Linear Fractional Programming

1
Linear Fractional Programming
2
What is LFP?

• Minimize
• Subject to
• p,q are n vectors, b is an m vector, A is an mn
  matrix, a,ß are scalar.

3
Lemma 11.4.1

• Let f(x)(ptxa)/(qtxß), and let S be a convex
  set such that qtxß?0 over S.
• Then f is both pseudoconvex and pseudoconcave
  over S.

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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.

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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.

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Solution Approach

• From the implications
• Search the extreme points until a Kuhn-Tucker
  point is reached.
• Direction
• If ?Kuhn-Tucker point, stop.
• Otherwise, -rjmax-ririlt0
• Increase nonbasic variable xj, adjust basic
  variables.
• Gilmore and Gomory(1963)
• Charnes and Cooper(1962)

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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.

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Gilmore and Gomory

• 2. Let rjmax-ririlt0, where rj is the ith
  component of rN.
• Determine the basic variable xB, to leave the
  basis by the minimum ratio test

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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 xk1. Replace
  k by k1, and go to step 1.

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ExampleGilmore and Gomory
min
s.t.
11
Iteration 1
x1 x2 x3 x4 x5 RHS
0 0 0 -
x3 -1 1 1 0 0 4
x4 0 1 0 1 0 6
x5 2 1 0 0 1 14
r 0 0 0 -
12
Computation of Iteration 1
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Iteration 2
x1 x2 x3 x4 x5 RHS
0 0 0 -
x3 0 1 0 11
x4 0 1 0 1 0 6
x1 1 0 0 7
r 0 0 0 -
14
Computation of Iteration 2
Optimal Solution x17, x20, min-12/11-1.09
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Charnes and Cooper

• Minimize
• Subject to

Minimize Subject to
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Example Charnes and Cooper

• Min
• s.t.

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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