OPERATIONS RESEARCH Michał Kulej Business Information Systems The development of the potential and academic programmes of Wrocław University of Technology Project co-financed by European Union within European Social Fund Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 1 / 41
Artificial Starting Solution The Big M Method The linear programming problem in which all constraints are ( ) with nonnegative right-hand sides offers a all-slack starting basic feasible solution. Problems with (=) and/or ( ) constraints need to use artificial variables to the beginning of simplex algorithm. There are two methods: the M-method(also called the Big M-method) and the Two-Phase method. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 2 / 41
Example Artificial Starting Solution The Big M Method max Z = 2x 1 + x 2 3x 3 x 1 + x 2 + x 3 6 2x 1 + x 2 = 14 x 1, x 2, x 3 0 After converting to standard form we have: max Z = 2x 1 + x 2 3x 3 x 1 + x 2 + x 3 s 1 = 6 2x 1 + x 2 = 14 x 1, x 2, x 3, s 1 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 3 / 41
Artificial Starting Solution Adding Artificial Variables The Big M Method The above system of equations is not in basic form - there is not a basic variable in each equation. If the equation i does not have a slack (or a variable that can play the role of a slack), an artificial variable, a i, is added to form a starting solution similar to the all-slack basic solution. However, because the artificial variables are not part of the original linear model, they are assigned a very high penalty in the objective function, thus forcing them (eventually) to equal zero in the optimal solution. In the considering example we add two artificial variables a 1 and a 2. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 4 / 41
Artificial Starting Solution The Big M Method Penalty Rule for Artificial Variables. Given M, a sufficiently large positive value (mathematically, M ), the objective coefficient of an artificial variable represents an appropriate penalty if: Artificial variable objective coefficient = { M, in maximization problems M, in minimization problems. So we have: max Z = 2x 1 + x 2 3x 3 Ma 1 Ma 2 a 1 +x 1 + x 2 + x 3 s 1 = 6 a 2 +2x 1 + x 2 = 14 x 1, x 2, x 3, s 1, a 1, a 2 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 5 / 41
Artificial Starting Solution The Big M Method Calculations The Big M Method First and second simplex tableau are as follows: c B BV a 1 a 2 x 1 x 2 x 3 s 1 Solution M a 1 1 0 1 1 1 1 6 6/1 = 6 M a 2 0 1 2 1 0 0 14 14/2 = 7 Z 0 0 3M 2 2M 1 M + 3 M 20M 2 x 1 1 0 1 1 1 1 6 M a 2 2 1 0 1 2 2 2 2/2 = 1 Z 3M + 2 0 0 M + 1 2M 1 2M 2 2M + 12 The optimal simplex tableau: c B BV a 1 a 2 x 1 x 2 x 3 s 1 Solution 1 1 2 x 1 0 2 1 2 0 0 7 1 0 s 1 1 2 0 1 2 1 1 1 Z M M + 1 0 0 3 0 14 Optimal solution is: x 1 = 7, s 1 = 1, x 2 = x 3 = 0 with Z = 14. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 6 / 41
Artificial Starting Solution The Big M Method No Feasible Solution - an Example The use of the penalty M will not force an artificial variable to zero level in the final simplex iteration if LPP does not have a feasible solution (i.e. the constraints are not consistent). In this case, the final simplex tableau will include at least one artificial variable at positive level. Solve the problem : The standard form: max Z = 2x 1 + 2x 2 6x 1 + 4x 2 24 x 1 5 x 1, x 2 0 max Z = 2x 1 + 2x 2 s 1 +6x 1 + 4x 2 = 24 x 1 s 2 = 5 x 1, x 2, s 1, s 2 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 7 / 41
Artificial Starting Solution The Big M Method Using The Big M Method to Solve the Problem It is enough to add only one artificial variable : max Z = 2x 1 + 2x 2 Ma 2 s 1 +6x 1 + 4x 2 = 24 a 2 +x 1 s 2 = 5 x 1, x 2, s 1, s 2 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 8 / 41
