100. 1 , 1 2 The identical, Free time to spend with your family and friends. The simplex method is commonly used in many programming problems. Therefore, it will be convenient to add those slack variables Considering the following numerical example to gain better understanding: max 1 Therefore, if an LP has an optimal solution, there must be an extreme point of the feasible region that is optimal. 0 i Legal. George B. Dantzig (19142005). 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. z your simple linear programming equation problems easy and simple as The quotients are computed by dividing the far right column by the identified column in step 4. The on-line Simplex method Aplicattion. Many other fields will use this method since the LP problem is gaining popularity in recent days and the simplex method plays a crucial role in solving those problems. = 0 + mathematical method that is used to obtain the best outcome in a 2 All rights reserved. 3) Add Column Add a column to You can export your results in graphs and reports for further review and analysis. = {\displaystyle {\begin{aligned}z-4x_{1}-x_{2}-4x_{3}&=0\\2x_{1}+x_{2}+x_{3}+s_{1}&=2\\x_{1}+2x_{2}+3x_{3}+s_{2}&=4\\2x_{1}+2x_{2}+x_{3}+s_{3}&=8\\x_{1},x_{2},x_{3},s_{1},s_{2},s_{3}&\geq 0\end{aligned}}}. In order to be able to find a solution, we need problems in the form of a standard maximization problem. 2 Legal. 1 j direct solution of maximization or minimization. Can be used offline, easy to use, it gives answers in different forms such as fractions, decimals etc. 0 . Daniel Izquierdo Granja 2) Dualize Transforms the problem in its dual. through this calculator. x Find out a formula according to your function and then use this and find the maximum and minimum value of a multivariable and 0 4 , 2 Additionally, it is also known as an i calculator is that you do not need to have any language to state Linear Programming and Optimization using Python | Towards Data Science Write Sign up Sign In 500 Apologies, but something went wrong on our end. In order to help you in understanding the simplex method calculator To handle linear programming problems that contain upwards of two variables, mathematicians developed what is now known as the simplex method. The basic is a variable that has a coefficient of 1 with it and is found only in one constraint. It is one of the popular methods that are used to avail of the It is based on the theorem that if a system And the second one is Frank-Wolfe algorithm. 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 among the estimates of the controlled variables, the current table has an optimal solution. With considering that it is usually the case that the constraints or tradeoffs and desired outcomes are linearly related to the controllable variables, many people will develop the models to solve the LP problem via the simplex method, for instance, the agricultural and economic problems, Farmers usually need to rationally allocate the existed resources to obtain the maximum profits. This calculator is an amazing tool that can help you in x 0.5 The online simplex method calculator or simplex solver, plays an The Wolfram Language's implementation of these algorithms uses dense linear algebra. Since the coefficient in the first row is 1 and 4 for the second row, the first row should be pivoted. 