Resources and environmental systems management under synchronic interval uncertainties

Abstract

Resources and environmental systems management (RESM) is challenged by the synchronic effects of interval uncertainties in the related practices. The synchronic interval uncertainties are misrepresented as random variables, fuzzy sets, or interval numbers in conventional RESM programming techniques including stochastic programming. This may lead to ineffectiveness of resources allocation, high costs of recourse measures, increased risks of unreasonable decisions, and decreased optimality of system profits. To fill the gap of few corresponding studies, a synchronic interval linear programming (SILP) method is proposed in this study. The proposition of interval sets and interval functions and coupling them with linear programming models lead to development of an SILP model for RESM. This enables incorporation of interval uncertainties in resource constraints and synchronic interval uncertainties in the programming objective into the optimization process. An analysis of the distribution-independent geometric properties of the feasible regions of SILP models results in proposition of constraint violation likelihoods. The tradeoff between system optimality and constraint violation is analyzed. The overall optimality of SILP systems under synchronic intervalness is quantified through proposition of integrally optimal solutions. Integration of these efforts leads to a violation-constrained interval integral method for optimization of RESM systems under synchronic interval uncertainties. Comparisons with selected existing methods reveal the effectiveness of SILP at eliminating negativity of synchronic intervalness, enabling risk management of and achieving overall optimality of RESM systems, and enhancing the reliability of optimization techniques for RESM problems. The exploited framework for analyzing synchronic interval uncertainties in RESM systems is helpful for addressing synchronisms of other uncertainties such as randomness or fuzziness and avoiding the resultant decision mistakes and disasters due to neglecting them.

Appendix

Lemma 1

Solutions under the conservative boundary are absolutely feasible, and that out of the optimistic boundary are infeasible.

Proof

Let R ⁻ = {X | A ⁺ X ≤ b ⁻ and X ≥ 0}, R ⁺ = {X | A ⁻ X ≤ b ⁺ and X ≥ 0}, and R(b, A) = {X | AX ≤ b; A ⁻ ≤ A ≤ A ⁺; X ≥ 0; b ⁻ ≤ b ≤ b ⁺; (b, A) ≠ (b ⁻, A ⁺) or (b ⁺, A ⁻); (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j ) > 0}. The lemma is equivalent with R ⁻ ⊂ R(b, A) ⊂ R ⁺. Let X = (x ₁,x ₂,…,x _n ) be any element in R ⁻, and A be a real vector satisfying A ⁻ ≤ A ≤ A ⁺. From definitions of R ⁻, we have A ⁺ X ≤ b ⁻, X ≥ 0, and AX ≤ A ⁺ X. Namely, AX ≤ A ⁺ X ≤ b ⁻ ≤ b where b ⁻ ≤ b ≤ b ⁺. Thus, X ∈ R(b, A), i.e. R ⁻⊆ R(b, A). Since (b, A) ≠ (b ⁻, A ⁺), R ⁻ absolutely belongs to R(b, A). Let X ^* = (x ^*₁ , x ^*₂ , …, x ^* _n ) be any vector in R(b, A) and satisfy X ^*∉ R ⁻. Similarly, we have X ^* ∈ R ⁺, i.e. R(b, A) ⊆ R ⁺. Accordingly, R(b, A) absolutely belongs to R ⁺ because (b, A) ≠ (b ⁺, A ⁻).□

Proposition 1

Let X be any solution in R _s where R _s = {X | X ≥ 0; A ⁺ X > b ⁻; A ⁻ X ≤ b ⁺; (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j ) > 0}. We have d _{X T} + d _{X L} > 0.