Artificial Starting Solution The Big M Method Calculations The Big M Method c B BV s 1 a 2 x 1 x 2 s 2 Solution 0 s 1 1 0 6 4 0 24 24/6 = 4 M a 2 0 1 1 0 1 5 5/1 = 5 Z 0 0 M 2 2 M 5M c B BV s 1 a 2 x 1 x 2 s 2 Solution 1 2 2 x 1 6 0 1 3 0 4 M a 2 1 6 1 0 2 3 1 1 1 Z 6 M + 1 2 3 0 0 3 M 2 3 M M + 8 The last simplex tableau is optimal - all Z - row coefficients are nonnegative. The artificial variable a 2 is basic variable with positive value so the problem is not consistent - it has no feasible solutions. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 9 / 41
Artificial Starting Solution The Two-Phase Method The Essence of The Two-Phase Method This method solves the linear programming problem in two phases: Phase I attempts to find a starting feasible solution and, if one is found, Phase II is used to solve the original problem. Phase I Put the problem in equation form, and add the necessary artificial variables to the constraints to get starting feasible basic solution. Next, find a basic solution of the resulting equations that, regardless of whether the problem is maximization or minimization, always minimizes the sum of the artificial variables. If minimum value of the sum is positive, the problem has no feasible solution, which ends the process. Otherwise, proceed to Phase II. Phase II Use the feasible solution from Phase I as starting feasible basic solution for original problem. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 10 / 41
Example Artificial Starting Solution The Two-Phase Method We use this method to solve the following problem: max Z = 4x 1 + x 2 3x 1 + x 2 = 3 4x 1 + 3x 2 6 x 1 + 2x 2 4 x 1, x 2 0 Using s 2 as a surplus in the second constraint and s 3 as a slack in the third constraint, the standard form of the problem is given as: max Z = 4x 1 + x 2 3x 1 + x 2 = 3 4x 1 + 3x 2 s 2 = 6 x 1 + 2x 2 + s 3 = 4 x 1, x 2, s 2, s 3 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 11 / 41
Artificial Starting Solution The Two-Phase Method The Model with Arificial Variables - Phase I The third equation has its basic variable (s 3 ) but the first and second do not. Thus we add the artificial variables a 1 and a 2 and we minimize the sum of the artificial variables a 1 + a 2. The resulting model is given as: min Z a = a 1 + a 2 3x 1 + x 2 +a 1 = 3 4x 1 + 3x 2 s 2 +a 2 = 6 x 1 + 2x 2 +s 3 = 4 x 1, x 2, s 2, s 3, a 1, a 2 0 Taking a 1, a 2, s 3 as a basic variables and using formula (2) for computing coefficients of Z a - row we obtain the first simplex tableau: c B BV x 1 x 2 s 2 a 1 a 2 s 3 Solution 1 a 1 3 1 0 1 0 0 3 1 a 2 4 3 1 0 1 0 6 0 s 3 1 2 0 0 0 1 4 Z a 7 4 1 0 0 0 9 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 12 / 41
Artificial Starting Solution The Two-Phase Method The Optimal Simplex Tableau of Phase I Now we use simplex algorithm for minimization problem and receive (after two iterations) the following optimum tableau: c B BV x 1 x 2 s 2 a 1 a 2 s 3 Solution 1 3 0 x 1 1 0 5 5 1 3 5 0 5 0 x 2 0 1 3 5 4 3 6 5 5 0 5 0 s 3 0 0 1 1 1 1 1 Z a 0 0 0 1 1 0 0 Because minimum Z a = 0. Phase I produces the basic feasible solution x 1 = 3 5, x 2 = 6 5, s 3 = 1. We can eliminate columns for a 1 and a 2 from tableau and move to Phase II. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 13 / 41
Phase II Artificial Starting Solution The Two-Phase Method After deleting the artificial columns, we write the original problem as max Z = 4x 1 + x 2 x 1 + 1 5 s 2 = 3 5 x 2 3 5 s 2 = 6 5 s 2 +s 3 = 1 x 1, x 2, s 2, s 3 0 Now we can begin the simplex algorithm with the following simplex tableau: c B BV x 1 x 2 s 2 s 3 Solution 4 x 1 1 0 1 3 5 0 5 1 x 2 0 1 3 5 0 6 5 0 s 2 0 0 1 1 1 1 18 Z 0 0 5 0 5 Because we are maximizing, this is the optimal tableau. The optimal solution is x 1 = 3 5, x 2 = 6 5 and Z = 18 5. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 14 / 41
