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Comment on “A nonlinear Lagrangian dual for integer programming”
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Operations Research Letters
Operations Research Letters 32 (2004) 197 – 198
www.elsevier.com/locate/dsw
Comment on “A nonlinear Lagrangian dual for integer programming” Paul A. Rubin Department of Management, The Eli Broad Graduate School of Management, Michigan State University, East Lansing, MI 48824-1122, USA Received 26 December 2002; received in revised form 11 March 2003; accepted 6 May 2003
Abstract We present a counterexample and correction to the contention by Xu and Li that the nonlinear Lagrangian dual problem they propose [Oper. Res. Lett. 30 (2002) 401] asymptotically has no duality gap. c 2003 Elsevier B.V. All rights reserved. Keywords: Integer programming; Nonlinear Lagrangian theory; Lagrangian relaxation; Dual method
1. Introduction
the problem dual to (1) is
Recently, Xu and Li [2] proposed a nonlinear Lagrangian dual for a general class of nonlinear integer programming problems. Using their notation, the primal problem is min{f(x): g(x) 6 0; x ∈ X }; n
(1) n
in which X ⊂ R is a =nite discrete set, f : R → R, g : Rn → Rm , and the feasible region S = {x ∈ X : g(x) 6 0} is presumed nonempty. Denote the optimal value of (1) by f∗ . For arbitrary p ¿ 0, they propose the nonlinear Lagrangian Lp (x; ) = m f(x) + (1= )exp( j=1 j gj (x)), where ∈ Rm p = {y ∈ Rm : y ¿ p}. Setting + Qp ( ) = min Lp (x; ); x∈X
PII of the original article: S0167-6377(02)00166-9 E-mail address: [email protected] (P.A. Rubin).
Dp = maxm Qp ( ): ∈Rp
(3)
They assert (Theorem 1, Asymptotic Strong Duality) that limp → ∞ Dp = f∗ , contending in the proof that for any p ¿ 1 there exists ∈ Rm p such that min Lp (x; ) ¿ min Lp (x; ):
x∈X \S
x∈S
(4)
Unfortunately, the Proof of Theorem 1 is Dawed: there is an implicit, and unwarranted, assumption that if the minimum = minx∈X \S maxj gj (x) of the worst single constraint violations among infeasible points occurs in constraint j, then every infeasible point violates constraint j. In fact, Theorem 1 is not correct, as we demonstrate below.
(2) 2. Counterexample Let X ={0; 1}2 ; f(x)=5−2x1 −x2 , g1 (x)=10x1 + 10x2 − 19 and g2 (x) = 10x1 − 10x2 − 9. Table 1 shows
c 2003 Elsevier B.V. All rights reserved. 0167-6377/$ - see front matter doi:10.1016/S0167-6377(03)00096-8
198
P.A. Rubin / Operations Research Letters 32 (2004) 197 – 198
Table 1 Counterexample x
f(x)
g1 (x)
g2 (x)
(0,0) (0,1) (1,0) (1,1)
5 4 3 2
−19 −9 −9 1
−9 −19 1 −9
the values of the three functions at every domain point. We =nd by inspection that S = {(0; 0); (0; 1)} and that (1) has optimal solution (0; 1) with objective value 4. Note that = 1 but that there is no constraint violated by both points in X \ S. Now consider any ∈ R2p for p ¿ 1. Obviously minx∈S L(x; ) ¿ minx∈S f(x) = 4. Outside S, we have L((1; 0); ) = 3 + (1= )exp(−9 1 + 2 ) and L((1; 1); ) = 2 + (1= )exp( 1 − 9 2 ). At least one of the two exponents −9 1 + 2 , 1 − 9 2 must be negative, since their sum is negative, so either L((1; 0); ) ¡ 3 + (1= ) 6 3 + 1=p ¡ 4 or L((1; 1); ) ¡ 2 + (1= ) 6 2 + 1=p ¡ 4. It follows that, for all ∈ R2p , minx∈X \S Lp (x; ) ¡ minx∈S Lp (x; ), contradicting (4). Moreover, for any optimal solution ∗ of (3), the corresponding primal solution x∗ (p) of (2) must be infeasible in (1). Finally, we see that Dp ¡ 3+1=p, and so limp → ∞ Dp 6 3 ¡ 4=f∗ , contradicting Theorem 1. A duality gap exists. 3. Implications In Section 4 et seq. of [2], Xu and Li consider only a single constraint, relying on the p-norm
surrogate constraint method of Li [1] to collapse multiple constraints down to one. Fortunately, Theorem 1 is valid for a single constraint. The correction to the proof is quite straightforward: is now de=ned by = minx∈X \S g(x) ¿ 0, where g(x) 6 0 is the single constraint, and = p in Eq. (20) of [2]. Thus, the combination of the proposed Lagrangian relaxation and the p-norm surrogate constraint method will produce a dual problem with no duality gap, at the cost of raising the severity of the nonlinearity (the relaxed problem now contains the exponential of a potentially high-degree polynomial of the nonlinear constraint functions) and possibly introducing some numerical stability issues. Useful areas for future research include determining which problem classes might produce computationally tractable duals using this method, and selecting (or developing) algorithms that will solve duals of this type eIciently. Acknowledgements The author acknowledges helpful comments from an anonymous reviewer. References [1] D. Li, Zero duality gap in integer programming: P-norm surrogate constraint method, Oper. Res. Lett. 25 (1999) 89–96. [2] Y. Xu, D. Li, A nonlinear Lagrangian dual for integer programming, Oper. Res. Lett. 30 (2002) 401–407.
