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Unformatted text preview: Table of Contents Chapter 2 (Linear Programming: Basic Concepts)
The Wyndor Glass Company Product Mix Problem (Section 2.1) Formulating the Wyndor Problem on a Spreadsheet (Section 2.2) The Algebraic Model for Wyndor (Section 2.3) The Graphical Method Applied to the Wyndor Problem (Section 2.4) Using the Excel Solver with the Wyndor Problem (Section 2.5) A Minimization Example—The Profit & Gambit Co. (Section 2.6) Introduction to Linear Programming (UW Lecture) 2.2 2.3–2.7 2.8 2.9–2.19 2.20–2.25 2.26–2.31 2.32–2.47 The Graphical Method and Properties of LP Solutions (UW Lecture) These slides are based upon a lecture introducing the basic concepts of linear programming and the Solver to firstyear MBA students at the University of Washington (as taught by one of the authors). The lecture is largely based upon a production problem using Lego building blocks. These slides are based upon a lecture introducing the graphical method and other concepts about linear programming solutions to firstyear MBA students at the University of Washington (as taught by one of the authors). 2.48–2.56 1 Wyndor Glass Co. Product Mix Problem Wyndor has developed the following new products: The company has three plants An 8foot glass door with aluminum framing. A 4foot by 6foot doublehung, woodframed window. Plant 1 produces aluminum frames and hardware. Plant 2 produces wood frames. Plant 3 produces glass and assembles the windows and doors. Questions:
1. 2. Should they go ahead with launching these two new products? If so, what should be the product mix? 2 Wyndor Glass Co. Product Mix Problem Summary of Assumptions Any product mix can be sold at unit profits of $300/door and $500/window Product requirements and available times by plant Summarizing in tabular form: Plant 1 produces aluminum frames and hardware. Each door requires 1 hour at plant 1. The plant has 4 hours per week available. Plant 2 produces wood frames. Each window requires 2 hours at plant 2. The plant has 12 hours per week available. Plant 3 produces glass and assembles the windows and doors. Each door requires 3 hours and each window requires 2 hours at plant 3. The plant has 18 hours per week available. Hours per week Plant Plant 1 Plant 2 Plant 3 Unit Profit Door 1 0 3 $300 Window 0 2 2 $500
3 Available 4 12 18 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). Doors $300 Windows $500 Hours Available 4 12 18 Unit Profit Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors Windows 4 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).
Doors $300 Windows $500 Hours Available 4 12 18 Unit Profit Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 0 Windows 0 Units Produced 5 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). Unit Profit Doors $300 Windows $500 Hours Available 4 12 18 Total Profit $800
G Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 1 Windows 1 Units Produced 11 Total Profit 12 =SUMPRODUCT(UnitProfit,UnitsProduced) 6 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 three consecutive cells. For example, if Quantity A <= Quantity B, put these three items (Quantity A, <=, Quantity B) in consecutive cells.
Unit Profit Doors $300 Windows $500 Hours Used 1 2 5 Hours Available 4 12 18 Total Profit $800 Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 1 Windows 1 <= <= <= Units Produced 5 6 7 8 9 E Hours Used =SUMPRODUCT(C7:D7,UnitsProduced) =SUMPRODUCT(C8:D8,UnitsProduced) =SUMPRODUCT(C9:D9,UnitsProduced) 7 A Trial Solution
Unit Profit Doors $300 Windows $500 Hours Used 4 6 18 Hours Available 4 12 18 Total Profit $2,700 Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 4 Windows 3 <= <= <= Units Produced The spreadsheet for the Wyndor problem with a trial solution (4 doors and 3 windows) entered into the changing cells. 8 Algebraic Model for Wyndor Glass Co.
