© Copyright 2004, Alan Marshall 1 Lecture 1 Linear Programming.

© Copyright 2004, Alan Marshall 1

Lecture 1Lecture 1

Linear Programming


AgendaAgenda

>Math Programming>Linear Programming

• Introduction• Exercise: Lego Enterprises• Terminology, Definitions• Possible Outcomes• Sensitivity Analysis


Math ProgrammingMath Programming

>Deals with resource allocation to maximize or minimize an objective subject to certain constraints

>Types:• Linear, Integer, Mixed, Nonlinear, Goal

>Relatively easy to solve using modern computing technology (potentially too easy!)


Our FocusOur Focus

>Linear, Integer (& Mixed Linear/Integer)

>Recognizing when linear/integer/mixed programming is appropriate

>Developing basic models>Computer solution

• Excel

>Interpreting results


Lego EnterprisesLego Enterprises

>Table profit is $16; Chair profit is $10>Table design

• 2 large blocks (side by side)• 2 small blocks (stacked under, centered)

>Chair design• 1 large block (seat)• 2 small blocks (back, bottom)

>Objective: select product mix to maximize profits using available resources


Understanding Lego ProblemUnderstanding Lego Problem

> Formulate as LP• Decision Variables, Objective Function,

Constraints

>Graph• Constraints, Objective function

>Find solution


LP FormulationLP Formulation

>Decision Variables• T = # of tables• C = # of chairs

>Objective• Maximize profit =

>Constraints• For large blocks:• For small blocks:




>Objective• Maximize profit: Z = 16T + 10C

>Constraints• For large blocks:• For small blocks:





>Constraints• For large blocks: 2T + 1C < 6• For small blocks:





>Constraints• For large blocks: 2T + 1C < 6• For small blocks: 2T + 2C < 8


Graphing Lego ExampleGraphing Lego Example

>Draw quadrant & axes• use T on x-axis and C on y-axis

>Add constraint lines• Find intercepts: set T to zero and solve

for C, set C to zero and solve for T

>Add profit equation• Select reasonable value

>Move profit equation outwards, as far as feasible



>Draw quadrant & axes• use T on x-axis and

C on y-axis



>Add constraint lines• Find intercepts: set

T to zero and solve for C, set C to zero and solve for T

>Large:• Tables: Max = 3• Chairs: Max = 6





>Large:• Tables: Max = 3• Chairs: Max = 6





>Small:• Tables: Max = 4• Chairs: Max = 4





>Small:• Tables: Max = 4• Chairs: Max = 4



>Add profit equation• Select reasonable

value

>40:• Tables: 40/16 = 2.5• Chairs: 40/10 = 4



>Add profit equation• Select reasonable

value

>40:• Tables: 40/16 = 2.5• Chairs: 40/10 = 4



>Move profit equation outwards, as far as feasible

>Solution: T = 2, C = 2

>Profit: 16(2)+10(2)=52


Characteristics Of LPsCharacteristics Of LPs

>Objective function and constraints are linear functions

>Constraint types are <, = , or > >Variables can assume any fractional

value>Decision variables are non-negative>Maximize or Minimize single objective



>Objective function and constraints are linear functions• Lego: All were linear trade-offs






>Constraint types are <, = , or > • Lego: All Constraints implied maximums

(<)

>Variables can assume any fractional value

>Decision variables are non-negative>Maximize or Minimize single objective





value• Lego: Fractional values can be viewed as

work-in-process at the end of the day

>Decision variables are non-negative>Maximize or Minimize single objective





value>Decision variables are non-negative

• Lego: Cannot produce negative amounts

>Maximize or Minimize single objective






• Lego: Maximizing Profit


Key DefinitionsKey Definitions

>Feasible solution: one that satisfies all constraints• can have many feasible solutions

>Feasible region: set of all feasible solutions

>Optimal solution: any feasible solution that optimizes the objective function• can have ties


Standard LP FormStandard LP Form

>All constraints expressed as equalities• use slack (<) or surplus (>) variables

>All variables are nonnegative>All variables appear on the left side of

the constraint functions>All constants appear on the right side

of the constraint functions>Formulate Lego problem in standard

form


Lego - Standard FormLego - Standard Form

>Maximize profit: Z = 16T + 10C>Subject to

• For large blocks: 2T + 1C + S1 = 6

• For small blocks: 2T + 2C + S2 = 8

• Non-negativities:, T, C, S1, S2 > 0

>Useful, because of the concept of slack and surplus• While we will not formulate this way in