Artificial Starting Solution The Two-Phase Method Removal of Artificial Variables after Phase I The removal of the artificial variables and their columns at the end of Phase I can take place only when they are all nonbasic. If one or more artificial variables are basic (at zero level) at the end of Phase I, then the following additional steps must be undertaken to remove them prior to start of Phase II. Step 1. Select a zero artificial variable to leave the basic solution (choosing the leaving variable) and designate its row as the pivot row. The entering variable can be any nonbasic variable with a nonzero(positive or negative) coefficient in the pivot row. Perform the associated simplex iteration. Step 2. Remove the column of the (just-leaving) artificial variable from the tableau. If all the zero artificial variables have been removed, go to Phase II. Otherwise, go back to Step 1. Remark All commercial packages use the two-phase method to solve linear programming problems. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 15 / 41
Sensitivity Analysis Changes in Objective Coefficients Changes in Objective Coefficients The sensitivity analysis is concerned with how changes in an parameters of the linear programming model affect the optimal solution. Let us consider the following problem (described in the basic form) of manufacturing two products W 1 and W 2 from two row materials S 1, S 2 : max Z = 3x 1 + 2x 2 [Maximizing the revenue] s 1 +2x 1 + x 2 = 100 [Limit on row material S 1 ] s 2 +x 1 + x 2 = 80 [Limit on row material S 2 ] s 3 +x 1 = 40 [Demand on W 1 ] x 1, x 2, s 1, s 2, s 3 0 The optimal solution of the model is x 1 = 20, x 2 = 60 with Z = 180. We want to determine the range of values of parameter (i.e. coefficient of objective function - unit price of product W 1 in our example ) over which the optimal solution remain unchanged. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 16 / 41
Sensitivity Analysis Changes in Objective Coefficients The Optimal Simplex Tableau of the Problem c B BV s 1 s 2 s 3 x 1 x 2 Solution 3 x 1 1-1 0 1 0 20 2 x 2-1 2 0 0 1 60 0 s 3-1 1 1 0 0 20 Z 1 1 0 0 0 180 Let us assumed that unit price of W 1 equals 3 + δ. Then we have: c B BV s 1 s 2 s 3 x 1 x 2 Solution 3 + δ x 1 1 1 0 1 0 20 2 x 2 1 2 0 0 1 60 0 s 3 1 1 1 0 0 20 Z 1 + δ δ + 1 0 0 0 180 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 17 / 41
Sensitivity Analysis Changes in Objective Coefficients Possibly Changes of Values of c 1 The optimal solution remains optimal as long as all the Z row coefficients are nonnegative, so we get: { 1 + δ 0 δ + 1 0 Solving these system inequalities we obtain that δ [ 1, 1]. Thus optimal solution will remain optimal for values of price of W 1 belonging to the interval [2,4]. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 18 / 41
Sensitivity Analysis Changes in Right-Hand Side Changes in Right-Hand Side Now we examine how the optimal solution to linear problem will changes if the right-hand side of a constraint is changed. For example we consider the constraint for row material S 1 (coefficient b 1 = 100). Changing in b i will leave Z - row of simplex tableau unchanged but will change the the solution column of the simplex tableau. If at least one coefficient becomes negative, the current solution is no longer feasible. So we look for the value of δ for which the following system equalities is consistent: 2x 1 + x 2 = 100 + δ x 1 + x 2 = 80 s 3 + x 1 = 40 x 1, x 2, s 1, s 2, s 3 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 19 / 41
Sensitivity Analysis Changes in Right-Hand Side Matrix Form of System Equations We describe this system in matrix form: 2 1 0 x 1 1 1 0 x 2 = 1 0 1 s 3 100 + δ 80 40 Multiplying both sides of this equations by the inverse matrix of the coefficients matrix we get: x 1 x 2 s 3 = 2 1 0 1 1 0 1 0 1 1 100 + δ 80 40 0 0 0 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 20 / 41
