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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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