Chap6 Duality Theory and Sensitivity Analysis

Chap6 Duality Theory and Sensitivity Analysis The rationale of duality theory Max 4x 1 + x 2 + 5x 3 + 3x 4 S.T. x 1 x 2 x 3 + 3x 4 1 5x 1 + x 2 + 3x 3 + 8x 4 55 x 1 + 2x 2 + 3x 3 5x 4 3 x 1 ~x 4 0 If we multiply a certain number (y 0) on both sides of any constraint, the result won t be affected. The arbitrary y is called the multiplier. y 1, y 2, and y 3 are the multipliers of constraints 1, 2, and 3, respectively. How can we tell the range of the optimal objective value? (Let Z * be the optimal value) Consider 4x 1 + x 2 + 5x 3 + 3x 4 5/3 (5x 1 + x 2 + 3x 3 + 8x 4 ) Z* 5/3(55) = 275/3 Similarly 4x 1 + x 2 + 5x 3 + 3x 4 constraint 2 + constraint 3 = (5x 1 + x 2 + 3x 3 + 8x 4 ) + ( x 1 + 2x 2 + 3x 3 5x 4 ) Z* < 55 + 3 = 58 That is (1) Z* 0.b 1 + 5/3.b 2 + 0.b 3 (2) Z* 0.b 1 + 1. b 2 + 1.b 3 In general, we want to determine the values of y i such that the following statement holds. Z* 4x 1 + x 2 + 5x 3 + 3x 4 y 1 (x 1 x 2 x 3 + 3x 4 ) + y 2 (5x 1 + x 2 + 3x 3 + 8x 4 ) + y 3 ( x 1 + 2x 2 + 3x 3 5x 4 ) y 1 + 55y 2 + 3y 3 Issue: which is the best estimate of the optimal value? Jin Y. Wang 6-1

So, we can conclude that (1) y 1, y 2, y 3 0 (2) 4 y 1 + 5y 2 y 3 1 y 1 + y 2 + 2y 3 5 y 1 + 3y 2 + 3y 3 3 3y 1 + 8y 2 5y 3 (3) Since we want to estimate Z* as best as possible, i.e. we want to have y 1 + 55y 2 + 3y 3 as less as possible. Therefore, we have Min y 1 + 55y 2 + 3y 3 The above formulation is called the Dual problem. The original formulation is called the Primal problem. Note: (1) Primal: coefficients of the objective function Dual: RHS (2) Primal: RHS Dual: coefficients of the objective function (3) Primal: coefficients of a variable in the functional constraint Dual: coefficients in a functional constraint (4) Max Min (5) Example of finding the dual problem Max 5x 1 + 4x 2 + 3x 3 S.T. 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x 3 11 3x 1 + 4x 2 + 2x 3 8 x 1, x 2, x 3 0 Matrix representation of the primal and dual problems Jin Y. Wang 6-2

Primal Problem (P) n Max Z = j c = 1 j x j Max Z = cx n S.T. = a j ij x j b 1 i, for i = 1, 2,, m S.T. Ax b x j 0, for j = 1, 2,, n x 0 Dual Problem (D) m Min W = j b = 1 i y i Min W = yb m S.T. = a i ij yi c 1 j, for j = 1, 2,, n S.T. ya c y i 0, for i = 1, 2,, m y 0 Weak duality property If x is a feasible primal solution, and y is a feasible dual solution, then cx yb <Proof> Corollary: If the primal is feasible and unbounded, the dual is infeasible and vice versa. Strong duality property If x * is an optimal solution for the primal problem and y * is an optimal solution for the dual problem, then cx * = y * b. A corollary for weak and strong duality property These two properties implies that cx < yb for feasible solutions if one or both of them are not optimal for their respective problems, whereas equality holds when both are optimal. Complementary solutions property Jin Y. Wang 6-3

