Lecture 7: Weak Duality

Transcription

1 EE 227A: Conve Optimization and Applications February 7, 2012 Lecture 7: Weak Duality Lecturer: Laurent El Ghaoui 7.1 Lagrange Dual problem Primal problem In this section, we consider a possibly non-conve optimization problem p := f 0 : f i 0, i = 1,, m, 7.1 where the functions f 0, f 1,..., f m We denote by D the domain of the problem which is the intersection of the domains of all the functions involved, and by X D its feasible set. We will refer to the above as the primal problem, and to the decision variable in that problem, as the primal variable. One purpose of Lagrange duality is to find a lower bound on a imization problem or an upper bounds for a maimization problem. Later, we will use duality tools to derive optimality conditions for conve problems Dual problem Lagrangian. To the problem we associate the Lagrangian L : R n R m R, with values L, λ := f 0 + λ i f i The variables λ R m, are called Lagrange multipliers. We observe that, for every feasible X, and every λ 0, f 0 is bounded below by L, λ: X, λ R m + : f 0 L, λ. 7.2 The Lagrangian can be used to epress the primal problem 7.1 as an unconstrained one. Precisely: p = L, λ, 7.3 ma λ 0, ν where we have used the fact that, for any vectors f R m, we have { 0 if f 0 ma λ 0 λt f = + otherwise. 7-1

2 Lagrange dual function. short the function We then define the Lagrange dual function dual function for gλ := L, λ. Note that, since g is the pointwise imum of affine functions L, is affine for every, it is concave. Note also that it may take the value. From the bound 7.2, by imizing over in the right-hand side, we obtain X, λ 0 : f 0 L, λ, = gλ, which, after imizing over the left-hand side, leads to the lower bound Lagrange dual problem. bound is p d, where λ R m +, ν R p : p gλ. The best lower bound that we can obtain using the above d = ma λ 0, ν gλ. We refer to the above problem as the dual problem, and to the vector λ R m as the dual variable. The dual problem involves the maimization of a concave function under conve sign constraints, so it is a conve problem. The dual problem always contains the implicit constraint λ dom g. We have obtained: Theorem 1 Weak duality. For the general possibly non-conve problem 7.1,weak duality holds: p d. Case with equality constraints. If equality constraints are present in the problem, we can represent them as two inequalities. It turns out that this leads to the same dual, as if we would directly use a single dual variable for each equality constraint, which is not restricted in sign. To see this, consider the problem We write the problem as p := f 0 : f i 0, i = 1,, m, h i = 0, i = 1,, p. p := f 0 : f i 0, i = 1,, m, h i 0, h i 0, i = 1,, p. Using a mulitplier ν ± i for the constraint ±h i 0, we write the associated Lagrangian as p p L, λ, ν +, ν = f 0 + λ i f i + ν + i h i + ν i h i = f 0 + λ i f i p ν i h i,

3 where ν := ν + ν does not have any sign constraints. Thus, inequality constraints in the original problem are associated with sign constraints on the corresponding multipliers, while the multipliers for the equality constraints are not eplicitly constrained Minima inequality Weak duality can also be obtained as a consequence of the following ima inequality, which is valid for any function φ of two vector variables, y, and any subsets X, Y: ma y Y X φ, y X ma y Y φ, y. 7.4 To prove this, start from, y : X φ, y ma φ, y Y y. and take the imum over X on the right-hand side, then the maimum over y Y on the left-hand side. Weak duality is indeed a direct consequence of the above. To see this, start from the unconstrained formulation 7.3, and apply the above inequality with φ = L the Lagrangian of the original problem, and y = λ the vector of Lagrange multipliers. Interpretation as a game. We can interpret the ima inequality result in the contet of a one-shot, zero-sum game. Assume that you have two players A and B, where A controls the decision variable, while B controls y. We assume that both players have full knowledge of the other player s decision, once it is made. The player A seeks to imize a payoff to player B L, y, while B seeks to maimize that payoff. The right-hand side in 7.4 is the optimal pay-off if the first player is required to play first. Obviously, the first player can do better by playing second, since then he or she knows the opponent s choice and can adapt to it. 7.2 Eamples Linear optimization problem Inequality form. Consider the LP in standard inequality form p = c T : A b, where A R m n, b R m, and the inequality in the constraint A b is interpreted component-wise. 7-3

