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A note on Lagrange multipliers in the multiple constraint case
Economics Letters 75 (2002) 141–145 www.elsevier.com / locate / econbase
A note on Lagrange multipliers in the multiple constraint case ´ * Manuel Besada, Miguel Miras ´ ´ , Universidade de Vigo, 36200 Vigo, Spain , Facultade de Economıa Departamento de Matematicas Received 22 April 2001; received in revised form 21 August 2001; accepted 1 October 2001
Abstract The well known reciprocal relationship among the Lagrange multiplier of the utility maximization problem subject to the budget constraint and the associated cost minimization problem has been recently extended to the multiple constraint case. A ‘dual’ minimization problem is attached to every constraint of the primal with the corresponding multipliers satisfying a reciprocal relationship. In general, the optimal solutions of the primal and dual problems are not the same. We prove that a maximum of the primal problem, that satisfies the second order sufficient conditions, is an extreme point of every dual problem. Furthermore, only when all the primal multipliers are strictly positive, the primal maximum is a minimum for all the dual problems. 2002 Elsevier Science B.V. All rights reserved. Keywords: Lagrange multipliers; Multiple constraints; Dual problems JEL classification: C61; D11
1. Introduction The classical problem of the household maximizing a utility function subject to an income constraint has been thoroughly studied. Provided that the problem is solvable, under the usual hypothesis on the utility (strict quasi-concavity, monotonicity, consumption set contained in the positive cone . . . ) and assuming that the money constraint is linear with positive prices, it is easy to establish that the optimal solution is unique and has strictly positive Lagrange multiplier. To this problem, called primal, one can associate a cost minimization problem, called dual. In fact, Silberberg (1990) proves that the second order sufficient conditions for the utility maximization and the cost minimization problems are equivalent: if one is satisfied, so is the other. Formally, in the single constraint case, a maximum of the primal problem with strictly positive multiplier that satisfies the * Corresponding author. Tel.: 134-9-868-12449; fax: 134-9-868-12401. ´ E-mail address: [email protected] (M. Miras). 0165-1765 / 02 / $ – see front matter PII: S0165-1765( 01 )00596-1
2002 Elsevier Science B.V. All rights reserved.
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
142
second order sufficient conditions is a minimum of the dual problem. Moreover, the Lagrange multipliers of both problems are reciprocal. Recently, Weber (1998) generalizes the former results to the multiple constraint case by associating with any maximization problem subject to m binding constraints (not necessarily linear), m different dual minimization problems; one for every constraint. He proves that any critical point of the Lagrangian of the primal problem, in particular the maximum, is a critical point of the Lagrangian of every dual problem and finds that the multipliers are reciprocally related. We extend Weber’s analysis by showing that the primal maximum is always an extreme point of every dual problem: a maximum if the associated primal multiplier is negative and a minimum if it is negative. Consequently, when all the Lagrange multipliers of the primal maximum are strictly positive, this point is a minimum for all the dual problems, with the multipliers related by the mentioned reciprocal relationships.
2. Preliminaries Let U :R n → R and g:R n → R m , g 5 ( g1 , . . . ,gm ), with m , n, be two continuously differentiable functions and denote by Dg(x) the Jacobian matrix of g at x. Consider the maximization problem Max U(x) s.t.: g j (x) 5 0, j 5 1, . . . ,m
(P) m
p
The Lagrangian for this primal problem is L(x) 5 U(x) 2 o j51 l j g j (x). Every constraint gk (x) 5 0, k [ h1, . . . ,mj, gives rise to a minimization dual problem Min gk (x)
H
0
U 2 U(x) 5 0 g j (x) 5 0, j ± k
s.t.: 0
(D k ) k
k
0
where U [ R is a fixed level of utility. The Lagrangian for (Dk ) is L (x) 5 gk (x) 1 m (U 2 U(x)) 1 o j ±k l jk g j (x). Whenever (P) is solvable and x 0 [ R n is an optimal solution of (P) we will take U 0 5 U(x 0 ). The main result in Weber (1998) states that if xo [ R n is a critical point of L, with non-zero multipliers, then x 0 is also a critical point of L k with multipliers given by
l kj 1 p ] ] m 5 p , l j 5 k ; k 5 1, . . . ,m, j ± k. lk m k
(1)
Note that the existence of degenerate constraints, those having zero as associated Lagrange multiplier, see Luenberger (1973), is ruled out explicitly.
