Explanation Given Information: The intake of used oil up to 50 million gallons of used oil and allocated to 3 refinery processes A, B and C, the electricity costs for process A amount to $ 150 , 000 per million gallon treated, while for processes B and C the cots are $ 100 , 000 and $ 50 , 000 respectively and process A can recover 60 % of the used oil, B can recover 55 % and C can recover 50 % . Assume the revenue is $ 4 million gallons of recovered oil and annual electrical budget is $ 3 million and C can handle up to 20 million gallons per year. Formula used: Following steps of the simplex method, Step 1: Use slack variables and write the inequalities in equation form. Step 2: Write the equations in an augmented matrix. Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Step 4: Select the pivot which is the smallest of the test ratios a b , where a is the entry in the right most column and b is the corresponding entry in the pivot column. Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Step 6: Repeat the above steps till all the entries in the bottom row are non negative. Use row operations to clear the column of the augmented matrix. Calculation: Let x represent amount of used oil allocated to refinery A, y represents amount of used oil allocated to refinery B and z represents the amount of used oil allocated to refinery C. Thus, the maximum revenues is given by, p = 4 × 60 100 x + 4 × 55 100 y + 4 × 50 100 z = 2.4 x + 2.2 y + 2.0 z = 24 x + 22 y + 20 z . And, the constraints are, x + y + z ≤ 50 3 x + 2 y + z ≤ 60 z ≤ 20 x ≥ 0 , y ≥ 0 , z ≥ 0 Consider the maximisation problem, Maximize p = 24 x + 22 y + 20 z Subject to x + y + z ≤ 50 3 x + 2 y + z ≤ 60 z ≤ 20 x ≥ 0 , y ≥ 0 , z ≥ 0 Step 1: Use slack variables and write the inequalities in equation form. Add the slack variables s , t and u . x + y + z + s = 50 3 x + 2 y + z + t = 60 z + u = 20 Step 2: Write the equations in an augmented matrix. The equations are, x + y + z + s = 50 3 x + 2 y + z + t = 60 z + u = 20 And, − 24 x − 22 y − 20 z + p = 0 Write the equations in augmented matrix, x y z s t u p s 1 1 1 1 0 0 0 50 t 3 2 1 0 1 0 0 60 u 0 0 1 0 0 1 0 20 p − 24 − 22 − 20 0 0 0 1 0 Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Since, the smallest negative number on the left side of the bottom row is − 24 . Thus, choose the left most entry and pivot the x column. Step 4: Select the pivot which is the smallest of the test ratios a b , where a is the entry in the right most column and b is the corresponding entry in the pivot column. Since, the test ratios are neither 0 nor negative. Thus, the test ratios are 50 1 = 50 and 60 3 = 20 . Since, 20 is smallest. Thus, the pivot is 3 in column x . x y z s t u p s 1 1 1 1 0 0 0 50 t 3 2 1 0 1 0 0 60 u 0 0 1 0 0 1 0 20 p − 24 − 22 − 20 0 0 0 1 0 Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Since, the pivot is t in column x and replace t with x . Thus, x is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations 3 R 1 − R 2 , R 4 + 8 R 2 , x y z s t u p s 0 1 2 3 − 1 0 0 90 x 3 2 1 0 1 0 0 60 u 0 0 1 0 0 1 0 20 p 0 − 6 − 12 0 8 0 1 480 Step 6: Repeat the above steps till all the entries in the bottom row are non negative

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 6.3, Problem 34E

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To calculate: The amount of used oil to be allocated to each process to maximize the revenues where the intake is up to 50 million gallons of used oil and allocated to 3 refinery processes A, B and C.

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Chapter 6.3, Problem 33E

Chapter 6.3, Problem 35E

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Finite Mathematics and Applied Calculus (MindTap Course List)

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