At each iteration, the simplex method simultaneously identifies a CPF solution x for the primal problem and a complementary solution y for the dual problem (in tabular form, the coefficients of the slack variables in row 0), where cx = yb. If x is not optimal for the primal problem, then y is not feasible for the dual problem. P: Max Z = 5x 1 + 4x 2 + 3x 3 S.T. 2x 1 + 3x 2 + x 3 5 4x 1 + x 2 + 2x 3 11 3x 1 + 4x 2 + 2x 3 8 x 1, x 2 x 3 0 D: Min W = 5y 1 +11y 2 + 8y 3 S.T. 2y 1 + 4y 2 + 3y 3 5 3y 1 + y 2 + 4y 3 4 y 1 + 2y 2 + 2y 3 3 y 1, y 2, y 3 0 Z - 5x 1-4x 2-3x 3 = 0 2x 1 + 3x 2 + x 3 + x 4 = 5 4x 1 + x 2 + 2x 3 + x 5 = 11 3x 1 4x 2 + 2x 3 + x 6 = 8 y 1 =, y 2 =, y 3 = Z - 1/2 x 1 + 2x 2 + 3/2x 6 = 12 1/2 x 1 + x 2 + x 4 + 1/2x 6 = 1 x 1 3x 2 + x 5 + x 6 = 3 3/2x 1 + 2x 2 + x 3 + 1/2x 6 = 4 y 1 =, y 2 =, y 3 = Z + 3x 2 + x 4 + x 6 = 13 x 1 + 2x 2 + 2x 4 x 6 = 2 5x 2 2x 4 + x 5 = 1 x 2 + x 3 3x 4 + 2x 6 = 1 y 1 =, y 2 =, y 3 = Recall that Jin Y. Wang 6-4

Iteration Basic Variable Z Any Z x B 1 0 Coefficient of: Right Side Original Variable Slack Variables c B B -1 A c B -1 A c B B -1 B -1 c B B -1 b B -1 b Let y (dual variables) = c B B -1 (the shadow price). Obviously, Z (=cx) = yb. The coefficients of original and slack variables in the objective row are ya c and y, respectively. Note that ya c represents the value of surplus variables in the dual. If the optimality hasn t yet reached, at least one coefficient in the objective row is < 0. Thus, at lease one constraint of the dual is violated. This also explains the strong duality property in some sense. Complementary optimal solutions property At the final iteration, the simplex method simultaneously identifies an optimal solution x * for the primal problem and a complementary optimal solution y * for the dual problem (the coefficients of the slack variables in row 0), where cx * = y * b In fact, y * = C B B -1 are the shadow prices. Symmetry property For any primal problem and its dual problem, all relationships between them must be symmetric because the dual of this dual problem is this primal problem. Duality theorem All possible relationships between the primal and dual problems If one problem has feasible solutions and a bounded objective function (and so has an optimal solution), then so does the other problem. If one problem has feasible solutions and an unbounded objective function (and so no optimal solution), then the other problem has no feasible solutions. Jin Y. Wang 6-5

If one problem has no feasible solutions, then the other problem has either no feasible solutions or an unbounded objective function. Dual Optimal Infeasible Unbounded Optimal Primal Infeasible Unbounded Example for primal infeasible dual infeasible Max 2x 1 x 2 Min y 1 2y 2 S.T. x 1 x 2 1 S.T. y 1 y 2 2 x 1 + x 2 2 y 1 + y 2 1 x 1, x 2 0 y 1, y 2 0 Applications of dual problem Dual problem can be solved directly by the simplex method. The number of functional constraints affects the computational effort of the simplex method far more than the number of variables does. It is useful for evaluating a proposed solution for the primal problem. Economic interpretation of duality Primal Problem (P) Z n j 1 n a j ij x j Max = = c j x j S.T. = b 1 i, for i = 1, 2,, m x j 0, for j = 1, 2,, n Dual Problem (D) W m j 1 m a i ij yi Min = = b i y i S.T. = c 1 j, for j = 1, 2,, n y i 0, for i = 1, 2,, m Jin Y. Wang 6-6

If resource i is not used for production, we rent it out with unit rental of y i. m a y i = 1 ij i means the rental income if we reduce the production amount of product j by one. m = a i ij yi c 1 j means the rental income be greater than the sales income. The objective of dual is to minimize the rental income in order to obtain the reasonable rental price. Primal-Dual relationships Dual is a LP problem and also has corner-point solutions. By using the augmented form, we can express these corner-point solutions as basic solutions. Because the functional constraints have the form, this augmented form is obtained by subtracting the surplus from the left-hand side of each constraint. Complementary basic solutions property Each basic solution in the primal problem has a complementary basic solution in the dual problem, where their respective objective function values are equal. Recall again: Coefficient of: Iteration Basic Variable Z Original Variable Slack Variables Right Side Any Z 1 c B B -1 A c c B B -1 c B B -1 b Given row 0 (objective row) of the simplex tableau for the primal basic solution, the complementary dual basic solution can be easily found. Primal Variable Associated Dual Variables (Original variable) x j (Surplus Variable) y m+j (j = 1, 2,, n) (Slack Variable) x n+i (Original Variable) y i (i = 1, 2,., m) Jin Y. Wang 6-7