3. Main results To analyze when the primal optimum is also a solution for all the dual problems, we look for a link between the corresponding second order sufficient conditions. For the sake of simplicity, and without
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
143
loss of generality, we consider the case of two constraints (m 5 2). Let us introduce some useful notations. For every r 5 1, . . . ,n, we will write
S
D
t
t
S
D
≠g j ≠g j ≠U ≠U =rU 5 ], . . . ,] , =r g j 5 ], . . . ,] , ≠x 1 ≠x r ≠x 1 ≠x r n
with =rU(x 0 ) and =r g j (x 0 ) representing their transpose column vectors at x 0 [ R . Let H(x 0 ) and H 1 (x 0 ) be the Hessian matrices of the Lagrangians L and L 1 , respectively, at x 0 and denote by Hr (x 0 ) and H r1 (x 0 ) the submatrices formed by taking their first r rows and first r columns, respectively. Then, the bordered Hessian matrices of (P) and (D1 ) and their principal minors can be written as 0 D(x 0 ) 5 0 =g1 (x 0 )t
1
0 0 =g2 (x 0 )t
0 D (x 0 ) 5 0 2=U(x 0 )t 1
1
0 Dr (x 0 ) 5 det 0 =r g1 (x 0 )t
1
0 0 =g2 (x 0 )t
1
2
2=U(x 0 ) =g2 (x 0 ) H 1 (x 0 )
0 0 =r g2 (x 0 )t
0 D (x 0 ) 5 det 0 2=rU(x 0 )t 1 r
=g1 (x 0 ) =g2 (x 0 ) H(x 0 )
2
=r g1 (x 0 ) =r g2 (x 0 ) , r 5 1, . . . n Hr (x 0 )
0 0 =r g2 (x 0 )t
2
2=rU(x 0 ) =r g2 (x 0 ) , r 5 1, . . . n 1 H (x 0 )
2
We can now prove that the primal maximum is an extreme point of every dual problem, but not necessarily a minimum. Theorem 1. Let U :R n → R, n . 2, and g:R n → R 2 be twice differentiable functions and xo [ R n such that the vectors h=g1 (xo ),=g2 (xo )j are linearly independent. Assume that xo is a maximum of the primal problem (P), with non-zero Lagrange multipliers l 1p , l 2p , that satisfies the second order sufficient conditions. Then for every k 5 1, 2, if l kp . 0, x 0 is a minimum of the dual problem (Dk ); whereas if l pk , 0, x 0 is a maximum of the dual problem (Dk ). Proof. We prove the result for the first dual problem. Obviously, the proof for the second dual problem is identical. By the Lagrange theorem, x 0 is a critical point of L, hence there exist unique l 1p , l p2 [ R such that =U(x 0 ) 2 l p1 =g1 (x 0 ) 2 l p2 =g2 (x 0 ) 5 0. Then, as l p1 and l p2 are non-zero, the vectors h=U(x 0 ),=g2 (x 0 )j are linearly independent and the rank condition for the problem (D1 ) is satisfied. In addition, x 0 is a critical point of the dual problem (D1 ) with multipliers given by relation (1), that is,
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
144
=g1 (x 0 ) 2 m 1 =U(x 0 ) 1 l 21 =g2 (x 0 ) 5 0. Furthermore, H 1 (x 0 ) 5 2 m 1 H(x 0 ); thus, it is easy to check that, for every r 5 1, . . . ,m, the following chain of equalities is valid, 0 0 2=rU(x 0 ) 0 =r g2 (x 0 ) D 5 0 t t 2=rU(x 0 ) =r g2 (x 0 ) 2 m 1 Hr (x 0 ) 0 0 2 l p1 =r g1 (x 0 ) 2 l p2 =r g2 (x 0 ) 0 =r g2 (x 0 ) 5 0 p t p t t 2 l 1 =r g1 (x 0 ) 2 l 2 =r g2 (x 0 ) =r g2 (x 0 ) 2 m 1 Hr (x 0 ) 0 0 2 l p1 =r g1 (x 0 ) 0 =r g2 (x 0 ) 5 0 1 r
* * *
*
*
2 l p1 =r g1 (x 0 )t =r g2 (x 0 )t 2 m 1 Hr (x 0 ) 0
*
0
*
2 m 1 =r g1 (x 0 )
*
5 ( m 1 )26 0 0 2 m 1 =r g2 (x 0 ) 5 (2m 1 )r24 Dr . 2 m 1 =r g1 (x 0 )t 2 m 1 =r g2 (x 0 )t 2 m 1 Hr (x 0 ) Since that xo is a maximum of (P) satisfying the second order sufficient conditions, thus, see Sydseater (1981), for every r 5 3, . . . ,n, Dr . 0 if r is even and Dr , 0 if r is odd. Suppose that l p1 . 0, or equivalently, m1 . 0. If r is even, (2m 1 )r 24 . 0 and Dr . 0, consequently D 1r . 0. If r is 1 1 r24 odd, D r . 0 because (2m ) , 0 and Dr , 0. Therefore, the second order sufficient conditions p 1 1 r 24 imply that xo is a minimum of (D1 ). Now, if l 1 , 0 then m , 0. In this case, (2m ) . 0 for all r 1 and the sign of D r coincides with the sign of Dr . It follows that x 0 is a maximum of (D1 ). h Naturally, Theorem 1 is valid if we have m , n constraints; the fact that we prove it for m 5 2 is just a matter of convenience. Analogous results can be stated if the primal problem is one of minimizing the objective function. Theorem 2. Let U :R n → R, n . 2, g:R n → R m , m , n, be twice differentiable functions and x 0 [ R n such that rank Dg(x 0 ) 5 m. If x 0 is a maximum of ( P), with strictly positive Lagrange multipliers, that satisfies the second order sufficient conditions, then, for every k 5 1, . . . ,m, x 0 is a minimum of the dual problem ( Dk ) and relation (1) holds. In summary, the primal–dual relationship extension to the multiple constraint case is completed: the primal maximum is a minimum of the kth-dual if the primal Lagrange multiplier associated with the restriction gk is positive and, moreover, the reciprocal expressions in (1) are valid.
