We use cookies to improve your experience on our site and to show you relevant advertising. Home; Math; Algebra; Find the value of X, Y and Z calculator to solve the 3 unknown variables X, Y and Z in a set of 3 equations. BYJUâS online linear programming calculator tool makes the calculations faster, and it displays the best optimal solution for the given objective functions with the system of linear constraints in a fraction of seconds. At this stage, no calculations are needed, just transfer the values ââfrom the preliminary stage to the corresponding table cells: We calculate the value of the objective function by elementwise multiplying the column Cb by the column P, adding the results of the products. Dual problem. Maximize 8x + 12y + 8z subject to x - 3y 56 4x +Z 16 y + 4z s 12 x20, y 20, 220 Write the dual problem. Dual simplex method calculator - Solve the Linear programming problem using Dual simplex method, step-by-step. Two-Phase Simplex Method Calculator. After this manipulation, the sign of inequality is reversed. Calculate: Define and solve a problem by using Solver / Example of a Solver evaluation . The number of variables in the basis is always constant, so it is necessary to choose which variable to derive from the basis, for which we calculate Q. LP Simplex and dual Simplex method choose . Enter the minimization problem and click the "Dual problem" button. Now in the constraint system it is necessary to find a sufficient number of basis variables. We use cookies to improve your experience on our site and to show you relevant advertising. Minimize 2x + y + 32 subject to the constraints below. Choose the right equivalent problem to solve. For the results of the calculations of the previous iteration, we remove the variable from the basis x8 and put in her place x2. Fill all cells with zeros corresponding to the variable that has just been entered into the basis: (The resolution element remains unchanged). 2 Applying the simplex method to the dual problem. Simplex Algorithm Calculator is an online application on the simplex algorithm and two phase method. There are many ways to write the code, so there are many equivalents to those requirements. A duplex communication system is a point-to-point system composed of two or more connected parties or devices that can communicate with one another in both directions. The solution for constraints equation with nonzero variables is called as basic variables. Complete, detailed, step-by-step description of solutions. We will now solve the dual of the example problem using the simplex method. Dual Problem for Standard Minimization. Thus, we have observed that, by solving (2), we can determine the shadow prices of (1) directly. By ⦠2) Dualize Transforms the problem in its dual. Learn more Hire us: Solving the traveling salesman problem using the branch and bound method. Linear Programming