Excel, we will still use these concepts


Possible LP OutcomesPossible LP Outcomes

>Unique optimal solution>Alternate optimal solutions>Unbounded problem>Infeasible problem


Example: Unique Optimal Soln Example: Unique Optimal Soln

>Solve graphically for the optimal solution:Max: z = 6x1 + 2x2

s.t. 4x1 + 3x2 > 12

x1 + x2 < 8

x1, x2 > 0


xx22

xx11

44xx11 + 3 + 3xx22 >> 12 12

xx11 + + xx22 << 8 8

33 88

44

88

Max 6Max 6xx11 + 2 + 2xx22

Example: Unique OptimalExample: Unique Optimal

>There is only one point in the feasible set that maximizes the objective function (x1 = 8, x2 = 0)


Example: Alternate Solutions Example: Alternate Solutions

>Solve graphically for the optimal solution:Max z = 6x1 + 3x2

s.t. 4x1 + 3x2 > 12

2x1 + x2 < 8

x1, x2 > 0


xx22

xx11

44xx11 + 3 + 3xx22 >> 12 12

22xx11 + + xx22 << 8 8

33 44

44

88

Max 6Max 6xx11 + 3 + 3xx22

Example: Alternate SolutionsExample: Alternate Solutions

>There are infinite points satisfying both constraints - objective function falls on a constraint line


Example: Infeasible ProblemExample: Infeasible Problem


s.t. 4x1 + 3x2 < 12

2x1 + x2 > 8

x1, x2 > 0


xx22

xx11

44xx11 + 3 + 3xx22 << 12 12

22xx11 + + xx22 >> 8 8

33 44

44

88

Example: Infeasible ProblemExample: Infeasible Problem

>No points satisfy both constraints• no feasible region, no optimal solution


Example: Unbounded ProblemExample: Unbounded Problem


s.t. x1 + x2 > 5

3x1 + x2 > 8

x1, x2 > 0


x2

x1

33xx11 + + xx22 >> 8 8

xx11 + + xx22 >> 5 5

Max 3Max 3xx11 + 4 + 4xx22

5

5

88

2.67

Example: Unbounded ProblemExample: Unbounded Problem

>objective function can be moved outward without limit; z can be increased infinitely


RECAPRECAP







FormulationFormulation

>Define decision variables: x1, x2, …

>Objective Function (max, min)>s.t., with constraints listed

• Variables on left side• Constants on right side• All variables nonnegative

>NB: “Standard Form” requires constraints stated as equalities• add slack/surplus variables


Possible LP OutcomesPossible LP Outcomes

>Unique optimal solution>Alternate optimal solutions>Unbounded problem>Infeasible problem


BreakBreak

15 Minutes


LP Models: Key QuestionsLP Models: Key Questions

>What am I trying to decide?>What is the objective?

• Is it to be minimized or maximized?

>What are the constraints?• Are they limitations or requirements?• Are they explicit or implicit?


ExampleExample

A chemical company makes and sells a product in 40-lb. and 80-lb. bags on a common production line. To meet anticipated orders, next week’s production should be at least 16,000 lbs. Profit contributions are $2 per 40-lb. bag, and $4 per 80-lb. bag. The packaging line operates 1500 minutes/week. 40-lb. bags require 1.2 min. of packaging time; 80-lb. bags require 3 min. The company has 6000 square feet of packaging material available. Each 40-lb. bag uses 6 square feet, and each 80-lb. bag uses 10 square feet. How many bags of each type should be produced?


Model DevelopmentModel Development

>What do we need to decide?What are our decision variables?



>What do we need to decide?x1 = number of 40-lb. bags to produce

x2 = number of 80-lb. bags to produce



>What is the objective?



>What is the objective?Maximize total profit



>What is the objective?Maximize total profitz = 2x1 + 4x2


Check Your Units!Check Your Units!