Proof

For any X ∈ R _s, we have X ≥ 0, A ⁺ X > b ⁻, A ⁻ X ≤ b ⁺, and (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j   − a ⁻ _j ) > 0. Since A ⁺ X − b ⁻ > 0, b ⁺ − A ⁻ X ≥ 0, (A ⁺)(A ⁺)^T > 0 and (A ⁻)(A ⁻)^T > 0, we have d _{X T} ≥ 0 and d _{X L} ≥ 0. Accordingly, we have d _{X T} + d _{X L} ≥ 0. If d _{X T}  = 0 and d _{X L}  = 0, we have (A ⁺ X − b ⁻)/((A ⁺)(A ⁺)^T) = 0 and (b ⁺ − A ⁻ X) / ((A ⁻)(A ⁻)^T) = 0, i.e. A ⁺ X = b ⁻ and A ⁻ X = b ⁺. Equivalently, ∑ _j=1,2,…,n (a ⁺ _j ·x _j ) = b ⁻, ∑ _j=1,2,…,n (a ⁻ _j ·x _j ) = b ⁺, and ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j )·x _j  = b ⁻ − b ⁺. Since ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j )·x _j ≥ 0 and b ⁻ − b ⁺ ≤ 0, equality ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j )·x _j  = b ⁻ − b ⁺ holds for any solution of non-negative decision variables x _j (j = 1, 2, …, n) if and only if b ⁺ = b ⁻ and a ⁺ _j  = a ⁻ _j for any j ∈ {1, 2, …, n}. As a result, (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j ) = 0, which contradicts with the given condition (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j ) > 0. Thus, it does not hold that d _{X T}  = d _{X L}  = 0.□

Theorem 1

For any i ∈ {1, 2, …, t}, the ith constraint in SILP model (1) is equivalent to inequality (2).

Proof

From formulations of CVL _i , we have inequality CVL _i ≤ CVL _imax is equivalent with [(A ⁺ _i X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)]/{[(A _i ⁺ X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)] + [(b _i ⁺ − A ⁻ _i X)/((A ⁻ _i )(A _i ⁻)^T)]} ≤ CVL _imax. Since X ∈ R _s, we have A ⁺ X > b ⁻, A ⁻ X ≤ b ⁺, and (b ⁺ − b ⁻) + ∑ _j=1,2,…,n (a ⁺ _j − a ⁻ _j ) > 0. Therefore, [(A ⁺ _i )(A ⁺ _i )^T(b _i ⁺ − A ⁻ _i X) + (A ⁻ _i )(A _i ⁻)^T(A ⁺ _i X − b ⁻ _i )] > 0. Besides, it holds for any i ∈ {1, 2, …, t} that (A ⁻ _i )(A _i ⁻)^T ≥ 0 and (A ⁺ _i )(A ⁺ _i )^T ≥ 0. Thus, we have inequality [(A _i ⁺ X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)]/{[(A _i ⁺ X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)] + [(b _i ⁺ − A ⁻ _i X)/((A ⁻ _i )(A _i ⁻)^T)]} ≤ CVL _imax is equivalent to [(A ⁺ _i X − b ⁻ _i ) / ((A ⁺ _i )(A _i ⁺)^T)] ≤ (CVL _imax){[(A ⁺ _i X − b ⁻ _i ) / ((A ⁺ _i )(A ⁺ _i )^T)] + [(b _i ⁺ − A ⁻ _i X) / ((A ⁻ _i )(A _i ⁻)^T)]}, (1 − CVL _imax)(A ⁺ _i X − b ⁻ _i )((A ⁻ _i )(A _i ⁻)^T) ≤ (CVL _imax)(b ⁺ _i − A ⁻ _i X)((A ⁺ _i )(A _i ⁺)^T), and then [(1 − CVL _imax)(A ⁻ _i )(A _i ⁻)^T A ⁺ _i + (CVL _imax)(A ⁺ _i )(A _i ⁺)^T A ⁻ _i ]X ≤ (CVL _imax)(A ⁺ _i ) (A ⁺ _i )^T b _i ⁺ + (1 − CVL _imax)(A ⁻ _i )(A ⁻ _i )^T b ⁻ _i .□

Proposition 2

For any i ∈ {1, 2, …, t}, assume X ∈ R _s and CVL _imax1 and CVL _imax2 are any two values of CVL _imax. If CVL _imax1 < CVL _imax2 and A _i (CVL _imax1)X ≤ b _i (CVL _imax1), then A _i (CVL _imax2)X ≤ b _i (CVL _imax2).