Let D = the number of doors to produce W = the number of windows to produce Maximize P = $300D + $500W subject to D≤4 2W ≤ 12 3D + 2W ≤ 18 and D ≥ 0, W ≥ 0. 9 Graphing the Product Mix
W 8
Product ion rat e (unit s pe r we ek) for windows 7 6 5 4 3 2 1 Origin A product mix of D = 4 and W = 6 (4, 6) A product mix of D = 2 and W = 3 (2, 3) 2 1 0 1 2 1 2 3 4 5 6 7 8 D Production rate (units per week) for doors 10 10 Graph Showing Constraints: D ≥ 0 and W ≥ 0 W
8 Producti on rate for wi ndows 6 4 2 0 2 4 6 8 D Production rate for doors 11 11 Nonnegative Solutions Permitted by D ≤ 4
W 8 D=4 Produc tion ra te for windows 6 4 2 0 2 6 4 Production rate for doors 8 D 12 12 Nonnegative Solutions Permitted by 2W ≤ 12
Production rate for windows W 8 2 W = 12 6 4 2 0 2 4 Production rate for doors 6 8 D 13 13 Boundary Line for Constraint 3D + 2W ≤ 18 Production rate for windows W
10 (0, 9)
8 1 _ (1, 7 ) 2 6 (2, 6) 3 D + 2 W = 18
1 _ (3, 4 ) 2 4 (4, 3)
2 1 _ (5, 1 ) 2 (6, 0)
0 2 4 6 8 D Production rate for doors 14 14 Changing RightHand Side Creates Parallel Constraint Boundary Lines Production rate for windows W
12 10 3D + 2W = 24
8 6 3D + 2W = 18
4 2 3D + 2W = 12 0 2 4 6 8 10 D Production rate for doors 15 15 Nonnegative Solutions Permitted by 3D + 2W ≤ 18 Production rate for windows W
10 8 6 3D + 2W = 18
4 2 0 2 4 6 8 D Production rate for doors 16 16 Graph of Feasible Region
Production rate for windows W 10 3 D + 2 W = 18 8 Plant 1
6 D=4 2 W =12 Plant 2
4 Feasible region 2 Plant 3 0 2 4 Production rate for doors 6 8 D 17 17 Objective Function (P = 1,500)
Production rate for windows 8 W 6 Feasible 4 P = 1500 = 300D + 500W region 2 0 2 4 Production rate for doors 6 8 D 18 18 Finding the Optimal Solution Production rate for w indow s
8 W P = 3600 = 300D+ 500W O ptim solution al (2, 6) P = 3000 = 300D+ 500W 6 4 F easible region P = 1500 = 300D+ 500W 2 0 2 4 6 8 10 Production rate for doors D 19 19 Summary of the Graphical Method Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint. Find the feasible region by determining where all constraints are satisfied simultaneously. Determine the slope of one objective function line. All other objective function lines will have the same slope. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line. A feasible point on the optimal objective function line is an optimal solution. 20 20 Identifying the Target Cell and Changing Cells Choose the “Solver” from the Tools menu. 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. 3 4 5 6 7 8 9 10 11 12 B C D E F G Unit Profit Plant 1 Plant 2 Plant 3 Doors $300 Windows $500 Hours Used 1 2 5 <= <= <= Hours Available 1 12 18 Total Profit $800 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 1 Windows 1 Units Produced 21 21 Adding Constraints To begin entering constraints, click the “Add” button to the right of the constraints window. Fill in the entries in the resulting Add Constraint dialogue box.