Acknowledgements We have benefited from the financial support of the Spanish Ministry of Education (grants PB98-0613-C02-01 and BEC2000-1388-C04-01 from DGICYT) and Xunta de Galicia (grant PGIDT00PXI30001PN).
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
145
References Luenberger, D.G., 1973. Introduction to Linear and Non-linear Programming. Addison-Wesley, Reading, Massachusetts. Silberberg, E., 1990. The Structure of Economics: A Mathematical Analysis. McGraw-Hill, New York. Sydseater, K., 1981. Topics in Mathematical Analysis for Economists. Academic Press Inc, London. Weber, C.E., 1998. A note on Lagrange multipliers with several binding constraints. Economics Letters 59, 71–75.
A note on Lagrange multipliers in the multiple constraint case ´ * Manuel Besada, Miguel Miras ´ ´ , Universidade de Vigo, 36200 Vigo, Spain , Facultade de Economıa Departamento de Matematicas Received 22 April 2001; received in revised form 21 August 2001; accepted 1 October 2001
Abstract The well known reciprocal relationship among the Lagrange multiplier of the utility maximization problem subject to the budget constraint and the associated cost minimization problem has been recently extended to the multiple constraint case. A ‘dual’ minimization problem is attached to every constraint of the primal with the corresponding multipliers satisfying a reciprocal relationship. In general, the optimal solutions of the primal and dual problems are not the same. We prove that a maximum of the primal problem, that satisfies the second order sufficient conditions, is an extreme point of every dual problem. Furthermore, only when all the primal multipliers are strictly positive, the primal maximum is a minimum for all the dual problems. 2002 Elsevier Science B.V. All rights reserved. Keywords: Lagrange multipliers; Multiple constraints; Dual problems JEL classification: C61; D11
1. Introduction The classical problem of the household maximizing a utility function subject to an income constraint has been thoroughly studied. Provided that the problem is solvable, under the usual hypothesis on the utility (strict quasi-concavity, monotonicity, consumption set contained in the positive cone . . . ) and assuming that the money constraint is linear with positive prices, it is easy to establish that the optimal solution is unique and has strictly positive Lagrange multiplier. To this problem, called primal, one can associate a cost minimization problem, called dual. In fact, Silberberg (1990) proves that the second order sufficient conditions for the utility maximization and the cost minimization problems are equivalent: if one is satisfied, so is the other. Formally, in the single constraint case, a maximum of the primal problem with strictly positive multiplier that satisfies the * Corresponding author. Tel.: 134-9-868-12449; fax: 134-9-868-12401. ´ E-mail address: [email protected] (M. Miras). 0165-1765 / 02 / $ – see front matter PII: S0165-1765( 01 )00596-1
2002 Elsevier Science B.V. All rights reserved.
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
142
second order sufficient conditions is a minimum of the dual problem. Moreover, the Lagrange multipliers of both problems are reciprocal. Recently, Weber (1998) generalizes the former results to the multiple constraint case by associating with any maximization problem subject to m binding constraints (not necessarily linear), m different dual minimization problems; one for every constraint. He proves that any critical point of the Lagrangian of the primal problem, in particular the maximum, is a critical point of the Lagrangian of every dual problem and finds that the multipliers are reciprocally related. We extend Weber’s analysis by showing that the primal maximum is always an extreme point of every dual problem: a maximum if the associated primal multiplier is negative and a minimum if it is negative. Consequently, when all the Lagrange multipliers of the primal maximum are strictly positive, this point is a minimum for all the dual problems, with the multipliers related by the mentioned reciprocal relationships.