4 The Lagrange function is and the corresponding dual function is The dual problem reads L, λ = c T + λ T A b gλ = L, λ = { b T λ if A T λ + c = 0 + otherwise. d = ma λ gλ = ma λ b T λ : λ 0, A T λ + c = 0. The dual problem is an LP in standard sign-constrained form, just as the primal problem was an LP in standard inequality form. Weak duality implies that c T + b T λ 0 for every, λ such that A b, A T λ = c. This property can be proven directly, by replacing c by A T λ in the left-hand side of the above inequality, and eploiting A b and λ 0. Standard form. We can also consider an LP in standard form: p = c T : A = b, 0. The equality constraints are associated with a dual variable ν that is not constrained in the dual problem. The Lagrange function is and the corresponding dual function is The dual problem reads L, λ, ν = c T λ T + ν T b A gλ = L, λ, ν = { b T ν if c = A T ν + λ + otherwise. d = ma λ 0, ν gλ, ν = ma ν b T ν : c A T ν. This is an LP in inequality form. 7-4

5 7.2.2 Minimum Euclidean distance problem Consider the problem of imizing the Euclidean distance to a given affine space: : A = b, 7.5 where A R p n, b R p. We assume that A is full row rank, or equivalently, AA T 0. The Lagrangian is and the Lagrange dual function is gν = L, ν = ν T A b, 1 L, ν = ν T A b. In this eample, the dual function can be computed analytically, using the optimality condition L, ν = + A T ν = 0. We obtain = A T ν, and The dual problem epresses as gν = 1 2 νt AA T ν b T ν. d = ma ν gν = ma ν 1 2 νt AA T ν b T ν. The dual problem can also be solved analytically, since it is unconstrained the domain of g is the entire space R p. We obtain ν = AA T 1 b, and We have thus obtained the bound p d. d = 1 2 bt AA T 1 b A non-conve boolean problem For a given matri W = W T 0, we consider the problem p = ma T W : 2 i 1, i = 1,..., n. Lagrange relaation. In this maimization problem, Lagrange duality will provide an upper bound on the problem. This is called a relaation, as we go above the true maimum, as if we d rela ignore constraints. 7-5

6 Lagrange dual. The Lagrangian writes L, λ = T W + n λ i 1 2 i = Tr D λ + T W D λ. where D λ := diagλ. To find the dual function, we need to maimize the Lagrangian with respect to the primal variable. We epress this problem as gλ = ma L, λ = t t :, t Tr D λ + T W D λ. The last inequality holds if and only if Dλ W 0 0 t Tr D λ 0. Hence the dual function is the optimal value of an SDP in one variable: Dλ W 0 gλ = t : 0. t 0 t Tr D λ We can solve this problem eplicitly: { Tr Dλ if D gλ = λ W otherwise. The dual problem involves imizing that is, getting the best upper bound the dual function over the variable λ 0: d = λ λ T 1 : diagλ W. The above is an SDP, in variable λ. Note that λ > 0 is automatically enforced by the PSD constraint. Geometric interpretation. The Lagrange relaation of the primal problem can be interpreted geometrically, as follows. For t > 0, λ > 0, consider the ellipsoids E t = { : T W t }, E λ = { : T D λ Tr D λ }. The primal problem amounts to find the smallest t 0 for which the ellipsoid E t contains the ball B := { : 1}. Note that for every λ > 0, E λ contains the ball B. To find an upper bound on the problem, we can find the smallest t for which there eist λ > 0 such that E t E λ. The latter condition is precisely diagλ W, t Tr D λ. 7-6

7 Figure 7.1. Geometric interpretation of dual problem in the boolean quadratic problem. In 2D the relaation turns out to be eact. 7.3 More on non-conve quadratic optimization The Boolean problem eaed previously is part of a general class of non-conve quadratic problems of the form p := ma q 0 : q i 0, i = 1,..., m, 7.6 where R n is the decision variable, and q i s are quadratic functions, of the form Lagrange relaation. have q i := T Q i + 2q T i + p i, i = 1,..., m, The idea is that if, for a given m-vector λ 0, and scalar t, we : q 0 λ i q i + t, then for every that is feasible for 7.6, the sum in the above is non-positive. Hence, q 0 t, so that t is an upper bound on our problem. The condition above is easy to check, as it involves a single quadratic function: indeed, it is equivalent to the LMI in t, λ: Q0 q 0 q T 0 r 0 Qi λ i q T i q i r i t. 7.7 Hence, the best upper bound that we can achieve using this approach is the SDP t,λ t : 7.7, λ