Unit Profit Doors $300 Windows $500 Hours Used 1 2 5 Hours Available 4 12 18 Total Profit $800 Plant 1 Plant 2 Plant 3 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 1 Windows 1 <= <= <= Units Produced 22 22 The Complete Solver Dialogue Box 23 23 Some Important Options Click on the “Options” button, and click in both the “Assume Linear Model” and the “Assume NonNegative” box. “Assume Linear Model” tells the Solver that this is a linear programming model. “Assume NonNegative” adds nonnegativity constraints to all the changing cells. 24 24 The Solver Results Dialogue Box 25 25 The Optimal Solution
B C D E F G 3 4 5 6 7 8 9 10 11 12 Unit Profit Plant 1 Plant 2 Plant 3 Doors $300 Windows $500 Hours Used 2 12 18 <= <= <= Hours Available 1 12 18 Total Profit $3,600 Hours Used Per Unit Produced 1 0 0 2 3 2 Doors 2 Windows 6 Units Produced 26 26 Wyndor Glass Co. Product Mix Problem Slides 226 have examined the Wyndor LP from two perspectives A spreadsheet model An algebraic model Formulated on an Excel spreadsheet. Analyzed and solved by Excel Solver. Obviously these two perspectives are tightly interrelated Formulated algebraically. Analyzed and solved by simple twodimensional graphical methods. The next slide summarizes the interrelationships between the spreadsheet LP model and its algebraic counterpart. 27 27 Components of a Linear Program Spreadsheet Model and Algebraic Formulation
Algebraic
Parameters (“fixed” constants) Decision Variables (decision maker chooses their values) Spreadsheet
Data Cells (constants in light blue shaded cells) Changing Cells (Solver chooses their values – numbers in yellow shaded cells) Objective Function Target Cell (Excel formula expressing the objective (summarizes the goal – e.g. max function [target] in terms of data and changing cell profit; min cost) values in dark orange shaded cell) Constraints (equations or inequalities that limit the allowable decision variable values; includes nonnegativity) Constraints (relationships – equations or inequalities – between output cells and data cells that limit the allowable values of changing cells; includes non negativity) NOTE: A spreadsheet LP model’s changing cells, target cell, and constraints are specified in the Solver Dialogue Box. Nonnegativity Constraints are checked in Options.
28 28 Four Assumptions of Linear Programming
1. Linearity: All functions – objective and constraints– are linear. This is often expressed in terms of the two fundamental properties of linear functions:
Proportionality: Effects are proportional to a decision variable’s value: E.g. for a decision variable D, the profit from D=10 is 10 x (Profit from D=1); resources required for D=10 are 10 x resources required for D=1 Additivity: Effects of two decision variables are the sum of the effects from each: E.g. the profit from D=1 and W=1 is (Profit from D=1) + (Profit from W=1); the resources required by D=1 and W=1 are (Resources required by D=1) + (Resources required by W=1) 1. Divisibility: Decision variables are infinitely divisible. Fractional values are allowed. 1. Certainty: Choices of decision variable values yield known, nonrandom outcomes. 1. Nonnegativity: Decision variables are nonnegative. (There are fancy ways around this. Specifically, any signfree d.v. can be replaced by the difference between two non negative variables. If X is signfree, replace it by X = RS, where R,S > 0) 29 29 The Profit & Gambit Co. Management has decided to undertake a major advertising campaign that will focus on the following three key products: The campaign will use both television and print media The general goal is to increase sales of these products. Management has set the following goals for the campaign: A spray prewash stain remover. A liquid laundry detergent. A powder laundry detergent. Question: how much should they advertise in each medium to meet the sales goals at a minimum total cost? Sales of the stain remover should increase by at least 3%. Sales of the liquid detergent should increase by at least 18%. Sales of the powder detergent should increase by at least 4%. 30 30 Profit & Gambit Co. Spreadsheet Model
3 4 5 6 7 8 9 10 11 12 13 14 B C D E F G Unit Cost ($millions) Television 1 Print Media 2 Increased Sales 3% 18% 8% Minimum Increase 3% 18% 4% Total Cost ($millions) 10 Stain Remover Liquid Detergent Powder Detergent Increase in Sales per Unit of Advertising 0% 1% 3% 2% 1% 4% Television 4 Print Media 3 >= >= >= Advertising Units 31 31 Algebraic Model for Profit & Gambit
Let TV = the number of units of advertising on television PM = the number of units of advertising in the print media Minimize Cost = TV + 2PM (in millions of dollars) subject to Stain remover increased sales: PM ≥ 3 Liquid detergent increased sales: 3TV + 2PM ≥ 18 Powder detergent increased sales: –TV + 4PM ≥ 4 and TV ≥ 0, PM ≥ 0. 32 32 Applying the Graphical Method
