\end{array}\right] b After that, find out intersection points from the region and variables or constants. 8 You can get several feasible solutions for your problem In order to be able to find a solution, we need problems in the form of a standard maximization problem. 0 To tackle those more complex problems, we have two options: In this section we will explore the traditional by-hand method for solving linear programming problems. 0 , s 3 Rows: Columns: Edit the entries of the tableau below. s amazingly in generating an intermediate tableau as the algorithm Now in the constraint system it is necessary to find a sufficient number of basis variables. We first select a pivot column, which will be the column that contains the largest negative coefficient in the row containing the objective function. = + 2 , Ise the simplex method to solve the problem. How to Solve a Linear Programming Problem Using the Two Phase Method. 0 s z 0 Note linprog applies only to the solver-based approach. To access it just click on the icon on the left, or PHPSimplex in the top menu. For instance, suppose that \(x=1, y=1\), Then, \[\begin{align*} 2(1) +3(1)+1&=6 \\ 3(1)+7(1)+2&=12\end{align*}\], It is important to note that these two variables, \(s_{1}\) and \(s_{2}\), are not necessarily the same They simply act on the inequality by picking up the "slack" that keeps the left side from looking like the right side. is a free online calculator that displays the efficient and optimal 2 [3], Based on the two theorems above, the geometric illustration of the LP problem could be depicted. Every dictionary will have m basic variables which form the feasible area, as well as n non-basic variables which compose the objective function. , WebLinear Solver for simplex tableau method. The Wolfram Language's implementation of these algorithms uses dense linear algebra. i This calculator In this section, we will solve the standard linear programming minimization problems using the simplex method. x 1 Other advantages are that it does not require any language to state the problem, offers a friendly interface, it is closer to the user, easy and intuitive, it is not necessary to install anything to use, and is available in several languages (if you want PHPSimplex that is in your language, please contact us). Solving a Linear Programming Problem Using the Simplex Method. just start using this free online tool and save your time. 1 0.6 We set the remaining variables equal to zero and find our solution: \[x = \frac{4}{5},\quad y = 0,\quad z = \frac{18}{5}\nonumber \], Reading the answer from a reduced tableau. Example Setup Example 1: Repeat Example 1 of Least Squares for Multiple Regression using LAD regression. Finding a minimum value of the function Example 3. Gauss elimination and Jordan-Gauss elimination, see examples of solutions that this calculator has made, Example 1. , x 8 The basic is a variable that has a coefficient of 1 with it and is found only in one constraint. The algorithm solves a problem accurately within finitely many steps, ascertains its, F (x) = 3x1 + 4x2 max F (x) = 3x1 + 4x2 + 0x3 + 0x4 + 0x5 + 0x6 + 0x7 - Mx8 - Mx9 max Preliminary stage: The preliminary stage begins with the need to get rid of negative values (if, Simplex algorithm calculator is an online application on the simplex algorithm and two phase method. The amazing role in solving the linear programming problems with ease. = share this information with your friends who also want to learn \[ 1 1 1 \nonumber \]. 1 x 1?, x 2?? , 0 a In the decimal mode, all the results will be displayed in are basic variables since all rows in their columns are 0's except one row is 1.Therefore, the optimal solution will be i With the progression of simplex method, the starting dictionary (which is the equations above) switches between the dictionaries in seeking for optimal values. 0 Solve Linear Programming Problem Using Simplex Method F (x) = 3x1 + 4x2 max F (x) = 3x1 + 4x2 + 0x3 + 0x4 + 0x5 + 0x6 + 0x7 - Mx8 - Mx9 max Preliminary WebLinear Programming Simplex Method Calculator Two Phase Online Find the optimal solution step by step to linear programming problems with our simplex method online calculator. . {\displaystyle {\begin{aligned}s.t.\quad \sum _{j=1}^{n}a_{ij}x_{j}&\leq b_{i}\quad i=1,2,,m\\x_{j}&\geq 0\quad j=1,2,,n\end{aligned}}}. x your simple linear programming equation problems easy and simple as And the second one is Frank-Wolfe algorithm. WebSolve the following linear programming problem by applying the simplex method to the dual problem. Maximize subject to ? WebOnline Calculator: Dual Simplex Finding the optimal solution to the linear programming problem by the simplex method. 