Linear Programming a geometrical preview

Transcrição

1 Linear Programming a geometrical preview ( DEGI) Operations Research 2012/ / 103

2 Linear Programming Learning Objectives Ability to formulate a linear program: define the decision variables (elements under control of the decision maker whose values determine the solution of the model); define the objective function as linear combinations of the decision variables (criterion the decision maker will use to evaluate alternative solutions to the problem); define the constraints of the problem as linear combinations of the decision variables (restrictions imposed upon the values of the decision variables by the caracteristics of the problem under study); Ability to represent graphically the decision space of a linear program. Ability to find, based on the graphical representation, the optimal solution of the linear program. Ability to determine, based on the graphical representation, the shadow price on a constraint: How much does the objective function increase if there is a unit increase in a constraint resource? Ability to determine, based on the graphical representation, over what range can a particular objective-function coefficient vary without changing the optimal solution. Ability to determine, based on the graphical representation, over what range can the right-hand-side of a constraint change in order to maintain the shadow price. ( DEGI) Operations Research 2012/ / 103

3 Production Planning at BA The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mould injection. BA recently won two new clients and intends to assign this production to one of its furnaces. The orders were of bottles of Sandeman Ruby Port (75cl) and one liter bottles of olive oil Oliveira da Serra. Both customers buy all the bottles that BA can produce. Due to differences in the production (number of cavities and different cycle times), each batch of portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30 hours. There are a total of 2000 hours available at the oven. Moreover, there are restrictions in the storage capacity of the bottles prior to their dispatch. The warehouse has 300m 3 available and each batch of portwine bottles occupies 6m 3 of warehouse space, while each batch of bottles of olive oil holds 5m 3. Finally we must also consider the capacity available in the decoration sector (e.g. labels, packaging), which is 200 hours. The portwine bottles spend 3 hours per batch and the olive oil bottles 5 hours per batch. Barbosa & Almeida wants to maximize the profit from these two orders, knowing that the profit per batch is 50e and 60e, respectively. ( DEGI) Operations Research 2012/ / 103

4 Production Planning at BA Linear Programming Model Decision variables x VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce; x A Number of batches of Azeite Oliveira da Serra bottles to produce. Objective: max Z = 50x VP + 60x A Subject to: 50x VP + 30x A 2000 (time in the oven) 6x VP + 5x A 300 (space in the warehouse) 3x VP + 5x A 200 (capacity in the decoration sector) x VP, x A 0 In this model x VP and x A may not be integer. ( DEGI) Operations Research 2012/ / 103

5 Production Planning at BA Graphical resolution Active and redundant constraints 80 X A 70 50X VP + 30X A = 2000 (restrição ativa) Sentido de crescimento de Z 60 Z = 50X VP + 60X A X VP = X VP + 5X A = 200 (restrição ativa) Tempo no Forno Espaço no armazém Capacidade na decoração Função objectivo X A =0 6X VP + 5X A = 300 X VP (restrição redundante) ( DEGI) Operations Research 2012/ / 103

6 Production Planning at BA Graphical resolution Optimal solution X A X VP + 30X A = Z = 50X VP + 60X A X VP + 5X A = X VP + 5X A = 300 X VP Tempo no Forno Capacidade na decoração { 50xVP + 30x A = x VP + 5x A = 200 Optimal solution: x VP = 25, x A = 25, Z = 2750 { xvp = 25 x A = 25 The optimal solution must be in a vertex of the solution space: The solutions inside the admissible region cannot be optimal because the profit may always increase by increasing x VP, x A or both. Espaço no armazém Função objectivo ( DEGI) Operations Research 2012/ / 103

7 Production Planning at BA Graphical resolution Relation between vertices and basic solutions 80 X A A X = (X VP; X A; Folga f ; Folga a ; Folga d ) A = (0; 40; 800; 100; 0) 50X VP + 30X A = 2000 B = (0; 0; 2000; 300; 200) C = (40; 0; 0; 60; 80) D = (25; 25; 0; 25; 0) U = (10; 20; 900; 140; 70) W = (50; 0; -500; 0; 50) Z = 50X VP + 60X A D U The vertices of the admissible region are admissible basic solutions. The other vertices correspond to basic, non-admissible solutions (negative variables). All the points inside the admissible region are non-basic solutions. A, B, C e D are admissible basic solutions; W is a basic, non-admissible solution; U is not a basic solution. 10 3X VP + 5X A = B C W X VP + 5X A = 300 X VP Tempo no forno Capacidade na decoração When moving from one vertex to another adjacent vertex one basic variable becomes non-basic and one non-basic variable becomes basic. Espaço no armazém Função objectivo ( DEGI) Operations Research 2012/ / 103

8 Production Planning at BA Graphical resolution (Sensitivity analysis Coefficients of the objective function) X A X VP + 30X A = 2000 Z = 50X VP + 60X A The optimal solution ramains unchanged while the slope of the objective function is maintained between the slope of the restriction time in the oven and the slope of the constaint decoration capacity X VP + 5X A = 200 Varying the slope of the objective function the optimal solution jumps from vertex to vertex X VP + 5X A = 300 X VP Tempo no Forno Capacidade na decoração Espaço no armazém Função objectivo ( DEGI) Operations Research 2012/ / 103

9 Production Planning at BA Graphical resolution (Sensitivity analysis Coefficients of the objective function) In order to calculate the variation of the coefficient of x VP (C VP x VP + 60x A ) we can write the two active constraints in the optimal solution with coefficient 60 in x A : 100x VP + 60x A 4000 (time in the oven) 36x VP + 60x A 2400 (decoration cap.) The coefficients of x VP in the two constraints will then be the limits for the variation of C VP such that the optimal solution does not change. 36 C VP C VP (initial value 50) 100 In order to calculate the variation of the coefficient of x A (50x VP + C A x A ) we can write the two active constraints in the optimal solution with coefficient 50 em x VP : 50x VP + 30x A 2000 (time in the oven) 50x VP x A (decoration cap.) The coefficients of x A in the two constraints will then be the limits for the variation of C A such that the optimal solution does not change. 30 C A C A (initial value 60) In both cases, the initial value of the coefficients is within the calculated range of variation. If the initial values were not contained in this interval, which ranges should be considered? ( DEGI) Operations Research 2012/ / 103