Proof

Let X be any vector in {X | X ∈ R _s; A _i (CVL _imax1)X ≤ b _i (CVL _imax1)}. Then X satisfies X ∈ R _s and A _i (CVL _imax1)X ≤ b _i (CVL _imax1). From Theorem 1, we have inequality A _i (CVL _imax1)X ≤ b _i (CVL _imax1) is equivalent with [(A ⁺ _i X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)]/{[(A ⁺ _i X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)] + [(b ⁺ _i − A ⁻ _i X)/((A ⁻ _i )(A ⁻ _i )^T)]} ≤ CVL _imax1. Since CVL _imax1 < CVL _imax2, so [(A ⁺ _i X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)]/{[(A ⁺ _i X − b ⁻ _i )/((A ⁺ _i )(A ⁺ _i )^T)] + [(b ⁺ _i − A ⁻ _i X)/((A ⁻ _i )(A ⁻ _i )^T)]} < CVL _imax2. Namely, X belongs to {X | X ∈ R _s and A _i (CVL _imax2)X ≤ b _i (CVL _imax2)}. No element in {X | X ∈ R _s and A _i (CVL _imax1)X ≤ b _i (CVL _imax1)} can satisfy A _i (CVL _imax2)X = b _i (CVL _imax2), because CVL _imax1 < CVL _imax2. Therefore, {X | X ∈ R _s and A _i (CVL _imax1)X ≤ b _i (CVL _imax1)} absolutely belongs to {X | X ∈ R _s and A _i (CVL _imax2)X ≤ b _i (CVL _imax2)}.□

Remark 1

For any CVL _imax ∈ [0, 1] and any i ∈ {1, 2, …, t}, L _ij (CVL _imax) ∈ [a ⁻ _ij , a ⁺ _ij ] and R _i (CVL _imax) ∈ [b ⁻ _i , b ⁺ _i ] where L _ij (CVL _imax) = [(1 − CVL _imax)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T]/[(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i )(A _i ⁺)^T] and R _i (CVL _imax) = [(CVL _imax)·b ⁺ _i ·(A ⁺ _i )(A _i ⁺)^T + (1 − CVL _imax)·b ⁻ _i ·(A ⁻ _i )(A _i ⁻)^T]/[(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i )(A _i ⁺)^T].

Proof

Based on formulations of L _ij (CVL _imax) and R _i (CVL _imax), it is equivalent to prove [(1 − CVL _imax)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T] ≥ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i ) (A _i ⁺)^T]·a ⁻ _ij , [(1 − CVL _imax)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T] ≤ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i ) (A _i ⁺)^T]·a ⁺ _ij , [(CVL _imax)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T + (1 − CVL _imax)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T] ≥ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i )(A _i ⁺)^T]·b ⁻ _i , and [(CVL _imax)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T + (1 − CVL _imax)·b ⁻ _i ·(A _i ⁻) (A ⁻ _i )^T] ≤ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i )(A _i ⁺)^T]·b _i ⁺. Namely, [(1 − CVL _imax)·a ⁺ _ij ·(A ⁻ _i )(A ⁻ _i )^T] ≥ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T]·a ⁻ _ij , [(CVL _imax)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T] ≤ [(CVL _imax)·(A ⁺ _i )(A _i ⁺)^T]·a ⁺ _ij , [(CVL _imax)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T] ≥ [(CVL _imax)·(A ⁺ _i )(A _i ⁺)^T]·b ⁻ _i , and [(1 − CVL _imax)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T] ≤ [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T]·b ⁺ _i . These inequalities hold because, for any i ∈ {1, 2, …, t} and any j ∈ {1, 2, …, n}, we have 1 − CVL _imax ≥ 0, CVL _imax ≥ 0, (A ⁻ _i )(A _i ⁻)^T ≥ 0, (A ⁺ _i )(A _i ⁺)^T ≥ 0, a ⁺ _ij ≥ a ⁻ _ij and b ⁺ _i ≥ b ⁻ _i .□

