EE/AA 578, Univ of Washington, Fall Duality

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1 7. Duality EE/AA 578, Univ of Washington, Fall 2016 Lagrange dual problem weak and strong duality geometric interpretation optimality conditions perturbation and sensitivity analysis examples generalized inequalities 7 1

2 Lagrangian standard form problem (not necessarily convex) minimize f 0 (x) subject to f i (x) 0, i = 1,...,m h i (x) = 0, i = 1,...,p variable x R n, domain D, optimal value p Lagrangian: L : R n R m R p R, with doml = D R m R p, L(x,λ,ν) = f 0 (x)+ m λ i f i (x)+ i=1 p ν i h i (x) i=1 weighted sum of objective and constraint functions λ i is Lagrange multiplier associated with f i (x) 0 ν i is Lagrange multiplier associated with h i (x) = 0 7 2

3 Lagrange dual function Lagrange dual function: g : R m R p R, g(λ,ν) = inf L(x,λ,ν) x D ( = inf x D f 0 (x)+ m λ i f i (x)+ i=1 ) p ν i h i (x) i=1 g is concave, can be for some λ, ν lower bound property: if λ 0, then g(λ,ν) p proof: if x is feasible and λ 0, then f 0 ( x) L( x,λ,ν) inf L(x,λ,ν) = g(λ,ν) x D minimizing over all feasible x gives p g(λ,ν) 7 3

4 Least-norm solution of linear equations dual function minimize x T x subject to Ax = b Lagrangian is L(x,ν) = x T x+ν T (Ax b) to minimize L over x, set gradient equal to zero: x L(x,ν) = 2x+A T ν = 0 = x = (1/2)A T ν plug in in L to obtain g: g(ν) = L(( 1/2)A T ν,ν) = 1 4 νt AA T ν b T ν a concave function of ν lower bound property: p (1/4)ν T AA T ν b T ν for all ν 7 4

5 Standard form LP minimize c T x subject to Ax = b, x 0 dual function Lagrangian is L is linear in x, hence L(x,λ,ν) = c T x+ν T (Ax b) λ T x g(λ,ν) = inf x L(x,λ,ν) = = b T ν +(c+a T ν λ) T x { b T ν A T ν λ+c = 0 otherwise g is linear on affine domain {(λ,ν) A T ν λ+c = 0}, hence concave lower bound property: p b T ν if A T ν +c 0 7 5

6 Equality constrained norm minimization dual function minimize x subject to Ax = b g(ν) = inf x ( x νt Ax+b T ν) = where v = sup u 1 u T v is dual norm of { b T ν A T ν 1 otherwise proof: follows from inf x ( x y T x) = 0 if y 1, otherwise if y 1, then x y T x 0 for all x, with equality if x = 0 if y > 1, choose x = tu where u 1, u T y = y > 1: x y T x = t( u y ) as t lower bound property: p b T ν if A T ν 1 7 6

7 Example: two-way partitioning minimize x T Wx subject to x 2 i = 1, i = 1,...,n a nonconvex problem; feasible set contains 2 n discrete points interpretation: partition {1,...,n} in two sets; W ij is cost of assigning i, j to the same set; W ij is cost of assigning to different sets dual function g(ν) = inf x (xt Wx+ i ν i (x 2 i 1)) = inf x (xt (W +diag(ν))x 1 T ν) { 1 = T ν W +diag(ν) 0 otherwise lower bound property: p 1 T ν if W +diag(ν) 0 example: ν = λ min (W)1 gives bound p nλ min (W) 7 7

8 The dual problem Lagrange dual problem maximize g(λ, ν) subject to λ 0 finds best lower bound on p, obtained from Lagrange dual function a convex optimization problem; optimal value denoted d λ, ν are dual feasible if λ 0, (λ,ν) domg often simplified by making implicit constraint (λ, ν) dom g explicit example: standard form LP and its dual (page 7 5) minimize c T x subject to Ax = b x 0 maximize b T ν subject to A T ν +c 0 7 8

