyright © 2013 pearson education, inc. publishing as prentice hall chapter topics model...
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2-1Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall
Chapter Topics
Model Formulation A Maximization Model Example Graphical Solutions of Linear Programming
Models A Minimization Model Example Irregular Types of Linear Programming
Models Characteristics of Linear Programming
Problems
Linear Programming: Formulation and Applications
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Objectives of business decisions frequently involve maximizing profit or minimizing costs.
Linear programming uses linear algebraic relationships to represent a firm’s decisions, given a business objective, and resource constraints.
Model formulation steps:Step 1 : Clearly define the problem: Objectives and decision to be made
Step 2 : Construct the objective function Step 3 : Formulate the constraints
Step 4: Solve the model
Linear Programming: An Overview
LP Model Formulation
Product mix problem - Beaver Creek Pottery Company
• How many bowls and mugs should be produced to maximize profits given labor and materials constraints?
Resource Requirements
Product Labor(Hr./Unit)
Clay(Lb./Unit)
Profit($/Unit)
Bowl 1 4 40
Mug 2 3 50
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LP Model Formulation
Resource 40 hrs of labor per dayAvailability: 120 lbs of clay
Decision B = number of bowls to produce per day
Variables: M = number of mugs to produce per day
Objective Maximize Z = $40B + $50MFunction: Where Z = profit per day
Resource 1B + 2M 40 hours of laborConstraints: 4B + 3M 120 pounds of clay
Non-Negativity B 0; M 0 Constraints:
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Complete Linear Programming Model:
Maximize Z = $40B + $50M
subject to: 1B + 2M 404B + 3M 120B, M 0
LP Model Formulation
Product mix problem - Beaver Creek Pottery Company
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A feasible solution does not violate any of the constraints:
Example: B = 5 bowls
M = 10 mugsZ = $40B + $50M= $700
Labor constraint check: 1(5) + 2(10) = 25
≤ 40 hours Clay constraint check: 4(5) + 3(10) = 70 ≤
120 pounds
Feasible Solutions
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An infeasible solution violates at least one of the constraints:
Example: B = 10 bowls
M = 20 mugs Z = $40B + $50M= $1400
Labor constraint check: 1(10) + 2(20) = 50 > 40 hours
Infeasible Solutions
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Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty).
Graphical methods provide visualization of how a solution for a linear programming problem is obtained.
Graphical Solution of LP Models
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Coordinate AxesGraphical Solution of Maximization Model (1 of 12)
Figure 2.2 Coordinates for graphical analysis
B is bowls
M is mugs
M
B
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Optimal SolutionGraphical Solution of Maximization Model (9 of 12)
Figure 2.10 Identification of optimal solution point
40M+50B
M
B
Product mix problem - Beaver Creek Pottery Company
• Gather information needed to conduct model building and analysis (steps 2 and 3)– Available capacity in each activity/process– How much capacity in each activity needed by each product– Profitability of each product
Given parameters (known data)Bowl Mug
Unit Profit $40 $50
Production Time (Hours ) Used Per Unit Produced
Resources Available
labor 1 2 40
clay 4 3 120
Complete Linear Programming Model:
Maximize Z = $40B + $50M
subject to: 1B + 2M 404B + 3M 120B, M 0
LP Model Formulation
Product mix problem - Beaver Creek Pottery Company
Developing a Spreadsheet Model
• Step #1: Data Cells– Enter all of the data for the problem on the spreadsheet.– Make consistent use of rows and columns.– It is a good idea to color code these “data cells” (e.g., light blue).
Beaver Creek Pottery Company
Bowl Mug
Unit Profit 40 50Resources
Resources Used Per Unit Produced Available
Labor 1 2 40
Clay 4 3 120
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Developing a Spreadsheet Model
• Step #2: Changing Cells– Add a cell in the spreadsheet for every decision that needs to be made.– If you don’t have any particular initial values, just enter 0 in each.– It is a good idea to color code these “changing cells” (e.g., yellow with border).
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Beaver Creek Pottery Company
Bowl Mug
Unit Profit 40 50
Resources
Resources Used Per Unit Produced Available
Labor 1 2 40
Clay 4 3 120
Units produced
Developing a Spreadsheet Model
• Step #3: Target Cell– Develop an equation that defines the objective of the model.– Typically this equation involves the data cells and the changing cells in order to
determine a quantity of interest (e.g., total profit or total cost).– It is a good idea to color code this cell (e.g., orange with heavy border).
