1 1 Slide Linear Programming (LP) Problem n A mathematical programming problem is one that seeks to...

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Linear Programming (LP) ProblemLinear Programming (LP) Problem

A A mathematical programming problemmathematical programming problem is one is one that seeks to maximize an objective function that seeks to maximize an objective function subject to constraints. subject to constraints.

If both the objective function and the If both the objective function and the constraints are linear, the problem is referred to constraints are linear, the problem is referred to as a as a linear programming problemlinear programming problem..

Linear functionsLinear functions are functions in which each are functions in which each variable appears in a separate term raised to variable appears in a separate term raised to the first power and is multiplied by a constant the first power and is multiplied by a constant (which could be 0).(which could be 0).

Linear constraintsLinear constraints are linear functions that are are linear functions that are restricted to be "less than or equal to", "equal restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.to", or "greater than or equal to" a constant.

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LP SolutionsLP Solutions

The The maximizationmaximization or or minimizationminimization of some of some quantity is the quantity is the objectiveobjective in all linear in all linear programming problems.programming problems.

A A feasible solutionfeasible solution satisfies all the problem's satisfies all the problem's constraints.constraints.

An An optimal solutionoptimal solution is a feasible solution that is a feasible solution that results in the largest possible objective results in the largest possible objective function value when maximizing (or smallest function value when maximizing (or smallest when minimizing).when minimizing).

A A graphical solution methodgraphical solution method can be used to can be used to solve a linear program with two variables.solve a linear program with two variables.

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Problem FormulationProblem Formulation

Problem formulation or modeling is the process Problem formulation or modeling is the process of translating a verbal statement of a problem of translating a verbal statement of a problem into a mathematical statement.into a mathematical statement.

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Guidelines for Model FormulationGuidelines for Model Formulation

Understand the problem thoroughly.Understand the problem thoroughly. Write a verbal description of the objective.Write a verbal description of the objective. Write a verbal description of each constraint.Write a verbal description of each constraint. Define the decision variables.Define the decision variables. Write the objective in terms of the decision Write the objective in terms of the decision

variables.variables. Write the constraints in terms of the decision Write the constraints in terms of the decision

variables.variables.

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Example 1: A Maximization ProblemExample 1: A Maximization Problem

LP FormulationLP Formulation

Max Max zz = 5 = 5xx11 + 7 + 7xx22

s.t. s.t. xx11 << 6 6

22xx11 + 3 + 3xx22 << 19 19

xx11 + + xx22 << 8 8

xx11, , xx22 >> 0 0

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Example 1: Graphical SolutionExample 1: Graphical Solution

Constraint #1 GraphedConstraint #1 Graphed

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22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

xx22

xx11

xx11 << 6 6

(6, 0)(6, 0)

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88

77

66

55

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22

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

(0, 6 (0, 6 1/31/3))

(9 (9 1/21/2, 0), 0)

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88

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

xx22

xx11

xx11 + + xx22 << 8 8

(0, 8)(0, 8)

(8, 0)(8, 0)

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Combined-Constraint GraphCombined-Constraint Graph

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

xx11 + + xx22 << 8 8

xx11 << 6 6

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Feasible Solution RegionFeasible Solution Region

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

FeasibleFeasibleRegionRegion

xx22

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Objective Function LineObjective Function Line

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

xx22

(7, 0)(7, 0)

(0, 5)(0, 5)

55xx11 + + 7x7x2 2 = 35= 3555xx11 + + 7x7x2 2 = 35= 35

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Optimal SolutionOptimal Solution

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

xx22

55xx11 + + 7x7x2 2 = 46= 4655xx11 + + 7x7x2 2 = 46= 46

Optimal SolutionOptimal SolutionOptimal SolutionOptimal Solution

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Summary of the Graphical Solution Summary of the Graphical Solution ProcedureProcedure

for Maximization Problemsfor Maximization Problems Prepare a graph of the feasible solutions for each Prepare a graph of the feasible solutions for each

of the constraints.of the constraints. Determine the feasible region that satisfies all Determine the feasible region that satisfies all

the constraints simultaneously..the constraints simultaneously.. Draw an objective function line.Draw an objective function line. Move parallel objective function lines toward Move parallel objective function lines toward

larger objective function values without entirely larger objective function values without entirely leaving the feasible region.leaving the feasible region.

