Computing the tolerance in multiobjective linear programming
Computing the tolerance in multiobjective linear programming Milan Hlad´ık Department of Applied Mathematics Charles University, Prague
Joint EUROPT-OMS Meeting 2007 July 4–7, Prague
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}.
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Definition I
Additive tolerance: any δ such that x ∗ remains efficient for all ˆ : |Cij − C ˆij | < δ; C
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Definition I
Additive tolerance: any δ such that x ∗ remains efficient for all ˆ : |Cij − C ˆij | < δ; C
I
Multiplicative tolerance: any δ such that x ∗ remains efficient ˆ : |C − C ˆ | < δ|G |. for all C
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Theorem (Additive tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the linear program max δ subject to DC T λ − |D|eδ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is an additive tolerance.
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Theorem (Additive tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the linear program max δ subject to DC T λ − |D|eδ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is an additive tolerance.
Theorem (Multiplicative tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the generalized linear fractional program max δ subject to DC T λ − δ|D||G |T λ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is a multiplicative tolerance.
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T .
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T . I
The linear program yields λ∗ ' (0.2343, 0, 0.7657), δ ∗ ' 2.0749.
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T . I
The linear program yields λ∗ ' (0.2343, 0, 0.7657), δ ∗ ' 2.0749.
I
The generalized linear fractional program (with G = C ) yields λ∗ ' (0.2665, 0, 0.7335), δ ∗ ' 0.2195.
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the linear program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , |D|k · e
inf
Dk · C T λ∗ . |D|k · e
k: Dkj 0
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the linear program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , |D|k · e
inf
Dk · C T λ∗ . |D|k · e
k: Dkj 0
2. For each i ∈ {1, . . . , s} with λ∗i = 0 set ∆+ ij := ∞ ∀j ∈ {1, . . . , n}, ∆− ij := ∞ ∀j ∈ {1, . . . , n}.
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Example Reconsider our ∗ δ ∆+ = δ ∗ δ∗
problem. The first step of Algorithm ∗ ∗ ∗ δ 2.2165 δ δ ∞ δ ∗ 2.2165 δ ∗ , ∆− = δ ∗ ∞ δ ∗ 2.2165 δ ∗ δ∗ ∞
where δ ∗ = 2.0749.
results in ∗ ∞ δ ∞ δ∗ , ∞ δ∗
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Example Reconsider our ∗ δ ∆+ = δ ∗ δ∗
problem. The first step of Algorithm ∗ ∗ ∗ δ 2.2165 δ δ ∞ δ ∗ 2.2165 δ ∗ , ∆− = δ ∗ ∞ δ ∗ 2.2165 δ ∗ δ∗ ∞
results in ∗ ∞ δ ∞ δ∗ , ∞ δ∗
where δ ∗ = 2.0749. The second step yields ∗ ∗ ∗ ∗ ∗ δ δ 2.2165 δ δ ∞ ∞ δ ∞ ∞ , ∆− = ∞ ∞ ∞ ∞ , ∆+ = ∞ ∞ δ ∗ δ ∗ 2.2165 δ ∗ δ∗ ∞ ∞ δ∗
Multiplicative tolerance matrix Compute matrices ∆+ and ∆− such that x ∗ detains efficiency for ˆij − Cij < |G |ij ∆+ . ˆ with −|G |ij ∆− < C every C ij ij
Multiplicative tolerance matrix Compute matrices ∆+ and ∆− such that x ∗ detains efficiency for ˆij − Cij < |G |ij ∆+ . ˆ with −|G |ij ∆− < C every C ij ij
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the generalized linear fractional program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , T ∗ |D|k · |G | λ
inf
Dk · C T λ∗ . T ∗ |D|k · |G | λ
k: Dkj 0
2. For each i ∈ {1, . . . , s} with λ∗i = 0 set ∆+ ij := ∞ ∀j ∈ {1, . . . , n}, ∆− ij := ∞ ∀j ∈ {1, . . . , n}.
Example Example Reconsider our results in ∗ δ ∆+ = δ ∗ δ∗
problem, G = C . The first step of Algorithm δ∗ δ∗ δ∗
∗ ∗ δ ∞ ∞ δ 0.2849 0.2849 δ ∗ , ∆− = δ ∗ ∞ ∞ δ ∗ , δ∗ ∞ ∞ δ∗ 0.2849 δ ∗ δ∗
and the second step yields ∗ ∗ ∗ ∗ ∗ δ δ 0.2849 δ δ ∞ ∞ δ ∞ ∞ , ∆− = ∞ ∞ ∞ ∞ , ∆+ = ∞ ∞ δ ∗ δ ∗ 0.2849 δ ∗ δ∗ ∞ ∞ δ∗ where δ ∗ = 0.2195.