2. Preliminaries Let U :R n → R and g:R n → R m , g 5 ( g1 , . . . ,gm ), with m , n, be two continuously differentiable functions and denote by Dg(x) the Jacobian matrix of g at x. Consider the maximization problem Max U(x) s.t.: g j (x) 5 0, j 5 1, . . . ,m
(P) m
p
The Lagrangian for this primal problem is L(x) 5 U(x) 2 o j51 l j g j (x). Every constraint gk (x) 5 0, k [ h1, . . . ,mj, gives rise to a minimization dual problem Min gk (x)
H
0
U 2 U(x) 5 0 g j (x) 5 0, j ± k
s.t.: 0
(D k ) k
k
0
where U [ R is a fixed level of utility. The Lagrangian for (Dk ) is L (x) 5 gk (x) 1 m (U 2 U(x)) 1 o j ±k l jk g j (x). Whenever (P) is solvable and x 0 [ R n is an optimal solution of (P) we will take U 0 5 U(x 0 ). The main result in Weber (1998) states that if xo [ R n is a critical point of L, with non-zero multipliers, then x 0 is also a critical point of L k with multipliers given by
l kj 1 p ] ] m 5 p , l j 5 k ; k 5 1, . . . ,m, j ± k. lk m k
(1)
Note that the existence of degenerate constraints, those having zero as associated Lagrange multiplier, see Luenberger (1973), is ruled out explicitly.
3. Main results To analyze when the primal optimum is also a solution for all the dual problems, we look for a link between the corresponding second order sufficient conditions. For the sake of simplicity, and without
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
143
loss of generality, we consider the case of two constraints (m 5 2). Let us introduce some useful notations. For every r 5 1, . . . ,n, we will write
S
D
t
t
S
D
≠g j ≠g j ≠U ≠U =rU 5 ], . . . ,] , =r g j 5 ], . . . ,] , ≠x 1 ≠x r ≠x 1 ≠x r n
with =rU(x 0 ) and =r g j (x 0 ) representing their transpose column vectors at x 0 [ R . Let H(x 0 ) and H 1 (x 0 ) be the Hessian matrices of the Lagrangians L and L 1 , respectively, at x 0 and denote by Hr (x 0 ) and H r1 (x 0 ) the submatrices formed by taking their first r rows and first r columns, respectively. Then, the bordered Hessian matrices of (P) and (D1 ) and their principal minors can be written as 0 D(x 0 ) 5 0 =g1 (x 0 )t
1
0 0 =g2 (x 0 )t
0 D (x 0 ) 5 0 2=U(x 0 )t 1
1
0 Dr (x 0 ) 5 det 0 =r g1 (x 0 )t
1
0 0 =g2 (x 0 )t
1
2
2=U(x 0 ) =g2 (x 0 ) H 1 (x 0 )
0 0 =r g2 (x 0 )t
0 D (x 0 ) 5 det 0 2=rU(x 0 )t 1 r
=g1 (x 0 ) =g2 (x 0 ) H(x 0 )
2
=r g1 (x 0 ) =r g2 (x 0 ) , r 5 1, . . . n Hr (x 0 )
0 0 =r g2 (x 0 )t
2
2=rU(x 0 ) =r g2 (x 0 ) , r 5 1, . . . n 1 H (x 0 )
2
We can now prove that the primal maximum is an extreme point of every dual problem, but not necessarily a minimum. Theorem 1. Let U :R n → R, n . 2, and g:R n → R 2 be twice differentiable functions and xo [ R n such that the vectors h=g1 (xo ),=g2 (xo )j are linearly independent. Assume that xo is a maximum of the primal problem (P), with non-zero Lagrange multipliers l 1p , l 2p , that satisfies the second order sufficient conditions. Then for every k 5 1, 2, if l kp . 0, x 0 is a minimum of the dual problem (Dk ); whereas if l pk , 0, x 0 is a maximum of the dual problem (Dk ). Proof. We prove the result for the first dual problem. Obviously, the proof for the second dual problem is identical. By the Lagrange theorem, x 0 is a critical point of L, hence there exist unique l 1p , l p2 [ R such that =U(x 0 ) 2 l p1 =g1 (x 0 ) 2 l p2 =g2 (x 0 ) 5 0. Then, as l p1 and l p2 are non-zero, the vectors h=U(x 0 ),=g2 (x 0 )j are linearly independent and the rank condition for the problem (D1 ) is satisfied. In addition, x 0 is a critical point of the dual problem (D1 ) with multipliers given by relation (1), that is,