8 The S-lemma. This mysterious name corresponds to a special case of non-conve quadratic optimization, where there is only a single constraint. Refer to appendi B of [BV] for more details. The problem bears the form ma q 0 : q 1 0, where both q 0, q 1 are arbitrary quadratic functions. The S-lemma states that if there eist a point R n such that q 1 < 0, then the Lagrange relaation is eact. The latter has the form of an SDP: t,λ t : Q0 q 0 q T 0 r 0 Q1 q λ 1 q1 T r t, λ This shows in particular that the apparently non-conve problem of finding the direction of maimal variance for a given covariance matri Σ is actually conve. Lagrange relaation for the problem ma : 2 =1 T Σ yields the dual problem check this! t t : ti Σ. From the S-lemma, the bound is eact. The S-lemma has many other applications. Eercises Figure 7.2. Localization problem with three range measurements in two dimensions. 7-8

9 1. Anchor localization. We are given anchor positions i R 3, and associated distances from these anchor points to an unknown object, R i, i = 1,..., m. The problem is to estimate a position of the object, and associated measure of uncertainty around the estimated point. Geometrically, the measurements imply that the object is located at the intersection of the m spheres of centers i and radiuses R i, i = 1,..., m. The main problem is to provide one point in the intersection located at some kind of center, and also a measure of the size of the intersection. In this problem, we seek to compute an outer spherical approimation to the intersection, that is, a sphere of center 0 and radius R 0, of imal volume, such that it contains the intersection. a First show how to find a point in the intersection, or detere it is empty, via SOCP. To simplify, and without loss of generality, we assume from now on that 0 is inside the intersection. This means that the vector z with components z i := R 2 i T i i, i = 1,..., m, is non-negative componentwise. b A first approach, which works well only in moderate dimensions 2D or 3D, simply entails gridding the boundary of the intersection. In 2D, we can parametrize the boundary eplicitly, as a curve, using an angular parameter. For each angular direction θ [0, 2π], we can easily find the point that is located on the boundary of the intersection, in the direction given by θ: we simply maimize t such that the point t cos θ, t sin θ is inside everyone of the spheres. There is an eplicit formula for the maimal value. One we have computed N points on the boundary, k, k = 1,..., N, we simply solve the SOCP 0,R 0 R 0 : R 0 0 k 2, k = 1,..., N. Compare the results with a uniform gridding of 13 and 63 points. Use the data X = 1, 2, 3 =, R T = R 1, R 2, R 3 = c Show that the optimal imum-volume spherical approimation can be obtained by solving R 2 0 = 0 ma { : i 2 2 R 2 i, i = 1,..., m }. d Using Lagrange relaation, show that an upper bound on the optimal radius can be obtained as F 0, y, with F 0, y := y i Ri 2 + ma y i i ,y 7-9

10 e Show that T y T z + Xy m if m F 0, y = y i 1 y i > 1, T y T z if m y i = 1, 0 = Xy, + otherwise. f Solve the problem via CVX, and compare your approimation with the gridding approach. 2. Reachable sets for discrete-time dynamical systems. Consider the discrete-time linear system t + 1 = At + Bpt, t = 0, 1, 2,... where A R n n and B R n np. We assume that the initial condition 0 is zero, while the signal p is considered to be noise, and is only known to be norm-bounded, precisely pt 2 1 for ever t 0. The goal of reachability analysis is to come up with bounds on the state at a certain time T. We seek a imum-volume sphere S that is guaranteed to contain T, irrespective of the values of the perturbation signal pt within its bounds. By applying the recursion, we can epress T as a linear combination of p0,..., pt 1: T 1 T = A t Bpt = Lp, t=0 where p = p0,..., pt 1, and L := [L0,..., LT 1], with Lt := A t B. a Show that a sufficient condition for S to contain the state vector at time T is T 1 λ 0 : p = p0,..., pt 1, p T L T Lp R0 2 + λtpt T pt 1. b Show how to compute the best approimation based on the condition above, via SDP. t=0 7-10