1 Convert this problem into our standard form: Minimize 3 x x 2 – 6 x 3 subject to 1 x x 2 – 3 x 3 ≥ x x 2 – 6 x 3 ≤ 5 7 x x 2 – 9 x 3 = -3 x 2, x 3 ≥ 0 How would this problem be input to your program?

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3 Because the b column is special and the z row is special, I chose to place them in column 0 and row 0 in my matrix for the Simplex algorithm. If you do this, the second phase of the project will probably be a little bit easier to implement.

4 public class Repeat_Add public static void main(String [ ] args) { int i, next_print; float x, y; double error; x= 1.0f / 3; // x= 1/3 y=0; next_print=3;

5 for (i=1; i <= ; i++) { y= y+x; error= y- (double) i / 3 ; if (i == next_print) { System.out.println(y + " should be " + (i/3) + " Error = " + error); next_print= next_print * 10; }

6 Errors resulting from repeatedly adding 1/3 (edited to make this more readable). 1.0 should be 1 Error = should be 10 Error = E should be 100 Error = E should be 1000 Error = should be Error = should be Error = should be Error = The “error” would be zero if the mathematics was 100% precise. But it is not due to floating point errors. This is why we compare values to some small epsilon instead of zero to test if they are zero or not.

7 LP problems can be: 1. infeasible, 2. unbounded, or 3. they have at least one basic (corresponding to a dictionary) optimal solution.

8 Another sample problem: Maximize 5 x x x 3 subject to x x 2 + x 3 ≤ 3 -x x 3 ≤ 2 2 x 1 - x x 3 ≤ 4 2 x x 2 - x 3 ≤ 2 x 1, x 2, x 3 ≥ 0

9 The initial dictionary: X4 = X X X3 X5 = X X X3 X6 = X X X3 X7 = X X X z = X X X3 The program output will be in dictionary format.

10 X1 enters. X7 leaves. z = After 1 pivot: X4 = X X X7 X5 = X X X7 X6 = X X X7 X1 = X X X z = X X X7

11 X3 enters. X6 leaves. z = After 2 pivots: X4 = X X X7 X5 = X X X7 X3 = X X X7 X1 = X X X z = X X X7

12 X2 enters. X5 leaves. z = After 3 pivots: X4 = X X X7 X2 = X X X7 X3 = X X X7 X1 = X X X z = X X X7 How could we argue that this must be the optimal solution?