Amount of print media advertising PM 10 Feasible region 8 6 4 PM = 3 2 TV + 4 PM = 4 4 2 0 2 3 TV + 2 PM = 18
10 4 6 8 Amount of TV advertising TV 33 33 The Optimal Solution
P M 1 0 C s =1 =T +2P ot 5 V M F a ibe es l r g on ei C s =1 =T +2P ot 0 V M 4 (4 ) ,3 o timl pa s lu n o tio 0 5 1 0 1T 5V A o n of T a ve tisin mu t Vd r g
34 34 Summary of the Graphical Method Draw the constraint boundary line for each constraint. Use the origin (or any point not on the line) to determine which side of the line is permitted by the constraint. Find the feasible region by determining where all constraints are satisfied simultaneously. Determine the slope of one objective function line. All other objective function lines will have the same slope. Move a straight edge with this slope through the feasible region in the direction of improving values of the objective function. Stop at the last instant that the straight edge still passes through a point in the feasible region. This line given by the straight edge is the optimal objective function line. A feasible point on the optimal objective function line is an optimal solution. 35 35 A Production Problem
Weekly supply of raw materials: 8 Small Bricks 6 Large Bricks Products: Table Profit = $20 / Table Chair Profit = $15 / Chair 36 36 Linear Programming Linear programming uses a mathematical model to find the best allocation of scarce resources to various activities so as to maximize profit or minimize cost. Let T = Number of tables to produce C = Number of chairs to produce Maximize Profit = ($20)T + ($15)C subject to 2T + C ≤ 6 large bricks 2T + 2C ≤ 8 small bricks and T ≥ 0, C ≥ 0. 37 37 Graphical Representation
Tables 5 4 2 Chairs + 2 Tables = 8 Small Bricks 3 2 Chairs + 2 Tables = 6 Large Bricks 1 1 2 3 4 5 6 Chairs 38 38 Components of a Linear Program Data Cells Changing Cells (“Decision Variables”) Target Cell (“Objective Function”) Constraints 39 39 Four Assumptions of Linear Programming Linearity Divisibility Certainty Nonnegativity 40 40 When is a Spreadsheet Model Linear? All equations (output cells) must be of the form = ax + by + cz + … where a, b, c are constants (data cells) and x, y, z are changing cells. Suppose C1:C6 are changing cells and D1:D6 are data cells. Which of the following can be part of an LP? SUMPRODUCT(D1:D6, C1:C6) SUM(C1:C6) C1 * SUM(C4:C6) SUMPRODUCT(C1:C3, C4:C6) IF(C1 > 3, 2*C3 + C4, 3*C3 + C5) IF(D1 > 3, C1, C2) MIN(C1, C2) MIN(D1, D2) * C1 ROUND(C1) 41 41 Why Use Linear Programming? Linear programs are easy (efficient) to solve The best (optimal) solution is guaranteed to be found (if it exists) Useful sensitivity analysis information is generated Many problems are essentially linear 42 42 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).
B Profit C Tables $20.00 D Chairs $15.00 E F G 3 4 5 6 7 8 Large Bricks Small Bricks Bill of Materials 2 1 2 2 Available 6 8 43 43 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).
3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables 0 Chairs 0 Available 6 8 Production Quantity: 44 44 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). 3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 0 Available 6 8 Total Profit $20.00 Production Quantity: 1 10 G Total Profit 11 =SUMPRODUCT(C4:D4,C11:D11) 45 45 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 three consecutive cells. For example, if Quantity A ≤ Quantity B, put these three items (Quantity A, ≤, Quantity B) in consecutive cells. Note the use of relative and absolute addressing to make it easy to copy formulas in column E.
3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 1 Total Used 3 4 <= <= Available 6 8 Total Profit $35.00 Production Quantity: 1 6 7 8 E Total Used =SUMPRODUCT(C7:D7,$C$11:$D$11) =SUMPRODUCT(C8:D8,$C$11:$D$11) 46 46 Defining the Target Cell Choose the “Solver” from the Tools menu. 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. 3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 1 Total Used 3 4 <= <= Available 6 8 Total Profit $35.00 Production Quantity: 1 47 47 Identifying the Changing Cells Enter all the changing cells in the “By Changing Cells” window. You may either drag the cursor across the cells or type the addresses. If there are multiple sets of changing cells, separate them by typing a comma. 3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 1 Total Used 3 4 <= <= Available 6 8 Total Profit $35.00 Production Quantity: 1 48 48 Adding Constraints To begin entering constraints, click the “Add” button to the right of the constraints window. Fill in the entries in the resulting Add Constraint dialogue box.