10 Production Planning at BA Graphical resolution (Sensitivity analysis RHS of the constraints) X A X VP + 30X A = b Z = 50X VP + 60X A Active constraint with another slope: the optimal value changes but the vertex does not change X VP + 5X A = 200 A change in the RHS value of a constraint translates the constraint into a paralel position X VP + 5X A = 300 X VP Tempo no Forno Capacidade na decoração Espaço no armazém Função objectivo ( DEGI) Operations Research 2012/ / 103

11 Production Planning at BA Graphical resolution (Sensitivity analysis RHS of the constraints) Reducing the number of available hours in the oven time (b 1 ) translates the line downwards. If the line surpasses the intersection between the constraint decoration capacity and x VP 0 (x VP, x A ) = (0, 40) the constraint decoration capacity will not be an active constraint any more and the optimal vertex changes. (0, 40) line oven time = b 1 b 1 = 1200 The optimal vertex will remain the same while: Increasing the number of available hours in the oven time (b 1 ) translates the line upwards. If the line surpasses the intersection between the constraint space in the warehouse and decoration capacity : (x VP, x A ) = (33 1 3, 20) the constraint oven time is not active anymore and the optimal vertex changes. (33 1, 20) line oven time = b 1 b 1 = b ( DEGI) Operations Research 2012/ / 103

12 Production Planning at BA Graphical resolution (Shadow prices 1 of the constraints) The shadow price of one constraint represents the variation in the objective function value if we increase the RHS of the constraint by one unit. Example: By how much would the profit increase if we could have an additional hour in the decoration? (increase the RHS of the contraint from 200 to 201) { 50xVP + 30x A = x VP + 5x A = 201 Optimal solution: x VP { xvp = x A = = , x A = , Z = With an additional hour in the decoration (an increase from 200 to 201) the value of the optimal solution increased from 2750 to The constraint decoration capacity has therefore a shadow price of The shadow price remains the same while the optimal vertex does not change (see the sensitivity analysis for the RHS of the constraints). The shadow price of a non-active constraint is zero. 1 The shadow prices are also called dual variables, marginal values or pi values. ( DEGI) Operations Research 2012/ / 103

13 Linear Programming Graphical resolution (Unique optimal solution) Objective: Z = 4 x 1 + x 2 0 x max Z = 4x 1 + x 2 Subject to: x 1 x 2 2 x 1 + 2x 2 8 x 1 - x 2 = 2 x 1 + 2x 2 = 8 x 1, x 2 0 x 1 ( DEGI) Operations Research 2012/ / 103

14 Linear Programming Graphical resolution (Non-unique optimal solution) x Z = x 1 - x Objective: - 28 max Z = x 1 x 2 30 Subject to: x 1 x 2 2 x 1 + 2x 2 8 x 1, x 2 0 x 1 - x 2 = 2 x 1 + 2x 2 = 8 x 1 ( DEGI) Operations Research 2012/ / 103

15 Linear Programming Graphical resolution (Unlimited solution) Z = 4 x 1 + x 2 0 x Objective: max Z = 4x 1 + x 2 x 1 - x 2 = 2 Subject to: x 1 x 2 2 x 1, x 2 0 x 1 ( DEGI) Operations Research 2012/ / 103

16 Linear Programming Graphical resolution (Without admissible solution) x 2 Objective: max Z Subject to: x 1 x 2 5 x 1 + 2x 2 8 x 1, x 2 0 x 1 - x 2 = -5 x 1 + 2x 2 = 8 x 1 ( DEGI) Operations Research 2012/ / 103

17 Mathematical Programming We consider only very special Mathematical Programming Models: All the variables take values in R or in Z. There is only one objective to minimize or to maximize. The objective and the constraints are linear functions of the decision variables. Linear Programming Models if all the variables take values in R. Integer Programming Models if all the variables take values in Z. ( DEGI) Operations Research 2012/ / 103

18 Mathematical Programming: fundamentals Objective: min f (X ) Subject to: g i (X ) 0 i {1,...,m} h i (X ) = 0 i {1,...,l} Admissible set All the points S R n that satisfy the constraints. Admissible solution Each X S is an admissible solution. Optimal solution X S f (X ) f (X ) X S ( DEGI) Operations Research 2012/ / 103

19 Mathematical Programming Non-linear programming example Curvas de nível de f x 2 Conjunto admissível (3, 6) Objective: min f (X ) = (x 1 3) 2 + (x 2 6) 2 Subject to: x1 2 + x 2 0 x 1 + x 2 6 x 1, x 2 0 Optimal solution: x 1 = 3 2, x 2 = 9 2, f = 9 2 x 1 x 2 = x 1 2 x 1 + x 2 = 6 ( DEGI) Operations Research 2012/ / 103

20 Simplex Method based on: Bradley, Hax, and Magnanti; Applied Mathematical Programming, Addison-Wesley, 1977 ( DEGI) Operations Research 2012/ / 103

21 Canonical Form Simplex Method Learning Objectives Ability to transform any linear program to a canonical form: with nonnegative variables by replacing each decision variable unconstrained in sign by a difference between two nonnegative variables. This replacement applies to all equations including the objective function. with equality constraints by changing inequalities to equalities by the introduction of slack and surplus variables. For inequalities, let the nonnegative surplus variable represent the amount by which the lefthand side exceeds the righthand side; for inequalities, let the nonnegative slack variable represent the amount by which the righthand side exceeds the lefthand side. with non-negative righthand-side coefficients by multiplying equations with a negative righthand side coefficient by -1. with one basic variable isolated in each constraint by adding a (nonnegative) artificial variable to any equation that does not have an isolated variable readily apparent, and construct the BIG M objective function. Simplex Method Ability to use the simplex method in tableau form to solve a linear program represented in the Canonical Form, i.e.: Ability to determine the starting point to initiate the simplex method. Ability to write the linear program in tableau form. Ability to use the improvement mechanism for moving from a point to another point with a better value of the objective function. Improvement Criterion In a maximization (minimization) problem, choose the nonbasic variable that has the most positive (negative) coefficient in the objective function of a canonical form. If that variable has a positive coefficient in some constraint, then a new basic feasible solution may be obtained by pivoting Ratio and Pivoting Criterion When improving a given canonical form by introducing variable x s into the basis, pivot in a constraint UNIVERSIDADE that DO PORTO gives the minimum ratio of righthand-side coefficient to the corresponding x s coefficient. Compute these ratios only for constraints that have a positive coefficient for x s. ( Ability to DEGI) detect termination criteria Operations to indicate Research when a solution has been obtained. 2012/ / 103