Calculator is a free online tool that displays the best optimal solution for the given constraints. The calculator given here can easily solve the problems related to the simplex method, two-phase method, and the graphical method as well. Egwald Web Services Domain Names Web Site Design Operations Research - Linear Programming - Dual Simplex Tableaux Generator ... Dual Solution: [y 1, y 2, y 3, ] = [26.571, 1.714, -15.714, ]; D = 1240. However, this Simplex algorithm does not exploit sparsity in the model. Solve either the original problem or its dual by the simplex method and then give the solutions to both. This element will allow us to calculate the elements of the table of the next iteration. Free solve for a variable calculator - solve the equation for different variables step-by-step This website uses cookies to ensure you get the best experience. Set up the dual problem. Hungarian method, dual simplex, matrix games, potential method, traveling salesman problem, dynamic programming Expression solver calculator The following expression solver calculator will evaluate math expressions with +, â , * ,and / signs. It optionally uses a dual Simplex method to solve LP subproblems in a mixed-integer (MIP) problem. If an inequality of the form â¤, then the compensating variable has the sign +, if the inequality of the form â¥, then the compensating variable has the sign -. If there are no basis variables in some restriction, then we add them artificially, and artificial variables enter the objective function with the coefficient -M if the objective function tends to max and M, if the objective function tends to min. We do not implement these annoying types of ads! In case of dual problem, these values are the optimal values of dual variables w 1 and w 2. It is a special case of mathematical programming. [+] Word Problems Video Playlist If you cannot find what you need, post your word problem in our calculator forum We use cookies to improve your experience on our site and to show you relevant advertising. u ⦠Set up initial Simplex table for the dual problem. P1 = (P1 * x3,1) - (x1,1 * P3) / x3,1 = ((525 * 5) - (2 * 700)) / 5 = 245; P2 = (P2 * x3,1) - (x2,1 * P3) / x3,1 = ((225 * 5) - (0 * 700)) / 5 = 225; P4 = (P4 * x3,1) - (x4,1 * P3) / x3,1 = ((75 * 5) - (0 * 700)) / 5 = 75; P5 = (P5 * x3,1) - (x5,1 * P3) / x3,1 = ((0 * 5) - (0 * 700)) / 5 = 0; x1,1 = ((x1,1 * x3,1) - (x1,1 * x3,1)) / x3,1 = ((2 * 5) - (2 * 5)) / 5 = 0; x1,3 = ((x1,3 * x3,1) - (x1,1 * x3,3)) / x3,1 = ((1 * 5) - (2 * 0)) / 5 = 1; x1,4 = ((x1,4 * x3,1) - (x1,1 * x3,4)) / x3,1 = ((0 * 5) - (2 * 0)) / 5 = 0; x1,5 = ((x1,5 * x3,1) - (x1,1 * x3,5)) / x3,1 = ((0 * 5) - (2 * 1)) / 5 = -0.4; x1,6 = ((x1,6 * x3,1) - (x1,1 * x3,6)) / x3,1 = ((0.5 * 5) - (2 * 2)) / 5 = -0.3; x1,7 = ((x1,7 * x3,1) - (x1,1 * x3,7)) / x3,1 = ((0 * 5) - (2 * 0)) / 5 = 0; x1,8 = ((x1,8 * x3,1) - (x1,1 * x3,8)) / x3,1 = ((-0.5 * 5) - (2 * -2)) / 5 = 0.3; x1,9 = ((x1,9 * x3,1) - (x1,1 * x3,9)) / x3,1 = ((0 * 5) - (2 * 0)) / 5 = 0; x2,1 = ((x2,1 * x3,1) - (x2,1 * x3,1)) / x3,1 = ((0 * 5) - (0 * 5)) / 5 = 0; x2,3 = ((x2,3 * x3,1) - (x2,1 * x3,3)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x2,4 = ((x2,4 * x3,1) - (x2,1 * x3,4)) / x3,1 = ((1 * 5) - (0 * 0)) / 5 = 1; x2,5 = ((x2,5 * x3,1) - (x2,1 * x3,5)) / x3,1 = ((0 * 5) - (0 * 1)) / 5 = 0; x2,6 = ((x2,6 * x3,1) - (x2,1 * x3,6)) / x3,1 = ((0 * 5) - (0 * 2)) / 5 = 0; x2,7 = ((x2,7 * x3,1) - (x2,1 * x3,7)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x2,8 = ((x2,8 * x3,1) - (x2,1 * x3,8)) / x3,1 = ((0 * 5) - (0 * -2)) / 5 = 0; x2,9 = ((x2,9 * x3,1) - (x2,1 * x3,9)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x4,1 = ((x4,1 * x3,1) - (x4,1 * x3,1)) / x3,1 = ((0 * 5) - (0 * 5)) / 5 = 0; x4,3 = ((x4,3 * x3,1) - (x4,1 * x3,3)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x4,4 = ((x4,4 * x3,1) - (x4,1 * x3,4)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x4,5 = ((x4,5 * x3,1) - (x4,1 * x3,5)) / x3,1 = ((0 * 5) - (0 * 1)) / 5 = 0; x4,6 = ((x4,6 * x3,1) - (x4,1 * x3,6)) / x3,1 = ((-0.5 * 5) - (0 * 2)) / 5 = -0.5; x4,7 = ((x4,7 * x3,1) - (x4,1 * x3,7)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x4,8 = ((x4,8 * x3,1) - (x4,1 * x3,8)) / x3,1 = ((0.5 * 5) - (0 * -2)) / 5 = 0.5; x4,9 = ((x4,9 * x3,1) - (x4,1 * x3,9)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x5,1 = ((x5,1 * x3,1) - (x5,1 * x3,1)) / x3,1 = ((0 * 5) - (0 * 5)) / 5 = 0; x5,3 = ((x5,3 * x3,1) - (x5,1 * x3,3)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x5,4 = ((x5,4 * x3,1) - (x5,1 * x3,4)) / x3,1 = ((0 * 5) - (0 * 0)) / 5 = 0; x5,5 = ((x5,5 * x3,1) - (x5,1 * x3,5)) / x3,1 = ((0 * 5) - (0 * 1)) / 5 = 0; x5,6 = ((x5,6 * x3,1) - (x5,1 * x3,6)) / x3,1 = ((0 * 5) - (0 * 2)) / 5 = 0; x5,7 = ((x5,7 * x3,1) - (x5,1 * x3,7)) / x3,1 = ((-1 * 5) - (0 * 0)) / 5 = -1; x5,8 = ((x5,8 * x3,1) - (x5,1 * x3,8)) / x3,1 = ((0 * 5) - (0 * -2)) / 5 = 0; x5,9 = ((x5,9 * x3,1) - (x5,1 * x3,9)) / x3,1 = ((1 * 5) - (0 * 0)) / 5 = 1; Maxx1 = ((Cb1 * x1,1) + (Cb2 * x2,1) + (Cb3 * x3,1) + (Cb4 * x4,1) + (Cb5 * x5,1) ) - kx1 = ((0 * 0) + (0 * 0) + (3 * 1) + (4 * 0) + (-M * 0) ) - 3 = 0; Maxx2 = ((Cb1 * x1,2) + (Cb2 * x2,2) + (Cb3 * x3,2) + (Cb4 * x4,2) + (Cb5 * x5,2) ) - kx2 = ((0 * 0) + (0 * 0) + (3 * 0) + (4 * 1) + (-M * 0) ) - 4 = 0; Maxx3 = ((Cb1 * x1,3) + (Cb2 * x2,3) + (Cb3 * x3,3) + (Cb4 * x4,3) + (Cb5 * x5,3) ) - kx3 = ((0 * 1) + (0 * 0) + (3 * 0) + (4 * 0) + (-M * 0) ) - 0 = 0; Maxx4 = ((Cb1 * x1,4) + (Cb2 * x2,4) + (Cb3 * x3,4) + (Cb4 * x4,4) + (Cb5 * x5,4) ) - kx4 = ((0 * 0) + (0 * 1) + (3 * 0) + (4 * 0) + (-M * 0) ) - 0 = 0; Maxx5 = ((Cb1 * x1,5) + (Cb2 * x2,5) + (Cb3 * x3,5) + (Cb4 * x4,5) + (Cb5 * x5,5) ) - kx5 = ((0 * -0.4) + (0 * 0) + (3 * 0.2) + (4 * 0) + (-M * 0) ) - 0 = 0.6; Maxx6 = ((Cb1 * x1,6) + (Cb2 * x2,6) + (Cb3 * x3,6) + (Cb4 * x4,6) + (Cb5 * x5,6) ) - kx6 = ((0 * -0.3) + (0 * 0) + (3 * 0.4) + (4 * -0.5) + (-M * 0) ) - 0 = -0.8; Maxx7 = ((Cb1 * x1,7) + (Cb2 * x2,7) + (Cb3 * x3,7) + (Cb4 * x4,7) + (Cb5 * x5,7) ) - kx7 = ((0 * 0) + (0 * 0) + (3 * 0) + (4 * 0) + (-M * -1) ) - 0 = M; Maxx8 = ((Cb1 * x1,8) + (Cb2 * x2,8) + (Cb3 * x3,8) + (Cb4 * x4,8) + (Cb5 * x5,8) ) - kx8 = ((0 * 0.3) + (0 * 0) + (3 * -0.4) + (4 * 0.5) + (-M * 0) ) - -M = M+0.8; Maxx9 = ((Cb1 * x1,9) + (Cb2 * x2,9) + (Cb3 * x3,9) + (Cb4 * x4,9) + (Cb5 * x5,9) ) - kx9 = ((0 * 0) + (0 * 0) + (3 * 0) + (4 * 0) + (-M * 1) ) - -M = 0; For the results of the calculations of the previous iteration, we remove the variable from the basis x1 and put in her place x6. Click on the "Pivot" button to perform the pivot operation. The preliminary stage begins with the need to get rid of negative values ââ(if any) in the right part of the restrictions. Usually the term "dual problem" refers to the Lagrangian dual problem but other dual problems are used â for example, the Wolfe dual problem and the Fenchel dual problem.The Lagrangian dual problem is obtained by forming the Lagrangian of a minimization problem by using nonnegative Lagrange multipliers to add the constraints to the objective function, and then solving for ⦠After unblocking website please refresh the page and click on find button again. Solve either the original problem or its dual by the simplex method, and then give the solutions to both. Click the "Initialize table" button. The 'interior-point-legacy' method is based on LIPSOL (Linear Interior Point Solver, ), which is a variant of Mehrotra's predictor-corrector algorithm , a primal-dual interior-point method.A number of preprocessing steps occur before the algorithm begins to iterate. The variables that are present in the basis are equal to the corresponding cells of the column P, all other variables are equal to zero. 10. Simplex Method: It is one of the solution method used in linear programming problems that involves two variables or a large number of constraint. subject to the following constraints. Dual Problem for Standard Minimization In a nutshell, we will reconstruct the minimization problem into a maximization problem by converting it into what we call a Dual Problem . Determine the dual problem. (1) This is different from Solving the dual problem ⦠3) Add Column Add a column to constraints matrix (and hence to costs vector). Home > Operation Research calculators > Dual Simplex method calculator. Since Problem (2) has a name, it is helpful to have a generic name for the original linear program. Linear Programming: It is a method used to find the maximum or minimum value for linear objective function. Maxx1 = ((Cb1 * x1,1) + (Cb2 * x2,1) + (Cb3 * x3,1) + (Cb4 * x4,1) + (Cb5 * x5,1) ) - kx1 = ((0 * 2) + (0 * 0) + (0 * 5) + (-M * 0) + (-M * 0) ) - 3 = -3; Maxx2 = ((Cb1 * x1,2) + (Cb2 * x2,2) + (Cb3 * x3,2) + (Cb4 * x4,2) + (Cb5 * x5,2) ) - kx2 = ((0 * 1) + (0 * 0) + (0 * 4) + (-M * 2) + (-M * 0) ) - 4 = -2M-4; Maxx3 = ((Cb1 * x1,3) + (Cb2 * x2,3) + (Cb3 * x3,3) + (Cb4 * x4,3) + (Cb5 * x5,3) ) - kx3 = ((0 * 1) + (0 * 0) + (0 * 0) + (-M * 0) + (-M * 0) ) - 0 = 0; Maxx4 = ((Cb1 * x1,4) + (Cb2 * x2,4) + (Cb3 * x3,4) + (Cb4 * x4,4) + (Cb5 * x5,4) ) - kx4 = ((0 * 0) + (0 * 1) + (0 * 0) + (-M * 0) + (-M * 0) ) - 0 = 0; Maxx5 = ((Cb1 * x1,5) + (Cb2 * x2,5) + (Cb3 * x3,5) + (Cb4 * x4,5) + (Cb5 * x5,5) ) - kx5 = ((0 * 0) + (0 * 0) + (0 * 1) + (-M * 0) + (-M * 0) ) - 0 = 0; Maxx6 = ((Cb1 * x1,6) + (Cb2 * x2,6) + (Cb3 * x3,6) + (Cb4 * x4,6) + (Cb5 * x5,6) ) - kx6 = ((0 * 0) + (0 * 0) + (0 * 0) + (-M * -1) + (-M * 0) ) - 0 = M; Maxx7 = ((Cb1 * x1,7) + (Cb2 * x2,7) + (Cb3 * x3,7) + (Cb4 * x4,7) + (Cb5 * x5,7) ) - kx7 = ((0 * 0) + (0 * 0) + (0 * 0) + (-M * 0) + (-M * -1) ) - 0 = M; Maxx8 = ((Cb1 * x1,8) + (Cb2 * x2,8) + (Cb3 * x3,8) + (Cb4 * x4,8) + (Cb5 * x5,8) ) - kx8 = ((0 * 0) + (0 * 0) + (0 * 0) + (-M * 1) + (-M * 0) ) - -M = 0; Maxx9 = ((Cb1 * x1,9) + (Cb2 * x2,9) + (Cb3 * x3,9) + (Cb4 * x4,9) + (Cb5 * x5,9) ) - kx9 = ((0 * 0) + (0 * 0) + (0 * 0) + (-M * 0) + (-M * 1) ) - -M = 0; Since there are negative values ââamong the estimates of the controlled variables, the current table does not yet have an optimal solution. The optimal solution is: w 1 = 3/8, w 2 = 3/4 z = 40 X 3/8 + 50 X 3/4= 105/2. To perform pivot operation. Next, you need to get rid of inequalities, for which we introduce compensating variables in the left-hand side of the inequalities. It was created by the American mathematician George Dantzig in 1947. Therefore, in the basis we introduce the variable with the smallest negative estimate. Using the calculator, we click "B" then enter Solution 1 Volume 15 Solution 1 Concentration 75 Solution 2 Concentration 95 Solution 3 Concentration 80. By browsing this website, you agree to our use of cookies. Each constraint must have one basis variable. Determine the dual problem. Problem (2) is called the dual of Problem (1). Primal to dual conversion calculator - Solve the Linear programming problem using Primal to dual conversion, step-by-step. By browsing this website, you agree to our use of cookies. Clicking "Calculate" we see the answer is: Volume of Solution 2 Needed 5. The best part about this calculator is that it can also generate the examples so that you can understand the method. Linear Programming - Dual Simplex Tableau Generator, solve a linear programming problem with the dual simplex algorithm. The algorithm solves a problem accurately within finitely many steps, ascertains its insolubility or a lack of bounds. Hungarian method, dual simplex, matrix games, potential method, traveling salesman problem, dynamic programming THE DUAL SIMPLEX METHOD. At the intersection of the line that corresponds to the variable that