Theorem 2

DCVL(CVL _imax) < DCVL(CVL _imax) if CVL _imax1 < CVL _imax2 where CVL _imax1 and CVL _imax2 are two levels of CVL _imax.

Proof

Due to the formulation of DCVL(CVL _imax), it is sufficient to prove that both F _ij (L _ij (CVL _imax)) and G _i (R _i (CVL _imax)) are monotonically decreasing with CVL _imax for any i ∈ {1, 2, …, t} and j ∈ {1, 2, …, n}. It is equivalent to prove that a) L _ij (CVL _imax1) ≥ L _ij (CVL _imax2) and b) R _i (CVL _imax1) ≤ R _i (CVL _imax2), since both F _ij (·) and G _i (·) are monotonically increasing functions.

(a). Since CVL _imax1 ≤ CVL _imax2 and a ⁻ _ij ≤ a ⁺ _ij , we have (CVL _imax1−CVL _imax2)·(a ⁻ _ij − a ⁺ _ij ) ≥ 0. Because (A ⁺ _i )(A _i ⁺)^T·(A ⁻ _i )(A _i ⁻)^T > 0, we have (CVL _imax1)·(1 − CVL _imax2)·a ⁻ _ij + (1 − CVL _imax1)·(CVL _imax2)·a ⁺ _ij ≥ (CVL _imax2)·(1 − CVL _imax1)·a ⁻ _ij + (1 − CVL _imax2)·(CVL _imax1)·a ⁺ _ij . Equivalently, we have (1 − CVL _imax1)·(1 − CVL _imax2)·a ⁺ _ij ·(A ⁻ _i )(A ⁻ _i )^T·(A _i ⁻)(A _i ⁻)^T + [(CVL _imax1)·(1 − CVL _imax2)·a ⁻ _ij + (1 − CVL _imax1)·(CVL _imax2)·a ⁺ _ij ]·(A ⁺ _i ) (A _i ⁺)^T·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(CVL _imax2)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T·(A _i ⁺)(A _i ⁺)^T ≥ (1 − CVL _imax1)·(1 − CVL _imax2)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T·(A _i ⁻)(A _i ⁻)^T + [(1 − CVL _imax1)·(CVL _imax2)·a ⁻ _ij + (CVL _imax1)·(1 − CVL _imax2)·a ⁺ _ij ]·(A ⁺ _i ) (A _i ⁺)^T·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(CVL _imax2)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T·(A _i ⁺)(A _i ⁺)^T, and furthermore (1 − CVL _imax1)·a ⁺ _ij · (A ⁻ _i )(A ⁻ _i )^T·(1 − CVL _imax2)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T·(1 − CVL _imax2)·(A ⁻ _i )(A ⁻ _i )^T + (1 − CVL _imax1)·a ⁺ _ij ·(A _i ⁻)(A _i ⁻)^T·(CVL _imax2)·(A ⁺ _i )(A ⁺ _i )^T + (CVL _imax1)·a ⁻ _ij ·(A ⁺ _i ) (A _i ⁺)^T·(CVL _imax2)·(A ⁺ _i )(A _i ⁺)^T ≥ (1 − CVL _imax2)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T·(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T·(1 − CVL _imax1)·(A ⁻ _i ) (A _i ⁻)^T + (1 − CVL _imax2)·a ⁺ _ij ·(A _i ⁻)(A _i ⁻)^T·(CVL _imax1)·(A ⁺ _i ) (A _i ⁺)^T + (CVL _imax2)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T·(CVL _imax1)·(A ⁺ _i ) (A _i ⁺)^T. Since [(1 − CVL _imax)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax)·(A ⁺ _i )(A _i ⁺)^T] ≥ 0 for any CVL _imax ∈ [0, 1] and any i ∈ {1, 2, …, t}, we have [(1 − CVL _imax1)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T]·[(1 − CVL _imax2)·(A ⁻ _i ) (A _i ⁻)^T + (CVL _imax2)·(A ⁺ _i ) (A _i ⁺)^T] ≥ [(1 − CVL _imax2)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·a ⁻ _ij ·(A ⁺ _i )(A _i ⁺)^T]·[(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(A ⁺ _i )(A _i ⁺)^T]. Thus, [(1 − CVL _imax1)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·a ⁻ _ij ·(A ⁺ _i ) (A _i ⁺)^T]/[(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(A ⁺ _i )(A _i ⁺)^T] ≥ [(1 − CVL _imax2)·a ⁺ _ij ·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·a ⁻ _ij ·(A ⁺ _i ) (A _i ⁺)^T]/[(1 − CVL _imax2)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·(A ⁺ _i )(A _i ⁺)^T]. It is equivalent to L _ij (CVL _imax1) ≥ L _ij (CVL _imax2) from the formulation of L _ij (CVL _imax).□

