Explanation Given information: The f = x + 2 y + 3 z subject to the constraints, x + z ≤ 20 x + y ≥ 30 y + z ≤ 20 Formula used: The mixed constraints by the use of Simplex method, (1) The problem is to maximize use f and to minimize use − f . (2) Make all the constraints to ≤ by multiply − 1 both the sides of the constraint and use slack variable to make simplex matrix. (3) Check the negative entry in the upper portion of the last column if entries are negative go to step 4 otherwise apply simplex method. (4) Select the negative entry in the same row and take that column as pivot column. In pivot column find all quotients above the line for the entries then find the pivot from the smallest positive quotient. Then after completing the operation go to step 3. Calculation: Consider the provided function, f = x + 2 y + 3 z And the constraints, x + z ≤ 20 x + y ≥ 30 y + z ≤ 20 Express all the constraints as ≤ constraints. Here, multiply x + y ≥ 30 by − 1 to obtain ≤ constraints. So, the constraints are, x + z ≤ 20 − x − y ≤ − 30 y + z ≤ 20 Slack variables are used to write the inequalities as equations. The equations after adding slack variables s 1 , s 2 , s 3 are, x + z + s 1 = 20 − x − y + s 2 = − 30 y + z + s 3 = 20 The simplex matrix is, [ x y z s 1 s 2 s 3 − f 1 0 1 1 0 0 0 20 − 1 − 1 0 0 1 0 0 − 30 0 1 1 0 0 1 0 20 1 2 3 0 0 0 1 0 ] The upper portion of last column has a negative value If values for the variables x , y , z , s 1 , s 2 , s 3 are computed then s 2 = − 30 is obtained. This violates the condition of simplex method that all the variables be non-negative. Choosing − 30 which is row two. In this row there are negative entries in column one and column two. Choosing the pivot column as column two. The pivot is determined from the quotients as, [ x y z s 1 s 2 s 3 − f 1 0 1 1 0 0 0 20 − 1 − 1 0 0 1 0 0 − 30 0 1 1 0 0 1 0 20 1 2 3 0 0 0 1 0 ] 20 / 0 − 30 / − 1 = 30 20 / 1 = 20 The least positive quotient is 20 . Hence the pivot row is row three and pivot column is column two. The intersection is 1 which the pivot as shown is, [ x y z s 1 s 2 s 3 − f 1 0 1 1 0 0 0 20 − 1 − 1 0 0 1 0 0 − 30 0 1 1 0 0 1 0 20 1 2 3 0 0 0 1 0 ] Pivoting with this entry gives the following matrix, [ x y z s 1 s 2 s 3 − f 1 0 1 1 0 0 0 20 − 1 0 1 0 1 1 0 − 10 0 1 1 0 0 1 0 20 1 0 1 0 0 − 2 1 − 40 ] The upper portion of last column has one negative value. If values for the variables x , y , z , s 1 , s 2 , s 3 are computed then it violates the condition of simplex method that all the variables be non-negative. Choosing − 10 which is row two. In this row there is negative entry in column one. Choosing this column as pivot column. The pivot is determined from the quotients as, [ x y z s 1 s 2 s 3 − f 1 0 1 1 0 0 0 20 − 1 0 1 0 1 1 0 − 10 0 1 1 0 0 1 0 20 1 0 1 0 0 − 2 1 − 40 ] 20 / 1 = 20 − 10 / − 1 = 10 20 / 0 The least positive quotient is 10

Mathematical Applications for the Management, Life, and Social Sciences

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ISBN: 9781337625340

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 4.5, Problem 19E

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