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Linear Fractional Programming

Linear Fractional Programming. 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. 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.

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Linear Fractional Programming

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  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)=(ptx+α)/(qtx+β), and let S be a convex set such that qtx+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, -rj=max{-ri:ri<=0} • Increase nonbasic variable xj, adjust basic variables. • Gilmore and Gomory(1963) • Charnes and Cooper(1962)

  7. Gilmore and Gomory(1963) • Initialization Step: • Find a starting basic feasible solution x1, • Form the corresponding tableau • Main Step • Compute • If , Stop. • Current xk is an optimal solution. • Otherwise, go to the step 2.

  8. Gilmore and Gomory 2. Let –rj=max{-ri:ri<=0}, where rj is the ith component of rN. Determine the basic variable xB, to leave the basis by the minimum ratio test:

  9. Gilmore and Gomory 3. Replace the variable xB, by the variable xj.Update the tableau corresponfing by pivoting at yrj. Let the current solution be xk+1. Replace k by k+1, and go to step 1.

  10. Example:Gilmore and Gomory:

  11. Iteration 1

  12. Computation of Iteration 1

  13. Iteration 2

  14. Computation of Iteration 2 Optimal Solution: x1=7, x2=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

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