(b). From the formulation of R _i (CVL _imax), we are going to prove [(CVL _imax1)·b ⁺ _i ·(A ⁺ _i )(A _i ⁺)^T + (1 − CVL _imax1)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T]/[(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(A ⁺ _i )(A _i ⁺)^T] ≤ [(CVL _imax2)·b ⁺ _i ·(A _i ⁺) (A _i ⁺)^T + (1 − CVL _imax2)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T]/[(1 − CVL _imax2)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·(A ⁺ _i )(A _i ⁺)^T]. That is, [(CVL _imax1)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T + (1 − CVL _imax1)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T]·[(1 − CVL _imax2)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·(A ⁺ _i ) (A ⁺ _i )^T] ≤ [(CVL _imax2)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T + (1 − CVL _imax2)·b ⁻ _i ·(A _i ⁻)(A ⁻ _i )^T]·[(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(A ⁺ _i )(A ⁺ _i )^T], or (CVL _imax1)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T·(1 − CVL _imax2)·(A ⁻ _i )(A ⁻ _i )^T + (1 − CVL _imax1)·b ⁻ _i ·(A _i ⁻) (A ⁻ _i )^T·(1 − CVL _imax2)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T·(CVL _imax2)·(A ⁺ _i )(A _i ⁺)^T + (1 − CVL _imax1)·b ⁻ _i ·(A _i ⁻) (A _i ⁻)^T·(CVL _imax2)·(A ⁺ _i )(A _i ⁺)^T ≤ [(CVL _imax2)·b ⁺ _i ·(A _i ⁺)(A _i ⁺)^T·(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (1 − CVL _imax2)·b ⁻ _i ·(A _i ⁻)(A _i ⁻)^T·(1 − CVL _imax1)·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax2)·b ⁺ _i ·(A ⁺ _i )(A _i ⁺)^T·(CVL _imax1)·(A ⁺ _i ) (A _i ⁺)^T + (1 − CVL _imax2)·b ⁻ _i ·(A ⁻ _i )(A _i ⁻)^T·(CVL _imax1)·(A ⁺ _i )(A ⁺ _i )^T. It is equivalent to [(CVL _imax1)·(1 − CVL _imax2)·b ⁺ _i + (1 − CVL _imax1)·b ⁻ _i ·(CVL _imax2)]·(A ⁺ _i )(A ⁺ _i )^T·(A ⁻ _i )(A ⁻ _i )^T + (1 − CVL _imax1)·(1 − CVL _imax2)·b ⁻ _i ·(A ⁻ _i )(A ⁻ _i )^T·(A ⁻ _i )(A ⁻ _i )^T + (CVL _imax1)·(CVL _imax2)·b ⁺ _i ·(A ⁺ _i )(A ⁺ _i )^T·(A ⁺ _i )(A ⁺ _i )^T ≤ [(1 − CVL _imax1)·(CVL _imax2)·b ⁺ _i + (CVL _imax1)·(1 − CVL _imax2)·b ⁻ _i ]·(A ⁺ _i )(A _i ⁺)^T·(A ⁻ _i )(A _i ⁻)^T + (1 − CVL _imax1)·(1 − CVL _imax2)·b ⁻ _i ·(A ⁻ _i )(A ⁻ _i )^T·(A ⁻ _i )(A _i ⁻)^T + (CVL _imax1)·(CVL _imax2)·b ⁺ _i ·(A ⁺ _i )(A _i ⁺)^T·(A ⁺ _i )(A _i ⁺)^T. Since CVL _imax1 ≤ CVL _imax2 and b ⁻ _i ≤ b ⁺ _i , we have [(CVL _imax1)·(1 − CVL _imax2)·b ⁺ _i + (1 − CVL _imax1)·b ⁻ _i ·(CVL _imax2)] − [(1 − CVL _imax1)·(CVL _imax2)·b ⁺ _i + (CVL _imax1)·(1 − CVL _imax2)·b ⁻ _i ] = (CVL _imax1−CVL _imax2)·(b ⁺ _i − b ⁻ _i ) and (CVL _imax1−CVL _imax2)·(b ⁺ _i − b ⁻ _i ) ≤ 0. At the meantime, (A ⁺ _i )(A ⁺ _i )^T·(A ⁻ _i )(A ⁻ _i )^T ≥ 0. Thus, we have R _i (CVL _imax1) ≤ R _i (CVL _imax2).□