3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 1 Total Used 3 4 <= <= Available 6 8 Total Profit $35.00 Production Quantity: 1 49 49 Some Important Options Click on the “Options” button, and click in both the “Assume Linear Model” and the “Assume NonNegative” box. “Assume Linear Model” tells the Solver that this is a linear programming model. “Assume NonNegative” adds nonnegativity constraints to all the changing cells. 50 50 The Solution After clicking “Solve”, you will receive one of four messages: “Solver found a solution. All constraints and optimality conditions are satisfied.” “Set cell values did not converge.” “Solver could not find a feasible solution.” “Conditions for Assume Linear Model are not satisfied.”
3 4 5 6 7 8 9 10 11 B Profit C Tables $20.00 D Chairs $15.00 E F G Large Bricks Small Bricks Bill of Materials 2 1 2 2 Tables Chairs 2 Total Used 6 8 <= <= Available 6 8 Total Profit $70.00 Production Quantity: 2 51 51 The Graphical Method for Solving LP’s Formulate the problem as a linear program Plot the constraints Identify the feasible region Draw an imaginary line parallel to the objective function (Z = a) Find the optimal solution 52 52 Example #1
Maximize Z = 3x1 + 5x2 subject to x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 53 53 Example #1
1 2 Maximize Z = 3x + 5x subject to
1 x2 10 9 x1>0 x ≤4 2x2 ≤ 12
1 2 Z = 30 8 7 6 5 4 2x2 = 12 Increasing Z x2>0 3x + 2x ≤ 18 and x1 ≥ 0, x2 ≥ 0. 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 1 1 2 3x +2x = 18 x =4 54 54 Example #1
1 2 Maximize Z = 3x + 5x subject to
1 x2 10 x1>0 x ≤4 2x2 ≤ 12
1 2 Z = 36 9 8 7 6 5 4 3x + 2x ≤ 18 and x1 ≥ 0, x2 ≥ 0. 2x2 = 12 Increasing Z
Optimal solution 3 2 1 1 2 3 4 5 6 7 8 9 10 x2>0 x1 1 1 2 3x +2x = 18 x =4 55 55 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10 2x1 – 3x2 ≤ 6 x 1 + x2 ≥ 6 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 56 56 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10 2x1 – 3x2 ≤ 6 x 1 + x2 ≥ 6 and
1 x2 10 9 8 7 6 5 4 3 2 1 1 2 1 x >0
1 2 x ≥ 0, x2 ≥ 0. All points (x ,x ) on or to the East of
1 the line x = 0 (the VERTICAL AXIS) 1
x >0
3
1 4 5 6 7 8 9 10 x1 x >0 57 57 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10 2x1 – 3x2 ≤ 6 x 1 + x2 ≥ 6 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 All points6(x1,x2) on or above (to the North) of line x25= 0 (the HORIZONTAL AXIS)
4 x2 > 0 3 2 1 1 x2 > 0 x2 > 0 2 3 4 5 6 7 8 9 10 x1 58 58 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10 2x1 – 3x2 ≤ 6 x 1 + x2 ≥ 6 and
1 2 x2 10 9 8 7 6 5 4 3 2 1 1 2 1 x >0
1 2 x ≥ 0, x ≥ 0. All points (x ,x ) satisfying the nonnegativity constraints form the first (NE) quadrant
2 x >0
x1 3 4 5 6 7 8 9 10 59 59 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10 2x1 – 3x2 ≤ 6 x 1 + x2 ≥ 6 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 x1 > 0 All p Nor oints ( x thea st) o 1,x2) o n f th e lin or ab e x1 ove ( + 2x to th e 2= 10
x2 > 0 3 4 5 6 7 8 9 10 x1 60 60 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10
1 2 x2 10 9 8 7 6 5 4 3 2 1 1 2 2x – 3x ≤ 6 x 1 + x2 ≥ 6 and x1 ≥ 0, x2 ≥ 0. x1 > 0 1 2 o ll p A in , (x ts o x) r no ov ab 1 to e(
2 the x =6 No ) es t w rth e f th o 2 ne li x 3 x2 > 0 3 4 5 6 7 8 9 10 x1 61 61 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10
1 2 x2 10 9 8 7 6 5 4 3 2 1 1 2 2x – 3x ≤ 6 x1 > 0 x2 > 0 3 4 5 6 7 8 9 10 x1 62 62 Example #2
Minimize Z = 15x1 + 20x2 subject to x1 +2x2 ≥ 10