Any feasible solution on the objective function Any feasible solution on the objective function line with the largest value is an optimal solution.line with the largest value is an optimal solution.

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Slack and Surplus VariablesSlack and Surplus Variables

A linear program in which all the variables are A linear program in which all the variables are non-negative and all the constraints are non-negative and all the constraints are equalities is said to be in equalities is said to be in standard formstandard form. .

Standard form is attained by adding Standard form is attained by adding slack slack variables variables to "less than or equal to" constraints, to "less than or equal to" constraints, and by subtracting and by subtracting surplus variablessurplus variables from from "greater than or equal to" constraints. "greater than or equal to" constraints.

Slack and surplus variables represent the Slack and surplus variables represent the difference between the left and right sides of the difference between the left and right sides of the constraints.constraints.

Slack and surplus variables have objective Slack and surplus variables have objective function coefficients equal to 0.function coefficients equal to 0.

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Example 1Example 1

Standard FormStandard Form

Max Max zz = 5 = 5xx11 + 7 + 7xx2 2 + 0+ 0ss1 1 + 0+ 0ss2 2 + 0+ 0ss33

s.t. s.t. xx11 + + ss11 = 6 = 6

22xx11 + 3 + 3xx22 + + ss22 = 19 = 19

xx11 + + xx22 + + ss3 3 = = 88

xx11, , xx2 2 , , ss11 , , ss22 , , ss33 >> 0 0

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Extreme Points and the Optimal SolutionExtreme Points and the Optimal Solution

The corners or vertices of the feasible region are The corners or vertices of the feasible region are referred to as the referred to as the extreme pointsextreme points..

An optimal solution to an LP problem can be An optimal solution to an LP problem can be found at an extreme point of the feasible region.found at an extreme point of the feasible region.

When looking for the optimal solution, you do not When looking for the optimal solution, you do not have to evaluate all feasible solution points.have to evaluate all feasible solution points.

You have to consider only the extreme points of You have to consider only the extreme points of the feasible region.the feasible region.

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The Five Extreme PointsThe Five Extreme Points

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1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

FeasibleFeasibleRegionRegion

1111 2222

3333

4444

5555

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Computer SolutionsComputer Solutions

Computer programs designed to solve LP Computer programs designed to solve LP problems are now widely available.problems are now widely available.

Most large LP problems can be solved with just a Most large LP problems can be solved with just a few minutes of computer time.few minutes of computer time.

Small LP problems usually require only a few Small LP problems usually require only a few seconds.seconds.

Linear programming solvers are now part of Linear programming solvers are now part of many spreadsheet packages, such as many spreadsheet packages, such as Microsoft Microsoft ExcelExcel..

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Interpretation of Computer OutputInterpretation of Computer Output

In this chapter we will discuss the following In this chapter we will discuss the following output:output:• objective function valueobjective function value• values of the decision variablesvalues of the decision variables• reduced costsreduced costs• slack/surplusslack/surplus

In Chapter 3 we will discuss how an optimal In Chapter 3 we will discuss how an optimal solution is affected by a:solution is affected by a:• change in a coefficient of the objective change in a coefficient of the objective

functionfunction• change in the right-hand side value of a change in the right-hand side value of a

constraintconstraint

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Example 1: Spreadsheet SolutionExample 1: Spreadsheet Solution

Partial Spreadsheet Showing Problem DataPartial Spreadsheet Showing Problem Data

A B C D12 Constraints X1 X2 RHS Values3 #1 1 0 64 #2 2 3 195 #3 1 1 86 Obj.Func.Coeff. 5 7

LHS CoefficientsA B C D

12 Constraints X1 X2 RHS Values3 #1 1 0 64 #2 2 3 195 #3 1 1 86 Obj.Func.Coeff. 5 7

LHS Coefficients

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Partial Spreadsheet Showing SolutionPartial Spreadsheet Showing Solution