>Always be sure that your units are consistent with the problem

>Our decision variable is measured in “Bags”>Our profit/objective function is in $

$bagsbag$

tObj.Fn.UniitDec.Var.Unf.Obj.Fn.Coe



>What are the constraints?



>What are the constraints?Aggregate production:Packaging time:Packaging materials:Nonnegativity:



>What are the constraints?Prod: 40x1 + 80x2 > 16,000

Time:Mat:NN:




Time: 1.2x1 + 3x2 < 1,500

Mat:NN:




Time: 1.2x1 + 3x2 < 1,500

Mat:6x1 + 10x2 < 6,000

NN:




Time: 1.2x1 + 3x2 < 1,500

Mat:6x1 + 10x2 < 6,000

NN: x1, x2 > 0


Check The Units!Check The Units!

>What are the constraints?Prod: 40x1 + 80x2 > 16,000lbs/bag x bags

Time: 1.2x1 + 3x2 < 1,500 min/bag x bags

Mat: 6x1 + 10x2 < 6,000 ft2/bag x bags

NN: x1, x2 > 0


Complete ModelComplete Model

x1 = no. of 40-lb. bags to produce

x2 = no. of 80-lb. bags to produce

Maximize z = 2x1 + 4x2

subject to 40x1 + 80x2 > 16,000

1.2x1 + 3x2 < 1,500

6x1 + 10x2 < 6,000

x1, x2 > 0


ExcelExcel

>Model Input• Basic model• Solver: identify objective function &

constraints

>Results• Answer Report• Sensitivity Report• Limits Report


Spreadsheet ModelSpreadsheet Model

A B C D E F G1 40-lb 80-lb23 DecVar 0 04 Constraint Constraint5 ObFnCoef 2 4 0 Amount Slack67 MinProd'n 40 80 0 >= 16000 160008 MachTime 1.2 3 0 <= 1500 15009 PackMat 6 10 0 <= 6000 6000


Cell FormulasCell Formulas

A B C D E F G1 40-lb 80-lb23 DecVar 0 04 Constraint Constraint5 ObFnCoef 2 4 =SUMPRODUCT(B5:C5,$B$3:$C$3) Amount Slack67 MinProd'n 40 80 =SUMPRODUCT(B7:C7,$B$3:$C$3) >= 16000 =ROUND(F7-D7,2)8 MachTime 1.2 3 =SUMPRODUCT(B8:C8,$B$3:$C$3) <= 1500 =ROUND(F8-D8,2)9 PackMat 6 10 =SUMPRODUCT(B9:C9,$B$3:$C$3) <= 6000 =ROUND(F9-D9,2)


Using SolverUsing Solver


Adding ConstraintsAdding Constraints


Note Assumptions ticked - essential!

Solver OptionsSolver Options


Answer ReportAnswer Report

Microsoft Excel 8.0e Answer Report

Target Cell (Max)Cell Name Original Value Final Value

$D$5 ObFnCoef 0 2200

Adjustable CellsCell Name Original Value Final Value

$B$3 DecVar 40-lb 0 500$C$3 DecVar 80-lb 0 300

ConstraintsCell Name Cell Value Formula Status Slack

$D$8 MachTime 1500 $D$8<=$F$8 Binding 0$D$9 PackMat 6000 $D$9<=$F$9 Binding 0$D$7 MinProd'n 44000 $D$7>=$F$7 Not Binding 28000


Sensitivity ReportSensitivity Report

Microsoft Excel 8.0e Sensitivity Report

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$3 DecVar 40-lb 500 0 2 0.4 0.4$C$3 DecVar 80-lb 300 0 4 1 0.666666667

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$D$8 MachTime 1500 0.666666667 1500 300 300$D$9 PackMat 6000 0.2 6000 1500 1000$D$7 MinProd'n 44000 0 16000 28000 1E+30


Optimal SolutionOptimal Solution

>Three parts: • decision variables• values of decision variables• value of objective function

>Decision variables:• basic (non-zero value),• non-basic (zero)• Basic variables are “in the solution”; non-

basic are not


Impact of Possible ChangesImpact of Possible Changes

>Change existing constraint• changes slope; may change size of

feasible region

>Add new constraint• may decrease feasible region (if binding)

>Remove constraint• may increase feasible region (if binding)

>Change objective• may change optimal solution


Sensitivity AnalysisSensitivity Analysis

>The next section will deal with the sensitivity analysis that can be done simply based on the reports generated, without rerunning the solution.

