Explanation Consider any nonstandard LP problem. Recall that a standard LP problem is the maximization problem of an objective function subject to some constraints with ≤ inequality. Thus, any LP problem which is not a maximization problem is nonstandard LP problem. Since, the basic solution of a standard LP problem is a solution where all the inactive variables, which are not pivots are 0 and the active variables are either 0 or non zero. Thus, value of all the variables in the basic solution of the standard LP problem is non negative. Since, in any nonstandard LP problem, one or more variables have negative value in the basic solution. Thus, the solution is not in the feasible region. Recall that Phase I involves calculations to make active variables have positive values. The computations are, star the rows for which the active variables have negative value then 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. Thus, Phase I makes the value of all the variables positive. For example, Consider the problem. Maximize p = x + y Subject to constraints, x + 2 y ≥ 6 − x + y ≤ 4 2 x + y ≤ 8 x ≥ 0 , y ≥ 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 and add the slack variables t and u . x + 2 y − s = 6 − x + + t = 4 2 x + y + u = 8 Thus, the equations are x + 2 y − s = 6 − x + + t = 4 2 x + y + u = 8 − x − y + p = 0 Write the equations in augmented matrix. x y s t u p s 1 2 − 1 0 0 0 6 t − 1 1 0 1 0 0 4 u 2 1 0 0 1 0 8 p − 1 − 1 0 0 0 1 0 Check the values of the active variables s t and u . s = 6 − 1 = − 6 And, t = 4 1 = 4 And, u = 8 1 = 8 Since, the value of s is negative, star the s row. Thus, the tableau is x y s t u p * s 1 2 − 1 0 0 0 6 t − 1 1 0 1 0 0 4 u 2 1 0 0 1 0 8 p − 1 − 1 0 0 0 1 0 Since, there are starred rows, use Phase I to proceed

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Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 5.4, Problem 53E

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The need for Phase I in a nonstandard LP problem.

Solution Summary: The author explains that Phase I is used to convert all the negative values of variables in the basic solution of a nonstandard LP problem.

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