Simplex Linear Programming I. Concept II. Model Template III. Class Example IV. Procedure V. Interpretation MAXIMIZATION METHOD Applied Management Science.

Slides:

Advertisements

Similar presentations

SIMPLEX METHOD.

Advertisements

Chapter 5: Linear Programming: The Simplex Method

Lecture 3 Linear Programming: Tutorial Simplex Method

Operation Research Chapter 3 Simplex Method.

SIMPLEX METHOD FOR LP LP Model.

Assignment (6) Simplex Method for solving LP problems with two variables.

Chapter 6 Linear Programming: The Simplex Method

The Simplex Method The geometric method of solving linear programming problems presented before. The graphical method is useful only for problems involving.

Dr. Sana’a Wafa Al-Sayegh

Computational Methods for Management and Economics Carla Gomes Module 6a Introduction to Simplex (Textbook – Hillier and Lieberman)

Optimization of thermal processes2007/2008 Optimization of thermal processes Maciej Marek Czestochowa University of Technology Institute of Thermal Machinery.

Linear Inequalities and Linear Programming Chapter 5

5.4 Simplex method: maximization with problem constraints of the form

Operation Research Chapter 3 Simplex Method.

Introduction to Management Science

Linear Programming: Computer Solution and Sensitivity Analysis

3-1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Linear Programming: Computer Solution and Sensitivity Analysis Chapter 3.

Stevenson and Ozgur First Edition Introduction to Management Science with Spreadsheets McGraw-Hill/Irwin Copyright © 2007 by The McGraw-Hill Companies,

LINEAR PROGRAMMING SIMPLEX METHOD.

Linear Programming - Standard Form

Learning Objectives for Section 6.2

Chapter 3 Linear Programming Methods 高等作業研究高等作業研究 ( 一 ) Chapter 3 Linear Programming Methods (II)

Chapter Topics Computer Solution Sensitivity Analysis

The Two-Phase Simplex Method LI Xiao-lei. Preview When a basic feasible solution is not readily available, the two-phase simplex method may be used as.

8. Linear Programming (Simplex Method) Objectives: 1.Simplex Method- Standard Maximum problem 2. (i) Greedy Rule (ii) Ratio Test (iii) Pivot Operation.

Simplex method (algebraic interpretation)

1 1 Slide Linear Programming (LP) Problem n A mathematical programming problem is one that seeks to maximize an objective function subject to constraints.

Chapter 6 Linear Programming: The Simplex Method Section 2 The Simplex Method: Maximization with Problem Constraints of the Form ≤

Topic III The Simplex Method Setting up the Method Tabular Form Chapter(s): 4.

1 1 Slide © 2000 South-Western College Publishing/ITP Slides Prepared by JOHN LOUCKS.

Kerimcan OzcanMNGT 379 Operations Research1 Linear Programming: The Simplex Method Chapter 5.

1 1 © 2003 Thomson  /South-Western Slide Slides Prepared by JOHN S. LOUCKS St. Edward’s University.

1 1 © 2003 Thomson  /South-Western Slide Slides Prepared by JOHN S. LOUCKS St. Edward’s University.

Public Policy Modeling Simplex Method Tuesday, October 13, 2015 Hun Myoung Park, Ph.D. Public Management & Policy Analysis Program Graduate School of International.

Business Mathematics MTH-367 Lecture 15. Chapter 11 The Simplex and Computer Solutions Methods continued.

3-1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Linear Programming: Computer Solution and Sensitivity Analysis Chapter 3.

Chapter 2 Introduction to Linear Programming n Linear Programming Problem n Problem Formulation n A Maximization Problem n Graphical Solution Procedure.

Chapter 6 Simplex-Based Sensitivity Analysis and Duality

4  The Simplex Method: Standard Maximization Problems  The Simplex Method: Standard Minimization Problems  The Simplex Method: Nonstandard Problems.

Mechanical Engineering Department 1 سورة النحل (78)

 Minimization Problem  First Approach  Introduce the basis variable  To solve minimization problem we simple reverse the rule that is we select the.

1 1 Slide © 2005 Thomson/South-Western Linear Programming: The Simplex Method n An Overview of the Simplex Method n Standard Form n Tableau Form n Setting.