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Beaver Creek Pottery Company
Bowl Mug
Unit Profit 40 50
Resources
Resources Used Per Unit Produced Available
Labor 1 2 40
Clay 4 3 120
Units produced total profit
Developing a Spreadsheet Model
• Step #4: Constraints– For any resource that is restricted, calculate the amount of that resource used in a
cell on the spreadsheet (an output cell).– Define the constraint in consecutive cells. For example, if Quantity A <= Quantity
B, put these three items (Quantity A, <=, Quantity B) in consecutive cells.
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Beaver Creek Pottery Company
Bowl Mug
Unit Profit 40 50Resources
Resources Used Per Unit Produced resources used AvailableLabor 1 2 < 40
Clay 4 3 < 120Bowl Mug
Units produced total profit
A Trial Solution
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Let’s do this problem in Excel….Pay attention to
Model layout and organizationData accuracySteps in the Solver Addin
Max or Minnon-negativity optionLinear optimization
Identifying the Target Cell and Changing Cells (Excel 2010)
• Choose the “Solver” from the Data tab.• Select the cell you wish to optimize in the “Set Target Cell” window.• Choose “Max” or “Min” depending on whether you want to maximize or minimize the
target cell.• Enter all the changing cells in the “By Changing Cells” window.
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Some Important Options (Excel 2007)
• Click on the “Options” button, and click in both the “Assume Linear Model” and the “Assume Non-Negative” box.– “Assume Linear Model” tells the Solver that this is a linear programming model.– “Assume Non-Negative” adds nonnegativity constraints to all the changing cells.
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The Complete Solver Dialogue Box (Excel 2007)
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Template for Resource-Allocation Problems
Activities
Unit Profit profit per unit of activityResources Resources
Used Available
SUMPRODUCTresource used per unit of activity (resource used per unit,
changing cells)
Total ProfitLevel of Activity changing cells SUMPRODUCT(profit per unit, changing cells)
<=
Con
stra
ints
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LP Model Formulation – Minimization (1 of 7)
Chemical Contribution
Brand Nitrogen (lb/ bag)
Phosphate (lb/ bag)
Super-gro 2 4
Crop-quick 4 3
Two brands of fertilizer available - Super-gro, Crop-quick.
Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.
Super-gro costs $6 per bag, Crop-quick $3 per bag.
Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?
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Decision Variables: x1 = bags of Super-grox2 = bags of Crop-quick
The Objective Function:Minimize Z = $6x1 + 3x2
Where: $6x1 = cost of bags of Super-Gro
$3x2 = cost of bags of Crop-Quick
Model Constraints:2x1 + 4x2 16 lb (nitrogen constraint)4x1 + 3x2 24 lb (phosphate constraint)x1, x2 0 (non-negativity constraint)
LP Model Formulation – Minimization (2 of 7)
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Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 16 4x2 + 3x2 24
x1, x2 0
Figure 2.16 Constraint lines for fertilizer model
Constraint Graph – Minimization (3 of 7)
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Figure 2.17 Feasible solution area
Feasible Region– Minimization (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 16 4x2 + 3x2 24
x1, x2 0
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Figure 2.18 The optimal solution point
Optimal Solution Point – Minimization (5 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 16 4x2 + 3x2 24
x1, x2 0
The optimal solution of a minimization problem is at the extreme point closest to the origin.
Template for Cost-Benefit Tradeoff Problems
Activities
Unit Cost cost per unit of activityBenefit Benefit
Achieved Needed
SUMPRODUCTbenefit achieved per unit of activity (benefit per unit,
changing cells)
Total CostLevel of Activity changing cells SUMPRODUCT(cost per unit, changing cells)
>=
Const
rain
ts
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Types of Functional Constraints
Type Form* Typical Interpretation Main Usage
Resource constraint LHS ≤ RHSFor some resource, Amount used ≤ Amount available
Resource-allocation problems and mixed problems
Benefit constraint LHS ≥ RHSFor some benefit, Level achieved ≥ Minimum Acceptable
Cost-benefit-trade-off problems and mixed problems
Fixed-requirement constraint
LHS = RHSFor some quantity, Amount provided = Required amount
Transportation problems and mixed problems
* LHS = Left-hand side (a SUMPRODUCT function). RHS = Right-hand side (a constant).
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Template for Mixed Problems
Activities
Unit Profit or Cost profit/cost per unit of activityResources Resources
Used Available
SUMPRODUCTresource used per unit of activity (resource used per unit,
changing cells)
Benefit BenefitAchieved Needed
SUMPRODUCTbenefit achieved per unit of activity (benefit per unit,
changing cells)
Total Profit or CostLevel of Activity changing cells
Const
rain
ts
SUMPRODUCT(profit/cost per unit, changing cells)
<=
>=
=
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