Download Solving a Diet Problem using the Simplex Method and more Study notes Calculus in PDF only on Docsity! 1 SOLUTION USING SIMPLEX METHOD If you would like to further talk about the solution methods in the classroom, you could show how to solve this problem using simplex method as follows: In our diet problem, we are looking at a “standard” minimization problem. What does the “standard minimization” mean? It means that the constraints are all linear; the constants are non-‐ negative; the inequality symbols are greater or equal to; the non-‐negativity constraints are placed on our variables; and, we’re minimizing. After recognizing that it’s a standard minimization problem, in order to apply the simplex method, we need to convert it to a standard maximization problem. To do that, we will start writing a matrix, which comes from the coefficients and constants of the constraints as well as the coefficients and constant of the objective. A = Then, we will find the transpose of this matrix by interchanging its rows and columns. AT = We can see that the rows of this matrix are the columns of the first matrix. This new matrix is a maximization problem so that the corresponding maximization problem is: Maximize w = 135.5 y1 + 45.5 y2 + 32.5 y3 + 1,000 y4 subject to the constraints 12 y1 + 3 y2 + 9 y3 + 171 y4 <= 2 33 y1 + 6 y2 + y3 + 150 y4 <= 1.5 y1 > 0, y2 >= 0, y3 >= 0, and y4 >= 0 This formulation is called the dual of the original problem. Since this is the standard form for a maximization problem, we can now apply the simplex method. To do that, the first step is to Step 1: Rewrite the objective formula as an equation: w -‐ 135.5 y1 – 45.5 y2 – 32.5 y3 -‐ 1,000 y4 = 0. We will call this the objective equation. Step 2: Remove the inequalities from the constraints. To do that, we will introduce two slack variables as m and n and add them to the constraints. We will call these the constraint equations: 2 w – 135.5 y1 + 45.5 y2 + 32.5 y3 + 1,000 y4 = 0 objective equation 12 y1 + 3 y2 + 9 y3 + 171 y4 + m= 2 constraint equation 33 y1 + 6 y2 + y3 + 150 y4 + n = 1.5 constraint equation Step 3: Let’s next put all of these in a tableau. RHS = right hand side Step 4: Our next step is to identify the pivot. To do that, we first find the pivot column. The pivot column is the column with the largest negative value in the objective equation. Let’s draw a line with the green marker around the pivot column. Next comes identifying the pivot row. This is done by dividing the values of the constraints by the value in the pivot column. The lowest value gives the pivot row. 2 divided by 171 = 0.012; 1.5 divided by 150 = 0.01. The pivot is where the pivot column and pivot row intersect. Let’s draw a line around the pivot with the red marker. Step 5: We now need to make the pivot equal to zero. To do this, we divide the pivot row by the pivot value. It’s best to keep the values as fractions. Step 6: Next we need to make the pivot column values into zeros. To do this, we add or subtract multiples of the pivot column. The pivot column value in the other constraint will go to zero if we add -‐171 times the pivot row. The pivot column in the objective function will go to zero if we add 1,000 times the pivot row. This then is the first iteration of the simplex method. y1 y2 y3 y4 m n w RHS values 12 3 9 171 1 0 0 2 33 6 1 150 0 1 0 1.5 -‐135.5 -‐45.5 -‐32.5 -‐1,000 0 0 1 0 y1 y2 y3 y4 m n w RHS values Ratios 12 3 9 171 1 0 0 2 2/171 33 6 1 150 0 1 0 1.5 1.5/150 -‐135.5 -‐45.5 -‐32.5 -‐1,000 0 0 1 0 y1 y2 y3 y4 m n w RHS values 12 3 9 171 1 0 0 2 33/150 6/150 1/150 1 0 1/150 0 1.5/150 -‐135.5 -‐45.5 -‐32.5 -‐1,000 0 0 1 0 y1 y2 y3 y4 m n w RHS values -‐25.6 -‐3.8 7.9 0 1 -‐1.1 0 0.3 0.2 0.04 0.007 1 0 0.007 0 0.01 -‐25.5 -‐25.5 -‐29.2 0 0 3.3 1 5