Chapter 4 Linear Programming: The Simplex Method

1 1 Slide © 2005 Thomson/South-Western Simplex-Based Sensitivity Analysis and Duality n Sensitivity Analysis with the Simplex Tableau n Duality.

1 1 Slide © 2000 South-Western College Publishing/ITP Slides Prepared by JOHN LOUCKS.

1 Simplex Method (created by George Dantzig in late 1940s) A systematic way of searching for an optimal LP solution BMGT 434, Spring 2002 Instructor: Chien-Yu.

Simplex Method for solving LP problems with two variables.

Simplex method : Tableau Form

Simplex Method Simplex: a linear-programming algorithm that can solve problems having more than two decision variables. The simplex technique involves.

 LP graphical solution is always associated with a corner point of the solution space.  The transition from the geometric corner point solution to the.

Business Mathematics MTH-367 Lecture 14. Last Lecture Summary: Finished Sec and Sec.10.3 Alternative Optimal Solutions No Feasible Solution and.

Department of Business Administration FALL -SPRING Management Science Operations Research by Ass. Prof. Sami Fethi Assoc. Prof. Sami Fethi © 2007 Pearson.

Decision Support Systems INF421 & IS Simplex: a linear-programming algorithm that can solve problems having more than two decision variables.

(iii) Simplex method - I D Nagesh Kumar, IISc Water Resources Planning and Management: M3L3 Linear Programming and Applications.

1 Introduction to Linear Programming Linear Programming Problem Linear Programming Problem Problem Formulation Problem Formulation A Simple Maximization.

GOOD MORNING CLASS! In Operation Research Class, WE MEET AGAIN WITH A TOPIC OF :

The Simplex Method The geometric method of solving linear programming problems presented before. The graphical method is useful only for problems involving.

Linear programming Simplex method.

Chapter 5 Simplex-Based Sensitivity Analysis and Duality

Chapter 4 Linear Programming: The Simplex Method

ENGM 631 Optimization Ch. 4: Solving Linear Programs: The Simplex Method.

ISS500E-Quantitiative Decision Tehniques

Chapter 3 The Simplex Method and Sensitivity Analysis

St. Edward’s University

Linear Programming SIMPLEX METHOD.

Linear Programming.

The Simplex Method The geometric method of solving linear programming problems presented before. The graphical method is useful only for problems involving.

Linear programming Simplex method.

Presentation transcript:

Simplex Linear Programming I. Concept II. Model Template III. Class Example IV. Procedure V. Interpretation MAXIMIZATION METHOD Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro, PhD Resourc e Planning a nd Allocation Management MGMT E-5050

Simplex Linear Programming * * DECISION VARIABLES MIGHT BE PRODUCTS BEING CONSIDERED FOR PRODUCTION Most real-life LP problems have more than two decision variables*, and thus are too large for the simple graphical solution procedure. In simplex LP problems the optimal solution will lie at a corner point of a multi-sided, multi- dimensional figure called an n - dimensional polyhedron that represents the feasible region. THE FEASIBLE REGION WOULD BE SIMILAR TO THE SURFACE OF A DIAMOND

Simplex Linear Programming  The simplex method examines the corner points in a systematic fashion, using basic algebraic concepts.  The same set of procedures is repeated time after time until an optimal solution is reached.  Each repetition, or iteration, increases the value of the objective function so that we are always moving closer to the optimal solution.

Simplex Linear Programming   The simplex method yields valuable economic information in addition to the optimal solution.  The manual computations must be mastered in order to successfully employ the software programs, and to interpret the computer printouts. WHY LEARN THE MANUAL COMPUTATIONS?

Problem Statement Black and White Color A firm produces two different types of television sets: Three resources are required to produce those televisions: Chassis24 units Labor160 hours Color Tubes10 units

Problem Statement TELEVISION RESOURCE REQUIREMENTS AND PROFIT MARGINS TelevisionChassis Labor HoursColor Tubes Unit Profit Black + White 150$6.00 Color 1101$15.00

Simplex Linear Programming Let X 1 = BLACK AND WHITE TELEVISIONS Let X 2 = COLOR TELEVISIONS Objective Function: Maximize Z = 6X X 2 subject to: 1X 1 + 1X 2 =< 24 chassis 5X X 2 =< 160 labor hours 0X 1 + 1X 2 =< 10 color tubes X 1, X 2 => 0 THE MODEL