A B C D89 X1 X2

10 5.0 3.01112 46.01314 Constraints Amount Used RHS Limits15 #1 5 <= 616 #2 19 <= 1917 #3 8 <= 8

Optimal Decision Variable Values

Maximized Objective Function

A B C D89 X1 X2

10 5.0 3.01112 46.01314 Constraints Amount Used RHS Limits15 #1 5 <= 616 #2 19 <= 1917 #3 8 <= 8

Optimal Decision Variable Values

Maximized Objective Function

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Interpretation of Computer OutputInterpretation of Computer Output

We see from the previous slide that:We see from the previous slide that:

• Objective Function Value = 46Objective Function Value = 46

• Decision Variable #1 (Decision Variable #1 (xx11) = 5) = 5

• Decision Variable #2 (Decision Variable #2 (xx22) = 3) = 3

• Slack in Constraint #1 = 1 (= 6 - 5)Slack in Constraint #1 = 1 (= 6 - 5)• Slack in Constraint #2 = 0 (= 19 - 19)Slack in Constraint #2 = 0 (= 19 - 19)• Slack in Constraint #3 = 0 (= 8 - 8)Slack in Constraint #3 = 0 (= 8 - 8)

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Reduced CostReduced Cost

The The reduced costreduced cost for a decision variable whose for a decision variable whose value is 0 in the optimal solution is the amount value is 0 in the optimal solution is the amount the variable's objective function coefficient the variable's objective function coefficient would have to improve (increase for would have to improve (increase for maximization problems, decrease for maximization problems, decrease for minimization problems) before this variable minimization problems) before this variable could assume a positive value. could assume a positive value.

The reduced cost for a decision variable with a The reduced cost for a decision variable with a positive value is 0.positive value is 0.

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Reduced CostsReduced Costs

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$8 X1 5.0 0.0 5 2 0.333333333$C$8 X2 3.0 0.0 7 0.5 2

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$B$13 #1 5 0 6 1E+30 1$B$14 #2 19 2 19 5 1$B$15 #3 8 1 8 0.333333333 1.666666667

Adjustable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$B$8 X1 5.0 0.0 5 2 0.333333333$C$8 X2 3.0 0.0 7 0.5 2

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease

$B$13 #1 5 0 6 1E+30 1$B$14 #2 19 2 19 5 1$B$15 #3 8 1 8 0.333333333 1.666666667

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Example 2: A Minimization ProblemExample 2: A Minimization Problem

LP FormulationLP Formulation

Min Min zz = 5 = 5xx11 + 2 + 2xx22

s.t. 2s.t. 2xx11 + 5 + 5xx22 >> 10 10

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

xx11, , xx22 >> 0 0

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Graph the ConstraintsGraph the Constraints

Constraint 1Constraint 1: When : When xx11 = 0, then = 0, then xx22 = 2; when = 2; when xx22 = = 0, 0, then then xx11 = 5. Connect (5,0) and (0,2). The ">" = 5. Connect (5,0) and (0,2). The ">" side is side is above this line.above this line.

Constraint 2Constraint 2: When : When xx22 = 0, then = 0, then xx11 = 3. But setting = 3. But setting xx11 to to 0 will yield 0 will yield xx22 = -12, which is not on the graph. = -12, which is not on the graph.

Thus, to get a second point on this line, set Thus, to get a second point on this line, set xx11 to any to any number larger than 3 and solve for number larger than 3 and solve for xx22: : when when xx11 = 5, = 5, then then xx22 = 8. Connect (3,0) and = 8. Connect (3,0) and (5,8). The ">" side is (5,8). The ">" side is to the right.to the right.

Constraint 3Constraint 3: When : When xx11 = 0, then = 0, then xx22 = 4; when = 4; when xx22 = = 0, 0, then then xx11 = 4. Connect (4,0) and (0,4). The ">" = 4. Connect (4,0) and (0,4). The ">" side is side is above this line.above this line.

