Duality Theory of Constrained Optimization

Transcription

1 Duality Theory of Constrained Optimization Robert M. Freund April, 2014 c 2014 Massachusetts Institute of Technology. All rights reserved. 1

2 2 1 The Practical Importance of Duality Duality is pervasive in nonlinear (and linear) optimization models in a wide variety of engineering and mathematical settings. Some elementary examples of models where duality is plays an important role are: Electrical networks. The current flows are primal variables and the voltage differences are the dual variables that arise in consideration of optimization (and equilibrium) in electrical networks. Economic markets. The primal variables are production levels and consumption levels, and the dual variables are prices of goods and services. Structural design. The tensions on the beams are primal variables, and the nodal displacements are the dual variables. Nonlinear (and linear) duality is extremely useful both as a conceptual as well as a computational tool. For example, dual problems and their solutions are used in connection with: (a) Identifying near-optimal solutions. A good dual solution can be used to bound the values of primal solutions, and so can be used to actually identify when a primal solution is optimal or near-optimal. (b) Proving optimality. Using a strong duality theorem, one can prove optimality of a primal solution by constructing a dual solution with the same objective function value. (c) Sensitivity analysis of the primal problem. The dual variable on a constraint represents the incremental change in the optimal solution value per unit increase in the right-hand-side (RHS) of the constraint. (d) Karush-Kuhn-Tucker (KKT) conditions. The optimal solution to the dual problem is a vector of KKT multipliers. (e) Convergence of algorithms. The dual problem is often used in the convergence analysis of algorithms. (f) Discovering and Exploiting Problem Structure. Quite often, the dual problem has some useful mathematical, geometric, or compu-

3 3 tational structure that can exploited in computing solutions to both the primal and the dual problem. 2 The Dual Problem 2.1 Constructing the Dual Problem Recall the basic constrained optimization model: OP : minimum x f(x) s.t. g 1 (x) 0 =. g m (x) 0 x X. In this model, we have f(x) : R n R and g i (x) : R n R, i = 1,..., m. We can always convert the constraints involving g i (x) to have the format g i (x) 0, and so we will now presume that our problem is of the form:

4 4 OP : z = minimum x f(x) s.t. g 1 (x) 0... g m (x) 0 x X. We will refer to X as the ground-set. Whereas in previous developments X was often assumed to be R n, in our study of duality it will be useful to presume that X itself is composed of a structural part of the feasible region of the optimization problem. A typical example is when the feasible region of our problem of interest is the intersection of k inequality constraints: F = {x R n : g i (x) 0, i = 1,..., k}. Depending on the structural properties of the functions g i ( ), we might wish re-order the constraints so that the first m of these k constraints (where of course m k) are kept explicitly as inequalities, and the last k m constraints are formulated into the ground-set Then it follows that: X := {x R n : g i (x) 0, i = m + 1,..., k}. F = {x R n : g i (x) 0, i = 1,..., m, } X, where now there are only m explicit inequality constraints. More generally, the ground-set X could be of any form, including the following:

5 5 (i) X = {x R n : x 0}, (ii) X = K where K is a convex cone, (iii) X = {x g i (x) 0, i = m + 1,..., m + k}, (iv) X = { x x Z n +}, (v) X = {x R n : Nx = b, x 0} where Nx = b models flows in a network, (vi) X = {x Ax b}, (vii) X = R n. We now describe how to construct the dual problem of OP. The first step is to form the Lagrangian function. For a nonnegative vector u, the Lagrangian function is defined simply as follows: m L(x, u) := f(x) + u T g(x) = f(x) + u i g i (x). i=1 The Lagrangian essentially takes the constraints out of the description of the feasible region of the model, and instead places them in the the objective function with multipliers u i for i = 1,..., m. It might be helpful to think of these multipliers as costs or prices or penalties. The second step is to construct the dual function L (u), which is defined as follows: L (u) := minimum x L(x, u) = minimum x f(x) + u T g(x) s.t. x X s.t. x X. (1)

6 6 Note that L (u) is defined for any given u. In order for the dual problem to be computationally viable, it will be necessary to be able to compute L (u) efficiently for any given u. Put another way, it will be necessary to be able to solve the optimization problem (1) efficiently for any given u. The third step is to write down the dual problem D, which is defined as follows: D : v = maximum u L (u) s.t. u Summary: Steps in the Construction of the Dual Problem Here is a summary of the process of constructing the dual problem of OP. The starting point is the primal problem: OP : z = minimum x f(x) s.t. g i (x) 0, i = 1,..., m x X. The dual problem D is constructed in the following three-step procedure: 1. Create the Lagrangian: L(x, u) := f(x) + u T g(x).

7 7 2. Create the dual function: L (u) := minimum x f(x) + u T g(x) s.t. x X. 3. Create the dual problem: D : v = maximum u L (u) s.t. u Example: the Dual of a Linear Problem Consider the linear optimization problem: LP : minimum x c T x s.t. Ax b x 0. We can rearrange the inequality constraints and re-write LP as: LP : minimum x c T x s.t. b Ax 0 x 0.

8 8 Let us define X := {x R n : x 0} and identify the linear inequality constraints b Ax 0 as the constraints g i (x) 0, i = 1,..., m. Therefore the Lagrangian for this problem is: L(x, u) = c T x + u T (b Ax) = u T b + (c A T u) T x. The dual function L (u) is defined to be: L (u) = min x X L(x, u) = min x 0 L(x, u). Let us now evaluate L (u). For a given value of u, L (u) is evaluated as follows: L (u) = min x 0 L(x, u) = min x 0 u T b + (c A T u) T x = u T b if A T u c if A T u c. The dual problem (D) is defined to be: and is therefore constructed as: (D) v = max u 0 L (u), (D) v = max u u T b s.t. A T u c u 0.

9 9 2.4 Remarks on Conventions involving ± It is often the case when constructing dual problems that functions such as L (u) takes the values of ± for certain values of u. Here we discuss conventions in this regard. Suppose we have an optimization problem: We could define the function: (P) : z = maximum x f(x) s.t. x S. f(x) if x S h(x) := if x / S. Then we can rewrite our problem as: (P) : z = maximum x h(x) s.t. x R n. Conversely, suppose that we have a function k( ) that takes on the value outside of a certain region S, but that k( ) is finite for all x S. Then the problem: (P) : z = maximum x k(x) s.t. x R n is equivalent to: (P) : z = maximum x k(x) s.t. x S. Similar logic applies to minimization problems over domains where the function values might take on the value +.