Simplex Linear Programming CONVERSION TO LINEAR EQUALITIES S 1 1X 1 + 1X 2 + 1S 1 = 24 chassis S 2 5X X 2 + 1S 2 = 160 labor hours S 3 0X 1 + 1X 2 + 1S 3 = 10 color tubes : S1, S2 =, S3 = COLOR TUBES SLACK VARIABLES: S1 = CHASSIS, S2 = LABOR HOURS, S3 = COLOR TUBES ADDING A SLACK VARIABLE TO EACH CONSTRAINT AND SETTING IT EQUAL TO THE RIGHT – HAND SIDE

Simplex Linear Programming CONVERSION TO LINEAR EQUALITIES SUITABLE FOR INCLUSION IN THE SIMPLEX MATRIX 00 1X 1 + 1X 2 + 1S 1 + 0S 2 + 0S 3 = 24 chassis 00 5X X 2 + 0S 1 + 1S 2 + 0S 3 = 160 labor hours 000 0X 1 + 1X 2 + 0S 1 + 0S 2 + 1S 3 = 10 color tubes PRESENCE REWRITE EACH CONSTRAINT TO REFLECT THE PRESENCE OR ABSENCE ABSENCE OF ALL VARIABLES IN THE PROBLEM

Simplex Linear Programming 1 st FEASIBLE SOLUTION TRADITIONALLY THE FIRST FEASIBLE SOLUTION PRODUCES NO PRODUCT AND HAS ALL RESOURCES INTACT X1X1X1X1 BLACK AND WHITE TVs 0 X2X2X2X2 COLOR TVs 0 S1S1S1S1 CHASSIS 24 S2S2S2S2 LABOR HOURS 160 S3S3S3S3 COLOR TUBES 10

The 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 Zj Cj-Zj THE BASIS OR MIX COLUMN SHOWS ALL VARIABLES GREATER THAN ZERO IN THE CURRENT SOLUTION

Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S S2S2S2S S3S3S3S Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX RIGHT-HAND SIDES ARE ENTERED INTO THE QUANTITYCOLUMN

Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 1X 1 + 1X 2 + 1S 1 + 0S 2 + 0S 3 = 24

Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 5X X 2 + 0S 1 + 1S 2 + 0S 3 = 160

Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 0X 1 + 1X 2 + 0S 1 + 0S 2 + 1S 3 = 10

Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S S2S2S2S S3S3S3S Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX

The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj Cj-Zj THE “Cj” or CONTRIBUTION MARGIN THE GROSS PROFIT PER UNIT FOR ALL VARIABLES IN THE PROBLEM. SURPLUS AND SLACK VARIABLES HAVE $0.00 Cj’s BY DEFINITION.

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj Cj-Zj

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS QUANTITY AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0. $0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj THE COMPUTED Zj ROW

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0. COMPUTING THE Cj – Zj ROW

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0. COMPUTING THE Cj – Zj ROW

Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. COMPUTING THE Cj – Zj ROW

The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. THE COMPLETED Cj - Zj ROW

The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0.

Simplex Linear Programming 1 st FEASIBLE SOLUTION TRADITIONALLY THE FIRST FEASIBLE SOLUTION PRODUCES NO PRODUCT AND HAS ALL RESOURCES INTACT X1X1X1X1 BLACK AND WHITE TVs0 X2X2X2X2 COLOR TVs 0 S1S1S1S1 CHASSIS 24 S2S2S2S2 LABOR HOURS 160 S3S3S3S3 COLOR TUBES 10

Simplex Linear Programming X 1 ( B+W TVs ) = 0 NOT IN THE BASIS X 2 ( Color TVs ) = 0 NOT IN THE BASIS Z ( Profit ) = $0.00 SINCE X 1 AND X 2 = 0 1 st FEASIBLE SOLUTION

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. WE BRING THE MOST PROFITABLE TELEVISION ( X2 ) INTO THE SOLUTION X2 BECOMES THE PIVOT COLUMN

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. WE COMPUTE EACH ROW’S RATIO OF QUANTITY DIVIDED BY ITS PIVOT COLUMN COEFFICIENT PIVOT COLUMN

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $0. S3S3S3S Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. THE LOWEST POSITIVE RATIO DENOTES THE PIVOT ROW PIVOT COLUMN PIVOTROW MEANS WE CAN ONLY PRODUCE TEN COLOR TVs THIS IS THE LIMITED RESOURCE