22 Simplex method: A systematic procedure for solving linear programs The method proceeds by moving from one feasible solution to another, at each step improving the value of the objective function and terminates after a finite number of such transitions. Two important characteristics of the simplex method: The method is robust it solves any linear program; it detects redundant constraints in the problem formulation; it identifies instances when the objective value is unbounded over the feasible region; it solves problems with one or more optimal solutions; the method is also self-initiating: it uses itself either to generate an appropriate feasible solution, as required, to start the method, or to show that the problem has no feasible solution. The method provides much more than just optimal solutions.. it indicates how the optimal solution varies as a function of the problem data (cost coefficients, constraint coefficients, and righthand-side data). it gives information intimately related with a linear program called the dual of the given problem: the simplex method automatically solves this dual problem along with the given problem. ( DEGI) Operations Research 2012/ / 103

23 Production Planning at BA Linear Programming Model Objective: max Z = 50x VP + 60x A Subject to: 50x VP + 30x A 2000 (time in the oven) 6x VP + 5x A 300 (space in the warehouse) 3x VP + 5x A 200 (capacity in the decoration sector) x VP, x A 0 Adding slack variables s 1, s 2 and s 3 : Objective: max 50x VP + 60x A Subject to: 50x VP +30x A +s 1 = x VP +5x A +s 2 = 300 3x VP +5x A +s 3 = 200 x VP, x A, s 1, s 2, s 3 0 ( DEGI) Operations Research 2012/ / 103

24 80 X A 70 Objective: max Z = 50x VP + 60x A X VP + 30X A = 2000 Subject to: 50x VP +30x A +s 1 = x VP +5x A +s 2 = 300 3x VP +5x A +s 3 = 200 x VP, x A, s 1, s 2, s 3 0 Basic (admissible) solution: Non-basic variables: Basic variables: { xvp = 0 x A = 0 Z = 0 s 1 = 2000 s 2 = 300 s 3 = Z = 50X VP + 60X A 3X VP + 5X A = X VP + 5X A = 300 Objective: maximize Z; choose the variable with the most positive coeficient in the objective function: x A s 3 is the first variable that turns zero when x A grows. x A grows from 0 until 40, a value such that s 3 = 0; s 1 and s 2 remain 0 X VP ( DEGI) Operations Research 2012/ / 103

25 80 X A Objective: max Z = 50x VP +60x A = 50x VP +60( x VP 1 5 x d ) = x VP 12s 3 Subject to: 32x VP +s 1 6s 3 = 800 3x VP +s 2 s 3 = x VP +x A s3 = 40 x VP, x A, s 1, s 2, s 3 0 Basic (admissible) solution: Non-basic variables: Basic variables: { xvp = 0 s 3 = 0 s 1 = 800 s 2 = 100 x A = 40 Z = X VP + 30X A = 2000 Z = 50X VP + 60X A 3X VP + 5X A = X VP + 5X A = 300 Objective: maximize Z; choose the variable with maximum coeficient in the objective function: x VP s 1 is the first variable that turns zero when x VP grows. x VP grows from 0 until 25, a value such that s 1 = 0; x A and s 2 remain 0 X VP ( DEGI) Operations Research 2012/ / 103

26 Objective: max Z = x VP 12s 3 = ( s s3) 12s3 = s1 8 s3 Subject to: x VP s s3 = s1 +s s3 = 25 +x A s1 80 s3 = 25 x VP, x A, s 1, s 2, s 3 0 Basic (admissible) solution: Non-basic variables: Basic variables: { s 1 = 0 s 3 = 0 x VP = 25 s 2 = 25 x A = 25 X A X VP + 30X A = 2000 Z = 50X VP + 60X A 3X VP + 5X A = X VP + 5X A = 300 This solution is optimal because an increase in s 1 or s 3 reduction of Z. X VP Z = 2750 ( DEGI) Operations Research 2012/ / 103

27 Simplex Method Steps of the algorithm Initial Basic Admissible Solution (BAS) : Obtain an initial Basic Admissible Solution (BAS). BAS is optimal? 2 Maximization if all the marginal costs are negative then the solution is optimal; Minimization if all the marginal costs are positive then the solution is optimal. If the BAS is not optimal then ITERATE. ITERATE : 1 Choose the variable that enters the basis: 3 Maximization the non-basic variable with the most positive marginal cost. Minimization the non-basic variable with the most negative marginal cost. 2 Choose the variable that enters the basis: divide the RHS of all the equations by the positive coefficients of the variable that will enter the basis; the variable that will leave the basis will be the basic variable of the equation with the lowest ratio; if all the coefficients of the variable that will enter the basis are negative or zero then the solution is non-limited. 3 Change the lines in the tableau in order to obtain coeficient equal to one for the pivot element and zero for all the other elements of the column, including the objective function. 4 BAS is optimal? 2 There are alternative optimal solutions when, in the optimal solution, one non-basic variable has a zero marginal cost. 3 If there is more than one variable that fulfills the criterion choose one arbitrarilly. ( DEGI) Operations Research 2012/ / 103

28 Production Planning at BA Simplex method Objective: max Z = 50x VP + 60x A Subject to: 50x VP + 30x A x VP + 5x A 300 3x VP + 5x A 200 x VP, x A 0 Objective: max Z = 50x VP + 60x A Subject to: 50x VP + 30x A + s 1 = x VP + 5x A + s 2 = 300 3x VP + 5x A + s 3 = 200 x VP, x A, s 1, s 2, s 3 0 ( DEGI) Operations Research 2012/ / 103

29 Production Planning at BA Simplex method x VP x A s 1 s 2 s 3 s s s Z x VP x A s 1 s 2 s s s x A Z x VP x A s 1 s 2 s 3 5 x VP s x A Z x A enters the basis because: max(50, 60) = 60 s 3 leaves the basis because: min( , 300 5, ) = x VP enters the basis because: max(14) = 14 s 1 leaves the basis because: min( , 100 3, 40 3 ) = All the coefficients of the objective function are 0. Optimal solution: (x VP, x A ) = (25, 25) and Z = 2750 What does s 2 = 25 mean? ( DEGI) Operations Research 2012/ / 103