Operations Research Letters
Operations Research Letters 32 (2004) 197 – 198
www.elsevier.com/locate/dsw
Comment on “A nonlinear Lagrangian dual for integer programming” Paul A. Rubin Department of Management, The Eli Broad Graduate School of Management, Michigan State University, East Lansing, MI 48824-1122, USA Received 26 December 2002; received in revised form 11 March 2003; accepted 6 May 2003
Abstract We present a counterexample and correction to the contention by Xu and Li that the nonlinear Lagrangian dual problem they propose [Oper. Res. Lett. 30 (2002) 401] asymptotically has no duality gap. c 2003 Elsevier B.V. All rights reserved. Keywords: Integer programming; Nonlinear Lagrangian theory; Lagrangian relaxation; Dual method
1. Introduction
the problem dual to (1) is
Recently, Xu and Li [2] proposed a nonlinear Lagrangian dual for a general class of nonlinear integer programming problems. Using their notation, the primal problem is min{f(x): g(x) 6 0; x ∈ X }; n
(1) n
in which X ⊂ R is a =nite discrete set, f : R → R, g : Rn → Rm , and the feasible region S = {x ∈ X : g(x) 6 0} is presumed nonempty. Denote the optimal value of (1) by f∗ . For arbitrary p ¿ 0, they propose the nonlinear Lagrangian Lp (x; ) = m f(x) + (1= )exp( j=1 j gj (x)), where ∈ Rm p = {y ∈ Rm : y ¿ p}. Setting + Qp ( ) = min Lp (x; ); x∈X
PII of the original article: S0167-6377(02)00166-9 E-mail address: [email protected] (P.A. Rubin).
Dp = maxm Qp ( ): ∈Rp
(3)
They assert (Theorem 1, Asymptotic Strong Duality) that limp → ∞ Dp = f∗ , contending in the proof that for any p ¿ 1 there exists ∈ Rm p such that min Lp (x; ) ¿ min Lp (x; ):
x∈X \S
x∈S
(4)
Unfortunately, the Proof of Theorem 1 is Dawed: there is an implicit, and unwarranted, assumption that if the minimum = minx∈X \S maxj gj (x) of the worst single constraint violations among infeasible points occurs in constraint j, then every infeasible point violates constraint j. In fact, Theorem 1 is not correct, as we demonstrate below.
(2) 2. Counterexample Let X ={0; 1}2 ; f(x)=5−2x1 −x2 , g1 (x)=10x1 + 10x2 − 19 and g2 (x) = 10x1 − 10x2 − 9. Table 1 shows
c 2003 Elsevier B.V. All rights reserved. 0167-6377/$ - see front matter doi:10.1016/S0167-6377(03)00096-8
198
P.A. Rubin / Operations Research Letters 32 (2004) 197 – 198
Table 1 Counterexample x
f(x)
g1 (x)
g2 (x)
(0,0) (0,1) (1,0) (1,1)
5 4 3 2
−19 −9 −9 1
−9 −19 1 −9
the values of the three functions at every domain point. We =nd by inspection that S = {(0; 0); (0; 1)} and that (1) has optimal solution (0; 1) with objective value 4. Note that = 1 but that there is no constraint violated by both points in X \ S. Now consider any ∈ R2p for p ¿ 1. Obviously minx∈S L(x; ) ¿ minx∈S f(x) = 4. Outside S, we have L((1; 0); ) = 3 + (1= )exp(−9 1 + 2 ) and L((1; 1); ) = 2 + (1= )exp( 1 − 9 2 ). At least one of the two exponents −9 1 + 2 , 1 − 9 2 must be negative, since their sum is negative, so either L((1; 0); ) ¡ 3 + (1= ) 6 3 + 1=p ¡ 4 or L((1; 1); ) ¡ 2 + (1= ) 6 2 + 1=p ¡ 4. It follows that, for all ∈ R2p , minx∈X \S Lp (x; ) ¡ minx∈S Lp (x; ), contradicting (4). Moreover, for any optimal solution ∗ of (3), the corresponding primal solution x∗ (p) of (2) must be infeasible in (1). Finally, we see that Dp ¡ 3+1=p, and so limp → ∞ Dp 6 3 ¡ 4=f∗ , contradicting Theorem 1. A duality gap exists. 3. Implications In Section 4 et seq. of [2], Xu and Li consider only a single constraint, relying on the p-norm
surrogate constraint method of Li [1] to collapse multiple constraints down to one. Fortunately, Theorem 1 is valid for a single constraint. The correction to the proof is quite straightforward: is now de=ned by = minx∈X \S g(x) ¿ 0, where g(x) 6 0 is the single constraint, and = p in Eq. (20) of [2]. Thus, the combination of the proposed Lagrangian relaxation and the p-norm surrogate constraint method will produce a dual problem with no duality gap, at the cost of raising the severity of the nonlinearity (the relaxed problem now contains the exponential of a potentially high-degree polynomial of the nonlinear constraint functions) and possibly introducing some numerical stability issues. Useful areas for future research include determining which problem classes might produce computationally tractable duals using this method, and selecting (or developing) algorithms that will solve duals of this type eIciently. Acknowledgements The author acknowledges helpful comments from an anonymous reviewer. References [1] D. Li, Zero duality gap in integer programming: P-norm surrogate constraint method, Oper. Res. Lett. 25 (1999) 89–96. [2] Y. Xu, D. Li, A nonlinear Lagrangian dual for integer programming, Oper. Res. Lett. 30 (2002) 401–407.