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. REAL VARIABLE X2 REPLACES SLACK VARIABLE S3 IN THE BASIS PIVOT COLUMN PIVOTROW TO PRODUCE TEN COLOR TVs MEANS THAT ALL COLOR TUBES MUST BE CONSUMED, MAKING S3 = 0, AND FORCING IT TO LEAVE THE BASIS

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. THE PIVOT NUMBER IS LOCATED AT THE INTERSECTION OF THE PIVOT ROW AND PIVOT COLUMN PIVOT COLUMN PIVOTROW

Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$0.$0.$0.$0.$0.$0. Cj-Zj $6.$15.$0.$0.$0. PIVOT COLUMN PIVOTROW THE PIVOT NUMBER MUST ALWAYS BE “1”

When the Pivot Number is Not “1” Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2S1S2S3 $0.1 $0.10 $15. X2X2X2X Zj$0.$0.$0.$0.$0.$0. Cj-Zj $15. PIVOT COLUMN PIVOTROW EXAMPLE THE ENTIRE PIVOT ROW MUST BE DIVIDED BY WHATEVER NUMBER NEEDED T0 FORCE A VALUE OF “1” FOR THE PIVOT NUMBER

When the Pivot Number is Not “1” Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0.1 $0.10 $15. X2X2X2X Zj$0.$0.$0.$0.$0.$0. Cj-Zj $15. PIVOT COLUMN PIVOTROW EXAMPLE

Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 1 st Row S1S Pivot Row S3S New 1 st Row S1S We subtract the pivot row from all other rows in the matrix ( except “Zj” and “Cj - Zj” ) in such a way as to force a “0” coefficient in the column above or below the pivot number. SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE FIRST ROW ( S 1 ) Pivot Number

Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 2 nd Row S2S Pivot Row S3S New 2 nd Row S2S2 0 SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE SECOND ROW ( S 2 ) HERE, THE PIVOT ROW MUST BE MULTIPLIED BY “10” IN ORDER TO FORCE THE REQUIRED ZERO COEFFICIENT IN THE NEW ROW Pivot Number

Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 2 nd Row S2S Pivot Row S3S New 2 nd Row S2S SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE SECOND ROW ( S 2 ) HERE, THE PIVOT ROW WAS MULTIPLIED BY “10” IN ORDER TO FORCE THE REQUIRED ZERO COEFFICIENT IN THE NEW ROW Pivot Number

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj Cj-Zj INSERTING THE TRANSFORMED ROWS INTO THE MATRIX

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150. Cj-Zj COMPUTING THE Zj ROW $0. $0. $150. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS QUANTITY AND ADD THE RESULTS

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0.S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS $0.

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj THE COMPLETED Zj ROW

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6. COMPUTING THE Cj - Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0. COMPUTING THE Cj - Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0. COMPUTING THE Cj - Zj ROW

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0. COMPUTING THE Cj - Zj ROW

Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. COMPUTING THE Cj - Zj ROW

The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE COMPLETED Cj – Zj ROW

The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15.

The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15.

Simplex Linear Programming 2 nd FEASIBLE SOLUTION X1X1X1X1 BLACK AND WHITE TVs 0 X2X2X2X2 COLOR TVs 10 S1S1S1S1 CHASSIS 14 S2S2S2S2 LABOR HOURS 60 S3S3S3S3 COLOR TUBES 0 TOTAL PROFIT = $150.00

Simplex Linear Programming 2 nd FEASIBLE SOLUTION Z ( total profit ) = $ X 1 ( B + W TVs ) = 0 NOT IN THE BASIS S 3 ( Color Tubes ) = 0 NOT IN THE BASIS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. WE BRING BLACK + WHITE TELEVISIONS (X1) INTO THE SOLUTION X1 BECOMES THE PIVOT COLUMN

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. WE COMPUTE EACH ROW’S RATIO OF QUANTITY DIVIDED BY ITS PIVOT COLUMN COEFFICIENT THE PIVOT COLUMN

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $0. S2S2S2S $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE LOWEST POSITIVE RATIO DENOTES THE PIVOT ROW THE PIVOT COLUMN PIVOTROW MEANS WE CAN ONLY PRODUCE 12 B+W TVs THIS IS THE LIMITED RESOURCE