30 Finding an initial Basic Admissible Solution (BAS) An important condition for the Simplex method is the availability of an initial Basic Admissible Solution in the canonical form. Sometimes this initial BAS is not evident or it may evn not exist (and sometimes there is no Basic Admissible Solution!). Solutions: Trial and error solve the system for different sets of variables, reduce it to the canonical form and test if the resulting solution is admissible. Using artificial variables. ( DEGI) Operations Research 2012/ / 103

31 Using artificial variables max n j=1 c j x j 1. Convert the LP problem to a normalized form. Subject to: n a ij x j = b i i {1,...,m} j=1 x j 0 j {1,...,n} max n j=1 c j x j 2. Examine each constraint and verify if there is some variable that can be basic for that constraint. If it does not exist, then sum an artificial variable y i, y i 0. Obs: The articial variables do not have any meaning in the original problem, that is why they are called artificial. They are only used to built an initial basis for the problem. Subject to: n a ij x j = b i i {1,...,k} j=1 n a ij x j + y i = b i i {k+1,...,m} j=1 x j 0 j {1,...,n} y i 0 i {k+1,...,m} ( DEGI) Operations Research 2012/ / 103

32 3. The artificial problem will only be equivalent to the original one if all the artificial variables have value zero. Objective all the artificial variables must leave the basis. 1 Two-fase method 2 Big M method ( DEGI) Operations Research 2012/ / 103

33 Big M method Assign the artificial variables a very high cost (M) (minimization problem) in the objective function. The simplex method will take care, by improving the objective function, to expel the arficicial variables from the basis artificial variables equal to zero. Subject to: min n j=1 c j x j + m i=k+1 My i n a ij x j = b i i {1,...,k} j=1 n a ij x j + y i = b i i {k+1,...,m} j=1 x j 0 j {1,...,n} y i 0 i {k+1,...,m} ( DEGI) Operations Research 2012/ / 103

34 Big M method example Objective: min Z = 3x 1 + x 2 + x 3 Subject to: x 1 2x 2 + x 3 + s 1 = 11 4x 1 + x 2 + 2x 3 s 2 = 3 2x 1 + x 3 = 1 x 1, x 2, x 3, s 1, s 2 0 Objective: min Z = 3x 1 + x 2 + x 3 + My 1 + My 2 Subject to: x 1 2x 2 + x 3 + s 1 = 11 4x 1 + x 2 + 2x 3 s 2 + y 1 = 3 2x 1 + x 3 + y 2 = 1 x 1, x 2, x 3, s 1, s 2, y 1, y 2 0 ( DEGI) Operations Research 2012/ / 103

35 Z = 3x 1 + x 2 + x 3 + M(3 + 4x 1 x 2 2x 3 + s 2 ) + M(1 + 2x 1 x 3 ) = 4M + ( 3 + 6M)x 1 + (1 M)x 2 + (1 3M)x 3 + Ms 2 x 1 x 2 x 3 s 1 s 2 y 1 y 2 s y y Z M M 3M 0 M 0 0 4M x 1 x 2 x 3 s 1 s 2 y 1 y 2 s y x Z M 0 0 M 0 3M M ( DEGI) Operations Research 2012/ / 103

36 x 1 x 2 x 3 s 1 s 2 y 1 y 2 s x x Z M M 0 x 1 x 2 x 3 s 1 s 2 1 x x x Z Optimal solution: (x 1, x 2, x 3, s 1, s 2 ) = (4, 1, 9, 0, 0) with Z = 2 ( DEGI) Operations Research 2012/ / 103

37 Some remarks The artificial variables are only used to serve as basic variables in a given equation. Once replaced in the base by the original variables, the artificial variables can be eliminated from the simplex tableau (eliminating the respective columns). If, in the optimal simplex tableau, some artificial variables have a value > 0, it means that the original problem has no admissible solution, and therefore it is an impossible problem. ( DEGI) Operations Research 2012/ / 103

38 Sensitivity Analysis based on: Bradley, Hax, and Magnanti; Applied Mathematical Programming, Addison-Wesley, 1977 Chapter 3, sections 3.1, 3.2, 3.3, 3.4 and 3.5 ( DEGI) Operations Research 2012/ / 103

39 Sensitivity Analysis We have already been introduced to sensitivity analysis via the geometry of a simple example. We saw that the values of the decision variables and those of the slack and surplus variables remain unchanged even though some coefficients in the objective function are varied. We also saw that varying the righthand-side value for a particular constraint alters the optimal value of the objective function in a way that allows us to impute a per-unit value, or shadow price, to that constraint. These shadow prices and the shadow prices on the implicit nonnegativity constraints, called reduced costs, remain unchanged even though some of the righthand-side values are varied. Since there is always some uncertainty in the data, it is useful to know over what range and under what conditions the components of a particular solution remain unchanged. Further, the sensitivity of a solution to changes in the data gives us insight into possible technological improvements in the process being modeled. For instance, it might be that the available resources are not balanced properly and the primary issue is not to resolve the most effective allocation of these resources, but to investigate what additional resources should be acquired to eliminate possible bottlenecks. Sensitivity analysis provides an invaluable tool for addressing such issues. ( DEGI) Operations Research 2012/ / 103

40 Sensitivity Analysis Learning Objectives Define shadow price of a constraint. Calculate shadow prices using only the initial and final simplex tableaux. Define reduced cost of a decision variable (activity). Calculate reduced costs using only the initial and final simplex tableaux. Price out an activity. Calculate the reduced cost of a variable using the shadow prices and the problem data. Calculate shadow prices using the problem data and the optimal basis. Determine how much the objective function coefficients can vary without changing the values of the decision variables in the optimal solution. Determine which variables will enter and leave the basis when the new cost coefficient reaches either of the extreme values of the range. Determine how much the right-hand-side values can vary such that the variables that constitute the basis remain the same. Determine which variables will enter and leave the basis when the new right-hand-side reaches either of the extreme values of the range. Identify alternative optimal solutions when they exist. Identify alternative optimal shadow prices when they exist. Analyze Solver (Microsoft Excel) sensitivity report and identify there shadow prices, reduced costs, and variation ranges for objective coefficients and right-hand-sides. ( DEGI) Operations Research 2012/ / 103