10 10 3 More Examples of Dual Constructions of Optimization Problems 3.1 The Dual of a Convex Quadratic Problem Consider the following convex quadratic optimization problem: QP : minimum x 1 2 xt Qx + c T x s.t. Ax b, where Q is symmetric and positive definite. To construct a dual of this problem, let us re-write the inequality constraints as b Ax 0 and dualize these constraints. We construct the Lagrangian: L(x, u) = 1 2 xt Qx + c T x + u T (b Ax) = u T b + (c A T u) T x xt Qx. The dual function L (u) is: L (u) = min x R n L(x, u) = min x R n u T b + (c A T u) T x xt Qx = u T b + min x R n(c A T u) T x xt Qx. For a given value of u, let x be the optimal value of x in this last expression. Then since the expression is a convex quadratic function of x, it follows that x must satisfy: whereby x is given by: 0 = x L( x, u) = (c A T u) + Q x,

11 11 x = Q 1 (c A T u). Substituting this value of x into the Lagrangian, we obtain: L (u) = L( x, u) = u T b 1 2 (c AT u) T Q 1 (c A T u). The dual problem (D) is defined to be: and is therefore constructed as: (D) v = max u 0 L (u), (D) v = max u u T b 1 2 (c AT u) T Q 1 (c A T u) s.t. u Remarks on Problems with Different Formats of Constraints Suppose that our problem has some inequality and equality constraints. Then just as in the case of linear optimization, we assign nonnegative dual variables u i to constraints of the form g i (x) 0, unrestricted dual variables u i to equality constraints g i (x) = 0, and non-positive dual variables u i to constraints of the form g i (x) 0. For example, suppose our problem is: OP : minimum x f(x) s.t. g i (x) 0, i L g i (x) 0, i G g i (x) = 0, i E x X.

12 12 Then the Lagrangian takes the form: L(x, u) := f(x) + u T g(x) = f(x) + i L u i g i (x) + i G u i g i (x) + i E u i g i (x), and the dual function L (u) is constructed as: L (u) := minimum x f(x) + u T g(x) s.t. x X. The dual problem is then defined to be: D : v = maximum u L (u) s.t. u i 0, i L u i 0, i G u i R, i E. For simplicity, we will typically presume that the constraints of OP are of the form g i (x) 0. This is only for simplicity of exposition, as all of the results discussed herein extend to the general case just described. 3.3 The Dual of a Log-Barrier Problem Consider the following logarithmic barrier problem:

13 13 BP : minimum x1,x 2,x 3 5x 1 + 7x 2 4x 3 3 ln(x j ) j=1 s.t. x 1 + 3x x 3 = 37 x 1 > 0, x 2 > 0, x 3 > 0. Let us define the ground-set as X = {x R n : x j > 0, j = 1,..., n}, and let us dualize on the single equality constraint. The Lagrangian function takes on the form: L(x, u) = 5x 1 + 7x 2 4x 3 3 ln(x j ) + u(37 x 1 3x 2 12x 3 ) j=1 = 37u + (5 u)x 1 + (7 3u)x 2 + ( 4 12u)x 3 3 ln(x j ). j=1 The dual function L (u) is constructed as: L (u) = min x X L(x, u) = min x>0 37u + (5 u)x 1 + (7 3u)x 2 + ( 4 12u)x 3 3 ln(x j ) = 37u + min x>0 (5 u)x 1 + (7 3u)x 2 + ( 4 12u)x 3 3 ln(x j ). j=1 j=1 Now notice that the optimization problem above separates into three univariate optimization problems of a linear function minus a logarithm term for each of the three positive variables x 1, x 2, and x 3. Examining x 1, it

14 14 holds that the minimization value will be if (5 u) 0, as we could set x 1 arbitrarily large. When (5 u) > 0, the problem of minimizing (5 u)x 1 ln(x 1 ) is a convex optimization problem whose solution is given by setting the first derivative with respect x 1 equal to zero. This means solving: (5 u) 1 x 1 = 0, or in other words, setting x 1 = Substituting this value of x 1, we obtain: 1 (5 u). ( ) 1 (5 u)x 1 ln(x 1 ) = 1 ln = 1 + ln(5 u). 5 u Using parallel logic for the other two variables, we arrive at: L (u) = 37u ln(5 u) + ln(7 3u) + ln( 4 12u) if u < 1/3 otherwise. Notice that L (u) is finite whenever 5 u > 0, 7 3u > 0, and 4 12u > 0, but that these three inequalities in u are equivalent to the single inequality u < 1/3. The dual problem (D) is defined to be: and is therefore constructed as: (D) v = max u R L (u),

15 15 (D) v = max u 37u ln(5 u) + ln(7 3u) + ln( 4 12u) s.t. u < 1/ The Dual of a Standard Conic Primal Problem We consider the following convex optimization problem in standard conic form: CP : z = minimum x c T x s.t. Ax = b x K, where K is a closed convex cone. Let K denote the dual cone of the closed convex cone K R n, defined by: K := {y R n y T x 0 for all x K}. Recall the following properties of K : Proposition 3.1 If K is a nonempty closed convex cone, then 1. K is a nonempty closed convex cone, and 2. (K ) = K. Let us consider K to be the ground-set (X = K) and we therefore dualize on the constraints Ax = b of CP, and consider the Lagrangian: L(x, u) := c T x + u T (b Ax).

16 16 The dual function is: It then follows that: L (u) := min x K L(x, u) = min x K ct x + u T (b Ax). L (u) = min x K c T x + u T (b Ax) = u T b + min x K (c A T u) T x b T u if c A T u K = if c A T u / K. We therefore write down the dual problem as: CD : v = maximum u,s b T u s.t. A T u + s = c s K. Remark 1 It is a straightforward exercise to show that the conic dual of CD is CP. 4 The Column Geometry of the Primal and Dual Problems In this section we develop a simple interpretation of the dual problem in the space of resources and costs, as follows. For a given x X, the primal problem specifies an array of resources and costs associated with x, namely:

17 17 s 1 ( ) s s 2 = z. s = m z g 1 (x) g 2 (x). g m (x) f(x) = ( ) g(x) f(x). We can think of each of these arrays as an array of resources and cost in R m+1. Define the set I: I := { (s, z) R m+1 there exists x X for which s g(x) and z f(x) }. This region is illustrated in Figure 1. z I -u H(u,α) s Figure 1: The set I. Proposition 4.1 If X is a convex set, and f( ), g 1 ( ),..., g m ( ) are convex functions on X, then I is a convex set.