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
144
=g1 (x 0 ) 2 m 1 =U(x 0 ) 1 l 21 =g2 (x 0 ) 5 0. Furthermore, H 1 (x 0 ) 5 2 m 1 H(x 0 ); thus, it is easy to check that, for every r 5 1, . . . ,m, the following chain of equalities is valid, 0 0 2=rU(x 0 ) 0 =r g2 (x 0 ) D 5 0 t t 2=rU(x 0 ) =r g2 (x 0 ) 2 m 1 Hr (x 0 ) 0 0 2 l p1 =r g1 (x 0 ) 2 l p2 =r g2 (x 0 ) 0 =r g2 (x 0 ) 5 0 p t p t t 2 l 1 =r g1 (x 0 ) 2 l 2 =r g2 (x 0 ) =r g2 (x 0 ) 2 m 1 Hr (x 0 ) 0 0 2 l p1 =r g1 (x 0 ) 0 =r g2 (x 0 ) 5 0 1 r
* * *
*
*
2 l p1 =r g1 (x 0 )t =r g2 (x 0 )t 2 m 1 Hr (x 0 ) 0
*
0
*
2 m 1 =r g1 (x 0 )
*
5 ( m 1 )26 0 0 2 m 1 =r g2 (x 0 ) 5 (2m 1 )r24 Dr . 2 m 1 =r g1 (x 0 )t 2 m 1 =r g2 (x 0 )t 2 m 1 Hr (x 0 ) Since that xo is a maximum of (P) satisfying the second order sufficient conditions, thus, see Sydseater (1981), for every r 5 3, . . . ,n, Dr . 0 if r is even and Dr , 0 if r is odd. Suppose that l p1 . 0, or equivalently, m1 . 0. If r is even, (2m 1 )r 24 . 0 and Dr . 0, consequently D 1r . 0. If r is 1 1 r24 odd, D r . 0 because (2m ) , 0 and Dr , 0. Therefore, the second order sufficient conditions p 1 1 r 24 imply that xo is a minimum of (D1 ). Now, if l 1 , 0 then m , 0. In this case, (2m ) . 0 for all r 1 and the sign of D r coincides with the sign of Dr . It follows that x 0 is a maximum of (D1 ). h Naturally, Theorem 1 is valid if we have m , n constraints; the fact that we prove it for m 5 2 is just a matter of convenience. Analogous results can be stated if the primal problem is one of minimizing the objective function. Theorem 2. Let U :R n → R, n . 2, g:R n → R m , m , n, be twice differentiable functions and x 0 [ R n such that rank Dg(x 0 ) 5 m. If x 0 is a maximum of ( P), with strictly positive Lagrange multipliers, that satisfies the second order sufficient conditions, then, for every k 5 1, . . . ,m, x 0 is a minimum of the dual problem ( Dk ) and relation (1) holds. In summary, the primal–dual relationship extension to the multiple constraint case is completed: the primal maximum is a minimum of the kth-dual if the primal Lagrange multiplier associated with the restriction gk is positive and, moreover, the reciprocal expressions in (1) are valid.
Acknowledgements We have benefited from the financial support of the Spanish Ministry of Education (grants PB98-0613-C02-01 and BEC2000-1388-C04-01 from DGICYT) and Xunta de Galicia (grant PGIDT00PXI30001PN).
´ / Economics Letters 75 (2002) 141 – 145 M. Besada, M. Miras
145
References Luenberger, D.G., 1973. Introduction to Linear and Non-linear Programming. Addison-Wesley, Reading, Massachusetts. Silberberg, E., 1990. The Structure of Economics: A Mathematical Analysis. McGraw-Hill, New York. Sydseater, K., 1981. Topics in Mathematical Analysis for Economists. Academic Press Inc, London. Weber, C.E., 1998. A note on Lagrange multipliers with several binding constraints. Economics Letters 59, 71–75.