9 weak duality: d p Weak and strong duality always holds (for convex and nonconvex problems) can be used to find nontrivial lower bounds for difficult problems for example, solving the SDP maximize 1 T ν subject to W +diag(ν) 0 gives a lower bound for the two-way partitioning problem on page 7 7 strong duality: d = p does not hold in general (usually) holds for convex problems conditions that guarantee strong duality in convex problems are called constraint qualifications 7 9

10 Slater s constraint qualification strong duality holds for a convex problem if it is strictly feasible, i.e., minimize f 0 (x) subject to f i (x) 0, i = 1,...,m Ax = b x intd : f i (x) < 0, i = 1,...,m, Ax = b also guarantees that the dual optimum is attained (if p > ) can be sharpened: (e.g., can replace int D with relative interior; linear inequalities do not need to hold with strict inequality, etc... ) there exist many other types of constraint qualifications 7 10

11 Inequality form LP primal problem minimize c T x subject to Ax b dual function g(λ) = inf x ( (c+a T λ) T x b T λ ) { b = T λ A T λ+c = 0 otherwise dual problem maximize b T λ subject to A T λ+c = 0, λ 0 from Slater s condition: p = d if A x b for some x in fact, p = d except when primal and dual are infeasible 7 11

12 primal problem (assume P S n ++) Quadratic program minimize x T Px subject to Ax b dual function g(λ) = inf x ( x T Px+λ T (Ax b) ) = 1 4 λt AP 1 A T λ b T λ dual problem maximize (1/4)λ T AP 1 A T λ b T λ subject to λ 0 from Slater s condition: p = d if A x b for some x in fact, p = d always 7 12

13 Economic (price) interpretation minimize f 0 (x) subject to f i (x) 0, i = 1,...,m. f 0 (x) is cost of operating firm at operating condition x; constraints give resource limits. suppose: can violate f i (x) 0 by paying additional cost of λ i (in dollars per unit violation), i.e., incur cost λ i f i (x) if f i (x) > 0 can sell unused portion of ith constraint at same price, i.e., gain profit λ i f i (x) if f i (x) < 0 total cost to firm to operate at x at constraint prices λ i : L(x,λ) = f 0 (x)+ m λ i f i (x) i=1 7 13

14 interpretations: dual function: g(λ) = inf x L(x,λ) is optimal cost to firm at constraint prices λ i weak duality: cost can be lowered if firm allowed to pay for violated constraints (and get paid for non-tight ones) duality gap: advantage to firm under this scenario strong duality: λ give prices for which firm has no advantage in being allowed to violate constraints

15 Min-max & saddle-point interpretation can express primal and dual problems in a more symmetric form: ( ) m { f0 (x) f f 0 (x)+ λ i f i (x) = i (x) 0, + otherwise, sup λ 0 i=1 so p = inf x sup λ 0 L(x,λ). also by definition, d = sup λ 0 inf x L(x,λ). weak duality can be expressed as supinf λ 0 x L(x,λ) inf sup L(x, λ) x λ 0 strong duality when equality holds. means can switch the order of minimization over x and maximization over λ 0. if x and λ are primal and dual optimal and strong duality holds, they form a saddle-point for the Lagrangian (converse also true). 7 15

16 Geometric interpretation for simplicity, consider problem with one constraint f 1 (x) 0 interpretation of dual function: g(λ) = inf (u,t) G (t+λu), where G = {(f 1(x),f 0 (x)) x D} t t G G λu + t = g(λ) p g(λ) u p d u λu+t = g(λ) is (non-vertical) supporting hyperplane to G hyperplane intersects t-axis at t = g(λ) 7 16

17 epigraph variation: same interpretation if G is replaced with A = {(u,t) f 1 (x) u,f 0 (x) t for some x D} t A λu + t = g(λ) p g(λ) u strong duality holds if there is a non-vertical supporting hyperplane to A at (0,p ) for convex problem, A is convex, hence has supp. hyperplane at (0,p ) Slater s condition: if there exist (ũ, t) A with ũ < 0, then supporting hyperplanes at (0,p ) must be non-vertical 7 17

