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. Since, the basic solution is not necessarily the optimal solution. Recall that Phase II uses the simplex method on the tableau obtained from the last step in the calculations in Phase I to obtain the optimal solution of the LP problem. Thus, Phase II is necessary to obtain the optimal solution of the LP problem. 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. 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. The first starred row is s and the largest positive number in the s row is 2. Thus, pivot the y column. The test ratios are 6 2 = 3 , 4 1 = 4 and 8 1 = 8 . Since, 3 is the smallest, pivot the entry 2 in y column. 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 Use row operations 2 R 2 − R 1 , 2 R 3 − R 1 and 2 R 4 + R 1 to clear the pivot column of the augmented matrix. x y s t u p y 1 2 − 1 0 0 0 6 t − 3 0 1 2 0 0 2 u 3 0 1 0 2 0 10 p − 1 0 − 1 0 0 2 6 Check the values of the active variables y t and u . y = 6 2 = 3 And, t = 2 2 = 1 And, u = 10 2 = 5 Since, all the values are positive, there are no starred rows

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