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1 Linear Programming Session 4-5Course: K0442 Quantitative MethodsYear: 20132Bina Nusantara University3Outline Todays Linear Programming Problem Problem Formulation A Maximization Problem Graphical Solution Procedure Extreme Points and the Optimal Solution Computer Solutions A Minimization Problem

3Linear Programming (LP) ProblemThe maximization or minimization of some quantity is the objective in all linear programming problems.All LP problems have constraints that limit the degree to which the objective can be pursued.A feasible solution satisfies all the problem's constraints.An optimal solution is a feasible solution that results in the largest possible objective function value when maximizing (or smallest when minimizing).A graphical solution method can be used to solve a linear program with two variables.4Linear Programming (LP) ProblemIf both the objective function and the constraints are linear, the problem is referred to as a linear programming problem.Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0).Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.5Problem FormulationProblem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.6Guidelines for Model FormulationUnderstand the problem thoroughly.Describe the objective.Describe each constraint.Define the decision variables.Write the objective in terms of the decision variables.Write the constraints in terms of the decision variables.7Example 1: A Maximization ProblemLP Formulation Max 5x1 + 7x2

s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8

x1, x2 > 088

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1 2 3 4 5 6 7 8 9 10 Example 1: Graphical SolutionConstraint #1 Graphed x2x1x1 < 6(6, 0)98

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1 2 3 4 5 6 7 8 9 10 Example 1: Graphical SolutionConstraint #2 Graphed2x1 + 3x2 < 19 x2x1(0, 6 1/3)(9 1/2, 0)10Example 1: Graphical SolutionConstraint #3 Graphed8

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1 2 3 4 5 6 7 8 9 10 x2x1x1 + x2 < 8(0, 8)(8, 0)11Example 1: Graphical SolutionCombined-Constraint Graph8

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1 2 3 4 5 6 7 8 9 10 2x1 + 3x2 < 19 x2x1x1 + x2 < 8x1 < 612Example 1: Graphical SolutionFeasible Solution Region8

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1 2 3 4 5 6 7 8 9 10 x1 x2FeasibleRegion

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1 2 3 4 5 6 7 8 9 10 Example 1: Graphical SolutionObjective Function Linex1 x2(7, 0)(0, 5)Objective Function5x1 + 7x2 = 3514Example 1: Graphical SolutionOptimal Solutionx1 x2Objective Function5x1 + 7x2 = 46Optimal Solution(x1 = 5, x2 = 3)15Summary of the Graphical Solution Procedurefor Maximization ProblemsPrepare a graph of the feasible solutions for each of the constraints.Determine the feasible region that satisfies all the constraints simultaneously..Draw an objective function line.Move parallel objective function lines toward larger objective function values without entirely leaving the feasible region.Any feasible solution on the objective function line with the largest value is an optimal solution.16Extreme Points and the Optimal SolutionThe corners or vertices of the feasible region are referred to as the extreme points.An optimal solution to an LP problem can be found at an extreme point of the feasible region.When looking for the optimal solution, you do not have to evaluate all feasible solution points.You have to consider only the extreme points of the feasible region.17Example 1: Graphical SolutionThe Five Extreme Points8

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1 2 3 4 5 6 7 8 9 10 x1FeasibleRegion12345 x218Computer SolutionsComputer programs designed to solve LP problems are now widely available.Most large LP problems can be solved with just a few minutes of computer time.Small LP problems usually require only a few seconds.Linear programming solvers are now part of many spreadsheet packages, such as Microsoft Excel.19Example 2: A Minimization ProblemLP Formulation

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10 4x1 - x2 > 12 x1 + x2 > 4

x1, x2 > 020Example 2: Graphical SolutionGraph the Constraints Constraint 1: When x1 = 0, then x2 = 2; when x2 = 0, then x1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 2: When x2 = 0, then x1 = 3. But setting x1 to 0 will yield x2 = -12, which is not on the graph. Thus, to get a second point on this line, set x1 to any number larger than 3 and solve for x2: when x1 = 5, then x2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. Constraint 3: When x1 = 0, then x2 = 4; when x2 = 0, then x1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.21Example 2: Graphical SolutionConstraints Graphed5

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1 1 2 3 4 5 6x24x1 - x2 > 12

x1 + x2 > 42x1 + 5x2 > 10x1Feasible Region22Example 2: Graphical SolutionGraph the Objective FunctionSet the objective function equal to an arbitrary constant (say 20) and graph it. For 5x1 + 2x2 = 20, when x1 = 0, then x2 = 10; when x2= 0, then x1 = 4. Connect (4,0) and (0,10).

Move the Objective Function Line Toward OptimalityMove it in the direction which lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined by the last two constraints.23Example 2: Graphical SolutionObjective Function Graphed5

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1 1 2 3 4 5 6x2Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 42x1 + 5x2 > 10x124Solve for the Extreme Point at the Intersection of the Two Binding Constraints

4x1 - x2 = 12 x1+ x2 = 4 Adding these two equations gives: 5x1 = 16 or x1 = 16/5. Substituting this into x1 + x2 = 4 gives: x2 = 4/5Example 2: Graphical Solution25Example 2: Graphical SolutionSolve for the Optimal Value of the Objective FunctionSolve for z = 5x1 + 2x2 = 5(16/5) + 2(4/5) = 88/5. Thus the optimal solution is x1 = 16/5; x2 = 4/5; z = 88/526Example 2: Graphical SolutionOptimal Solution5

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1 1 2 3 4 5 6x2Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 42x1 + 5x2 > 10

