OPS 301 Module B and Additional Topics in Linear Programming Dr. Steven Harrod

Topics Definition of Linear Programming Steps to Formulate a Problem Applications Mathematical Requirements Linear functions Convex, non-negative feasible region Single objective Steps to Formulate a Problem Solution Techniques Graphical Simplex

Linear Programming A mathematical tool to make multiple decisions affecting a single outcome or objective Guaranteed to find the optimal solution (best possible set of decisions) relative to the stated objective and linear assumptions This slide provides some reasons that capacity is an issue. The following slides guide a discussion of capacity.

LP Applications Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm’s profit Determining the best diet for persons or animals based on any combination of nutrition, cost, taste, etc. Determining production, routes, and storage locations that will minimize total shipping cost This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

Requirements of an LP Problem Linear Acceptable: X1, 2X2, 9 Not acceptable: X2, X3, |X|, √X Convex A line drawn between any two points on the region will not travel outside the region Single Objective This slide can be used to frame a discussion of capacity. Points to be made might include: - capacity definition and measurement is necessary if we are to develop a production schedule - while a process may have “maximum” capacity, many factors prevent us from achieving that capacity on a continuous basis. Students should be asked to suggest factors which might prevent one from achieving maximum capacity.

Convex and Linear X2 Feasible region X1 – 80 – 60 – 40 – 20 – 80 – 60 – 40 – 20 – Feasible region | | | | | | | | | | | 0 20 40 60 80 100 X1

Linear, But Not Convex X2 Feasible region X1 – 80 – 60 – 40 – 20 – 80 – 60 – 40 – 20 – Feasible region | | | | | | | | | | | 0 20 40 60 80 100 X1

Convex, But Not Linear X2 Feasible region X1 – 80 – 60 – 40 – 20 – 80 – 60 – 40 – 20 – Feasible region | | | | | | | | | | | 0 20 40 60 80 100 X1

Formulating an LP What is the goal? What do you control? “Objective Function” Maximize – make as large as possible Minimize – make as small as possible What do you control? Called “Decision Variables” Must be non-negative What are the limits? Resources not to exceed (<=) Minimum requirements to meet (>=) Items that must be used/served exactly (=) Constraints are labeled “s.t.” for “subject to”

Example Young MBA Erica Cudahy may invest up to $1,000. She can invest her money in stocks and loans. Each dollar invested in stocks yields 10¢ profit, and each dollar invested in a loan yields 15¢ profit. At least 30% of all money invested must be in stocks, and at least $400 must be in loans. source: Winston, Operations Research, 2004

Example What is the goal? Maximize profit. What do we control? Dollar amount invested each in stocks and loans. What are the limits? At least $400 invested in loans 30% invested in stocks $1000 maximum available to invest

Mathematical Formulation Label our decision variables X1 dollar amount invested in stocks X2 dollar amount invested in loans Write objective function the value resulting from our decisions 0.10 X1 + 0.15 X2

Mathematical Formulation Write Limits $1000 available to invest X1 + X2 <= 1000 at least 30% of all money in stocks 0.3 (X1 + X2) <= X1 Arrange all variables on left side – 0.7 X1 + 0.3 X2 <=0 at least $400 in loans X2 >=400

Complete Formulation max 0.10 X1 + 0.15 X2 s.t. X1 + X2 <= 1000

Graphical Solution Can be used when there are two decision variables Plot the constraint equations at their limits by converting each equation to an equality Identify the feasible solution space Create an iso-profit line based on the objective function Move this line outwards until the optimal point is identified

Graphical Solution

Graphical Example max 7 X1 + 5 X2 s.t. 4X1 + 3X2 ≤ 240 2X1 + 1X2 ≤ 100

Graphical Solution 2X1 + 1X2 ≤ 100 4X1 + 3X2 ≤ 240 X2 Feasible region – 80 – 60 – 40 – 20 – 2X1 + 1X2 ≤ 100 4X1 + 3X2 ≤ 240 Feasible region | | | | | | | | | | | 0 20 40 60 80 100 X1 Figure B.3

Graphical Solution Iso-Profit Line Solution Method $210 = 7X1 + 5X2 – 80 – 60 – 40 – 20 – | | | | | | | | | | | 0 20 40 60 80 100 Number of Watch TVs Number of Walkmans X1 X2 Assembly (constraint B) Electronics (constraint A) Feasible region Figure B.3 Choose a possible value for the objective function $210 = 7X1 + 5X2 Solve for the axis intercepts of the function and plot the line X2 = 42 X1 = 30