is derived from the basis, and the column that corresponds to the variable that is entered into the basis, is the resolving element. If the calculator did not compute something or you have ⦠All other cells remain unchanged. Complete, detailed, step-by-step description of solutions. Please add atozmath.com to your ad blocking whitelist or disable your adblocking software. Without using the calculator: You have 15 liters of 75% antifreeze. X+ y +z 212 4x + y x 20.720, 249 220 Determine the dual problem. more. By browsing this website, you agree to our use of cookies. Each equation has containing the unknown variables X, Y and Z. All other cells remain unchanged. Compensating variables are included in the objective function of the problem with a zero coefficient. ]+ denotes the projection on the nonnegative orthant Rm +, α precondition: Add solver: Load the Solver Add-in in Excel. All other cells remain unchanged. Solving an equivalent (dual) problem solves the original (primary) problem. Solve the Linear programming problem using, This site is protected by reCAPTCHA and the Google. Solve for (comma-separated): Leave empty for automatic determination, or specify variables like x,y . (The data from the previous iteration is taken as the initial data). We've detected that you are using AdBlock Plus or some other adblocking software which is preventing the page from fully loading. For what the corresponding restrictions are multiplied by -1. 1) Restart The screen back in the default problem. Problem (1) has come to be called the primal. Learn more The elements of the Q column are calculated by dividing the values ââfrom column P by the value from the column corresponding to the variable that is entered in the basis: We deduce from the basis the variable with the least positive value of Q. As programmers, we transform requirements into code. Click on the "Find pivot" button to locate the pivot element. P1 = (P1 * x3,6) - (x1,6 * P3) / x3,6 = ((245 * 0.4) - (-0.3 * 140)) / 0.4 = 350; P2 = (P2 * x3,6) - (x2,6 * P3) / x3,6 = ((225 * 0.4) - (0 * 140)) / 0.4 = 225; P4 = (P4 * x3,6) - (x4,6 * P3) / x3,6 = ((75 * 0.4) - (-0.5 * 140)) / 0.4 = 250; P5 = (P5 * x3,6) - (x5,6 * P3) / x3,6 = ((0 * 0.4) - (0 * 140)) / 0.4 = 0; x1,1 = ((x1,1 * x3,6) - (x1,6 * x3,1)) / x3,6 = ((0 * 0.4) - (-0.3 * 1)) / 0.4 = 0.75; x1,2 = ((x1,2 * x3,6) - (x1,6 * x3,2)) / x3,6 = ((0 * 0.4) - (-0.3 * 0)) / 0.4 = 0; x1,3 = ((x1,3 * x3,6) - (x1,6 * x3,3)) / x3,6 = ((1 * 0.4) - (-0.3 * 0)) / 0.4 = 1; x1,4 = ((x1,4 * x3,6) - (x1,6 * x3,4)) / x3,6 = ((0 * 0.4) - (-0.3 * 0)) / 0.4 = 0; x1,5 = ((x1,5 * x3,6) - (x1,6 * x3,5)) / x3,6 = ((-0.4 * 0.4) - (-0.3 * 0.2)) / 0.4 = -0.25; x1,6 = ((x1,6 * x3,6) - (x1,6 * x3,6)) / x3,6 = ((-0.3 * 0.4) - (-0.3 * 0.4)) / 0.4 = 0; x1,8 = ((x1,8 * x3,6) - (x1,6 * x3,8)) / x3,6 = ((0.3 * 0.4) - (-0.3 * -0.4)) / 0.4 = 0; x1,9 = ((x1,9 * x3,6) - (x1,6 * x3,9)) / x3,6 = ((0 * 0.4) - (-0.3 * 0)) / 0.4 = 0; x2,1 = ((x2,1 * x3,6) - (x2,6 * x3,1)) / x3,6 = ((0 * 0.4) - (0 * 1)) / 0.4 = 0; x2,2 = ((x2,2 * x3,6) - (x2,6 * x3,2)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x2,3 = ((x2,3 * x3,6) - (x2,6 * x3,3)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x2,4 = ((x2,4 * x3,6) - (x2,6 * x3,4)) / x3,6 = ((1 * 0.4) - (0 * 0)) / 0.4 = 1; x2,5 = ((x2,5 * x3,6) - (x2,6 * x3,5)) / x3,6 = ((0 * 0.4) - (0 * 0.2)) / 0.4 = 0; x2,6 = ((x2,6 * x3,6) - (x2,6 * x3,6)) / x3,6 = ((0 * 0.4) - (0 * 0.4)) / 0.4 = 0; x2,8 = ((x2,8 * x3,6) - (x2,6 * x3,8)) / x3,6 = ((0 * 0.4) - (0 * -0.4)) / 0.4 = 0; x2,9 = ((x2,9 * x3,6) - (x2,6 * x3,9)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x4,1 = ((x4,1 * x3,6) - (x4,6 * x3,1)) / x3,6 = ((0 * 0.4) - (-0.5 * 1)) / 0.4 = 1.25; x4,2 = ((x4,2 * x3,6) - (x4,6 * x3,2)) / x3,6 = ((1 * 0.4) - (-0.5 * 0)) / 0.4 = 1; x4,3 = ((x4,3 * x3,6) - (x4,6 * x3,3)) / x3,6 = ((0 * 0.4) - (-0.5 * 0)) / 0.4 = 0; x4,4 = ((x4,4 * x3,6) - (x4,6 * x3,4)) / x3,6 = ((0 * 0.4) - (-0.5 * 0)) / 0.4 = 0; x4,5 = ((x4,5 * x3,6) - (x4,6 * x3,5)) / x3,6 = ((0 * 0.4) - (-0.5 * 0.2)) / 0.4 = 0.25; x4,6 = ((x4,6 * x3,6) - (x4,6 * x3,6)) / x3,6 = ((-0.5 * 0.4) - (-0.5 * 0.4)) / 0.4 = 0; x4,8 = ((x4,8 * x3,6) - (x4,6 * x3,8)) / x3,6 = ((0.5 * 0.4) - (-0.5 * -0.4)) / 0.4 = 0; x4,9 = ((x4,9 * x3,6) - (x4,6 * x3,9)) / x3,6 = ((0 * 0.4) - (-0.5 * 0)) / 0.4 = 0; x5,1 = ((x5,1 * x3,6) - (x5,6 * x3,1)) / x3,6 = ((0 * 0.4) - (0 * 1)) / 0.4 = 0; x5,2 = ((x5,2 * x3,6) - (x5,6 * x3,2)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x5,3 = ((x5,3 * x3,6) - (x5,6 * x3,3)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x5,4 = ((x5,4 * x3,6) - (x5,6 * x3,4)) / x3,6 = ((0 * 0.4) - (0 * 0)) / 0.4 = 0; x5,5 = ((x5,5 * x3,6) - (x5,6 * x3,5)) / x3,6 = ((0 * 0.4) - (0 * 0.2)) / 0.4 = 0; x5,6 = ((x5,6 * x3,6) - (x5,6 * x3,6)) / x3,6 = ((0 * 0.4) - (0 * 0.4)) / 0.4 = 0; x5,8 = ((x5,8 * x3,6) - (x5,6 * x3,8)) / x3,6 = ((0 * 0.4) - (0 * -0.4)) / 0.4 = 0; x5,9 = ((x5,9 * x3,6) - (x5,6 * x3,9)) / x3,6 = ((1 * 0.4) - (0 * 0)) / 0.4 = 1; Maxx1 = ((Cb1 * x1,1) + (Cb2 * x2,1) + (Cb3 * x3,1) + (Cb4 * x4,1) + (Cb5 * x5,1) ) - kx1 = ((0 * 0.75) + (0 * 0) + (0 * 2.5) + (4 * 1.25) + (-M * 0) ) - 3 = 2; Maxx5 = ((Cb1 * x1,5) + (Cb2 * x2,5) + (Cb3 * x3,5) + (Cb4 * x4,5) + (Cb5 * x5,5) ) - kx5 = ((0 * -0.25) + (0 * 0) + (0 * 0.5) + (4 * 0.25) + (-M * 0) ) - 0 = 1; Maxx6 = ((Cb1 * x1,6) + (Cb2 * x2,6) + (Cb3 * x3,6) + (Cb4 * x4,6) + (Cb5 * x5,6) ) - kx6 = ((0 * 0) + (0 * 0) + (0 * 1) + (4 * 0) + (-M * 0) ) - 0 = 0; Maxx8 = ((Cb1 * x1,8) + (Cb2 * x2,8) + (Cb3 * x3,8) + (Cb4 * x4,8) + (Cb5 * x5,8) ) - kx8 = ((0 * 0) + (0 * 0) + (0 * -1) + (4 * 0) + (-M * 0) ) - -M = M; Since there are no negative values ââamong the estimates of the controlled variables, the current table has an optimal solution. 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