linear fractional programming. what is lfp? minimize subject to p,q are n vectors, b is an m vector,...
TRANSCRIPT
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.
xq
xpt
t
0
x
bAx
Lemma 11.4.1
Let f(x)=(ptx+α)/(qtx+β), and let S be a convex set such that qtx+β0 over S.
Then f is both pseudoconvex and pseudoconcave over S.
Implications of lemma 11.4.1Since 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.
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.
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)
NBxfxfr tB
tN
tN
1)()( 0Nr
Gilmore and Gomory(1963)
Initialization Step: Find a starting basic feasible solution x1, Form the corresponding tableauMain Step
1. Compute – If , Stop.
Current xk is an optimal solution.– Otherwise, go to the step 2.
NBxfxfrrr tB
tN
tN
11121 )()(),(
0Nr
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:
}0:{min1
ijy
b
miyb y
ij
i
rj
r
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.
Example:Gilmore and Gomory:
min
s.t.
43
22
21
21
xx
xx
0,
142
6
4
21
21
2
21
xx
xx
x
xx
x2
0
1
2
3
4
5
6
7(2,6) (4,6)
(7,0)(0,0)
(0,4)
x1
Iteration 1x1 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 -
)( 1xf16
10
16
2
16
10
16
2
Computation of Iteration 1
2
4
1
1
xp
xqt
t
)0,0,0,16
2,
16
10()( txf
)16
2,
16
10()( t
N xf )0,0,0()( tB xf
)16
2,
16
10(
12
10
11
)0,0,0()16
2,
16
10(
)()(),( 11121
NBxfxfrrr tB
tN
tN
leavexyy
b
enterxrrrrrr
r
rrrr
ijij
i
N
tB
5
1154321
543
}2
14),
0
6(),
1
4{(}0:min{
),16
10(},,,,max{
0
)0,0,0(),,(
Iteration 2x1 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 -
)( 1xf121
10
121
47
121
52
2
3
2
1
2
1
2
1
121
5
Computation of Iteration 2
12
11
2
2
xp
xqt
t
)0,0,0,121
47,
121
10()( 2 txf
)121
47,
121
10()( t
N xf )0,0,0()( tB xf
)121
5,
121
52(01)
121
10,0,0()0,
121
47(
)()(),(
21
21
21
23
12252
NBxfxfrrr tB
tN
tN
.,0 StoprN
Optimal Solution: x1=7, x2=0, min=-12/11=-1.09
Charnes and Cooper
Minimize
Subject to
xq
xpt
t
0
x
bAx
Minimize
Subject to
zyp t
0
0
1
0
z
y
zyq
bzAyt
zxy
xqz
t
1
Example: Charnes and Cooper
Min
s.t.
zyy 22 21
0,,
143
0142
06
04
21
21
21
2
21
zyy
zyy
zyy
zy
zyy
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
711 z
yx
022 z
yx