The m basic variables for the primal problem are required to have a coefficient of zero in the objective row, which thereby requires the m associated dual variables to be zero, i.e., nonbasic variables for the dual problem. The values of the remaining n (basic) variables will be the simultaneous solution to the system of equations. The dual solution read from row 0 must also be a basic solution. Complementary slackness property Primal Variable Associated Dual Variable Basic Nonbasic (m variables) Nonbasic Basic (n variables) If one of (primal) variables is a basic variable (> 0), then the corresponding one (dual) must be a nonbasic variable (= 0). Wyndor Example: Max Z = 3x 1 + 5x 2 Min W = 4y 1 + 12y 2 + 18y 3 S.T. x 1 4 S.T. y 1 + 3y 3 3 2x 2 12 2y 2 + 2y 3 5 3x 1 + 2x 2 18 x 1, x 2 0 y 1, y 2, y 3 0 Max Z = 3x 1 + 5x 2 Min W = 4y 1 + 12y 2 + 18y 3 S.T. x 1 + x 3 = 4 S.T. y 1 + 3y 3 y 4 = 3 2x 2 + x 4 = 12 2y 2 + 2y 3 y 5 = 5 3x 1 + 2x 2 + x 5 = 18 Given x = (4, 6, 0, 0, -6), we know that x 1, x 2, and x 5 are basic variables. The coefficients of these variables in row 0 are zero. Thus, y 4, y 5, and y 3 are nonbasic variables (=0). y 1 = 3 2y 2 = 5 Jin Y. Wang 6-8

Given x = (0, 6, 4, 0, 6), we have Let x 1 * ~x n * be primal feasible solution, y 1 * ~y m * be dual feasible solution. Then, they are both optimal if and only if (1) x j * > 0 implies m i= 1 * a y = c. ij i j n (2) a j= 1 ij * x < b implies y * i = 0. j i Ex: Max 5x 1 + 6x 2 + 9x 3 + 8x 4 Min 5y 1 + 3y 2 S.T. x 1 + 2x 2 + 3x 3 + x 4 5 S.T. y 1 + y 2 5 x 1 + x 2 + 2x 3 + 3x 4 3 2y 1 + y 2 6 x 1,, x 4 0 3y 1 + 2y 2 9 y 1 + 3y 2 8 y 1, y 2 0 Ex: Max 8x 1 9x 2 + 12x 3 + 4x 4 + 11x 5 Jin Y. Wang 6-9

S.T. 2x 1 3x 2 + 4x 3 + x 4 + 3x 5 1 x 1 + 7x 2 + 3x 3 2x 4 + x 5 1 5x 1 + 4x 2 6x 3 + 2x 4 + 3x 5 22 x 1,.., x 5 0 Complementary optimal basic solutions property Each optimal basic solution in the primal problem has a complementary optimal basic solution in the dual problem, where their respective objective functions are equal. The coefficients in the primal row 0 are all 0 All dual variables (including decision and surplus variables) are all 0. Nonnegative surplus variables imply that functional constraints are satisfied. Nonnegativity constraints are satisfied. How to write the dual if the primal is not in the standard form Max 3x 1 + x 2 x 3 S.T. x 1 + x 2 + 4x 3 5 x 1 x 2 7x 3 = 7 x 1, x 2 0, x 3 is free Jin Y. Wang 6-10

Summary Primal (Dual) Dual (Primal) Max Min i th constraint is (sensible) i th variable 0 (sensible) i th variable 0 (sensible) i th constraint is (sensible) i th constraint is = (odd) i th variable is unsigned (odd) i th variable is unsigned (odd) i th constraint is = (odd) i th constraint is (bizarre) i th variable 0 (bizarre) i th variable 0 (bizarre) i th constraint is (bizarre) For constraint in primal, we can convert to by multiplying both sides by -1. For negative variable x (< 0) in primal, we can use another variable, whose value is (-1)*x, to replace x. Ex: Max 5x 1 + 4x 2 S.T. x 1 x 2 5 3x 1 + x 2 2 x 1 + 2x 2 = 9 x 1 0, x 2 0 The radiation therapy example Min Z = 0.4x 1 + 0.5x 2 Jin Y. Wang 6-11

