1. Linear Fractional Programming

2. What is LFP? Minimize Subject to p,q are n vectors, b is an m vector, A is an m*n matrix, α,β are scalar.

3. Lemma 11.4.1 Let f(x)=(p t x+α)/(q t x+β), and let S be a convex set such that q t x+β  0 over S. Then f is both pseudoconvex and pseudoconcave over S.

4. Implications of lemma 11.4.1 Since f is both pseudoconvex and pseudoconcave over S, then by Theorem 3.5.11, it is also quasiconvex, quasiconcave, strictly quasiconvex, and strictly quasiconcave. Since f is both pseudoconvex and pseudoconcave, the by theorem 4.3.7, a point satisfying the kuhn-Tucker conditions for a minimization problem is also a global minimum over S. Likewise, a point satisfying the kuhn-Tucker conditions for a maximization problem is also a global maximum over S.

5. Implications of lemma 11.4.1(cont.) Since f is strictly quasiconvex and strictly quasiconcave, then by Theorem 3.5.6, a local minimum is also a global minimum over S. Likewise, a local maximum is also a global maximum over S. Since f is quasiconvex and quasiconcave, if the feasible region is bounded, then by theorem 3.5.3, the f has a minimum at an extreme point of the feasible region and also has a maximum at an extreme point of the feasible region.

6. Solution Approach From the implications: Search the extreme points until a Kuhn-Tucker point is reached. Direction:  If  Kuhn-Tucker point, stop.  Otherwise, -r j =max{-r i :r i <=0}  Increase nonbasic variable x j, adjust basic variables. Gilmore and Gomory(1963) Charnes and Cooper(1962)

7. Gilmore and Gomory(1963) Initialization Step: Find a starting basic feasible solution x 1, Form the corresponding tableau Main Step 1. Compute – If, Stop. Current x k is an optimal solution. – Otherwise, go to the step 2.

8. Gilmore and Gomory 2. Let – r j =max{-r i :r i <=0}, where r j is the ith component of r N. Determine the basic variable x B, to leave the basis by the minimum ratio test:

9. Gilmore and Gomory 3. Replace the variable x B, by the variable x j.Update the tableau corresponfing by pivoting at y rj. Let the current solution be x k+1. Replace k by k+1, and go to step 1.

10. Example:Gilmore and Gomory: min s.t.

11. Iteration 1 x1x1 x2x2 x3x3 x4x4 x5x5 RHS 000- x3x3 11004 x4x4 010106 x5x5 2100114 r000-

12. Computation of Iteration 1

13. Iteration 2 x1x1 x2x2 x3x3 x4x4 x5x5 RHS 000- x3x3 01011 x4x4 010106 x1x1 1007 r000-

14. Computation of Iteration 2 Optimal Solution: x 1 =7, x 2 =0, min=-12/11=-1.09

15. Charnes and Cooper Minimize Subject to Minimize Subject to

16. Example: Charnes and Cooper Min s.t.

17. Solved by Lingo Global optimal solution found at iteration: 6 Objective value: -1.090909 Variable Value Reduced Cost Y1 0.6363636 0.000000 Y2 0.000000 4.727273 Z 0.9090909E-01 0.000000