18 18 Now define: H u,α := {(s, z) R m+1 : u T s + z = α}. We say that H u,α is a lower support of I if Let us also define the line: u T s + z α for all (s, z) I. L := {(s, z) R m+1 : s = 0}. Note that H u,α L = {(0, α)}. Now consider the problem of determining the hyperplane H u,α that is a lower support of I and whose intersection with L is the highest, that is, whose value α is maximized. This problem is:

19 19 maximum u,α α s.t. H u,α is a lower support of I = maximum u,α α s.t. u T s + z α for all (s, z) I = maximum u 0,α α s.t. u T s + z α for all (s, z) I = maximum u 0,α α s.t. u T g(x) + f(x) α for all x X

20 20 = maximum u 0,α α s.t. L(x, u) α for all x X = maximum u 0,α α s.t. min x X L(x, u) α = maximum u 0,α α s.t. L (u) α = maximum u L (u) s.t. u 0. This last expression is exactly the dual problem (D), whose value is given by v. We summarize this observation as: Remark 1 The dual problem corresponds to finding a hyperplane H u,α that is a lower support of I, and whose intersection with L is the highest. This highest value is exactly the value of the dual problem, namely v.

21 21 5 The Dual is a Concave Maximization Problem Theorem 5.1 The dual function L (u) is a concave function on its domain S = {u R m : u 0}. Proof: Let u 1 0 and u 2 0 be two values of the dual variables, and let u = λu 1 + (1 λ)u 2, where λ [0, 1]. Then L (u) = min x X f(x) + u T g(x) = min x X λ [ f(x) + u T 1 g(x)] + (1 λ) [ f(x) + u T 2 g(x)] λ [ min x X f(x) + u T 1 g(x)] + (1 λ) [ min x X (f(x) + u T 2 g(x)] = λl (u 1 ) + (1 λ)l (u 2 ). Therefore L (u) is a concave function. 6 Weak Duality Let z and v be the optimal values of the primal and the dual problems: OP : z = minimum x f(x) s.t. g i (x) 0, i = 1,..., m x X, D : v = maximum u L (u) s.t. u 0.

22 22 Theorem 6.1 Weak Duality Theorem: If x is feasible for OP and ū is feasible for D, then f( x) L (ū). In particular, z v. Proof: If x is feasible for OP and ū is feasible for D, then Therefore z v. f( x) f( x) + ū T g( x) min x X f(x) + ūt g(x) = L (ū). Corollary 6.1 If x is feasible for OP and ū 0 is feasible for D, and f( x) = L (ū), then x and ū are optimal solutions of OP and D, respectively. Corollary 6.2 If z =, then D has no feasible solution. Corollary 6.3 If v = +, then OP has no feasible solution. A strong duality theorem cannot necessarily be established for a general nonlinear optimization problem without adding assumptions regarding convexity and the existence of a Slater point. This will be developed in Section 8. 7 Saddlepoint Optimality Criteria The pair ( x, ū) is called a saddlepoint of the Lagrangian L(x, u) if x X, ū 0, and L( x, u) L( x, ū) L(x, ū) for all x X and u 0. Theorem 7.1 A pair ( x, ū) is a saddlepoint of the Lagrangian if and only if the following three conditions are satisfied:

23 23 (i) L( x, ū) = L (ū), (ii) g( x) 0, and (iii) ū T g( x) = 0. Furthermore, ( x, ū) is a saddlepoint of L(x, u) if and only if the following two conditions are satisfied: (I) x and ū are, respectively, optimal solutions of OP and D, and (II) there is no duality gap, i.e., z = f( x) = L (ū) = v. Proof: Suppose that ( x, ū) is a saddlepoint. By definition of the saddlepoint, condition (i) must be true. Also by definition of the saddlepoint, for any u 0, f( x) + ū T g( x) f( x) + u T g( x). This implies that g( x) 0, since otherwise the above inequality can be violated by picking u 0 appropriately. This establishes (ii). Taking u = 0, we get ū T g( x) 0; hence, ū T g( x) = 0, establishing (iii). Also, note that from the above observations that x and ū are feasible for OP and D, respectively, and f( x) = L( x, ū) = L (ū), which implies that they are optimal solutions of their respective problems, and there is no duality gap, thus establishing (I) and (II). Suppose now that x and ū 0 satisfy conditions (i), (ii), and (iii). Then L( x, ū) L(x, ū) for all x X by (i). Furthermore, for any u 0 we have: L( x, ū) = f( x) f( x) + u T g( x) = L( x, u). Thus ( x, ū) is a saddlepoint of L(x, u). Finally, suppose x and ū are optimal solutions of OP and D with no duality gap. Primal and dual feasibility implies that L (ū) L( x, ū) = f( x) + ū T g( x) f( x). Since there is no duality gap, equality holds throughout, implying conditions (i), (ii), and (iii), and hence ( x, ū) is a saddlepoint of L(x, u).

24 24 Remark 2 Note that up until this point, we have made no assumptions about the functions f(x), g 1 (x),..., g m (x) or the structure of the set X. Suppose that our constrained optimization problem OP is a convex program, namely f(x), g 1 (x),..., g m (x) are convex functions, and that X = R n : OP : z = minimum x f(x) s.t. g(x) 0 x R n. Suppose that x together with ū satisfy the KKT conditions: (i) f( x) + ū T g( x) = 0 (ii) g( x) 0 (iii) ū 0 (iv) ū T g( x) = 0. Then because f(x), g 1 (x),..., g m (x) are convex functions, condition (i) is equivalent to L( x, ū) = min x X L(x, ū) = L (ū) for X = R n. Therefore the KKT conditions imply conditions (i), (ii), and (iii) of Theorem 7.1, and so we have: Theorem 7.2 Suppose X = R n, and f(x), g 1 (x),..., g m (x) are convex differentiable functions, and x together with ū satisfy the KKT conditions. Then x and ū are optimal solutions of OP and D, with no duality gap.