1 2 FEASIBLE REGION:
1 2 x2 10 9 8 7 6 5 4 3 2 1 1 2 2x – 3x ≤ 6 x1 > 0 All points (x ,x ) that simultaneously satisfy ALL CONSTRAINTS x2 > 0 3 4 5 6 7 8 9 10 x1 63 63 Example #2
1 2 Minimize Z = 15x + 20x subject to x1 +2x2 ≥ 10
1 2 x2 10 9 8 7 6 2x – 3x ≤ 6
1 2 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 The objective function Z=15x +20x is a line with slope (15/20) =  ¾. This particular line passes thru
1 2 Decreasing Z (x ,x ) = (10,0) so OV, the Objective function Value is OV = Z = 15∙10 + 20∙0 = 150
64 64 Example #2
1 2 Minimize Z = 15x + 20x subject to x1 +2x2 ≥ 10
1 2 x2 10 9 8 7 6 5 2x – 3x ≤ 6 Decreasing Z 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 The lowest OV line that intersects with the feasible region passes 65 65 Example #3
Maximize Z = x1 + x2 subject to x1 +2x2 = 8 x1 – x 2 ≤ 0 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 66 66 Example #3
Maximize Z = x1 + x2 subject to x1 +2x2 = 8
1 2 x2 10 x –x ≤0 and x1 ≥ 0, x2 ≥ 0.
Increasing Z 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 67 67 Example #3
Maximize Z = x1 + x2 subject to x1 +2x2 = 8
1 2 x2 10 9 8 7 6 5 4 x –x ≤0 and x1 ≥ 0, x2 ≥ 0. Increasing Z This feasible region contains a single point, 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 68 68 Properties of Linear Programming Solutions An optimal solution must lie on the boundary of the feasible region. There are exactly four possible outcomes of linear programming: If an LP model has one optimal solution, it must be at a corner point. If an LP model has many optimal solutions, at least two of these optimal solutions are at corner points. A unique optimal solution is found. An infinite number of optimal solutions exist. No feasible solutions exist. The objective function is unbounded (there is no optimal solution). 69 69 Example #4 (Multiple Optimal Solutions)
Minimize Z = 6x1 + 4x2 subject to x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 70 70 Example #4 (Multiple Optimal Solutions)
1 2 Minimize Z = 6x + 4x subject to x1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 Increasing Z (2,6) Note: Objective function and third constraint have same slope. All points along edge from (4,3) to (2,6) are optimal. (4,3) 71 71 Example #5 (No Feasible Solution)
Maximize Z = 3x1 + 5x2 subject to x1 ≥ 5 x2 ≥ 4 3x1 + 2x2 ≤ 18 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 72 72 Example #5 (No Feasible Solution)
Maximize Z = 3x1 + 5x2 subject to x1 ≥ 5
2 x2 10 x ≥4
1 2 9 8 3x + 2x ≤ 18 and x1 ≥ 0, x2 ≥ 0. 7 6 5 4 3 2 1 1 2 3 4 5 Note: The feasible region is empty! No point satisfies all constraints simultaneously.
6 7 8 9 10 x1 73 73 Example #6 (Unbounded Solution)
Maximize Z = 5x1 + 12x2 subject to x1 ≤ 5 2x1 –x2 ≤ 2 and x1 ≥ 0, x2 ≥ 0. x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 74 74 Example #6 (Unbounded Solution)
Maximize Z = 5x1 + 12x2 subject to x1 ≤ 5 2x1 –x2 ≤ 2 and x1 ≥ 0, x2 ≥ 0.
Note: Z can be made arbitrarily x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 Increasing Z 75 75 The Simplex Method Algorithm
1. 2. Start at a feasible corner point (often the origin). Check if adjacent corner points improve the objective function:
a) b) If so, move to adjacent corner and repeat step 2. If not, current corner point is optimal. Stop.
x2 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 76 76 ...
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