Sensitivity Analysis The Inverse Matrix of Basis Changes in Right-Hand Side The inverse matrix we could obtain from the optimal simplex tableau. It is consisting of the columns in the optimal tableau that correspond to the initial basic variables BV = {s 1, s 2, s 3 }(taking in the same order): 2 1 0 1 1 0 1 0 1 1 = 1 1 0 1 2 0 1 1 1 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 21 / 41
Sensitivity Analysis Changes in Right-Hand Side Hence we get: 1 1 0 1 2 0 1 1 1 100 + δ 80 40 0 0 0 For the current set of basic variable (basis) to remain optimal we require that the following system of inequalities holds: 20 + δ 0 60 δ 0 20 δ 0 The solution remains optimal (basic variable x 1, x 2, s 3 ) as long as δ [ 20, 20] or if between 80 and 120 row material S1 (b 1 [80, 120], where the current amount is b 1 = 100) are available. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 22 / 41
Dual Problem Definition Of the Dual Problem Associated with any linear programming problem is another linear problem, called the dual problem (Dual in short). Now we explain how to find the dual problem to the given LPP, we discuss the economic interpretation of the dual problem and we discuss the relation that exist between an LPP (called Primal) and its dual problem. We consider the LPP with normal (canonical) form : max Z = c 1 x 1 + c 2 x 2 + + c n x n a 11 x 1 + a 12 x 2 + + a 1n x n b 1 a 21 x 1 + a 22 x 2 + + a 2n x n b 2... a m1 x 1 + a m2 x 2 + + a mn x n b m x j 0 (j = 1, 2,...,n) The original problem is called the primal. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 23 / 41
Dual Problem Definition Of the Dual Problem Dual Problem of LPP if Primal is in Canonical Form The dual problem is defined as follows: min W = b 1 y 1 + b 2 y 2 + + b m y m a 11 y 1 + a 21 y 2 + + a m1 y m c 1 a 12 y 1 + a 22 y 2 + + a m2 y m c 2... a 1n y 1 + a 2n y 2 + + a mn y m c n y i 0 (i = 1, 2,...,m) Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 24 / 41
Dual Problem Definition Of the Dual Problem The Construction of the Dual Problem from the Primal Problem max Z min W (x 1 0) (x 2 0) (x n 0) x 1 x 2 x n (y 1 0) y 1 a 11 a 12 a 1n b 1 (y 2 0) y 2 a 21 a 22 a 2n b 2...... (y m 0) y m a m1 a m2 a mn b m c 1 c 2 c n Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 25 / 41
Example Dual Problem Definition Of the Dual Problem The Furniture Company STYLE manufactures tables and chairs. A table sells for $160, and chair sells for $200. The demand for tables and chairs is unlimited. The manufacture of each type of furniture require labor, lumber, and inventory space. The amount of each resource needed to make tables and chairs and daily limit of available resources is given in the following table: Resources needed Amount of Resource Table Chair resource available(hours) Labor(hours) 2 4 40 Lumber((feet) 3 ) 18 18 216 Inventory space((feet) 2 ) 24 12 240 STYLE wants to maximize total revenue. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 26 / 41
Dual Problem Definition Of the Dual Problem Primal and Dual of the Example Problem Primal problem max z = 160x 1 + 200x 2 2x 1 + 4x 2 40 18x 1 + 18x 2 216 24x 1 + 12x 2 240 x 1, x 2 0. Dual problem min w = 40y 1 + 216y 2 + 40y 3 2y 1 + 18y 2 + 24y 3 160 4y 1 + 18y 2 + 12y 3 200 y 1, y 2, y 3 0. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 27 / 41
Dual Problem Interpretation of Dual for STYLE Economic Interpretation of Duality Suppose there is an entrpreneur who wants to purchase all of Style s resourses i.e. 40 hours of labor, 210 (feet) 3 of lumber and 240(feet) 2 of inventory space. Then he must determine the price he is willing to pay for a unit of each of STYLE s resources. Let y 1, y 2 i y 3 be the price for one hour of labor, one cubic feet of lumber and one square feet of inventory space. We show that the resource prices should be determine by solving the dual problem. The total price the entrepreneur must pay for the resources is 40y 1 + 216y 2 + 240y 3 and since he wish to minimize the cost of purchasing the resources he wants to: minimize W = 40y 1 + 216y 2 + 240y 3. But he must set resource prices high enough to induce STYLE to sell its resources. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 28 / 41