25 25 8 Strong Duality for Convex Optimization Problems In this section we consider the following convex problem: OPE : z = minimum x f(x) s.t. g L (x) 0 g E (x) = 0 x X, where we assume here that f(x) and g i (x) are convex functions for i L, g i (x) are linear functions (and so can be written as g i (x) = b i A i x) for i E, and X is a convex set. We use the notation g L (x) for the vector of values of g i (x) for i L and similarly g E (x) for the vector of values of g i (x) for i E. With this notation we have: g E (x) = 0 Ax = b. We further assume with no loss of generality that the matrix A has full row rank. (If A does not have full row rank, we can eliminate dependent rows to yield a new system with full row rank.) It follows from the full row rank of A that if A T u = 0, then u = 0. Note that the standard conic optimization problem CP is a special case of OPE; this follows by setting f(x) := c T x, X := K, and having a vacuous set of convex inequalities (more formally, L := ). Herein we explore and state sufficient conditions that will guarantee that the dual problem will have an optimal solution with no duality gap. For completeness, let us construct the dual problem for this format. We form the Lagrangian: L(x, u) := f(x) + u T g(x) = f(x) + u T Lg L (x) + u T Eg L (x), where u L and u E are vectors of multipliers indexed over the inequality constraints and equality constraints, respectively. Then the dual function L (u)

26 26 is: L (u) := minimum x f(x) + u T g(x) s.t. x X. The dual problem is then defined to be: D : v = maximum u L (u) s.t. u L 0 u E R E, where E denotes the number of indices in the index set E. 8.1 The Perturbation Function We now introduce the concept of perturbing the RHS vector of the optimization problem OPE. We perturb the RHS of OPE by an amount y = (y L, y E ), and obtain the new problem OPE y defined as: OPE y : z(y) = minimum x f(x) s.t. g L (x) y L g E (x) = y E x X. We call OPE y the perturbed primal problem and z(y) the perturbation function. Note that our original problem OPE is just OPE 0 and that z = z(0).

27 27 Next define: Y := {(y L, y E ) : there exists x X for which g L (x) y L and g E (x) = y E }. Lemma 8.1 Y is a convex set, and z(y) is a convex function whose domain is Y. Proof: Let y 1, y 2 Y and let y 3 = λy 1 +(1 λ)y 2 where λ [0, 1]. Let x 1 be feasible for OPE y 1 and x 2 be feasible for OPE y 2, and let x 3 = λx 1 +(1 λ)x 2. Then x 3 be feasible for OPE y 3, whereby y 3 Y, proving that Y is convex. To show that z(y) is a convex function, let y 1, y 2, y 3 be as given above. The aim is to show that z(y 3 ) λz(y 1 ) + (1 λ)z(y 2 ). First assume that z(y 1 ) and z(y 2 ) are finite. For any ε > 0, there exist x 1 and x 2 feasible for OPE y 1 and OPE y 2, respectively, and f(x 1 ) z(y 1 )+ε and f(x 2 ) z(y 2 )+ε. Now let x 3 = λx 1 + (1 λ)x 2. Then x 3 is feasible for OPE y 3, and z(y 3 ) f(x 3 ) λf(x 1 ) + (1 λ)f(x 2 ) λ(z(y 1 ) + ε) + (1 λ)(z(y 2 ) + ε) = λz(y 1 ) + (1 λ)z(y 2 ) + ε. As this is true for any ε > 0, it follows that z(y 3 ) λz(y 1 ) + (1 λ)z(y 2 ), and so z(y) is convex. If z(y 1 ) and/or z(y 2 ) is not finite, the proof goes through with appropriate modifications. The perturbation function z(y) is illustrated in Figure 2. One immediate consequence of the convexity of z( ) is the existence of a subgradient of z(y) when y inty : Corollary 8.1 For every ȳ inty, there exists a subgradient of z( ) at y = ȳ. 8.2 Slater Points and Strong Duality A vector x 0 is called a Slater point for OPE if x 0 satisfies: g L (x 0 ) < 0,

28 28 z -u z(y) 0 H(u,α) y Figure 2: The perturbation function z(y). g E (x 0 ) = 0, which is the same as Ax 0 = b, and x intx. Note that a Slater point is feasible for OPE and satisfies all constraints strictly in the sense of being interior, except for the equality constraints which by definition must be satisfied at equality. Our main strong duality theorem, stated below, asserts that if OPE has a Slater point, then strong duality holds and the dual problem attains its optimum. Theorem 8.1 (Strong Duality Theorem) Consider the problem OPE, where f(x) and g i (x) are convex functions for i L, g i (x) are linear functions for i E, and X is a convex set. If z is finite and if OPE has a Slater point, then: (i) D attains its optimum at some ū = (ū L, ū E ).

29 29 (ii) ū is an optimal solution of D if and only if ( ū) z(0), where z(y) is the perturbation function. (iii) z = v. (iv) Let ū be any dual optimal solution. The set of primal optimal solutions (if any) is characterized as the intersection S 1 S 2, where: S 1 = {x X g L (x) 0, g E (x) = 0, and ū T g(x) = 0}, and S 2 = {x X L(x, ū) = L (ū)}. Proof: Define the following set: S := {(w L, w E, t) there exists x X for which g L (x) w L, g E (x) = w E, and f(x) < z + t}, and notice that S is a nonempty convex set. Also notice that (w L, w E, t) = (0, 0, 0) / S, and so there is a hyperplane separating (0, 0, 0) from S. This means that there exists (ū, θ) = (ū L, ū E, θ) 0 and a scalar α for which and ū T L w L + ū T E w E + θt α for all (w L, w E, t) S, ū T L 0 + ūt E 0 + θ 0 α. The second system above implies that α 0. This together with the definition of S implies that the first system can be rewritten as: ū T L(g L (x)+s L )+ū T Eg E (x)+θ(f(x) z +δ) 0 for all x X, s L 0, and δ > 0. This then implies that ū L and θ 0. If θ > 0 then we can assume (by re-scaling if necessary) that θ = 1, and then setting s L = 0 we obtain: ū T Lg L (x) + ū T Eg E (x) + f(x) z δ for all x X and δ > 0. This in turn implies that L (ū) z δ for all δ > 0, and hence L (ū) z. Furthermore, ū L 0 implies that ū is feasible for D, whereby v L (ū) z. But from weak duality we know that v z, and hence v = L (ū) = z and ū solves D. This proves (i) and (iii) in the case that θ > 0. We now show, by contradiction, that indeed θ > 0, which will complete the proofs of (i) and (iii).