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Constraints GraphedConstraints Graphed

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1 2 3 4 5 61 2 3 4 5 6 1 2 3 4 5 61 2 3 4 5 6

xx22xx22

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

22xx11 + 5 + 5xx22 >> 10 1022xx11 + 5 + 5xx22 >> 10 10

xx11xx11

Feasible RegionFeasible Region

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Graph the Objective FunctionGraph the Objective Function

Set the objective function equal to an Set the objective function equal to an arbitrary constant (say 20) and graph it. For 5arbitrary constant (say 20) and graph it. For 5xx11 + 2+ 2xx22 = 20, when = 20, when xx11 = 0, then = 0, then xx22 = 10; when = 10; when xx22= = 0, then 0, then xx11 = 4. Connect (4,0) and (0,10). = 4. Connect (4,0) and (0,10).

Move the Objective Function Line Toward Move the Objective Function Line Toward OptimalityOptimality

Move it in the direction which lowers its Move it in the direction which lowers its value (down), since we are minimizing, until it value (down), since we are minimizing, until it touches the last point of the feasible region, touches the last point of the feasible region, determined by the last two constraints.determined by the last two constraints.

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Objective Function GraphedObjective Function Graphed

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1 2 3 4 5 6 1 2 3 4 5 6

xx22xx22Min Min zz = 5 = 5xx11 + 2 + 2xx22

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

Min Min zz = 5 = 5xx11 + 2 + 2xx22

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

22xx11 + 5 + 5xx22 >> 10 1022xx11 + 5 + 5xx22 >> 10 10

xx11xx11

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Solve for the Extreme Point at the Intersection of the Solve for the Extreme Point at the Intersection of the Two Binding ConstraintsTwo Binding Constraints

44xx11 - - xx22 = 12 = 12

xx11+ + xx22 = 4 = 4

Adding these two equations gives: Adding these two equations gives:

55xx11 = 16 or = 16 or xx11 = 16/5. = 16/5.

Substituting this into Substituting this into xx11 + + xx22 = 4 gives: = 4 gives: xx22 = 4/5 = 4/5 Solve for the Optimal Value of the Objective FunctionSolve for the Optimal Value of the Objective Function

Solve for Solve for zz = 5 = 5xx11 + 2 + 2xx22 = 5(16/5) + 2(4/5) = 88/5. = 5(16/5) + 2(4/5) = 88/5.

Thus the optimal solution is Thus the optimal solution is

xx11 = 16/5; = 16/5; xx22 = 4/5; = 4/5; z z = 88/5= 88/5

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Optimal SolutionOptimal Solution

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1 2 3 4 5 6 1 2 3 4 5 6

xx22xx22Min Min zz = 5 = 5xx11 + 2 + 2xx22

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

Min Min zz = 5 = 5xx11 + 2 + 2xx22

44xx11 - - xx22 >> 12 12

xx11 + + xx22 >> 4 4

22xx11 + 5 + 5xx22 >> 10 10

Optimal: Optimal: xx11 = 16/5 = 16/5

xx22 = 4/5 = 4/5

22xx11 + 5 + 5xx22 >> 10 10

Optimal: Optimal: xx11 = 16/5 = 16/5

xx22 = 4/5 = 4/5xx11xx11

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Partial Spreadsheet Showing Problem DataPartial Spreadsheet Showing Problem Data

A B C D12 Constraints X1 X2 RHS3 #1 2 5 104 #2 4 -1 125 #3 1 1 46 Obj.Func.Coeff. 5 2

LHS CoefficientsA B C D

12 Constraints X1 X2 RHS3 #1 2 5 104 #2 4 -1 125 #3 1 1 46 Obj.Func.Coeff. 5 2

LHS Coefficients

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Partial Spreadsheet Showing FormulasPartial Spreadsheet Showing Formulas

A B C D9

10 X1 X211 Dec.Var.Values1213 =B6*B11+C6*C111415 Constraints Amount Used Amount Avail.16 #1 =B3*$B$11+C3*$C$11 >= =D317 #2 =B4*$B$11+C4*$C$11 >= =D418 #3 =B5*$B$11+C5*$C$11 >= =D5