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. REAL VARIABLE X1 REPLACES SLACK VARIABLE S2 IN THE BASIS THE PIVOT COLUMN PIVOTROW TO PRODUCE 12 B+W TVs MEANS THAT ALL REMAINING LABOR HOURS MUST BE CONSUMED, MAKING S2 = 0, AND FORCING IT TO LEAVE THE BASIS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT NUMBER IS LOCATED AT THE INTERSECTION OF THE PIVOT ROW AND PIVOT COLUMN THE PIVOT COLUMN PIVOTROW

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT COLUMN PIVOTROW THE PIVOT NUMBER MUST ALWAYS BE “1”

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT COLUMN PIVOTROW DIVIDE THE ENTIRE PIVOT ROW BY “5”

Toward 3 rd Feasible Solution ROW TRANSFORMATION Old 1 st Row S1S Pivot Row X1X New 1 st Row S1S SUBTRACT THE PIVOT ROW ( X 1 ) FROM THE FIRST ROW ( S 1 ) HERE, THERE WAS NO NEED TO MULTIPLY THE PIVOT ROW BY ANY NUMBER IN ORDER TO FORCE A ZERO COEFFICIENT BELOW THE PIVOT NUMBER Pivot Number

Toward 3 rd Feasible Solution ROW TRANSFORMATION Old 3 rd Row X2X Pivot Row X1X New 3 rd Row X2X SUBTRACT THE PIVOT ROW ( X 1 ) FROM THE THIRD ROW ( X 2 ) HERE, THERE WAS NO NEED TO TRANSFORM ROW “3” BECAUSE IT ALREADY HAD A ZERO COEFFICIENT BELOW THE PIVOT NUMBER Pivot Number

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj Cj-Zj INSERTING THE TRANSFORMED ROWS INTO THE MATRIX

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222. Cj-Zj COMPUTING THE Zj ROW $0. $72. $150. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS QUANTITY AND ADD THE RESULTS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6. Cj-Zj COMPUTING THE Zj ROW $0. $6. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0.S $6.X $15. X2X2X2X Zj$222.$6.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2 Cj-Zj COMPUTING THE Zj ROW $0. $1.2 $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj COMPUTING THE Zj ROW $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS -$12.

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj THE COMPLETED Zj ROW

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0.$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0.$0.$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2$3.0 Cj-Zj$0.$0.$0.-$1.2 COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 THE COMPLETED Cj – Zj ROW

The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0

The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0

The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 THERE ARE NO POSITIVE NUMBERS IN THE Cj – Zj ROW THEOPTIMALSOLUTION

Simplex Linear Programming 3 rd AND OPTIMAL FEASIBLE SOLUTION X 1 = 12 X 2 = 10 S 1 = 2 S 2 = 0 S 3 = 0 Z = $ twelve Produce twelve black and white televisions ten Produce ten color televisions two There are two chassis left over no There are no labor hours left over no There are no color tubes left over $ Total profit realized is $ A RECORD SHOULD BE KEPT OF WHAT EACH VARIABLE REPRESENTS

The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S $6. X1X1X1X $15. X2X2X2X Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2 -$3.0 -$3.0 THE SLACK VARIABLE SHADOW PRICES ALWAYS FOUND IN THE Cj – Zj ROW

Simplex Linear Programming 3 rd AND OPTIMAL FEASIBLE SOLUTION SHADOW PRICES S 1 = $0.00 S 2 = - $1.20 S 3 = - $3.00 WILLING TO PAY WE WOULD BE WILLING TO PAY ZERO DOLLARS UP TO ZERO DOLLARS FOR ADDITIONAL CHASSIS WOULD BE WILLING TO PAY WE WOULD BE WILLING TO PAY $1.20 UP TO $1.20 FOR ADDITIONAL LABOR HOURS WE WOULD BE WILLING TO PAY $3.00 UP TO $3.00 FOR ADDITIONAL COLOR TUBES THIS PARTICULAR SIMPLEX METHOD DISPLAYS NEGATIVE VALUES FOR POSITIVE SHADOW PRICES. THIS IS CALLED THE “MIRROR EFFECT”.

Simplex Linear Programming under QM for WINDOWS

Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions

Template

Simplex Linear Programming Maximization