30 30 Suppose that θ = 0. Then setting s L = 0 we obtain: ū T Lg L (x) + ū T Eg E (x) 0 for all x X. Substituting in the value of the Slater point x = x 0 yields: ū T Lg L (x 0 ) + ū T Eg E (x 0 ) 0. But g E (x 0 ) = 0 and g L (x 0 ) < 0, which when combined with ū L 0 implies that ū L = 0. Therefore we have: ū T Eg E (x) 0 for all x X. In particular, since x 0 intx, it follows that for some ε > 0 that x 0 +d X whenever d ε, whereby ū T Eg E (x 0 + d) 0 whenver d ε. We can write this out in matrix form as ū T E (b Ax0 Ad) 0 whenever d ε. However, Ax 0 = b which implies that u T EAd 0 for any d satisfying d ε and hence u T E Ad 0 for any d. Setting d := AT u E yields A T u E 2 = u T E Ad 0 which implies that AT u E = 0, which implies that that u E = 0 as a consequence of the assumption that A has full row rank. But then we have (ū, θ) = (ū L, ū E, θ) = 0, which contradicts the fact that this triplet must be nonzero. This completes the proofs of (i) and (iii). We now show (ii), starting with ( ) of (ii). Suppose ū is optimal for D. Then from (iii) we have min {f(x) + x X ūt g(x)} = L (ū) = z = z(0). Thus, f(x) + ū T g(x) z(0) for every x X. For any given fixed y, if x X and g L (x) y L and g E (x) = y E, then because ū L 0 it holds that Thus for fixed y, f(x) + ū T y f(x) + ū T g(x) z(0). ū T y + z(y) = min x f(x) + ū T y z(0), s.t. g L (x) y L g E (x) = y E x X,

31 31 that is, z(y) z(0) ū T (y 0), and so ū z(0), which proves ( ) of (ii). To show ( ) of (ii), let ū z(0). Then z(y) z(0) ū T (y 0), for y Y. Noting that z(y) is non-increasing in y i for i L, let e i denote the i th unit vector for i L, and observe: z(0) z(e i ) z(0) ū T (e i 0) = z(0) ū i, which shows that ū i 0 for i L, and hence ū L 0. Thus ū is feasible for D. Next, for any x X, let ỹ = g( x). Then which implies that that is, Thus z(g( x)) = min x f(x) f( x) s.t. g L (x) g L ( x) g E (x) = g E ( x) x X f( x) z(g( x)) z(0) ū T (g( x) 0), z(0) f( x) + ū T g( x). z = z(0) min x X {f( x) + ūt g( x)} = L (ū). By weak duality, however, z L (ū), so z = L (ū) and so ū is optimal for D. This shows ( ) of (ii), completing the proof of (ii). We now show (iv). Suppose x S 1 S 2. Then ū L 0, x satisfies L (ū) = L( x, ū), and ū T g( x) = 0, whereby ( x, ū) satisfy the conditions of Theorem 7.1 and hence x is optimal for the primal problem OPE. Conversely, suppose x is optimal for the primal problem. Then z = v, and so again

32 32 invoking Theorem 7.1 we have g L ( x) 0, g E ( x) = 0, L (ū) = L( x, ū), and ū T g( x) = 0. Therefore x S 1 S 2. Let us see how Theorem 8.1 specializes to the case of conic optimization. Recall that the conic optimization primal problem format is: CP : z = minimum x c T x s.t. Ax = b x K, where K is a closed convex cone. In Section 3.4 we derived the conic dual problem: CD : v = maximum u,s b T u s.t. A T u + s = c s K, where K is the dual cone of K. Note that CP is a special case of OPE with a vacuous set of inequality constraints and K as the ground-set, i.e., X = K. A Slater point for CP then is a point x 0 that satisfies: Theorem 8.1 then implies: Ax 0 = b and x 0 intk. Corollary 8.2 (Strong Duality for CP) If CP has a Slater point, then v = z and the conic dual CD attains its optimum. We can also define a Slater point for CD to be a point (u 0, s 0 ) that satisfies: A T u 0 + s 0 = c and s 0 intk. By recasting CD in the format of CP, one can prove:

33 33 Corollary 2 (Strong Duality for CD) If CD has a Slater point, then v = z and the primal attains its optimum. Finally, these two corollaries combine to yield: Corollary 3 (Strong Duality for CP and CD) If CP and CD each have a Slater point, then v = z and the primal and dual problems attain their optima. 8.3 An Example of a (Conic) Convex Problem with Finite Duality Gap Consider the problem: P : z = minimum x1,x 2,x 3 x 1 s.t. x 2 +x 3 = 0 x 1 10 (x 1, x 2 ) x 3, where denotes the Euclidean norm. Note that this problem is a conic problem with K = Q 3 = {(x 1, x 2, x 3 ) : (x 1, x 2 ) x 3 }, i.e., K is the second-order cone in R 3, and so can be formatted as: P : z = minimum x1,x 2,x 3 x 1 s.t. x 2 +x 3 = 0 x 1 10 (x 1, x 2, x 3 ) Q 3. Note that P is feasible (set x 1 = x 2 = x 3 = 0), and that the first and third constraints combine to force x 1 = 0 in any feasible solution, whereby

34 34 z = 0. Also note that any primal feasible solution is of the form x = (0, β, β) for some β 0. The ground-set X for this problem is X = K = Q 3. We form the Lagrangian by using multipliers u 1, u 2 on the two linear constraints and remembering that X = K = {(x 1, x 2, x 3 ) (x 1, x 2 ) x 3 }, the Lagrangian is: L(x 1, x 2, x 3, u 1, u 2 ) = x 1 +u 1 ( x 2 +x 3 )+u 2 ( x 1 10) = 10u 2 +(1 u 2, u 1, u 1 ) T (x 1, x 2, x 3 ). Then L (u 1, u 2 ) = min 10u 2 + (1 u 2, u 1, u 1 ) T (x 1, x 2, x 3 ). (x 1,x 2 ) x 3 It is straightforward to show that L (u 1, u 2 ) is given by: 10u 2 if (1 u 2, u 1 ) u 1 L (u 1, u 2 ) = if (1 u 2, u 1 ) > u 1. We therefore can write the dual problem as: D : v = maximum u1,u 2 10u 2 s.t. (1 u 2, u 1 ) u 1 u 1 R, u 2 0. It is then straightforward to see that the only feasible solution to D will have u 2 = 1, whereby v = 10 < 0 = z. Furthermore, the set of optimal solutions of D is comprised of those vectors (u 1, u 2 ) = (α, 1) for all α 0. Hence both P and D attain their optima and there is a finite duality gap z v = 0 ( 10) = 10. Because there is a duality gap for this problem, the hypotheses of Theorem 8.1 cannot all hold. Let us examine this further. The interior of the primal ground-set X = K = Q 3 is intk = {(x 1, x 2, x 3 ) (x 1, x 2 ) < x 3 }. As discussed above, the only feasible solutions of P are of the form x = (x 1, x 2, x 3 ) = (0, β, β) for β 0, and for all such feasible solutions we have