Theorem 3

If the necessarily optimal solution exists for SILP-2 model (3), the integrally optimal solution equals to the necessarily optimal solution.

Proof

Let X _opt = (x _1opt, x _2opt, …, x _nopt) be the integrally optimal solution of model (3). From Definition 6, we have $\int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{j} } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right)$ is maximized when X = X _opt. Namely, there does not exist another feasible solution X ^′ = (x ^′₁ , x ^′₂ , …, x ^′ _n ) such that $\int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{j}^{\prime } } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right) \, \quad > \, \int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{{j{\text{opt}}}} } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right)$. Assume X _opt is not the necessarily optimal solution that exists for SILP-2 model (3). From the definition of necessarily optimal solution, there must exist another vector of feasible solutions, assumed as X ^″ = (x ^″₁ , x ^″₂ , …, x ^″ _n ), such that $\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{j}^{\prime \prime } } \right]} \, \ge \, \sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{{j{\text{opt}}}} } \right]}$ for all combinations of d _k ∈ [d ⁻ _k , d ⁺ _k ] (k = 1, 2, …, r) and $\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{j}^{\prime \prime } } \right]} > \sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{{j{\text{opt}}}} } \right]}$ for at least one combination of d _k ∈ [d ⁻ _k , d ⁺ _k ] (k = 1, 2, …, r). Thus, we have $\int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{j}^{\prime \prime } } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right) > \int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{{j{\text{opt}}}} } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right)$. This is contradictory to the maximization of $\int \ldots \int \left\{ {\sum\nolimits_{j = 1}^{n} {\left[ {\left[ {g_{j} \left( {d_{1} ,d_{2} , \ldots ,d_{r} } \right) \, + h_{j} } \right]x_{{j{\text{opt}}}} } \right]} } \right\}d\left( {d_{1} } \right) \ldots d\left( {d_{r} } \right)$. Therefore, the integrally optimal solution is also the necessarily optimal solution.□