Optimal: x1 = 16/5 x2 = 4/5x127Feasible RegionThe feasible region for a two-variable linear programming problem can be nonexistent, a single point, a line, a polygon, or an unbounded area.Any linear program falls in one of three categories:is infeasible has a unique optimal solution or alternate optimal solutionshas an objective function that can be increased without boundA feasible region may be unbounded and yet there may be optimal solutions. This is common in minimization problems and is possible in maximization problems.28Sensitivity Analysis and Interpretation of SolutionIntroduction to Sensitivity AnalysisGraphical Sensitivity Analysis29In the previous chapter we discussed:objective function valuevalues of the decision variables

In this chapter we will discuss:changes in the coefficients of the objective functionchanges in the right-hand side value of a constraintIntroduction to Sensitivity Analysis30Introduction to Sensitivity AnalysisSensitivity analysis (or post-optimality analysis) is used to determine how the optimal solution is affected by changes, within specified ranges, in:the objective function coefficientsthe right-hand side (RHS) valuesSensitivity analysis is important to a manager who must operate in a dynamic environment with imprecise estimates of the coefficients. Sensitivity analysis allows a manager to ask certain what-if questions about the problem.31Example 1LP FormulationMax 5x1 + 7x2

s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8

x1, x2 > 032Example 1Graphical Solution2x1 + 3x2 < 19 x2x1x1 + x2 < 8Max 5x1 + 7x2x1 < 6Optimal Solution: x1 = 5, x2 = 38

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11 2 3 4 5 6 7 8 9 1033Objective Function CoefficientsLet us consider how changes in the objective function coefficients might affect the optimal solution.The range of optimality for each coefficient provides the range of values over which the current solution will remain optimal.Managers should focus on those objective coefficients that have a narrow range of optimality and coefficients near the endpoints of the range.34Example 1Changing Slope of Objective Functionx1FeasibleRegion12345 x2Coincides withx1 + x2 < 8constraint line8

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11 2 3 4 5 6 7 8 9 10Coincides with2x1 + 3x2 < 19constraint lineObjective functionline for 5x1 + 7x235Range of OptimalityGraphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.Slope of an objective function line, Max c1x1 + c2x2, is -c1/c2, and the slope of a constraint, a1x1 + a2x2 = b, is -a1/a2.36Example 1Range of Optimality for c1 The slope of the objective function line is -c1/c2. The slope of the first binding constraint, x1 + x2 = 8, is -1 and the slope of the second binding constraint, 2x1 + 3x2 = 19, is -2/3.Find the range of values for c1 (with c2 staying 7) such that the objective function line slope lies between that of the two binding constraints: -1 < -c1/7 < -2/3 Multiplying through by -7 (and reversing the inequalities): 14/3 < c1 < 737Example 1Range of Optimality for c2 Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints: -1 < -5/c2 < -2/3

Multiplying by -1: 1 > 5/c2 > 2/3Inverting : 1 < c2/5 < 3/2

Multiplying by 5: 5 < c2 < 15/238Example 1Summary of Range of Optimality for c1 and c2 14/3 < c1 < 7 or 4 2/3 < c1 < 7

Since the current value of c1 is 5, c1 is allowed to increase by 2 and decrease by 1/3

< c2 < 15/2 or 5 < c2 < 7 1/2

Since the current value of c2 is 7, c2 is allowed to increase by 1/2 and decrease by 2

39Right-Hand SidesLet us consider how a change in the right-hand side for a constraint might affect the feasible region and perhaps cause a change in the optimal solution.The improvement in the value of the optimal solution per unit increase in the right-hand side of a constraint is called the dual price of the constraintThe range of feasibility is the range over which the dual price is applicable.As the RHS increases, other constraints will become binding and limit the change in the value of the objective function.40Dual PriceAlgebraically, a dual price is determined by adding +1 to the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints. The dual price is equal to the difference in the values of the objective functions between the new and original problems.The dual price for a nonbinding constraint is 0.A negative dual price indicates that the objective function will not improve if the RHS is increased.Shadow price ( or dual value) = dual price for maximize problems and negative of dual price for minimize problems41Relevant Cost and Sunk CostA resource cost is a relevant cost if the amount paid for it is dependent upon the amount of the resource used by the decision variables. E.g. hourly labor cost for piece workers Relevant costs are reflected in the objective function coefficients. A resource cost is a sunk cost if it must be paid regardless of the amount of the resource actually used by the decision variables. E.g. hourly cost for salaried employees Sunk resource costs are not reflected in the objective function coefficients.42Cautionary Note onthe Interpretation of Dual PricesResource cost is sunkThe dual price is the maximum amount you should be willing to pay for one additional unit of the resource.Resource cost is relevantThe dual price is the maximum premium over the normal cost that you should be willing to pay for one unit of the resource, since the dual price is the amount the value of the resource exceeds its cost.43Example 1Dual PricesConstraint 1: Since x1 < 6 is not a binding constraint, its dual price is 0.Constraint 2: Change the RHS value of the second constraint to 20 and resolve for the optimal point determined by the last two constraints 2x1 + 3x2 = 20 and x1 + x2 = 8. The solution is x1 = 4, x2 = 4, z = 48. Hence, the dual price = znew - zold = 48 - 46 = 2.44Example 1Dual PricesConstraint 3: Change the RHS value of the third constraint to 9 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9.

The solution is: x1 = 8, x2 = 1, z = 47. The dual price is znew - zold = 47 - 46 = 1.45ReferenceDavid R. Anderson [et.al]. (2008). Quantitative Methods for Business. 11. SOWES. New York. ISBN: 10: 0324651813.Bina Nusantara University46Bina Nusantara University47Thank You