35 35 (x 1, x 2 ) = β = x 3, so indeed there is no primal feasible solution that lies in the interior of the ground-set X. Therefore the primal problem has no Slater point, and hence the conclusions of Theorem 8.1 do not apply to this problem instance. We can formulate the perturbed problem and the perturbation function z(y) := z(y 1, y 2 ) as the value of: P (y1,y 2 ) : z(y 1, y 2 ) := minimum x1,x 2,x 3 x 1 s.t. x 2 +x 3 = 0 + y 1 x y 2 (x 1, x 2, x 3 ) Q 3. Let us examine the perturbation function z(y) in detail. Note that when y 1 < 0, then any feasible solution of P (y1,y 2 ) has x 3 = x 2 + y 1 < x 2 x 2, whereby P (y1,y 2 ) is infeasible and so z(y) = +. When y 1 = 0 and y 2 10, then x 1 = 0 in any feasible solution, and indeed (x 1, x 2, x 3 ) = (0, 0, 0) is feasible and hence optimal and z(y) = 0. When y 1 = 0 and y 2 < 10, then x 1 > 0 in any feasible solution, but then x 3 > x 2 = x 3 which is not possible, hence P (y1,y 2 ) is infeasible and so z(y) = +. Last of all, when y 1 > 0, note that any feasible solution will have x 1 10 y 2, but also (x 1, x 2, x 3 ) = ( 10 y 2, (10+y 2) 2 y1 2 2y 1, (10+y 2) 2 +y1 2 2y 1 ) is feasible and hence optimal and z(y) = 10 y 2. Therefore the perturbation function z(y) can be summarized as follows: + if y 1 < 0 z(y 1, y 2 ) = 0 if y 1 = 0, y if y 1 = 0, y 2 < y 2 if y 1 > 0. Here we see that the domain of z(y) is Y := {y = (y 1, y 2 ) : y 1 = 0, y 2 10} {y = (y 1, y 2 ) : y 1 > 0}. The point y = (0, 0) / inty, and indeed z(y) has no subgradient at y = (0, 0).

36 36 9 Duality Strategies 9.1 Dualizing Bad Constraints Suppose we wish to solve: OP : minimum x f(x) s.t. g(x) 0 Nx g. Suppose that optimization over the constraints Nx g is easy, but that the addition of the constraints g(x) 0 makes the problem much more difficult. This can happen, for example, when the constraints Nx g are network constraints, and when g(x) 0 are non-network constraints in the model. Let and re-write OP as: X = {x Nx g} OP : minimum x f(x) s.t. g(x) 0 x X. The Lagrangian is: L(x, u) = f(x) + u T g(x),

37 37 and the dual function is: L (u) := minimum x f(x) + u T g(x) s.t. x X, which is L (u) := minimum x f(x) + u T g(x) s.t. Nx g. Notice that L (u) is easy to evaluate for any value of u if, say, f(x) and g i (x) are linear or convex and separable, i = 1,..., m, and so in this case we can attempt to solve OP by designing an algorithm to solve the dual problem: D : maximum u L (u) s.t. u Dualizing A Large Problem into Many Small Problems Suppose we wish to solve: OP : minimum x 1,x 2 (c1 ) T x 1 +(c 2 ) T x 2 s.t. B 1 x 1 +B 2 x 2 d A 1 x 1 b 1 A 2 x 2 b 2

38 38 Notice here that if it were not for the constraints B 1 x 1 + B 2 x 2 d that we would be able to separate the problem into two separate problems. Let us dualize on these constraints. Let: and re-write OP as: X = { (x 1, x 2 ) A 1 x 1 b 1, A 2 x 2 b 2} OP : minimum x 1,x 2 (c1 ) T x 1 +(c 2 ) T x 2 s.t. B 1 x 1 +B 2 x 2 d (x 1, x 2 ) X The Lagrangian is: L(x, u) = (c 1 ) T x 1 + (c 2 ) T x 2 + u T (B 1 x 1 + B 2 x 2 d) = u T d + ((c 1 ) T + u T B 1 )x 1 + ((c 2 ) T + u T B 2 )x 2, and the dual function is: L (u) = minimum x 1,x 2 ut d + ((c 1 ) T + u T B 1 )x 1 + ((c 2 ) T + u T B 2 )x 2 s.t. (x 1, x 2 ) X which can be re-written as: L (u) = u T d + minimum A 1 x 1 b 1 ((c1 ) T + u T B 1 )x 1 + minimum A 2 x 2 b 2 ((c2 ) T + u T B 2 )x 2

39 39 Notice once again that L (u) is easy to evaluate for any value of u, and so we can attempt to solve OP by designing an algorithm to solve the dual problem: D : maximum u L (u) s.t. u Illustration of Lagrange Duality in Discrete Optimization In order to suggest the computational power of duality theory and to illustrate duality constructs in discrete optimization, let us consider the simple constrained shortest path problem portrayed in Figure Figure 3: A constrained shortest path problem. The objective in this problem is to find the shortest (least cost) path from node 1 to node 6 subject to the constraint that the chosen path uses

40 40 at least four arcs. In an applied context the network might contain millions of nodes and arcs and billions of paths. For example, in one application setting that can be solved using the ideas considered in this discussion, each arc would have both a travel cost and a positive utility with, say, the scenic beauty of the route (which for our example is 1 for each arc). The objective is to find the least cost path between a given pair of nodes from all those paths whose travel utility is at least 4 units. Table 1 lists all paths from node 1 to node 6 in the graph pictured in Figure 3. Of the seven different paths from node 1 to node 6, notice that there are exactly four different paths that use at least four arcs; these are path , path , path , and path Since path has the least cost of these four paths (at a cost of 5), this path is the optimal solution of the constrained shortest path problem. Path Cost Number of Arcs Table 1: Paths from node 1 to node 6. Now suppose that we have available an efficient algorithm for solving (unconstrained) shortest path problems (indeed, such algorithms are readily available in practice), and that we wish to use this algorithm to solve the given problem. Let us first formulate the constrained shortest path problem as an integer program:

41 41 CSP P : z = min xij x 12 +x 13 +x 23 +x 24 +3x 34 +2x 35 +x 54 +x 46 +3x 56 s.t. 4 x 12 x 13 x 23 x 24 x 34 x 35 x 54 x 46 x 56 0 x 12 +x 13 = 1 x 12 x 23 x 24 = 0 x 13 +x 23 x 34 x 35 = 0 x 24 +x 34 +x 54 x 46 = 0 x 35 x 54 x 56 = 0 x 46 +x 56 = 1 x ij {0, 1} for all x ij. In this formulation, x ij is a binary variable indicating whether or not arc i j is chosen as part of the optimal path. The first constraint in this formulation specifies that the optimal path must contain at least four arcs, which is written as a rearrangement of the expression: x 12 + x 13 + x 23 + x 24 + x 34 + x 35 + x 54 + x 46 + x The remaining constraints are the familiar constraints describing a shortest path problem. They indicate that one unit of flow is available at node 1 and must be sent to node 6; nodes 2, 3, 4, and 5 act as intermediate nodes that neither generate nor consume flow. Since the decision variables x ij must be integer, any feasible solution to these constraints traces out a path from node 1 to node 6. The objective function determines the cost of any such path.