>While this can be useful, mastery of this material is not important for the course, or programme as you can always simply run the model again with the changes

>However, we will look at this briefly


Sensitivity AnalysisSensitivity Analysis

>Used to determine how optimal solution is affected by changes, within specified ranges: objective function or RHS coefficients (only 1 at a time)

>Important to managers who must operate in a dynamic environment with imprecise estimates of coefficients

>Sensitivity analysis allows us to ask certain what-if questions


Objective Function CoefficientsObjective Function Coefficients

>If an objective function coefficient changes, slope of objective function line changes. At some threshold, another corner point may become optimal.

>Question: How much can objective coefficient change without changing optimal corner point?


x2

x1

current optimal solution

new optimal solution

Geometric IllustrationGeometric Illustration









Reduced Costs Reduced Costs (Objective Function Coefficients)(Objective Function Coefficients)

>Reduced cost for decision variable not in solution (current value is 0) is amount variable's objective function coefficient would have to improve (increase for max, decrease for min) before variable could enter solution

>Reduced cost for decision variable in solution is 0









Objective Function RangesObjective Function Ranges

>Interval within which original solution remains optimal (same decision variables in solution) while keeping all other data constant

>Within this range, associated reduced cost is valid

>Value of the objective function might change in this range of optimality









RHS Coefficient ChangesRHS Coefficient Changes

>When a right-hand-side value changes, the constraint moves parallel to itself

>Question: How is the solution affected, if at all?

>Two cases:• constraint is binding• constraint is nonbinding


x2

x1

optimal solution

Binding constraints

Binding constraints have zero slackNonbinding constraints have positive slack

Nonbinding constraint

Geometric IllustrationGeometric Illustration


Binding ConstraintsBinding Constraints

Microsoft Excel 8.0e Answer Report

Target Cell (Max)Cell Name Original Value Final Value

$D$5 ObFnCoef 0 2200

Adjustable CellsCell Name Original Value Final Value

$B$3 DecVar 40-lb 0 500$C$3 DecVar 80-lb 0 300

ConstraintsCell Name Cell Value Formula Status Slack

$D$8 MachTime 1500 $D$8<=$F$8 Binding 0$D$9 PackMat 6000 $D$9<=$F$9 Binding 0$D$7 MinProd'n 44000 $D$7>=$F$7 Not Binding 28000


Tightening & Relaxing ConstraintsTightening & Relaxing Constraints

>Tightening a constraint means to make it more restrictive; i.e. decreasing the RHS of a less than constraint, or increasing the RHS of a greater constraint. This compresses the feasible region.

>Relaxing a constraint means to make it less restrictive; i.e., expand the feasible region.


x2

x1

Original optimal solution

New optimal solution (z decreases)

Effect of Tightening a ConstraintEffect of Tightening a Constraint


x2

x1

optimal solution

Effect of Relaxing a ConstraintEffect of Relaxing a Constraint









Dual Prices (RHS Coefficients)Dual Prices (RHS Coefficients)

>Amount objective function will improve per unit increase in constraint RHS value

>Reflects value of an additional unit of resource (if resource cost is sunk); reflects extra value over normal cost of resource (when resource cost is relevant)

>Always 0 for nonbinding constraint (positive slack or surplus at optimal solution)
















RHS RangesRHS Ranges

>As long as the constraint RHS coefficient stays within this range, the associated dual price is valid

>For changes outside this range, must resolve


Shadow vs Dual PricesShadow vs Dual Prices

>Shadow Price: Amount objective function will change per unit increase in RHS value of constraint

>For maximization problems, dual prices and shadow prices are the same

>For minimization problems, shadow prices are the negative of dual prices


Next ClassNext Class

>We will look at the two handout exercises• To be posted on the website, with

solution files

>Decision Theory>In Lecture 3, we will do additional

problems in both Linear Programming and Decision Theory

© Copyright 2004, Alan Marshall 1 Lecture 1 Linear Programming.

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t large

t small

c constraints

tables c

large blocks

small blocks

solution slide