Decision Variables

Minimized Objective Function

A B C D9

10 X1 X211 Dec.Var.Values1213 =B6*B11+C6*C111415 Constraints Amount Used Amount Avail.16 #1 =B3*$B$11+C3*$C$11 >= =D317 #2 =B4*$B$11+C4*$C$11 >= =D418 #3 =B5*$B$11+C5*$C$11 >= =D5

Decision Variables


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Partial Spreadsheet Showing SolutionPartial Spreadsheet Showing Solution

A B C D9

10 X1 X211 Dec.Var.Values 3.20 0.8001213 17.6001415 Constraints Amount Used Amount Avail.16 #1 10.4 >= 1017 #2 12 >= 1218 #3 4 >= 4

Decision Variables


A B C D9

10 X1 X211 Dec.Var.Values 3.20 0.8001213 17.6001415 Constraints Amount Used Amount Avail.16 #1 10.4 >= 1017 #2 12 >= 1218 #3 4 >= 4

Decision Variables


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Feasible RegionFeasible Region

The feasible region for a two-variable linear The feasible region for a two-variable linear programming problem can be nonexistent, a single programming problem can be nonexistent, a single point, a line, a polygon, or an unbounded area.point, a line, a polygon, or an unbounded area.

Any linear program falls in one of three categories:Any linear program falls in one of three categories:• is infeasible is infeasible • has a unique optimal solution or alternate has a unique optimal solution or alternate

optimal solutionsoptimal solutions• has an objective function that can be increased has an objective function that can be increased

without boundwithout bound A feasible region may be unbounded and yet there A feasible region may be unbounded and yet there

may be optimal solutions. This is common in may be optimal solutions. This is common in minimization problems and is possible in minimization problems and is possible in maximization problems.maximization problems.

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Special CasesSpecial Cases

Alternative Optimal SolutionsAlternative Optimal Solutions

In the graphical method, if the objective function In the graphical method, if the objective function line is parallel to a boundary constraint in the line is parallel to a boundary constraint in the direction of optimization, there are direction of optimization, there are alternate alternate optimal solutionsoptimal solutions, with all points on this line , with all points on this line segment being optimal.segment being optimal.

InfeasibilityInfeasibility

A linear program which is overconstrained so A linear program which is overconstrained so that no point satisfies all the constraints is said that no point satisfies all the constraints is said to be to be infeasibleinfeasible..

UnboundedUnbounded

(See example on upcoming slide.)(See example on upcoming slide.)

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Example: Infeasible ProblemExample: Infeasible Problem

Solve graphically for the optimal solution:Solve graphically for the optimal solution:

Max Max zz = 2 = 2xx11 + 6 + 6xx22

s.t. 4s.t. 4xx11 + 3 + 3xx22 << 1212

22xx11 + + xx22 >> 8 8

xx11, , xx22 >> 0 0

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Example: Infeasible ProblemExample: Infeasible Problem

There are no points that satisfy both constraints, There are no points that satisfy both constraints, hence this problem has no feasible region, and hence this problem has no feasible region, and no optimal solution. no optimal solution.

xx22

xx11

44xx11 + 3 + 3xx22 << 12 12

22xx11 + + xx22 >> 8 8

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44

88

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Example: Unbounded ProblemExample: Unbounded Problem

Solve graphically for the optimal solution:Solve graphically for the optimal solution:

Max Max z z = 3 = 3xx11 + 4 + 4xx22

s.t. s.t. xx11 + + xx22 >> 5 5

33xx11 + + xx22 >> 8 8

xx11, , xx22 >> 0 0

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Example: Unbounded ProblemExample: Unbounded Problem

The feasible region is unbounded and the The feasible region is unbounded and the objective function line can be moved parallel to objective function line can be moved parallel to itself without bound so that itself without bound so that zz can be increased can be increased infinitely. infinitely.

xx22

xx11

33xx11 + + xx22 >> 8 8

xx11 + + xx22 >> 5 5

Max 3Max 3xx11 + 4 + 4xx22

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2.672.67

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