41 Production Planning at BA with Favaios Bottles The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mould injection. BA recently won two new clients and intends to assign this production to one of its furnaces. The orders were of bottles of Sandeman Ruby Port (75cl), one liter bottles of olive oil Oliveira da Serra and one liter bottles of Favaios Moscatel. The customers buy all the bottles that BA can produce. Due to differences in the production (number of cavities and different cycle times), each batch of portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30 hours and a batch of favaios takes 50 hours to produce. There are a total of 2000 hours available at the oven. Moreover, there are restrictions in the storage capacity of the bottles prior to their dispatch. The warehouse has 300m 3 available and each batch of portwine bottles occupies 6m 3 of warehouse space, each batch of bottles of olive oil holds 5m 3 and each batch of favaios holds 6m 3. Finally we must also consider the capacity available in the decoration sector (e.g. labels, packaging), which is 200 hours. The portwine bottles spend 3 hours per batch, the olive oil bottles 5 hours per batch and the favaios bottles also 5 hours per batch. Barbosa & Almeida wants to maximize the profit from these two orders, knowing that the profit per batch is 50e 60e and 20e, respectively. ( DEGI) Operations Research 2012/ / 103

42 Production Planning at BA with Favaios Bottles Decision variables x VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce; x A Number of batches of Azeite Oliveira da Serra bottles to produce; x F Number of batches of Favaios bottles to produce. Objective: max Z = 50x VP + 60x A + 20x F Subject to: 50x VP +30x A +50x F 2000 (time in the oven) 6x VP +5x A +6x F 300 (space in the warehouse) 3x VP +5x A +5x F 200 (capacity in the decoration sector) x VP, x A, x F 0 In this model x VP, x A and x F may not be integer. ( DEGI) Operations Research 2012/ / 103

43 Production Planning at BA Producing Favaios Bottles x VP x A x F s 1 s 2 s 3 s s s Z x VP x A x F s 1 s 2 s 3 s s x A Z x VP x A x F s 1 s 2 s x VP s x A Z x A enters the basis because: max(50, 60, 20) = 60 s 3 leaves the basis because: min( , 300 5, ) = x VP enters the basis because: max(14) = 14 s 1 leaves the basis because: min( , 100 3, 40 3 ) = = 25 5 All the coefficients of the objective function are 0. Optimal solution: (x VP, x A, x F ) = (25, 25, 0) and Z = 2750 ( DEGI) Operations Research 2012/ / 103

44 Some remarks about the sensitivity analysis We wish to analyze the effect on the optimal solution of changing various elements of the problem data without re-solving the linear program or having to remember any of the intermediate tableaux generated in solving the problem by the simplex method. The type of results that can be derived in this way are conservative, in the sense that they provide sensitivity analysis for changes in the problem data small enough so that the same decision variables remain basic, but not for larger changes in the data. ( DEGI) Operations Research 2012/ / 103

45 Shadow Prices Reduced Costs ( DEGI) Operations Research 2012/ / 103

46 Production Planning at BA Producing Favaios Bottles x VP x A x F s 1 s 2 s x VP s x A Z The complete optimal solution for the problem is (x VP, x A, x F, s 1, s 2, s 3 ) = (25, 25, 0, 0, 25, 0) and Z = 2750 The binding constraints in this optimal solution are therefore: the first one (time in the oven, s 1 = 0); the third one (capacity in the decoration sector, s 3 = 0); and x F = 0. 50x VP +30x A +50x F = x VP +5x A +5x F = 200 x F = 0 (x VP, x A, x F ) = (25, 25, 0) Z = 2750 ( DEGI) Operations Research 2012/ / 103

47 Shadow Price Definition The shadow price associated with a particular constraint is the change in the optimal value of the objective function per unit increase in the righthand-side value of that constraint, all other problem data remaining unchanged. ( DEGI) Operations Research 2012/ / 103

48 Production Planning at BA Producing Favaios Bottles Shadow Prices What will be the increase of the profit per unit increase in the oven time? 50x VP +30x A +50x F = x VP +5x A +5x F = 200 x F = 0 50x VP +30x A = x VP +5x A = 200 x F = 0 50x VP +30x A = x VP +30x A = 1200 x F = 0 x VP = x A = x F = 0 (x VP, x A, x F ) = ( , , 0) Z = = Profit increase: = 7 16 x VP x A x F s 1 s 2 s 3 x VP s x A Z ( DEGI) Operations Research 2012/ / 103

49 Production Planning at BA Producing Favaios Bottles Shadow Prices What will be the increase of the profit per unit increase in the decoration capacity? 50x VP +30x A +50x F = x VP +5x A +5x F = 201 x F = 0 50x VP +30x A = x VP +5x A = 201 x F = 0 50x VP +30x A = x VP +30x A = 1206 x F = 0 x VP = (x VP, x A, x F ) = ( , , 0) Z = = Profit increase: = x VP x A x F s 1 s 2 s 3 x VP s x A x A = Z x F = 0 ( DEGI) Operations Research 2012/ / 103

50 Each constraint of a problem has as shadow price. The shadow price for a constraint is the negative of the coefficient of the appropriate slack (or artificial) variable in the objective function of the final tableau. ( DEGI) Operations Research 2012/ / 103

51 Reduced Cost Definition The reduced cost associated with the nonnegativity constraint for each variable is the shadow price of that constraint (i.e., the corresponding change in the objective function per unit increase in the lower bound of the variable). ( DEGI) Operations Research 2012/ / 103

52 Production Planning at BA Producing Favaios Bottles Reduced costs What will be the reduction of the profit if we must produce at least one batch of favaios bottles? 50x VP +30x A +50x F = x VP +5x A +5x F = 200 x F = 1 50x VP +30x A +50 = x VP +5x A +5 = 200 x F = 1 50x VP +30x A = x VP +5x A = 195 x F = 1 x VP = 195/8 x A = 195/8 x F = 1 (x VP, x A, x F ) = ( 195 8, 195 8, 1) Z = Profit reduction: , 25 = 48, 75 = = 2701, 25 x VP x A x F s 1 s 2 s 3 x VP s x A Z ( DEGI) Operations Research 2012/ / 103

53 Production Planning at BA Producing Favaios Bottles Reduced costs What will be the reduction of the profit if we must produce at least one batch of favaios bottles? x VP x A x F s 1 s 2 s 3 s s s Z x VP x A x F s 1 s 2 s x VP s x A Z We can compute the reduced cost of x F in a different way by computing the oportunity cost for producing 1 batch of favaios bottles: = = As the revenue of 1 batch of favaios bottles is 20, the profit will change by = ( DEGI) Operations Research 2012/ / 103