42 42 Notice that if we eliminate the first constraint from this problem, then the problem becomes an unconstrained shortest path problem that can be solved by our available algorithm. Let us define the ground-set X by all constraints except the first constraint. Then X is the set of solutions of: x 12 +x 13 = 1 x 12 x 23 x 24 = 0 x 13 +x 23 x 34 x 35 = 0 x 24 +x 34 +x 54 x 46 = 0 x 35 x 54 x 56 = 0 (2) x 46 +x 56 = 1 x ij {0, 1} for all x ij. Then the ground-set X defines the set of paths from node 1 to node 6. Let us then dualize on the first (the complicating) constraint by moving it to the objective function multiplied by the Lagrange multiplier u. The objective function then becomes: minimum xij c ij x ij + u, i,j 4 i,j and collecting the terms, the modified problem becomes: x ij

43 43 L (u) = min xij (1 u)x 12 + (1 u)x 13 + (1 u)x 23 +(1 u)x 24 + (3 u)x 34 + (2 u)x 35 +(1 u)x 54 + (1 u)x 46 + (3 u)x u s.t. x 12 +x 13 = 1 x 12 x 23 x 24 = 0 x 13 +x 23 x 34 x 35 = 0 x ij {0, 1} for all x ij. x 24 +x 34 +x 54 x 46 = 0 x 35 x 54 x 56 = 0 x 46 +x 56 = 1 Let L (u) denote the optimal objective function value of this problem, for any given multiplier u. Notice that for a fixed value of u, 4u is a constant and the problem becomes a shortest path problem with modified arc costs given by c ij = c ij u. Note that for a fixed value of u, this problem can be solved by our available algorithm for the shortest path problem. Also notice that for any fixed value of u, L (u) is a lower bound on the optimal objective value z to the constrained shortest path problem defined by the original formulation. To obtain the best lower bounds, we need to solve the Lagrangian dual problem: D : v = maximum u L (u) s.t. u 0.

44 44 To relate this problem to the earlier material on duality, the ground-set X is the set of feasible paths, namely the solutions of (2). In this instance, X is a finite (discrete) set. That is, X is the set of paths connecting node 1 to node 6. Suppose that we plot the cost z and the number of arcs r used in each of these paths, see Figure 4. z (cost) conv(s) z* = 5 v* = r (number of arcs) Figure 4: Resources and costs for the constrained shortest-path problem. We obtain the seven dots in this figure corresponding to the seven paths listed in Table 1. Now, because our original problem was formulated as a minimization problem, the geometric dual is obtained by finding the line that lies on or below these dots and that has the largest intercept with the feasibility line. Note from Figure 4 that the optimal value of the dual is v = 4.5, whereas the optimal value of the original primal problem is z = 5. In this example, because the cost-resource set is not convex (it is simply the seven dots), we encounter a duality gap. In general, integer optimization models will usually have such duality gaps, but they are typically not very large in practice. Nevertheless, the dual bounding information provided by v z can be used in branch and bound procedures to devise efficient algorithms for solving integer optimization models. Let us summarize. The original constrained shortest path problem might be very difficult to solve directly (at least when there are hundreds of thousands of

45 45 nodes and arcs). Duality theory permits us to relax (eliminate) the complicating constraint and to solve a much easier Lagrangian shortest path problem with modified arc costs c ij = c ij u. Solving for any u generates a lower bound L (u) on the value z to the original problem. Finding the best lower bound max u 0 L (u) requires that we develop an efficient procedure for solving for the optimal u. Duality theory permits us to replace a single and difficult problem (in this instance, a constrained shortest-path problem) with a sequence of much easier problems (in this instance, a number of unconstrained shortest path problems), one for each value of u encountered in our search procedure. Practice on a wide variety of problem applications has shown that the dual problem and subsequent branch and bound procedures can be extremely effective in many applications. To conclude this discussion, note from Figure 4 that the optimal value for u in the dual problem is u = 1.5. Also note that if we subtract u = 1.5 from the cost of every arc in the network, then the shortest path (with respect to the modified costs) has cost 1.5. Both paths and are shortest paths for this modified problem. Adding 4u = = 6 to the modified shortest path cost of 1.5 gives a value of = 4.5 which equals v. Note that if we permitted the flow in the original problem to be split among multiple paths (i.e., formulate the problem as a linear optimization problem instead of as an integer optimization problem), then the optimal solution would send one half of the flow on both paths and and incur a cost of: = 4.5. In general, the value of the dual problem is always identical to the value of the original (primal) problem if we permit convex combinations of primal solutions. This equivalence between convexification and dualization is easy to see geometrically. 11 Duality Theory Exercises 1. Consider the problem (P) min x f(x) = x s.t. g(x) = x 0 x X = [0, ). Formulate the dual of this problem.

46 46 2. Consider the problem (P) min x c T x s.t. b Ax 0 x 0. a. Formulate a dual based on g(x) = (ḡ(x), g(x)) = (b Ax, x) and X = R n. b. How does the dual derived in part (a.) compare with the dual derived in the notes Section 2.3? 3. Consider the two equivalent problems (P 1 ) min x x (P 2 ) min x 1 2 xt x b Ax 0 b Ax 0 x R n x R n. Derive the dual problems (D 1 ) and (D 2 ) of (P 1 ) and (P 2 ). relation between (D 1 ) and (D 2 )? 4. Let P be given as a function of the parameter y: What is the P y : z(y) = min x 1 + 5x 2 7x 3 s.t. x 1 3x 2 + x 3 y, x X = {x R 3 : x j = 0 or 1, j = 1, 2, 3}. Construct the dual D y as a function of the parameter y. Graph the function z(y) as a function of y. For what values of y is there a duality gap? 5. Consider the problem (P) min x c T x xt Qx s.t. b Ax 0. where Q is symmetric and positive semidefinite. a. What is the dual? b. Show that the dual of the dual is the primal.