S.T. 0.3x 1 + 0.1x 2 2.7 0.5x 1 + 0.5x 2 = 6 0.6x 1 + 0.4x 2 6 x 1, x 2 0 Dual simplex method (section 7.1) Max 8x 1 8x 2 9x 3 Min y 1 8y 2 2y 3 S.T. x 1 + x 2 + x 3 1 S.T. y 1 2y 2 + y 3 8 2x 1 4x 2 x 3 8 y 1 4y 2 y 3 8 x 1 x 2 x 3 2 y 1 y 2 y 3 9 x 1 ~ x 3 0 y 1 ~ y 3 0 Note that the current primal does not have a feasible solution. However, the dual does have one. Working on the dual is one way of tackling this situation. Another way is working the complementary counterpart on the primal. Convert the dual to standard form and add slack variables (currently, the basic solution is feasible). W + y 1 8y 2 2y 3 = 0 y 1 + 2y 2 y 3 + y 4 = 8 y 1 + 4y 2 + y 3 + y 5 = 8 y 1 + y 2 + y 3 + y 6 = 9 Add slack variables to the primal (currently, the basic solution is infeasible). Z + 8x 1 + 8x 2 + 9x 3 = 0 x 1 + x 2 + x 3 + x 4 = 1 2x 1 4x 2 x 3 + x 5 = 8 x 1 x 2 x 3 + x 6 = 2 In the dual problem, pick y 3 to enter and pick y 5 to leave. Jin Y. Wang 6-12

The coefficients of the dual objective function correspond to the current value of primal s basic variable. The dual current values of basic variables correspond to the coefficients of the primal objective function. The dual entering column corresponds to the primal leaving row. In dual: Find a nonbasic variable with a negative coefficient in row 0 (positive coefficient in algebraic form) to enter. In Primal: Find a basic variable with a negative current value to leave. In dual: Find the variable corresponding to the entering variable with a positive coefficient and smallest ratio to leave. In Primal: Find the variable with a negative coefficient in the leaving row and smallest ratio to enter. For dual, pick y 3 to enter and pick y 5 to leave. W y 1 0y 2 + 2y 5 = 16 2y 1 + 6y 2 + y 4 + y 5 = 16 y 1 + 4y 2 + y 3 + y 5 = 8 0y 1 3y 2 y 5 + y 6 = 1 For primal, pick x 6 to leave, x 2 to enter. Z + 16x 1 + x 3 + 8x 6 = 16 2x 1 + 0x 3 + x 4 + x 6 = 1 6x 1 + 3x 3 + x 5 4x 6 = 0 x 1 + x 2 + x 3 x 6 = 2 Conditions for adopting the dual simplex method Jin Y. Wang 6-13

The prerequisite for using the dual simplex is the dual has initial feasible solution. Thus, the primal coefficient of row 0 must be nonnegative. What if both primal and dual do not have a feasible initial solution? Big M or two-phase method. Another example for dual simplex Z + 4x 2 + x 4 + x 5 = 12 x 1 3x 2 + 11x 4 x 5 = 4 x 2 + x 3 3x 4 + 2x 5 = 8 Jin Y. Wang 6-14

Sensitivity Analysis Max Z = cx Ax = b x 0 Recalled that we can rewrite it as the following Max Z = c B x B + c N x N S.T. Bx B + Nx N = b x B, x N 0 Max Z + (c B B -1 N - c N )x N = c B B -1 b x B = B -1 b B -1 Nx N What happens to the optimal solution if (1) b is changed (2) c is changed (3) A is changed (4) add a new variable (5) add a new constraint b is changed to b Max cx S.T. Ax = b x 0 In this case, the original optimal basic solution is no more feasible, since b is changed to b. In order to make sure that the basic won t change, we need to check B -1 b. If B -1 b 0 The basic won t change, we are at optimality since (c B B -1 N - c N 0) Otherwise (B -1 b < 0) The new basic solution is infeasible. That is, the basic change. Need to solve the problem by using dual simplex. Ex: Max Z = 3x 1 + 5x 2 Jin Y. Wang 6-15

S.T. x 1 + x 3 = 4 2x 2 +x 4 = 12 3x 1 + 2x 2 +x 5 = 18 x * = 2 6, x B = 2 x x x 3 2 1 x4, x N = x5 b changes from 4 12 18 to 4 15 18 b changes from 4 12 18 to 4 24 18 The final tableau can be obtained based on the given information. Based on Z + (c B B -1 N - c N )x N = c B B -1 b and x B = B -1 b B -1 Nx N or recall the fundamental insight table. Jin Y. Wang 6-16