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Let us denote C c := 12 .(C + C ) and consider the MOLP problem max C c x.
x∈M
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Let us denote C c := 12 .(C + C ) and consider the MOLP problem max C c x.
x∈M
Let (λ∗ , δ ∗ ) be an optimal solution to the linear (or generalized linear fractional) program.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ is defined as before.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ midpoint matrix is 0 10 C c = 0 10 10 10
is defined as before. The 0 80 10 20 . 10 10
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ midpoint matrix is 0 10 C c = 0 10 10 10
is defined as before. The 0 80 10 20 . 10 10
Then λ∗ ' (0.2343, 0, 0.7657) (or λ∗ ' (0.2665, 0, 0.7335)), DC I λ∗ > 0 and hence x ∗ is necessarily efficient.
The End.
Joint EUROPT-OMS Meeting 2007 July 4–7, Prague
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}.
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Definition I
Additive tolerance: any δ such that x ∗ remains efficient for all ˆ : |Cij − C ˆij | < δ; C
Introduction Consider the multiobjective linear program max Cx,
x∈M
where M := {x ∈ Rn : Ax ≤ b}. Let x ∗ be an efficient solution.
Definition I
Additive tolerance: any δ such that x ∗ remains efficient for all ˆ : |Cij − C ˆij | < δ; C
I
Multiplicative tolerance: any δ such that x ∗ remains efficient ˆ : |C − C ˆ | < δ|G |. for all C
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Theorem (Additive tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the linear program max δ subject to DC T λ − |D|eδ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is an additive tolerance.
Computing the tolerances Normal cone of M at the point x ∗ N (x ∗ ) := {x ∈ Rn : Dx ≥ 0}. T For a nondegenerate basic solution, D = (A−1 B ) .
Theorem (Additive tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the linear program max δ subject to DC T λ − |D|eδ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is an additive tolerance.
Theorem (Multiplicative tolerance, Hlad´ık 2007) Let (λ∗ , δ ∗ ) be an optimal solution to the generalized linear fractional program max δ subject to DC T λ − δ|D||G |T λ ≥ 0, λ, δ ≥ 0, e T λ = 1. Then δ ∗ is a multiplicative tolerance.
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T .
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T . I
The linear program yields λ∗ ' (0.2343, 0, 0.7657), δ ∗ ' 2.0749.
Example Example (Hansen, Labb´e, Wendell 1989) Consider
0 10 0 80 C = 0 10 10 20 , 10 10 10 10 M = {x ∈ R4 : 4x1 + 9x2 + 7x3 + 10x4 ≤ 6000, x1 + x2 + 3x3 + 40x4 ≤ 4000, x ≥ 0}, and the basic solution x ∗ ' (1333.33, 0, 0, 66.67)T . I
The linear program yields λ∗ ' (0.2343, 0, 0.7657), δ ∗ ' 2.0749.
I
The generalized linear fractional program (with G = C ) yields λ∗ ' (0.2665, 0, 0.7335), δ ∗ ' 0.2195.
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the linear program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , |D|k · e
inf
Dk · C T λ∗ . |D|k · e
k: Dkj 0
Additive tolerance matrix Compute upper matrix ∆+ and lower matrix ∆− such that x ∗ ˆ with −∆− < C ˆ − C < ∆+ . detains efficiency for every C
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the linear program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , |D|k · e
inf
Dk · C T λ∗ . |D|k · e
k: Dkj 0
2. For each i ∈ {1, . . . , s} with λ∗i = 0 set ∆+ ij := ∞ ∀j ∈ {1, . . . , n}, ∆− ij := ∞ ∀j ∈ {1, . . . , n}.
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Example Reconsider our ∗ δ ∆+ = δ ∗ δ∗
problem. The first step of Algorithm ∗ ∗ ∗ δ 2.2165 δ δ ∞ δ ∗ 2.2165 δ ∗ , ∆− = δ ∗ ∞ δ ∗ 2.2165 δ ∗ δ∗ ∞
where δ ∗ = 2.0749.