47 47 6. Consider the program (P ) min x f(x) s.t. g i (x) 0, i = 1,..., m x X, where f( ) and g i ( ) are convex functions and the set X is a convex set. Consider the matrix below: Property of D v finite v finite v finite v finite Property of P attained not attained attained not attained v = + v = v = z v = z v < z v < z z finite Slater point z finite no Slater point z = z = (infeasible) Prove by citing relevant theorems from the notes, etc., that the following cases cannot occur: 2, 3, 4, 5, 7, 8, 9, 11, 13, 15, 17, 18, 19, and Let X R n, Y R m, f(x, y) : (X Y ) R. A saddlepoint ( x, ȳ) of f(x, y) is a point ( x, ȳ) X Y such that f( x, y) f( x, ȳ) f(x, ȳ) for all x X and y Y. Prove the equivalence of the following three results: (a) There exists a saddlepoint ( x, ȳ) of f(x, y). (b) There exists x X and ȳ Y for which max y Y f( x, y) = min x x f(x, ȳ). (c) min x X {max y Y f(x, y)} = max y Y {min x X f(x, y)}.

48 48 8. Consider the logarithmic barrier problem: P (θ) : minimize x c T x θ n ln(x j ) j=1 s.t. Ax = b x > 0. Compute a dual of P (θ) by dualizing on the constraints Ax = b. What is the relationship between this dual and the dual of the original linear program (without the logarithmic barrier function)? 9. Consider the program P (γ) : minimize x n ln(x j ) j=1 s.t. Ax = b c T x = γ x > 0. Compute a dual of P (γ) by dualizing on the constraints Ax = b and c T x = γ. What is the relationship between the optimality conditions of P (θ) in the previous exercise and P (γ)? Consider the program P (δ) : minimize x n ln(x j ) ln(δ c T x) j=1 s.t. Ax = b c T x < δ x > 0. Compute a dual of P (δ) by dualizing on the constraints Ax = b. What is the relationship between the optimality conditions of P (δ) and P (θ) and P (γ) of the previous two exercises? 10. Consider the conic form problem CP of convex optimization CP : z = minimum x c T x s.t. Ax = b x K,

49 49 and the associated conic dual problem: CD : v = maximum u,s b T u s.t. A T u + s = c s K, where K is a closed convex cone. Show that the dual of the dual is the primal, that is, that the conic dual of CD is CP. 11. Prove Corollary 2, which asserts that the existence of Slater point for the conic dual problem guarantees strong duality and that the primal attains its optimum. 12. Consider the following minimax problems: min x X max y Y ϕ(x, y) and max y Y min x X ϕ(x, y) where X and Y are nonempty compact convex sets in R n and R m, respectively, and ϕ(x, y) is convex in x for fixed y, and is concave in y for fixed x. (a) Show that min x X max y Y ϕ(x, y) max y Y min x X ϕ(x, y) in the absence of any convexity/concavity assumptions on X, Y, and/or ϕ(, ). (b) Show that f(x) := max y Y ϕ(x, y) is a convex function in x and that g(y) := min x X ϕ(x, y) is a concave function in y. (c) Use a separating hyperplane theorem to prove: min x X max y Y ϕ(x, y) = max y Y min x X ϕ(x, y). 13. Let X and Y be nonempty sets in R n, and let f( ), g( ) : R n R. Consider the following conjugate functions f ( ) and g ( ) defined as follows: and f (u) := min x X {f(x) ut x}, g (u) := max x X {g(x) ut x}. (a) Construct a geometric interpretation of f ( ) and g ( ). (b) Show that f ( ) is a concave function on X := {u f (u) > }, and g ( ) is a convex function on Y := {u g (u) < + }.

50 50 (c) Prove the following weak duality theorem between the conjugate primal problem min{f(x) g(x) x X Y } and the conjugate dual problem max{f (u) g (u) u X Y }: min{f(x) g(x) x X Y } max{f (u) g (u) u X Y }. (d) Now suppose that f( ) is a convex function, g( ) is a concave function, intx inty, and min{f(x) g(x) x X Y } is finite. Show that equality in part (13c) holds true and that max{f (u) g (u) u X Y } is attained for some u = u. (e) Consider a standard inequality constrained nonlinear optimization problem using the following notation: OP : minimum x f(x) s.t. ḡ 1 (x) 0,. ḡ m (x) 0, x X. By suitable choices of f( ), g( ), X, and Y, formulate this problem as an instance of the conjugate primal problem min{f(x) g(x) x X Y }. What is the form of the resulting conjugate dual problem max{f (u) g (u) u X Y }? 14. Consider the following problem: z = min x 1 + x 2 s.t. x x 2 2 = 4 2x 1 x 2 4. (a) Formulate the Lagrange dual of this problem by incorporating both constraints into the objective function via multipliers u 1, u 2. (b) Compute the gradient of L (u) at the point ū = (1, 2). (c) Starting from ū = (1, 2), perform one iteration of the steepest ascent method for the dual problem. In particular, solve the following problem where d = L (ū): max α L (ū + α d) s.t. ū + α d 0 α 0.

51 Prove that Remark 1 is true. 16. Consider the following very general conic problem: GCP : z = minimum x c T x s.t. Ax b K 1 x K 2, where K 1 R m and K 2 R n are each a closed convex cone. Derive the following conic dual for this problem: GCD : v = maximum y b T y s.t. c A T y K 2 y K 1, and show that the dual of GCD is GCP. How is the conic format of Subection 3.4 a special case of GCP? 17. Consider the example of Subsection 8.3 for which there is finite duality gap. (a) In this example the primal problem is a convex problem. Why is there nevertheless a duality gap? What hypotheses are absent that otherwise would guarantee strong duality? (b) The text in Subsection 8.3 reads It is straightforward to show that L (u 1, u 2 ) is given by: 10u 2 if (1 u 2, u 1 ) u 1 L (u 1, u 2 ) = if (1 u 2, u 1 ) > u 1. Give a formal demonstration of this straightforward statement. 18. For a given scalar γ, the γ-level set H γ of the dual problem D is given by: H γ := {u : u 0, L (u) γ}. Show that if OP has a feasible solution x 0 satisfying g(x 0 ) < 0, then the level sets of D must be bounded, i.e., H γ is a bounded set for all γ.

52 Consider the optimization problem P : minimize f(x) x s.t. g(x) 0 x X, and its Lagrangian relaxation: L (u) = min x X f(x) + ut g(x). Let v = max u 0 L (u) denote the optimal value of the Lagrangian dual problem. Suppose that the variables x are partitioned as x = (ξ, w) and that f(x), g(x), and X are of the form: f(x) = f(ξ, w) = c T ξ + d(w) g(x) = g(ξ, w) = b Aξ + h(w) X = {(ξ, w) ξ 0}. That is, ξ models the linear portion of the problem and w together with the functions d(w) and h(w) model the nonlinear portion of the problem. Show for this problem that { { L b (u) = T u + min w d(w) + u T h(w) } if A T u c if A T u c.