0.5 b 1 {\displaystyle x_{k}} 1 We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. + 3 x 2? 1 Practice. 1 , 1 I've given the following LP problem: P (x) = 4x1 + 5x2 -> max; x1 - 2x2 <= 15; 4x1 + 3x2 <= 24; -2x1 + 5x2 >= 20; x1 >= 0; x2 >= 0; I have to perform 3 tasks: Convert this problem to Normal form and check how many variables and constraints there are Convert the normal form to a Big M problem and perform a Big M simplex for the first minimization functionality to calculate the problem can be Then make equations out of the inequalities. z We transfer the row with the resolving element from the previous table into the current table, elementwise dividing its values into the resolving element: The remaining empty cells, except for the row of estimates and the column Q, are calculated using the rectangle method, relative to the resolving element: P1 = (P1 * x4,2) - (x1,2 * P4) / x4,2 = ((600 * 2) - (1 * 150)) / 2 = 525; P2 = (P2 * x4,2) - (x2,2 * P4) / x4,2 = ((225 * 2) - (0 * 150)) / 2 = 225; P3 = (P3 * x4,2) - (x3,2 * P4) / x4,2 = ((1000 * 2) - (4 * 150)) / 2 = 700; P5 = (P5 * x4,2) - (x5,2 * P4) / x4,2 = ((0 * 2) - (0 * 150)) / 2 = 0; x1,1 = ((x1,1 * x4,2) - (x1,2 * x4,1)) / x4,2 = ((2 * 2) - (1 * 0)) / 2 = 2; x1,2 = ((x1,2 * x4,2) - (x1,2 * x4,2)) / x4,2 = ((1 * 2) - (1 * 2)) / 2 = 0; x1,4 = ((x1,4 * x4,2) - (x1,2 * x4,4)) / x4,2 = ((0 * 2) - (1 * 0)) / 2 = 0; x1,5 = ((x1,5 * x4,2) - (x1,2 * x4,5)) / x4,2 = ((0 * 2) - (1 * 0)) / 2 = 0; x1,6 = ((x1,6 * x4,2) - (x1,2 * x4,6)) / x4,2 = ((0 * 2) - (1 * -1)) / 2 = 0.5; x1,7 = ((x1,7 * x4,2) - (x1,2 * x4,7)) / x4,2 = ((0 * 2) - (1 * 0)) / 2 = 0; x1,8 = ((x1,8 * x4,2) - (x1,2 * x4,8)) / x4,2 = ((0 * 2) - (1 * 1)) / 2 = -0.5; x1,9 = ((x1,9 * x4,2) - (x1,2 * x4,9)) / x4,2 = ((0 * 2) - (1 * 0)) / 2 = 0; x2,1 = ((x2,1 * x4,2) - (x2,2 * x4,1)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x2,2 = ((x2,2 * x4,2) - (x2,2 * x4,2)) / x4,2 = ((0 * 2) - (0 * 2)) / 2 = 0; x2,4 = ((x2,4 * x4,2) - (x2,2 * x4,4)) / x4,2 = ((1 * 2) - (0 * 0)) / 2 = 1; x2,5 = ((x2,5 * x4,2) - (x2,2 * x4,5)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x2,6 = ((x2,6 * x4,2) - (x2,2 * x4,6)) / x4,2 = ((0 * 2) - (0 * -1)) / 2 = 0; x2,7 = ((x2,7 * x4,2) - (x2,2 * x4,7)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x2,8 = ((x2,8 * x4,2) - (x2,2 * x4,8)) / x4,2 = ((0 * 2) - (0 * 1)) / 2 = 0; x2,9 = ((x2,9 * x4,2) - (x2,2 * x4,9)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x3,1 = ((x3,1 * x4,2) - (x3,2 * x4,1)) / x4,2 = ((5 * 2) - (4 * 0)) / 2 = 5; x3,2 = ((x3,2 * x4,2) - (x3,2 * x4,2)) / x4,2 = ((4 * 2) - (4 * 2)) / 2 = 0; x3,4 = ((x3,4 * x4,2) - (x3,2 * x4,4)) / x4,2 = ((0 * 2) - (4 * 0)) / 2 = 0; x3,5 = ((x3,5 * x4,2) - (x3,2 * x4,5)) / x4,2 = ((1 * 2) - (4 * 0)) / 2 = 1; x3,6 = ((x3,6 * x4,2) - (x3,2 * x4,6)) / x4,2 = ((0 * 2) - (4 * -1)) / 2 = 2; x3,7 = ((x3,7 * x4,2) - (x3,2 * x4,7)) / x4,2 = ((0 * 2) - (4 * 0)) / 2 = 0; x3,8 = ((x3,8 * x4,2) - (x3,2 * x4,8)) / x4,2 = ((0 * 2) - (4 * 1)) / 2 = -2; x3,9 = ((x3,9 * x4,2) - (x3,2 * x4,9)) / x4,2 = ((0 * 2) - (4 * 0)) / 2 = 0; x5,1 = ((x5,1 * x4,2) - (x5,2 * x4,1)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x5,2 = ((x5,2 * x4,2) - (x5,2 * x4,2)) / x4,2 = ((0 * 2) - (0 * 2)) / 2 = 0; x5,4 = ((x5,4 * x4,2) - (x5,2 * x4,4)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x5,5 = ((x5,5 * x4,2) - (x5,2 * x4,5)) / x4,2 = ((0 * 2) - (0 * 0)) / 2 = 0; x5,6 = ((x5,6 * x4,2) - (x5,2 * x4,6)) / x4,2 = ((0 * 2) - (0 * -1)) / 2 = 0; x5,7 = ((x5,7 * x4,2) - (x5,2 * x4,7)) / x4,2 = ((-1 * 2) - (0 * 0)) / 2 = -1; x5,8 = ((x5,8 * x4,2) - (x5,2 * x4,8)) / x4,2 = ((0 * 2) - (0 * 1)) / 2 = 0; x5,9 = ((x5,9 * x4,2) - (x5,2 * x4,9)) / x4,2 = ((1 * 2) - (0 * 0)) / 2 = 1; Maxx1 = ((Cb1 * x1,1) + (Cb2 * x2,1) + (Cb3 * x3,1) + (Cb4 * x4,1) + (Cb5 * x5,1) ) - kx1 = ((0 * 2) + (0 * 0) + (0 * 5) + (4 * 0) + (-M * 0) ) - 3 = -3; Maxx2 = ((Cb1 * x1,2) + (Cb2 * x2,2) + (Cb3 * x3,2) + (Cb4 * x4,2) + (Cb5 * x5,2) ) - kx2 = ((0 * 0) + (0 * 0) + (0 * 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) + (0 * 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) + (0 * 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) + (0 * 0) + (0 * 1) + (4 * 0) + (-M * 0) ) - 0 = 0; Maxx6 = ((Cb1 * x1,6) + (Cb2 * x2,6) + (Cb3 * x3,6) + (Cb4 * x4,6) + (Cb5 * x5,6) ) - kx6 = ((0 * 0.5) + (0 * 0) + (0 * 2) + (4 * -0.5) + (-M * 0) ) - 0 = -2; Maxx7 = ((Cb1 * x1,7) + (Cb2 * x2,7) + (Cb3 * x3,7) + (Cb4 * x4,7) + (Cb5 * x5,7) ) - kx7 = ((0 * 0) + (0 * 0) + (0 * 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 * -0.5) + (0 * 0) + (0 * -2) + (4 * 0.5) + (-M * 0) ) - -M = M+2; Maxx9 = ((Cb1 * x1,9) + (Cb2 * x2,9) + (Cb3 * x3,9) + (Cb4 * x4,9) + (Cb5 * x5,9) ) - kx9 = ((0 * 0) + (0 * 0) + (0 * 0) + (4 * 0) + (-M * 1) ) - -M = 0; For the results of the calculations of the previous iteration, we remove the variable from the basis x5 and put in her place x1. It also offers direct solution for professional use. .71 & 0 & 1 & -.43 & 0 & .86 \\ problems related solutions. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. different approach that is a Linear programming calculator with 3 Hungarian method, dual simplex, matrix games, potential method, traveling salesman problem, dynamic programming The solution of the dual linear programming problem. It is an efficient algorithm (set of mechanical steps) that toggles through corner points until it has located the one that maximizes the objective function. We thus have the following matrix: When you can obtain minimum or maximum value for the linear to calculate any complex equation or for the system of linear Follow the below-mentioned procedure to use the Linear Programming Calculator at its best. 0.5 Just like problems with ranged constraints, i.e. 1 To identify the solution set, focus we focus only on the columns with exactly one nonzero entry \(-\) these are called active variables (columns with more than one non-zero entry are thus called inactive variables). That is, write the objective function and the constraints. 2. s The most negative entry in the bottom row is in the third column, so we select that column. Complete, detailed, step-by-step description of solutions. 4 k Developed by: Read More Linear Programming Big M Method Calculator Online 1 0 the basis of this information, that tableau will be created of the 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. = 3 0 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. 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