results in ∗ ∞ δ ∞ δ∗ , ∞ δ∗
Proof and example Sketch of proof. ˆ of C are determined so that the The possible deviations C ˆ T λ∗ > 0 remains true. inequality D C
Example Reconsider our ∗ δ ∆+ = δ ∗ δ∗
problem. The first step of Algorithm ∗ ∗ ∗ δ 2.2165 δ δ ∞ δ ∗ 2.2165 δ ∗ , ∆− = δ ∗ ∞ δ ∗ 2.2165 δ ∗ δ∗ ∞
results in ∗ ∞ δ ∞ δ∗ , ∞ δ∗
where δ ∗ = 2.0749. The second step yields ∗ ∗ ∗ ∗ ∗ δ δ 2.2165 δ δ ∞ ∞ δ ∞ ∞ , ∆− = ∞ ∞ ∞ ∞ , ∆+ = ∞ ∞ δ ∗ δ ∗ 2.2165 δ ∗ δ∗ ∞ ∞ δ∗
Multiplicative tolerance matrix Compute matrices ∆+ and ∆− such that x ∗ detains efficiency for ˆij − Cij < |G |ij ∆+ . ˆ with −|G |ij ∆− < C every C ij ij
Multiplicative tolerance matrix Compute matrices ∆+ and ∆− such that x ∗ detains efficiency for ˆij − Cij < |G |ij ∆+ . ˆ with −|G |ij ∆− < C every C ij ij
Algorithm Let (λ∗ , δ ∗ ) be an optimal solution to the generalized linear fractional program. 1. For each i ∈ {1, . . . , s} and j ∈ {1, . . . , n} set ∆+ ij ∆− ij
:= :=
inf
Dk · C T λ∗ , T ∗ |D|k · |G | λ
inf
Dk · C T λ∗ . T ∗ |D|k · |G | λ
k: Dkj 0
2. For each i ∈ {1, . . . , s} with λ∗i = 0 set ∆+ ij := ∞ ∀j ∈ {1, . . . , n}, ∆− ij := ∞ ∀j ∈ {1, . . . , n}.
Example Example Reconsider our results in ∗ δ ∆+ = δ ∗ δ∗
problem, G = C . The first step of Algorithm δ∗ δ∗ δ∗
∗ ∗ δ ∞ ∞ δ 0.2849 0.2849 δ ∗ , ∆− = δ ∗ ∞ ∞ δ ∗ , δ∗ ∞ ∞ δ∗ 0.2849 δ ∗ δ∗
and the second step yields ∗ ∗ ∗ ∗ ∗ δ δ 0.2849 δ δ ∞ ∞ δ ∞ ∞ , ∆− = ∞ ∞ ∞ ∞ , ∆+ = ∞ ∞ δ ∗ δ ∗ 0.2849 δ ∗ δ∗ ∞ ∞ δ∗ where δ ∗ = 0.2195.
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Let us denote C c := 12 .(C + C ) and consider the MOLP problem max C c x.
x∈M
Application to interval MOLP An interval MOLP problem max Cx,
x∈M
where C varies in C I = {C ⊆ Rs×n : C ≤ C ≤ C }.
Definition I
A vector x ∈ M is possibly efficient if it is efficient for some C ∈ CI.
I
A vector x ∈ M is necessarily efficient if it is efficient for every C ∈ C I .
Let us denote C c := 12 .(C + C ) and consider the MOLP problem max C c x.
x∈M
Let (λ∗ , δ ∗ ) be an optimal solution to the linear (or generalized linear fractional) program.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ is defined as before.
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ midpoint matrix is 0 10 C c = 0 10 10 10
is defined as before. The 0 80 10 20 . 10 10
Sufficient condition Theorem The vector x ∗ is necessarily efficient if DC I λ∗ > 0.
Example Consider the interval matrix [−1, 1] [8, 12] [−1, 1] [75, 85] C I = [−1, 1] [8, 12] [8, 12] [17, 23] [8, 12] [8, 12] [8, 12] [8, 12] The feasible set M and its point x ∗ midpoint matrix is 0 10 C c = 0 10 10 10
is defined as before. The 0 80 10 20 . 10 10
Then λ∗ ' (0.2343, 0, 0.7657) (or λ∗ ' (0.2665, 0, 0.7335)), DC I λ∗ > 0 and hence x ∗ is necessarily efficient.
The End.