Iteration Basic Variable Z Any Z 1 x B 0 Coefficient of: Right Side Original Variable Slack Variables c B B -1 A c B -1 A c B B -1 B -1 c B B -1 b B -1 b Basic Final tableau after RHS changing Coefficient of: Right Side Variable x 1 x 2 x 3 x 4 x 5 Z x 3 x 2 x 1 Apply dual simplex. We pick x 1 to leave and x 4 to enter. Basic Coefficient of: Variable x 1 x 2 x 3 x 4 x 5 Right Side Z x 3 x 2 x 4 Allowable range for RHS given that the basic won t change Max Z = 3x 1 + 5x 2 S.T. x 1 +x 3 = 4 2x 2 +x 4 = 12 3x 1 + 2x 2 +x 5 = 18 b changes from 4 12 18 to 4 12 + Δ. We need to guarantee B -1 b 0 18 Simultaneous changes in Right-Hand Sides Jin Y. Wang 6-17

Adopt the 100 percent rule. Calculate for each change percentage of the allowable change (increase or decrease) for that right-hand side to remain within its allowable range. If the sum of the percentage changes does not exceed 100 percent, the shadow prices definitely will still be valid (basic remains the same). If the sum does exceed 100 percent, then we cannot be sure. Example (Wyndor) Constraint Shadow Price Current RHS Allowable Increase Allowable Decrease Plant 1 0 4 Infinite 2 Plant 2 1.5 12 6 6 Plant 3 1 18 6 6 4 4 Change b from 12 to 15. 18 15 b 2 : percentage of allowable increase = b 3 : percentage of allowable decrease = Conclusion: c is changed to c Supposed c is changed to c Max c x Ax = b x 0 Notice that the optimal solution computed before is still. In order to check the optimality, look at the objective row and be sure. That is, if the above inequality holds, we are still at optimality. Otherwise, use (revised) simplex method starting with x * as initial feasible solution. Ex: Max Z = 3x 1 + 5x 2 Jin Y. Wang 6-18

S.T. x 1 + x 3 = 4 2x 2 + x 4 = 12 3x 1 + 2x 2 + x 5 = 18 At the final iteration, we have If c changes from [ 3 5] to [ 5] 4. How much can c change and still have the same optimal solution? The coefficients of nonbasic variables are changed Let c B B -1 N - c N remains 0 Example 1: x B = [ x 1, x 2, x 3 ] T, x N = [x 4, x 5, x 6 ] T, c B = [1 2 1 ], c N = [2 3 3] 1 3 1 2 0 0 B = 0 1 4, N = 1 0 1 0 2 2 2 1 0 Jin Y. Wang 6-19

The coefficients of basic variables are changed same reasoning. Example 2 (Wyndor): At the final iteration, x B = [x 3, x 2, x 1 ] T, x N = [x 4, x 5 ] T, c B = [0 5 3], c N = [0 0], B = 0 2 0, N = 1 0 0 2 3 0 1 How much can we change c 2 given the original optimal is still optimal? 1 0 1 0 0 Simultaneous changes on objective function coefficients Variable Value Reduced cost Current coefficient Allowable increase Allowable decrease x 1 0 4.5 3 4.5 Infinite x 2 9 0 5 Infinite 3 Apply the 100 percent rule. In the Wyndor example, if c 1 is increased by 1.5, then c 2 can be decreased by as much as. If c 1 is increased by 3, then c 2 can only be decreased by as much as. Reduce cost For any nonbasic decision variable x j, the associated coefficient in row 0 is referred to as the reduced cost for x j. It is the minimum amount by which the unit cost of activity j would have to be reduced to make it worthwhile to undertake activity j (increase x j from zero). Jin Y. Wang 6-20

Interpreting c j as the unit profit of activity j, the reduce cost for x j is the maximum allowable increase in c j to keep the current BF solution optimal. A is changed to A Max cx S.T. A x = b x 0 Similar to the procedures when b is changed; but a bit more complex. Change the coefficients of nonbasic variables. Current optimal basic solution is still feasible. Only need to check the optimality. An easy way: If the complementary dual basic solution y * still satisfies the single dual constraint that has changed, then the original optimal solution remains optimal. Or use our usual way of checking whether c B B -1 N - c N 0. Change the coefficients of basic variables Check the feasibility first. If feasible, perform the optimality check. If infeasible, compute the coefficients of row 0 and reoptimize with the dual simplex or two phase/big M methods. Sometimes, coefficients of c and A may be changed simultaneously. The solutions techniques are similar as discussed before. Let s go to the examples directly. Jin Y. Wang 6-21

