Explanation Given Information: The cost of production in Texas plant, For each desk costs is $36 and for each computer table is $60. But the plant can produce at most 120 units per day. The cost of production in Louisiana plant, For each desk costs is $42 and for each computer table is $57. But the plant can produce at most 150 units per day. Also, the number of computer tables produces by the plant must be at least 10 more than the number of desks. And, the order is for 130 desks and 130 computer tables. Formula used: The Simplex Method for mixed Constraints: Step 1: If the provided problem is to minimize the function f , then maximize − f . Step 2: Make the constraints ≤ by multiplying − 1 to both sides of any ≥ constraints. Step 3: Examine the upper portion of the last column for any negative entries in the simplex matrix. a. Use the simplex method if there are no negative entries. b. Go to step 4 if there are negative entries. Step 4: Follow the following procedure if there are negative entries in the upper portion of the matrix: a. Search for any negative entry in the row and select that column as the pivot column. b. Find the pivot from the smaller positive quotient c. Complete the pivot operations and return to the step 3. Calculation: Assume the following variables, x 1 = Number of desks made in Texas x 2 = Number of tables made in Texas x 3 = Number of desk made in Louisiana x 4 = Number of computer tables made in Louisiana The production cost of each desk in Texas is $36 and $42 at Louisiana. Similarly, the production cost of each computer table in Texas is $60 and $57 at Louisiana. Therefore, the total cost is given by, Cost = 36 x 1 + 60 x 2 + 42 x 3 + 57 x 4 The objective is to minimize the production cost. Thus, the objective function is, Minimize Z = 36 x 1 + 60 x 2 + 42 x 3 + 57 x 4 The constraints are as follows, x 1 + x 2 ≤ 120 x 3 + x 4 ≤ 150 x 1 + x 3 ≥ 130 x 2 + x 4 ≥ 130 And, x 2 − x 1 ≥ 10 Now, convert this to maximization problem. Thus, the function that has to be maximized is, − Z = − 36 x 1 − 60 x 2 − 42 x 3 − 57 x 4 The modified constraints are; x 1 + x 2 ≤ 120 x 3 + x 4 ≤ 150 − x 1 − x 3 ≤ − 130 − x 2 − x 4 ≤ 130 And, − x 2 + x 1 ≤ 10 Rewrite these constraints as equalities using slack variables, x 1 + x 2 + s 1 = 120 x 3 + x 4 + s 2 = 150 − x 1 − x 3 + s 3 = − 130 − x 2 − x 4 + s 4 = − 130 The constraints as equalities using slack variables are; − x 2 + x 1 + s 5 = − 10 x 1 , x 2 , x 3 , x 4 ≥ 0 Thus, the simplex matrix is; [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 s 4 s 5 − Z 1 1 0 0 1 0 0 0 0 0 120 0 0 1 1 0 1 0 0 0 0 150 − 1 0 − 1 0 0 0 1 0 0 0 − 130 0 − 1 0 − 1 0 0 0 1 0 0 − 130 1 − 1 0 0 0 0 0 0 1 0 − 10 36 60 42 57 0 0 0 0 0 1 0 ] The upper portion of the last column has negative values. If values for the variables x , y , s 1 , s 2 , s 3 , s 4 , s 5 are computed then s 3 = − 130 are obtained. This violates the condition of simplex method that all the variables be non-negative. Choosing s 3 = − 130 which is row 3. In this row, there is negative entry in column 1. Choosing the pivot column as column 1. The pivot is determined from the quotients as, − 130 − 1 = 130 Thus, the pivot is − 1 in column x 1 . [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 s 4 s 5 − Z 1 1 0 0 1 0 0 0 0 0 120 0 0 1 1 0 1 0 0 0 0 150 − 1 0 − 1 0 0 0 1 0 0 0 − 130 0 − 1 0 − 1 0 0 0 1 0 0 − 130 1 − 1 0 0 0 0 0 0 1 0 − 10 36 60 42 57 0 0 0 0 0 1 0 ] Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Since, the pivot is s 3 in column x 1 and replace s 3 with x 1 . Thus, x 1 is the active variable. Use row operations to clear the pivot column of the augmented matrix. First, use the row operation − R 3 → R 3 then − R 3 + R 1 → R 1 , − R 3 + R 5 → R 5 and − 36 R 3 + R 6 → R 6 . [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 s 4 s 5 − Z 0 1 − 1 0 1 0 1 0 0 0 − 10 0 0 1 1 0 1 0 0 0 0 150 1 0 1 0 0 0 − 1 0 0 0 130 0 − 1 0 − 1 0 0 0 1 0 0 − 130 0 − 1 − 1 0 0 0 1 0 1 0 − 140 0 60 6 57 0 0 36 0 0 1 − 4680 ] The upper portion of the last column has negative values. If values for the variables x , y , s 1 , s 2 , s 3 , s 4 , s 5 are computed then s 5 = − 140 are obtained. This violates the condition of simplex method that all the variables be non-negative. Choosing s 5 = − 140 which is row 5. In this row, there is negative entry in column 2. Choosing the pivot column as column 2. The pivot is determined from the quotients as, − 140 − 1 = 140 and − 130 − 1 = 130 Thus, the pivot is − 1 in column x 2 . [ x 1 x 2 x 3 x 4 s 1 s 2 s 3 s 4 s 5 − Z 0 1 − 1 0 1 0 1 0 0 0 − 10 0 0 1 1 0 1 0 0 0 0 150 1 0 1 0 0 0 − 1 0 0 0 130 0 − 1 0 − 1 0 0 0 1 0 0 − 130 0 − 1 − 1 0 0 0 1 0 1 0 − 140 0 60 6 57 0 0 36 0 0 1 − 4680 ] Clear the pivot column using the pivot and label the pivot row with the label of the pivot column

Mathematical Applications for the Management, Life, and Social Sciences

12th Edition

ISBN: 9781337625340

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 4, Problem 46RE

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To calculate: The quantity of each type that should be produced at each location so that the cost of the order is minimum and also find the minimum cost.

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Chapter 4, Problem 45RE

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The quantity of each type that should be produced at each location so that the cost of the order is minimum and also find the minimum cost.

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To calculate: The quantity of each type that should be produced at each location so that the cost of the order is minimum and also find the minimum cost.

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