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.
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Acknowledgements
This research was supported by the National Key Research and Development Plan (2016YFC0502800, 2016YFA0601502), the Natural Sciences Foundation (51520105013, 51679087), the 111 Project (B14008) and the Natural Science and Engineering Research Council of Canada. We are very grateful to the editor and two anonymous peer reviewers who provided many constructive comments on how to improve our manuscript.
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Appendix
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.□
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Cheng, G., Huang, G., Dong, C. et al. Resources and environmental systems management under synchronic interval uncertainties. Stoch Environ Res Risk Assess 32, 435–456 (2018). https://doi.org/10.1007/s00477-017-1445-5
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DOI: https://doi.org/10.1007/s00477-017-1445-5