Example 1: Changes in the coefficients of a nonbasic variable Given the following LP problem and the final tableau. Max Z = 3x 1 + 5x 2 S.T. x 1 + x 3 = 4 2x 2 + x 4 = 24 3x 1 + 2x 2 + x 5 = 18 Basic Coefficient of: Right Side Variable x 1 x 2 x 3 x 4 x 5 Z 9/2 0 0 0 5/2 45 x 3 1 0 1 0 0 4 x 2 3/2 1 0 0 1/2 9 x 4-3 0 0 1-1 6 Changes are made to the coefficients of x 1. Let c 1 = 4 and a 31 = 2. Applying the dual theory. Or compute the value of c B B -1 N - c N. Example 2: Changes in the coefficients of a basic variable Use the same problem as in Example 1. Changes are made to the coefficients of x 2. Let c 2 = 3, a 22 = 3, and a 32 = 4. Compute the new current basic solution. Optimality check. Jin Y. Wang 6-22

Reoptimize using revised simplex method, if needed. Add x n+1 Add a new variable (x n+1 ) to the objective function and each constraint Assume the value of this new variable be zero. That is, let the new solution * x 1 be.. * x n 0 This new solution is still feasible. Check optimality If c B B -1 N - c N 0 done, this new solution is the optimal solution. If c B B -1 N - c N < 0 basic changed adopt (revised) simplex starting from the above solution. Example: Max x 1 + x 2 S.T. 2x 1 + x 2 + x 3 = 4 x 1 + 2x 2 + x 4 = 4 x 1 ~ x 4 0 4 / 3 4 / 3 x* = add x 5 0 0 Jin Y. Wang 6-23

Max x 1 + x 2 + 2x 5 S.T. 2x 1 + x 2 + x 3 + 3x 5 = 4 x 1 + 2x 2 + x 4 3x 5 = 4 x 1 ~x 5 0 Keep x 1, x 2 as basic variables, x 3, x 4, and x 5 as nonbasic variables. 4 / 3 4 / 3 x = 0, check c B B -1 N - c N 0 0 Jin Y. Wang 6-24

Add a new constraint If new slack 0 (current constraint satisfies the new constraint) done, at optimality If the current optimal solution satisfies the new constraint, then it would still be the optimal solution. The reason is that a new constraint can only eliminate some previously feasible solutions without adding any new ones. If new slack 0 (current constraint does not satisfy the new constraint) basis changed Introduce this new constraint into the final simplex tableau, where the usual additional variable (slack) is designated to be the new basic variable for this new row. Apply elementary operations to convert to proper form resolve the problem by dual simplex. Example: Max Z = 3x 1 + 5x 2 S.T. x 1 4 2x 2 24 3x 1 + 2x 2 18 x 1, x 2 0 Add a new constraint 2x 1 + 3x 2 24. The previous optimal solution (0, 9) violates the new constraint. So the previous optimal solution is no longer feasible. Add the new constraint to the current final simplex tableau. Basic Variable Revised final tableau Coefficient of: x 1 x 2 x 3 x 4 x 5 x 6 RHS Z 9/2 0 0 0 5/2 0 45 x 3 1 0 1 0 0 0 4 x 2 3/2 1 0 0 1/2 0 9 x 4-3 0 0 1-1 0 6 x 6 2 3 0 0 0 1 24 Convert to proper form Jin Y. Wang 6-25

Basic Variable Converted to proper form Coefficient of: x 1 x 2 x 3 x 4 x 5 x 6 RHS Z 9/2 0 0 0 5/2 0 45 x 3 1 0 1 0 0 0 4 x 2 3/2 1 0 0 1/2 0 9 x 4-3 0 0 1-1 0 6 x 6-5/2 0 0 0-3/2 1-3 Apply the dual simplex Basic Variable New final tableau Coefficient of: x 1 x 2 x 3 x 4 x 5 x 6 RHS Z 1/3 0 0 0 0 5/3 40 x 3 1 0 1 0 0 0 4 x 2 3/2 1 0 0 0 1/3 8 x 4-4/3 0 0 1 0-2/3 8 x 6 5/3 0 0 0 1-2/3 2 Jin Y. Wang 6-26