Variables that are unrestricted in sign

1. Variables that are unrestricted in sign LI Xiao-lei

2. Preview • In solving LPs with the simplex algorithm, we used the ratio test. • The ratio test depended on the fact that any feasible point required all variables to be nonnegative. • If some variables are allowed to be unrestricted in sign (urs), the ratio test and therefore the simplex algorithm are no longer valid.

3. Preview • For each urs variable xi, we begin by defining two new variables xi’ and xi’’. Then , substitute xi’-xi’’ for xi in each constraint and in the objective function. Also add the sign restrictions xi’≥0 and xi’’≥0. since all variables are now required to be nonnegative, we can proceed with the simplex.

4. Preview For any basic feasible solution, each urs variable xi must fall into one of the following three cases: • Case 1 xi’>0 and xi’’=0 This case occurs if a bfs has xi>0. in this case, xi=xi’-xi’’=xi’. Thus, xi=xi’. For example, if xi=3 in a bfs, this will be indicated by xi’=3 and xi’’=0. • Case 2 xi’=0 and xi’’>0 This case occurs if xi<0. Since xi=xi’-xi’’, we obtain xi=-xi’’. For example, if xi=-5 in a bfs, we will have xi’=0 and xi’’=5. then xi=0-5=-5. • Case 3 xi’=xi’’=0 In this case, xi=0-0=0.

5. Example 7 • A baker has 30 oz of flour and 5 packages of yeast. Baking a loaf of bread requires 5 oz of flour and 1 package of yeast. Each loaf of bread can be sold for 30¢. The baker may purchase additional flour at 4¢/oz or sell leftover flour at the same price. Formulate and solve an LP to help the baker maximize profits (revenues-costs).

6. Example 7 Solution Define x1= number of loaves of bread baked x2= number of ounces by which flour supply is increased by cash transactions Therefore, x2>0 means that x2 oz of flour were purchased, and x2<0 means that –x2 ounces of flour were sold. x2=0 means no flour was bought or sold. x1≥0 and x2 is urs.

7. Example 7 The appropriate LP is max z =30x1-4x2 s.t. 5x1≤30+x2 (flour constraint) x1≤ 5 (Yeast constraint) x1≥0, x2 urs since x2 is urs, we substitute x2’-x2’’ for x2 in the objective function and constraints. max z =30x1-4x2’+4x2’’ s.t. 5x1≤30+x2’-x2’’ (flour constraint) x1≤ 5 (Yeast constraint) x1,x2’,x2’’≥0

8. Example 7 Transforming the objective function to row 0 form and adding slack variables to the two constraints, we obtain the initial tableau. Note: no matter how many pivots we make, the x2’ column will always be the negative of the x2’’ column.

9. Example 7 x1 enters the basis in row 2. x2’’ enters the basis in row 1.

10. Example 7 The optimal solution to the baker’s problem is z=170, x1=5, x2’’=5, x2’=0,s1=s2=0. Thus, the baker can earn a profit of 170¢ by baking 5 loaves of bread. Since x2=x2’-x2’’=0-5=-5, the baker should sell 5 oz of flour.