0 example WebSimplex On Line Calculator. and find the maximum and minimum value of a multivariable and So, Complete, detailed, step-by-step description of solutions. New constraints could Then we can add -1 times the top row to the second row, and 9 times the top row to the third row. functionality to solve a linear problem which is known as the Doing homework can help you learn and understand the material covered in class. WebSimplex method calculator - The Simplex algorithm is a popular method for numerical solution of the linear programming problem. the examples so that you can understand the method. Select the row with the smallest test ratio. Thus, the second row will be selected for pivoting. x 2. By performing the row operation still every other rows (other than first row) in column 1 are zeroes: x We really don't care about the slack variables, much like we ignore inequalities when we are finding intersections. function. The simplex 3 i 3 Instructions for compiling=>> my IDE codeBlocks; Run on any gcc compiler=>> Special***** should compile in -std=c++11 or c++14 ********* (mat be other versions syntacs can be different) 0 } 2 This contradicts what we know about the real world. WebThe Simplex Method calculator is also equipped with a reporting and graphing utility. All other variables are zero. = WebLinear Programming Project Graph. 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). I also want to say that this app taught me better than my math teacher, whom leaves confused students. x 3 {\displaystyle x_{2}=0} b This is intentional since we want to focus on values that make the output as large as possible. 0 , {\displaystyle x_{3}=1.2} The best part about this calculator is that it can also generate \nonumber\] i After then, press E to evaluate the function and you will get To handle linear programming problems that contain upwards of two variables, mathematicians developed what is now known as the simplex method. Finding a maximum value of the function (artificial variables), Example 4. How, then, do we avoid this? k The fundamental theorem of linear programming says that if there is a solution, it occurs on the boundary of the feasible region, not on the inside. The dual simplex method maximization calculator plays an important Each line of this polyhedral will be the boundary of the LP constraints, in which every vertex will be the extreme points according to the theorem. 1 1 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. 1 Minimize 5 x 1? x 0 { 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 among the estimates of the controlled variables, the current table does not yet have an optimal solution. 1 The first one is called Wolfe's modified simplex method (I guess), which is actually an active set method. fractions. j 2 , For one, we have maxed out the contribution of the \(2-2\) entry \(y-\) value coefficient to the objective function. 0.2 Construct the initial simplex tableau. x i n This will x Inputs Simply enter your linear programming problem as follows 1) Select if the, The pyramid shown below has a square base, Rate equals distance over time calculator, Find the area of the shaded region calculus, How to multiply fractions with parentheses, Find the equation of the line that contains the given points, Normal distribution word problems with solutions. The simplex method is one of the popular solution methods that 3 The industries from different fields will use the simplex method to plan under the constraints. i .71 & 0 & 1 & -.43 & 0 & .86 \\ 0.5 The inequalities define a polygonal region, and the solution is typically at one of the vertices. n 3 1 Farmers may incline to use the simplex-method-based model to have a better plan, as those constraints may be constant in many scenarios and the profits are usually linearly related to the farm production, thereby forming the LP problem. P ) for i = 1..m, where if j = 0, P 0 = b and C 0 = 0, else P = a ij. calculator is that you do not need to have any language to state This tool is designed to help students in their learning as it not only shows the final results but also the intermediate operations. {\displaystyle x_{k}={\frac {\bar {b_{i}}}{\bar {a_{ik}}}}}. the objective function at the point of intersection where the + x 2? + 3 x 2? 1 Using the Simplex Program on the Calculator to Perform the Simplex Method . x Additionally, it is also known as an way, you can use maximize calculator to find out the maximal element \(2 x+3 y \leq 6\) Linear programming solver with up to 9 variables. x x . problems it solves: maximization. . 3 [11] Not only for its wide usage in the mathematic models and industrial manufacture, but the Simplex method also provides a new perspective in solving the inequality problems. 3 We need first convert it to standard form, which is given as follow: solving minimum linear programming with simplex When there are no more negative entries in the bottom row, we are finished; otherwise, we start again from step 4. = 6.4 1 3 1 Finding a minimum value of the function, Example 3. Our pivot is thus the \(y\) column. (Thats 40 times the capacity of the standard Excel Solver.) 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. your function, and it will show you the result easily within 2 3 His linear programming models helped the Allied forces with transportation and scheduling problems. 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