54 General Discussion Fundamental relationship between shadow prices, reduced costs, and the problem data. Shadow x 1... x n s 1... s m price s 1 a a 1n b 1 y s m a m1... a mn b m y m Z c 1... c n At the final tableau: Z c 1... c n c n+1... c n+m z 0 =... = y 1... y m Recall that, at each iteration of the simplex method, the objective function is transformed by subtracting from it a multiple of the row in which the pivot was performed. Consequently, the final form of the objective function could be obtained by subtracting multiples of the original constraints from the original objective function. Consider first the final objective coefficients associated with the original basic variables: s 1,..., s m. Let π 1,..., π m be the multiples of each row that are subtracted from the original objective function to obtain its final form. Since s i appears only in the ith constraint and has a +1 coefficient we should have c n+i = 0 1π i = π i = y i. Thus the shadow prices are the multiples π i. Since these multiples can be used to obtain every objective coefficient in the final form, the reduced cost c j of variable x j is given by c j = c j m i=1 a ij y i (j = 1... n). Since c j = 0 for the m basic variables: 0 = c j m i=1 a ij y i (for j basic). This is a system of m equations in m unknowns that uniquely determines the values of the shadow prices y i. The current value of the objective function is z 0 = m i=1 b i y i, z 0 = m i=1 b i y i... ( DEGI) Operations Research 2012/ / 103

55 Variation in the objective coefficients How much the objective-function coefficients can vary without changing the values of the decision variables in the optimal solution? We will make the changes one at a time, holding all other coefficients and righthand-side values constant. ( DEGI) Operations Research 2012/ / 103

56 Production Planning at BA Producing Favaios Bottles Variation in the objective coefficient of x F Suppose that we increase the objective function coefficient of x F in the original problem formulation by c F : In applying the simplex method, multiples of the rows were subtracted from the objective function to yield the final system of equations. Therefore, the objective function in the final tableau will remain unchanged except for the addition of c F x F. x VP x A x F s 1 s 2 s 3 s s s Z c F x VP x A x F s 1 s 2 s x VP s x A Z c F x F will become a candidate to enter the basis, only when its objective-function coefficient is positive. The optimal solution remains unchanged as long as: c F 0 c F ( DEGI) Operations Research 2012/ / 103

57 Production Planning at BA Producing Favaios Bottles Variation in the objective coefficient of x VP x VP x A x F s 1 s 2 s 3 s s s Z c VP x VP x A x F s 1 s 2 s x VP s x A Z c VP x VP x A x F s 1 s 2 s x VP s x A Z c VP 1 32 c VP c VP 25 c VP The current solution remains unchanged while: c VP 0 ( c VP 78) c VP 0 ( c VP 14) c VP 0 ( c VP 50) Taking the most limiting inequalities, the bounds on c VP are: 14 c VP c new VP c new VP 100 ( DEGI) Operations Research 2012/ / 103

58 General Discussion Ranges of the cost coefficients in the optimal solution. Shadow x 1... x n s 1... s m price s 1 a a 1n b 1 y s m a m1... a mn b m y m Z c 1... c n At the final tableau: Z c 1... c n c n+1... c n+m z 0 If x j is a non-basic variable at the final tableau with objective function coefficient c j that was changed to c new j = c j + c j, then the current solution remains unchanged so long as the new reduced cost c new j remains nonnegative, that is, c new j. = c j + c j m i=1 a ij y i = c j + c j 0. The range on the variation of the objective-function coefficient of a nonbasic variable is then given by: < c j < c j ; ( < c j + c j = c new j < c j c j ). If x r is a basic variable at the final tableau with objective function coefficient c r that was changed to c new r = c r + c r, then c new r = c r + c r m i=1 a ij y i = c r + c r. Since x r is a basic variable, c r = 0. So, to recover a canonical form with c new r = 0, we subtract c r times the rth constraint in the final tableau from the final form of the objective function, obtaining new reduced costs for all nonbasic variables: c new j = c j a rj c r, where a rj is the coefficient of variable x j in the rth constraint in the final tableau. For all basic variables c new j = 0 and the current basis remains optimal if c new j 0. The range on the variation of the objective-function coefficient is: Max j { cj a rj a rj > 0 } c r Min j { cj a rj a rj < 0 }.. ( DEGI) Operations Research 2012/ / 103

59 Variations on the righthand-side values ( DEGI) Operations Research 2012/ / 103

60 Production Planning at BA Producing Favaios Bottles Variations on the righthand-side values Varying the righthand-side value of a constraint that is non-binding in the optimal solution: the warehouse constraint. x VP x A x F s 1 s 2 s 3 s s b 2 s Z x VP x A x F s 1 s 2 s x VP s b 2 5 x A Z We add an amount b 2 to the righthand side of the warehouse constraint. s 2 was basic in the initial tableau and is basic in the optimal tableau s 2 = 300+ b 2 s 2 = 25 + b 2 In order to keep the solution feasible: 25 + b 2 0 b 2 25 b new 2 = b ( DEGI) Operations Research 2012/ / 103

61 Production Planning at BA Producing Favaios Bottles Variations on the righthand-side values Varying the righthand-side value of a constraint that is binding in the optimal solution: the decoration constraint. We add an amount b 3 to the righthand side of the decoration constraint. This is equivalent to decreasing the value of the slack variable s 3 by b 3, that is substituting s 3 by s 3 b 3 in the original problem formulation. s 3, which is zero in the final solution, is changed to s 3 = b 3 x VP = b3 0 that is b3 3 s 2 = b3 0 that is b3 7 x A = b3 0 that is b b b new x VP x A x F s 1 s 2 s 3 s s s b 3 Z x VP x A x F s 1 s 2 s 3 x VP b 3 s b 3 x A b 3 Z b 3 For b new 3 < ( b 3 < 80) the basic variable x A becomes negative. What variable should enter the basis to take its place? In order for the new basis to be an optimal solution, the entering variable must be chosen so that the reduced costs are not allowed to become positive. ( DEGI) Operations Research 2012/ / 103