18 Complementary slackness assume strong duality holds, x is primal optimal, (λ,ν ) is dual optimal f 0 (x ) = g(λ,ν ) = inf x ( f 0 (x )+ f 0 (x ) f 0 (x)+ m λ if i (x)+ i=1 m λ if i (x )+ i=1 ) p νih i (x) i=1 p νih i (x ) i=1 hence, the two inequalities hold with equality x minimizes L(x,λ,ν ) λ i f i(x ) = 0 for i = 1,...,m (known as complementary slackness): λ i > 0 = f i (x ) = 0, f i (x ) < 0 = λ i =

19 Karush-Kuhn-Tucker (KKT) conditions the following four conditions are called KKT conditions (for a problem with differentiable f i, h i ): 1. primal constraints: f i (x) 0, i = 1,...,m, h i (x) = 0, i = 1,...,p 2. dual constraints: λ 0 3. complementary slackness: λ i f i (x) = 0, i = 1,...,m 4. gradient of Lagrangian with respect to x vanishes: f 0 (x)+ m λ i f i (x)+ i=1 p ν i h i (x) = 0 i=1 from page 7 18: if strong duality holds and x, λ, ν are optimal, then they must satisfy the KKT conditions 7 19

20 KKT conditions for convex problem if x, λ, ν satisfy KKT for a convex problem, then they are optimal: from complementary slackness: f 0 ( x) = L( x, λ, ν) from 4th condition (and convexity): g( λ, ν) = L( x, λ, ν) hence, f 0 ( x) = g( λ, ν) if Slater s condition is satisfied: x is optimal if and only if there exist λ, ν that satisfy KKT conditions recall that Slater implies strong duality, and dual optimum is attained generalizes optimality condition f 0 (x) = 0 for unconstrained problem 7 20

21 example: water-filling (assume α i > 0) minimize n i=1 log(x i +α i ) subject to x 0, 1 T x = 1 x is optimal iff x 0, 1 T x = 1, and there exist λ R n, ν R such that λ 0, λ i x i = 0, 1 x i +α i +λ i = ν if ν < 1/α i : λ i = 0 and x i = 1/ν α i if ν 1/α i : λ i = ν 1/α i and x i = 0 determine ν from 1 T x = n i=1 max{0,1/ν α i} = 1 interpretation n patches; level of patch i is at height α i flood area with unit amount of water resulting level is 1/ν 1/ν x i i α i 7 21

22 Perturbation and sensitivity analysis (unperturbed) optimization problem and its dual minimize f 0 (x) subject to f i (x) 0, i = 1,...,m h i (x) = 0, i = 1,...,p maximize g(λ, ν) subject to λ 0 perturbed problem and its dual min. f 0 (x) s.t. f i (x) u i, i = 1,...,m h i (x) = v i, i = 1,...,p max. g(λ,ν) u T λ v T ν s.t. λ 0 x is primal variable; u, v are parameters p (u,v) is optimal value as a function of u, v we are interested in information about p (u,v) that we can obtain from the solution of the unperturbed problem and its dual 7 22

23 global sensitivity result assume strong duality holds for unperturbed problem, and that λ, ν are dual optimal for unperturbed problem apply weak duality to perturbed problem: p (u,v) g(λ,ν ) u T λ v T ν = p (0,0) u T λ v T ν sensitivity interpretation if λ i large: p increases greatly if we tighten constraint i (u i < 0) if λ i small: p does not decrease much if we loosen constraint i (u i > 0) if ν i large and positive: p increases greatly if we take v i < 0; if ν i large and negative: p increases greatly if we take v i > 0 if ν i small and positive: p does not decrease much if we take v i > 0; if ν i small and negative: p does not decrease much if we take v i <

24 local sensitivity: if (in addition) p (u,v) is differentiable at (0,0), then λ i = p (0,0) u i, ν i = p (0,0) v i proof (for λ i ): from global sensitivity result, p (0,0) u i = lim tց0 p (te i,0) p (0,0) t λ i hence, equality p (0,0) u i = lim tր0 p (te i,0) p (0,0) t λ i p (u) for a problem with one (inequality) constraint: u = 0 u p (u) p (0) λ u 7 24

25 Duality and problem reformulations equivalent formulations of a problem can lead to very different duals reformulating the primal problem can be useful when the dual is difficult to derive, or uninteresting common reformulations introduce new variables and equality constraints make explicit constraints implicit or vice-versa transform objective or constraint functions e.g., replace f 0 (x) by φ(f 0 (x)) with φ convex, increasing 7 25