Graphical Solution $210 = $7X1 + $5X2 X2 (0, 42) (30, 0) X1 – 80 – 80 – 60 – 40 – 20 – $210 = $7X1 + $5X2 (0, 42) (30, 0) | | | | | | | | | | | 0 20 40 60 80 100 X1 Figure B.4

Graphical Solution $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 – 80 – 60 – 40 – 20 – $350 = $7X1 + $5X2 $280 = $7X1 + $5X2 $210 = $7X1 + $5X2 $420 = $7X1 + $5X2 | | | | | | | | | | | 0 20 40 60 80 100 X1 Figure B.5

Optimal solution point Graphical Solution X2 – 80 – 60 – 40 – 20 – Maximum profit line Optimal solution point (X1 = 30, X2 = 40) $410 = $7X1 + $5X2 | | | | | | | | | | | 0 20 40 60 80 100 X1 Figure B.6

Corner-Point Method 2 3 1 4 X2 X1 – 80 – 60 – 40 – 20 – 80 – 60 – 40 – 20 – 2 3 | | | | | | | | | | | 0 20 40 60 80 100 1 X1 Figure B.7 4

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Solve for the intersection of two constraints 2X1 + 1X2 ≤ 100 (assembly time) 4X1 + 3X2 ≤ 240 (electronics time) 4X1 + 3X2 = 240 - 4X1 - 2X2 = -200 + 1X2 = 40 4X1 + 3(40) = 240 4X1 + 120 = 240 X1 = 30 Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Corner-Point Method The optimal value will always be at a corner point Find the objective function value at each corner point and choose the one with the highest profit Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0 Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400 Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350 Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410

Beyond 2 Variables, Matrix Algebra To solve problems of more than two variables with more complex feasible regions (decimal and fractional values) requires computers and matrix algebra Matrix algebra allows easy calculation of corner points Simplex method finds optimal corner point

Matrix Algebra max 7 X1 + 5 X2 s.t. 4X1 + 3X2 ≤ 240 2X1 + 1X2 ≤ 100 Define 4 matrices:

Matrix Formulation max cx s.t. Ax <=b x>=0

Linear System Solution IF Ax=b, and A is “Square” (number of rows equals number of columns) DEFINE the inverse of A: A-1 THEN x=A-1b Use spreadsheet to calculate inverse How to convert Ax<=b to Ax=b?

Slack Variables “S” max 7 X1 + 5 X2 + 0 S1 + 0 S2 s.t. {one of X1, X2, S1, S2} = 0 X1, X2, S1, S2 >=0

Instructions for Slack Add one new slack variable to each constraint Slack variables added to objective with zero coefficient Square matrix by choosing some variables equal to zero

Slack Variables “S” New matrices:

Solution Method Type all of these matrices into Excel In each row of ? Enter a single 1 Fill the rest of the row with zeros Do not put 1 in same column on multiple rows Do not leave cells blank, zeros are required Label matrix “x” your corner point

Solution Method Select the cells of the corner point Enter formula =mmult(minverse(A matrix cells), b matrix cells) Keep all cells selected Hold keys ctrl & shift, then press Enter. If successful, number values should display in corner point Pick a new cell Label “Objective Value” Enter formula =mmult(c matrix cells, corner point cells) Again hold keys ctrl & shift, then press enter

Finding Corner Points Select different combinations of variables to set to zero by moving 1 value within row Round corner points to 3 decimals (#.###) Ignore corner points found with negative values Objective value calculated is value given displayed choices or decisions for variables Given all combinations, all possible corner points, the optimal solution will be one of these

Simplex Method Realistic problems quickly grow to have more corner points than humanly possible to examine Simplex method starts from a simple point and jumps from point to point until reaching optimal, then stops. Does not usually need to cover a large number of points Solution can be proven optimal without listing all corner points Developed by George Dantzig in the late 1940s Most computer-based LP packages use the simplex method

Simplex Spreadsheet Simplex spreadsheet provided for you Follow instructions on spreadsheet Simplex method uses additional matrix algebra to calculate a directional indicator to next best corner point Simplex also uses matrix algebra to identify stopping point

Conclusion Understand Solution Techniques Use of LP Mathematical Requirements Steps to Formulate a Problem Solution Techniques Graphical Simplex Matrix Algebra Foundation of Advanced Work