62 x VP x A x F s 1 s 2 s 3 x VP b 3 s b 3 x A b 3 Z b 3 The final tableau for b new 3 = 120 ( b 3 = 80) will be: x VP x A x F s 1 s 2 s 3 x VP s x A Z For b new 3 < 120 the basic variable x A becomes negative. What variable should enter the basis to take its place? In order for the new basis to be an optimal solution, the entering variable must be chosen so that the reduced costs are not allowed to become positive. Regardless of which variable enters the basis, the entering variable will be isolated in row 3 of the final tableau to replace x A, which leaves the basis. To isolate the entering variable, we must perform a pivot operation, and a multiple, say t, of row 3 in the final tableau will be subtracted from the objective-function row. x VP x A x F s 1 s 2 s 3 x VP s x A Z t t 5 +t t 16 5 t 0 ( DEGI) Operations Research 2012/ / 103

63 x VP x A x F s 1 s 2 s x VP s x A Z t t 5 8 +t t 5 16 t t t t 5 16 t 0 that is t that is t 5 0 that is t that is t 30 0 t 70 3 Since the coefficient of s 1 is most constraining on t, s 1 will enter the basis. Note that the range on the righthand-side value and the variable transitions that would occur if that range were exceeded by a small amount are easily computed. However, the pivot operation actually introducing s 1 into the basis and eliminating x A need not be performed. ( DEGI) Operations Research 2012/ / 103

64 General Discussion Variations in the Righthand-Side values. x 1... x n s 1... s m s 1 a a 1n b s m a m1... a mn b m Z c 1... c n At the final tableau: x 1... x n s 1... s m s 1 a a 1n β β 1m b 1... s m a m1... a mn β m1... β mm b m Z c 1... c n c n+1... c n+m z 0 As this is a canonical form, a ij and β ij in the final tableau will be structured so that one basic variable is isolated in each contraint. We can change the coefficient b k in the kth righthand-side by b k with all the other data held fixed, simply by substituting s k b k for s k in the initial tableau. To see how this change affects the updated righthand-side coefficients, we make the same substitution in the final tableau. The terms β ik s k become β ik (s k b k ) = β ik s k β ik b k. As β ik b k is a constant, we move it to the righthand side to give modified righthand-side values b i + β ik b k for (i = 1, 2,..., m).... ( DEGI) Operations Research 2012/ / 103

65 As long as all of these values are nonnegative, the basis specified by the final tableau remains optimal, since the reduced costs have not been changed. Consequently, the current basis is optimal { whenever b i } + β ik b k 0 for { (i = 1, 2,..., } m) or equivalently, Max bi i β β ik > 0 b k Min bi i β ik β ik < 0 ik When b k reaches either its upper or lower bound any further increase (or decrease) in its value makes one of the updated righthand sides, say b r + β rk b k negative. At this point the basic variable in row r leaves the basis, and we must pivot in row r of the final tableau to find the variable to be introduced in its place. Since pivoting subtracts a multiple t of the pivot row from the objective equation, the new objective equation has coefficients c j ta rj (j = 1, 2,..., n). For the basis to be optimal, each of these coefficients must be nonpositive. Since c j = 0 for the basic variable being dropped and since its coefficient in costraint r is a rj = 1, we must have t 0. For any nonnegative t, the updated coefficient c j ta rj for any other variable remains nonpositive if a rj 0. { Consequently we need only to consider a rj < 0, and t is given by: cj } t = Min j a a rj < 0 rj The index u giving this minimum has c u ta ru = 0, and the corresponding variable x u can become the new basic variable in row r by pivoting on a ru. Note that the pivot is made on a negative coefficient. ( DEGI) Operations Research 2012/ / 103

66 Alternative optimal solutions As in the case of the objective function and righthand-side ranges, the final tableau of the linear program tells us something conservative about the possibility of alternative optimal solutions. If all reduced costs of the nonbasic variables are strictly negative (positive) in a maximization (minimization) problem, then there is no alternative optimal solution, because introducing any variable into the basis at a positive level would reduce (increase) the value of the objective function. If one or more of the reduced costs are zero, there may exist alternative optimal solutions. ( DEGI) Operations Research 2012/ / 103

67 Alternative shadow prices Independent of the question of whether or not alternative optimal solutions exist in the sense that different values of the decision variables yield the same optimal value of the objective function, there may exist alternative optimal shadow prices. If all righthand-side values in the final tableau are positive, then there do not exist alternative optimal shadow prices. If one or more of these values are zero, then there may exist alternative optimal shadow prices. ( DEGI) Operations Research 2012/ / 103

68 Alternative Shadow Prices Example x VP x A x F s 1 s 2 s 3 x VP s x A Z Since the righthand-side value in row 3 is zero, it is possible to take x A (basic variable in row 3) out of the basis as long as there is a variable to introduce into the basis. The candidates to be introduced into the basis are x F, s 1 and s 3, the one that corresponds to: Min j { cj a rj a rj < 0 } { = Min } s 1 will enter the basis. ( DEGI) Operations Research 2012/ / 103

69 Duality in Linear Programming based on: Bradley, Hax, and Magnanti; Applied Mathematical Programming, Addison-Wesley, 1977 Chapter 4, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6 and 4.7 ( DEGI) Operations Research 2012/ / 103

70 Duality in Linear Programming In the preceding chapter on sensitivity analysis, we saw that the shadow-price interpretation of the optimal simplex multipliers is a very useful concept. First, these shadow prices give us directly the marginal worth of an additional unit of any of the resources. Second, when an activity is priced out using these shadow prices, the opportunity cost of allocating resources to that activity relative to other activities is determined. Duality in linear programming is essentially a unifying theory that develops the relationships between a given linear program and another related linear program stated in terms of variables with this shadow-price interpretation. The importance of duality is twofold: First, fully understanding the shadow-price interpretation of the optimal simplex multipliers can prove very useful in understanding the implications of a particular linear-programming model. Second, it is often possible to solve the related linear program with the shadow prices as the variables in place of, or in conjunction with, the original linear program, thereby taking advantage of some computational efficiencies. ( DEGI) Operations Research 2012/ / 103

71 Duality in Linear Programming Learning Objectives Write the dual problem of a linear programming problem. Understand the: weak duality property, optimality property, unboundedness property, strong duality property. Verify that the complementary slackness conditions hold for a given problem. Apply the dual simplex method. Apply the parametric primal-dual algorithm. ( DEGI) Operations Research 2012/ / 103

