Page 135 - IJOCTA-15-1

Duality for robust multi-dimensional vector variational control problem under invexity
subject to ξ(t, σ, ζ, σ κ , ζ κ , ω, ϱ)=1, the invexity of function-
als follows as below :
2b+σ−7≦0, b∈[1, 2]
−5ω =0, µ∈[0, 1]
µσ t 1
2
1 3 (i) To show that R Ψ 1 (., ¯ a 1 )dν = R (σ +
σ(0, 0)= , σ(1, 1)= , Ω Ω
2 2 ¯ a 1 ω)dν is invex w.r.t. η and ξ at (ζ, ϱ).
1 Consider
h i
a 1 ∈ 0, , a 2 ∈[0, 3], [t 1 , t 2 ]∈[0, 0]×[1, 1]. Z Z
2
A 1 = Ψ 1 (Λ, σ κ , ¯a 1 )dν − Ψ 1 (Π, ζ κ , ¯a 1 )dν
Thus, the set of feasible solutions for (RUVP1) will
Ω Ω
be Z
− η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)(Ψ 1 ) σ (Π, ζ κ , ¯a 1 )dν
−5ω =0,
Ω
T 1 ={(σ, ω)∈K ×Υ : σ ≤3, µσ t 1
Z
1 3
σ(0, 0)= , σ(1, 1)= , t∈Ω, b∈[1, 2], µ∈[0, 1]}. − D κ η((t, σ, ζ, σ κ , ζ κ , ω, ϱ))(Ψ 1 ) σ κ (Π, ζ κ ¯a 1 )dν
2 2 Ω
To obtain a feasible point (σ, ω), sufficiency con- Z
dition (2) gives − ξ(t, σ, ζ, σ κ , ζ κ , ω, ϱ)(Ψ 1 ) ω (Π, ζ κ , ¯a 1 )dν
Ω
χ 1 5τ Z 1 Z Z
2
ω = − . = σ + ω dν − 2.35dν − 8ωσdν
4χ 2 2χ 2 Ω 2 Ω Ω
∗
Solving (2 ), we get Z 1
− dν
2
χ 1 5τ + Ω
σ =5 − t 1 +d 1 , d 1 ∈R . Z 1 2 3 Z
4χ 2 2χ 2 = 3t 1 + 2 + 40 dt 1 dt 2 − 2.35dt 1 dt 2
4
1
1
∗
Taking χ 1 = , χ 2 = , (3 ) gives d 1 = 1 2 and τ = Ω Z Z Ω 1
2
2
1 − 8ωσdt 1 dt 2 − dt 1 dt 2
50 . Thus, Ω Ω 2
1 3
3t 1 =0.0875≧0
(σ, ω)= + ,
4 2 20 and
is obtained as the robust feasible solution to Z Z
A 2 = Ψ 2 (Λ, σ κ , ¯a 2 )dν − Ψ 2 (Π, ζ κ , ¯a 2 )dν
(RUVP1), which is shown graphically in figure 1. Ω Ω
Z
Now the Wolfe type dual to the problem (RUVP1) − η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)(Ψ 2 ) σ (Π, ζ κ , ¯a 2 )dν
is given by : Ω
Z
Z
1
2
(WD1) max ζ + ϱ+λ(2b+ζ −7)+ − D κ η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)(Ψ 2 ) σ κ (Π, ζ κ , ¯a 2 )dν
(ζ,ϱ) Ω 2 Ω Z
2
−5ϱ), 3−ϱ +λ(2b+ζ −7)+ − ξ(t, σ, ζ, σ κ , ζ κ , ω, ϱ)(Ψ 2 ) ω (Π, ζ κ , ¯a 2 )dν
τ(µζ t 1
Ω

−5ϱ) dν Z Z 74 Z
τ(µζ t 1 2
= 3−ω dν − dν + 2ξωdν
Ω Ω 25 Ω
subject to Z Z Z
= 2.9775dt 1 dt 2 − 2.96dt 1 dt 2 + 0.4dt 1 dt 2
∗
=0 (4 )
Ω Ω Ω
χ 1 (2ζ +λ)+χ 2 λ+λ−(τ) t 1
1
∗
χ 1 ( −5τ)+χ 2 (−2ϱ−5τ)−5τ =0 (5 )
2 =0.4175≧0
R
1 3 Therefore, the invexity of Ω Ψ i (., ¯ a i )dν,i=
∗
ζ(0, 0)= , ζ(1, 1)= , (6 ) 3 1
2 2 1, 2 at (ζ, ϱ)= − , holds.
2 5
2 Z
X T
∗
χ≧0, χ i =1, λ≧0. (7 ) (ii) To show that λ θ(., ., b)dν is invex
Ω
i=1 w.r.t. η and ξ at (ζ, ϱ).
Solving dual constraints, using χ=
Z Z

T
T
(χ 1 , χ 2 )= 2 1 , τ = 1 , λ=1, we get (ζ, ϱ)= A 3 = λ θ(Λ, σ κ , b)dν − λ θ(Π, ζ κ , b)dν
,
3 3 50
3 1
− , . Hence, (ζ, ϱ, χ, λ, τ, ¯ a, b, µ)= Z Ω Ω
2 5
T
1
3 1 2 1
− , , , , 1, 1 , , 3, 2, 1 is a robust feasible − η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)λ θ σ (Π, ζ κ , b)dν
2 5 3 3 50 2 2
solution to (WD1). Z Ω
T
To validate Theorem 3, the invexity of function- − D κ η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)λ θ σ κ (Π, ζ κ b)dν
Ω
3 1
als is to be explored at (ζ, ϱ)= − , w.r.t. Z
2 5 − T
η and ζ. Taking η(t, σ, ζ, σ κ , ζ κ , ω, ϱ)=4ω and ξ(t, σ, ζ, σ κ , ζ κ , ω, ϱ)λ θ ω (Π, ζ κ , b)dν
Ω
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