Dual Problem Interpretation of Dual for STYLE Economic Interpretation of Duality For example, he must offer STYLE at lest $160 for a combination of resources that includes 2 hours of labor, 18 cubic feet of lumber and 24 square feet of inventory space, because STYLE could use these resources to manufacture table that can be sold for $160. Since he is offering 2y 1 + 18y 2 + 24y 3 for the resources used to produce table, he must choose y 1, y 2, y 3 to satisfy 2y 1 + 18y 2 + 24y 3 160. Similar reasoning for the chair gives: 4y 1 + 18y 2 + 12y 3 200. The sign restrictions y 1, y 2, y 3 0 must also hold. So we see that the solution to the dual of STYLE problem does yield prices for labor, lumber and inventory space. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 29 / 41
Dual Problem Finding the Dual to any LPP Economic Interpretation of Duality We illustrate how to find the dual problem on the following example: max z = 2x 1 + x 2 min w = 2y 1 + 3y 2 + y 3 x 1 + x 2 = 2 y 1 - Unrestricted 2x 1 x 2 3 y 2 0 x 1 x 2 1 y 3 0 x 1 0 y 1 + 2y 2 + y 3 2 x 2 Unrestricted y 1 y 2 y 3 = 1. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 30 / 41
Dual Problem Rules for Construction of Dual Economic Interpretation of Duality The general conclusion from the preceding example is that the variables and constraints in the primal and dual problems are defined by rules in the following table: Maximization problem Minimization problem Constraints Variables 0 0 = Unrestricted Variables Constraints 0 0 Unrestricted = Table: Rules for construction the Dual Problem Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 31 / 41
Dual Problem Primal-Dual Relationships The Key Relationships between the Primal and Dual Theorem The dual of the dual problem yields the original primal problem. Theorem (Weak duality property) If we choose any feasible solution to the primal and any feasible solution to the dual, the W value for the feasible dual solution will be at least as large as the Z value for the feasible primal solution. Let x = [x 1, x 2,...,x n ] T be any feasible solution to the primal and y = [y 1, y 2,...,y m ] be any feasible solution to the dual. Then (Z value for x) (W value for y) From this theorem results two properties: Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 32 / 41
Dual Problem Primal-Dual Relationships Property If x = ( x 1, x 2,..., x n ) and ȳ = (ȳ 1, ȳ 2,...,ȳ m ) are feasible solutions of primal problem and dual problem respectively such that Z = c 1 x 1 + c 2 x 2 +... + c n x n = b 1 ȳ 1 + b 2 ȳ 2 +... + b m ȳ m = W, then x must be optimal solution for primal problem and ȳ must be optimal solution for dual problem. Property If the primal(dual) is unbounded, then the dual(primal) problem is infeasible. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 33 / 41
Dual Problem Primal-Dual Relationships Duality Theorem Theorem The following are only possible relations between the primal and dual problems: 1 If one problem has feasible solutions and a bounded objective function(and so has optimal solution), then so does the other problem. 2 If one problem has feasible solutions and an unbounded objective function(and so no optimal solution), then the other problem has no feasible solutions. 3 If one problem has no feasible solutions, then the other problem has either no feasible solution or an unbounded objective function. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 34 / 41