72 A Preview of Duality ( DEGI) Operations Research 2012/ / 103

73 Production Planning at BA The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mould injection. BA recently won two new clients and intends to assign this production to one of its furnaces. The orders were of bottles of Sandeman Ruby Port (75cl), one liter bottles of olive oil Oliveira da Serra and one liter bottles of Favaios Moscatel. The customers buy all the bottles that BA can produce. Due to differences in the production (number of cavities and different cycle times), each batch of portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30 hours and a batch of favaios takes 50 hours to produce. There are a total of 2000 hours available at the oven. Moreover we must also consider the capacity available in the decoration sector (e.g. labels, packaging), which is 200 hours. The portwine bottles spend 3 hours per batch, the olive oil bottles 5 hours per batch and the favaios bottles also 5 hours per batch. Barbosa & Almeida wants to maximize the profit from these two orders, knowing that the profit per batch is 50e 60e and 20e, respectively. ( DEGI) Operations Research 2012/ / 103

74 Production Planning at BA Decision variables x VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce; x A Number of batches of Azeite Oliveira da Serra bottles to produce; x F Number of batches of Favaios bottles to produce. Objective: max Z = 50x VP + 60x A + 20x F Subject to: 50x VP +30x A +50x F 2000 (time in the oven) 3x VP +5x A +5x F 200 (capacity in the decoration sector) x VP, x A, x F 0 In this model x VP, x A and x F may not be integer. ( DEGI) Operations Research 2012/ / 103

75 Production Planning at BA Shadow prices x VP x A x F s O s D s O s D Z At the final tableau: x VP x A x F s O s D x VP 1 0 x A The shadow prices, y O for the oven capacity and y D for the decoration capacity can be determined from the final tableau as the negative of the reduced costs associated with the slack variables s O and s D. Thus these shadow prices are y O = 7 16 and y D = Z ( DEGI) Operations Research 2012/ / 103

76 Economic properties of the shadow prices associated with the resources Shadow x 1... x n s 1... s m price s 1 a a 1n b 1 y s m a m1... a mn b m y m Z c 1... c n At the final tableau: Z c 1... c n c n+1... c n+m z 0 Reduced costs in terms of shadow prices:. c j = c j m i=1 a ij y i (j = 1... n) a ij is the amount of resource i used per unit of activity j, and y i is the imputed value of that resource. The term m i=1 a ij y i is the total value of the resources used per unit of activity j and is thus the marginal resource cost per unit of activity j. If we think of the objective coefficients c j as being marginal revenues, the reduced costs c j = c j m i=1 a ij y i (j = 1... n).. are simply net marginal revenues (i.e., marginal revenue minus marginal cost). ( DEGI) Operations Research 2012/ / 103

77 Production Planning at BA Reduced cost (net marginal revenue) for each type of bottle (decision variable) x VP x A x F s O s D s O s D Z At the final tableau: x VP x A x F s O s D x VP x A Z Shadow price for the oven capacity: y O = Shadow price for the decoration capacity: y D = Net marginal revenue for the production of portwine bottles: c VP = c VP m i=1 a ivpy i = 50 ( ) = 0 Net marginal revenue for the production of azeite bottles: c A = c A m i=1 a iay i = 60 ( ) = 0 Net marginal revenue for the production of favaios bottles: c F = c F m i=1 a if y i = 20 ( ) = For the basic variables the reduced costs are zero. The marginal revenue is equal to the marginal cost for these activities. For all nonbasic variables the reduced costs are negative. The marginal revenue is less than the marginal cost for these activities, so they should not be pursued. ( DEGI) Operations Research 2012/ / 103

78 Production Planning at BA Shadow prices (opportunity costs) associated with the consumption of the resources (constraints) x VP x A x F s O s D s O s D Z At the final tableau: x VP x A x F s O s D x VP x A Z If we value the total resources at the shadow prices, we find their value is exactly equal to the optimal value of the objective function: = 2750 ( DEGI) Operations Research 2012/ / 103

79 Production Planning at BA Can shadow prices be determined directly? For a maximization decision problem, the shadow prices must satisfy the requirement that marginal revenue be less than or equal to marginal cost for all activities. The shadow prices must be nonnegative since they are associated with less-than-or-equal-to constraints in a maximization decision problem. Imagine that BA does not own its productive capacity but has to rent it. In this case the shadow prices could be seen as rent rates. Objective: max Z = 50x VP + 60x A + 20x F Subject to: 50x VP + 30x A + 50x F x VP + 5x A + 5x F 200 x VP, x A, x F 0 Objective: min V = 2000y O + 200y D Subject to: 50y O + 3y D 50 30y O + 5y D 60 50y O + 5y D 20 y O, y D 0 ( DEGI) Operations Research 2012/ / 103

80 Production Planning at BA Primal and Dual Decision Problems PRIMAL Decision Problem: Objective: max Z = 50x VP + 60x A + 20x F Subject to: 50x VP + 30x A + 50x F 2000 Optimal solution: x VP = 25 x A = 25 x F = 0 Z = 2750 with shadow prices: y O = 7 16 y D = x VP + 5x A + 5x F 200 x VP, x A, x F 0 DUAL Decision Problem: Objective: min V = 2000y O + 200y D Subject to: Optimal solution: y O = 7 16 y D = V = 2750 with shadow prices: x VP = 25 x A = 25 x F = 0 50y O + 3y D 50 30y O + 5y D 60 50y O + 5y D 20 y O, y D 0 ( DEGI) Operations Research 2012/ / 103

81 Production Planning at BA Requirements of the optimality conditions of the Simplex Method The reduced costs of the basic variables are zero. If a decision variable of the primal is positive, then the corresponding constraint in the dual must hold with equality. The variables with negative reduced costs are nonbasic variables (zero valued) If a constraint holds as a strict inequality, then the corresponding decision variable must be zero. If a shadow price is positive, then the corresponding constraint must hold with equality. If x VP 0 then c VP = 50 (50ŷ O + 3ŷ D ) = 0 If x A 0 then c A = 60 (30ŷ O + 5ŷ D ) = 0 If c F = 20 (50ŷ O + 5ŷ D ) < 0 then x F = 0 If ŷ O > 0 then 50 x VP + 30 x A + 50 x F = 2000 If ŷ D > 0 then 3 x VP + 5 x A + 5 x F = 200 ( DEGI) Operations Research 2012/ / 103