26 Introducing new variables and equality constraints minimize f 0 (Ax+b) dual function is constant: g = inf x L(x) = inf x f 0 (Ax+b) = p we have strong duality, but dual is quite useless reformulated problem and its dual minimize f 0 (y) subject to Ax+b y = 0 maximize b T ν f 0(ν) subject to A T ν = 0 dual function follows from g(ν) = inf (f 0(y) ν T y +ν T Ax+b T ν) x,y { f = 0 (ν)+b T ν A T ν = 0 otherwise 7 26

27 norm approximation problem: minimize Ax b minimize y subject to y = Ax b can look up conjugate of, or derive dual directly (see page 7 4) g(ν) = inf x,y ( y +νt y ν T Ax+b T ν) { b = T ν +inf y ( y +ν T y) A T ν = 0 otherwise { b = T ν A T ν = 0, ν 1 otherwise dual of norm approximation problem maximize b T ν subject to A T ν = 0, ν

28 Implicit constraints LP with box constraints: primal and dual problem minimize c T x subject to Ax = b 1 x 1 maximize b T ν 1 T λ 1 1 T λ 2 subject to c+a T ν +λ 1 λ 2 = 0 λ 1 0, λ 2 0 reformulation with box constraints made implicit minimize f 0 (x) = subject to Ax = b { c T x 1 x 1 otherwise dual function g(ν) = inf 1 x 1 (ct x+ν T (Ax b)) = b T ν A T ν +c 1 dual problem: maximize b T ν A T ν +c

29 Problems with generalized inequalities minimize f 0 (x) subject to f i (x) Ki 0, i = 1,...,m h i (x) = 0, i = 1,...,p Ki is generalized inequality on R k i definitions are parallel to scalar case: Lagrange multiplier for f i (x) Ki 0 is vector λ i R k i Lagrangian L : R n R k 1 R k m R p R, is defined as L(x,λ 1,,λ m,ν) = f 0 (x)+ m λ T i f i (x)+ i=1 p ν i h i (x) i=1 dual function g : R k 1 R k m R p R, is defined as g(λ 1,...,λ m,ν) = inf x D L(x,λ 1,,λ m,ν) 7 29

30 lower bound property: if λ i K i 0, then g(λ 1,...,λ m,ν) p proof: if x is feasible and λ K i 0, then f 0 ( x) f 0 ( x)+ m λ T i f i ( x)+ i=1 inf x D L(x,λ 1,...,λ m,ν) = g(λ 1,...,λ m,ν) p ν i h i ( x) i=1 minimizing over all feasible x gives p g(λ 1,...,λ m,ν) dual problem weak duality: p d always maximize g(λ 1,...,λ m,ν) subject to λ i K i 0, i = 1,...,m strong duality: p = d for convex problem with constraint qualification (for example, Slater s: primal problem is strictly feasible) 7 30

31 primal SDP (F i,g S k ) Semidefinite program minimize c T x subject to x 1 F 1 + +x n F n G Lagrange multiplier is matrix Z S k Lagrangian L(x,Z) = c T x+tr(z(x 1 F 1 + +x n F n G)) dual function g(z) = inf x L(x,Z) = dual SDP { tr(gz) tr(fi Z)+c i = 0, i = 1,...,n otherwise maximize tr(gz) subject to Z 0, tr(f i Z)+c i = 0, i = 1,...,n p = d if primal SDP is strictly feasible ( x with x 1 F 1 + +x n F n G) 7 31

32 A nonconvex problem with strong duality minimize x T Ax+2b T x subject to x T x 1 A 0, hence nonconvex dual function: g(λ) = inf x (x T (A+λI)x+2b T x λ) unbounded below if A+λI 0 or if A+λI 0 and b R(A+λI) minimized by x = (A+λI) b otherwise: g(λ) = b T (A+λI) b λ dual problem and equivalent SDP: maximize b T (A+λI) b λ subject to A+λI 0 b R(A+λI) maximize t [ λ A+λI b subject to b T t ] 0 strong duality although primal problem is not convex (not easy to show; if interested, see appendix B in book) 7 32

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