Dual Problem Primal-Dual Relationships Interpretation of Optimal Dual Decision Variables Now we can give an interpretation of the dual problem for maximization problem. We know that for optimal solutions x of primal problem and ȳ of dual problem the following equality holds: Z = c 1 x 1 + c 2 x 2 +... + c n x n = b 1 ȳ 1 + b 2 ȳ 2 +... + b m ȳ m = W So each b i ȳ i can be interpreted as the contribution to profit by having b i units of resource i available for the primal problem. Thus, the optimal dual variable ȳ i (it is called shadow price) is interpreted as the contribution to profit per unit of resource i(i = 1, 2,...,m). Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 35 / 41
Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem Reading Optimal Dual from Z-row of Optimal Simplex Tableau for STYLE After solving primal problem by simplex method we could read the optimal dual solution from the optimal simplex tableau. Let us consider the LPP for firm STYLE. The optimal simplex tableau is as follows: Table: Optimal simplex tableau s 1 s 2 s 3 x 1 x 2 1 200 x 2 2 1 18 0 0 1 8 160 x 1 1 1 2 9 0 1 0 4 0 s 3 6-2 1 0 0 48 20 Z 20 3 0 0 0 2240 The basic variables are ZB = {x 2, x 1, s 3 } and the basis is 1 4 2 0 B = 18 18 0. Inverse matrix for B is: B 1 2 1 18 0 = 1 1 2 9 0 12 24 1 6 2 1 Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 36 / 41
Dual Problem The Optimal Solution of Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem (y 1, y 2, y 3 ) we can compute using matrix B 1 as follows: (y 1, y 2, y 3 ) = c B B 1 = (200, 160, 0) 1 1 2 2 1 18 0 1 9 0 6 2 1 = (20, 20 3, 0). where the vector c B contains the coefficients of objective function corresponding the basic variables. The optimal dual solution are coefficients of variables s 1, s 2, s 3 of Z row optimal simplex tableau. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 37 / 41
Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem Reding Dual Solution from Optimal Simplex Tableau If the primal problem is any form, then the optimal dual solution may be read from Z row optimal simplex tableau by using the following rules: Optimal value of dual variable y i if constraint i is a ( ) equal to coefficient of s i in Z row optimal simplex tableau. Optimal value of dual variable y i if constraint i is a ( ) equal to -(coefficient of e i in Z row optimal simplex tableau), where e i is surplus variable. Optimal value of dual variable y i if constraint i is an equality (=) equal to (coefficient of a i in Z row optimal simplex tableau)-m, where a i is artificial variable. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 38 / 41
Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem Example max z = 3x 1 + 2x 2 + 5x 5 x 1 + 3x 2 + 2x 3 15 2x 2 x 3 5 2x 1 + x 2 5x 3 = 10 x 1, x 2, x 3 0. To solve the problem we use the M method: max Z = 3x 1 + 2x 2 + 5x 5 Ma 2 Ma 3 x 1 + 3x 2 + 2x 3 + s 1 = 15 2x 2 x 3 e 2 + a 2 = 5 2x 1 + x 2 5x 3 + a 3 = 10 x 1, x 2, x 3, s 1, a 2, a 3 0. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 39 / 41
Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem The Big-M Method, the Last Simplex Tableau x 1 x 2 x 3 s 1 e 2 a 2 a 3 4 5 5 x 3 0 0 1 23 23 5 23 2 23 2 2 x 2 0 1 0 23 9 9 23 23 1 23 9 17 3 x 1 1 0 0 23 23 17 7 23 23 Z 0 0 0 51 23 The dual problem has the following form: 58 23 M 58 23 M + 9 23 15 23 65 23 120 23 565 23 min W = 15y 1 + 5y 2 + 10y 3 y 1 + 2y 3 3 3y 1 + 2y 2 + y 3 2 2y 1 y 2 5y 3 5 y 1 0, y 2 0, y 3 Unrestricted. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 40 / 41
Optimal Dual Solution Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem Reading optimal dual solution from optimal simplex tableau we get: The first constraint is inequality so y 1 = 51 23 (coefficient of Z row for column s 1 ). The second constraint is inequality so y 2 = 58 23 (- coefficient of Z row for column e 2 ). The third constraint is equality = so y 3 = 9 23 (coefficient of Z row for column a 3 minus M). Optimal value of objective function of dual problem is W = 15 51 58 23 + 5( 23 ) + 10 9 23 = 565 23 and equals optimal value of objective function of the primal problem. Michał Kulej (Wrocł Univ. of Techn.) OPERATIONS RESEARCH BIS 41 / 41