Explanation Given Information: The provided constraints are 5 x − y + w ≥ 1000 z + w ≤ 2000 x + y ≥ 500 x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0 Formula used: To convert the minimization problem to the maximization problem, multiply the objective function by − 1 . To compute solution of general linear programming problems, use slack and surplus variables to covert inequalities into equations and star the rows for which the active variables have negative value. If there are starred rows, use Phase I to proceed. Phase I: To remove the stars that is to make active variables have positive values. Start from the first starred row and locate the largest positive number and pivot the column, use test ratios and the smallest test ratio corresponds to a pivot and clear the pivot column. Check again and star the rows for which the active variables have negative value and repeat the process until there are no starred rows and move to Phase II. Phase II: Use the simplex method to solve the obtained standard maximization tableau. Choose the smallest negative number on the left side of the bottom row and pivot the column. 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. Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Repeat till all the entries in the bottom row are non negative. If there are no negative entries in the bottom row of the standard maximization tableau, then the obtained tableau represents the solution to the problem. Use row operations to clear the column of the augmented matrix. Calculation: Consider the objective function c = 5 x + y + z + w of the minimization problem. Recall that to convert the minimization problem to the maximization problem, multiply the objective function by − 1 . − c = − ( 5 x + y + z + w ) p = − 5 x − y − z − w Thus, the corresponding maximization problem is Maximize p = − x − y − z − w subject to 5 x − y + w ≥ 1000 z + w ≤ 2000 x + y ≥ 500 x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0 Recall that to compute solution of general linear programming problems, use slack and surplus variables to covert inequalities into equations and star the rows for which the active variables have negative value. Subtract the surplus variable s , t and add the slack variable u . 5 x − y + w − s = 1000 z + w + t = 2000 x + y + u = 500 Thus, the equations are 5 x − y + w − s = 1000 z + w + t = 2000 x + y + u = 500 5 x + y + z + w + p = 0 Write the equations in augmented matrix. x y z w s t u p s 5 − 1 0 1 − 1 0 0 0 1000 t 0 0 1 1 0 1 0 0 2000 u 1 1 0 0 0 0 1 0 500 p 5 1 1 1 0 0 0 1 0 Check the values of the active variables s and t . s = 1000 − 1 = − 1000 And, t = 2000 1 = 2000 And, u = 500 1 = 500 Since, the value of s is negative, star the s row. Thus, the tableau is x y z w s t u p * s 5 − 1 0 1 − 1 0 0 0 1000 t 0 0 1 1 0 1 0 0 2000 u 1 1 0 0 0 0 1 0 500 p 5 1 1 1 0 0 0 1 0 θ Since, there are starred rows, use Phase I to proceed

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ISBN: 9781337280426

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5.4, Problem 18E

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To calculate: The minimum value of c=5x+y+z+w subject to constraint,

5x−y+w≥1000z+w≤2000x+y≥500x≥0,y≥0,z≥0,w≥0

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Chapter 5.4, Problem 17E

Chapter 5.4, Problem 19E

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The minimum value of c = 5 x + y + z + w subject to constraint, 5 x − y + w ≥ 1000 z + w ≤ 2000 x + y ≥ 500 x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0

Solution Summary: The author explains how to calculate the minimum value of c=5x+y+z+w subject to constraint.

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To calculate: The minimum value of c=5x+y+z+w subject to constraint,

5x−y+w≥1000z+w≤2000x+y≥500x≥0,y≥0,z≥0,w≥0

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Chapter 5 Solutions

The minimum value of c = 5 x + y + z + w subject to constraint, 5 x − y + w ≥ 1000 z + w ≤ 2000 x + y ≥ 500 x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0

Solution Summary: The author explains how to calculate the minimum value of c=5x+y+z+w subject to constraint.

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To calculate: The minimum value of c=5x+y+z+w subject to constraint, 5x−y+w≥1000z+w≤2000x+y≥500x≥0,y≥0,z≥0,w≥0

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Chapter 5 Solutions

To calculate: The minimum value of c=5x+y+z+w subject to constraint,

5x−y+w≥1000z+w≤2000x+y≥500x≥0,y≥0,z≥0,w≥0