Solution to EE 617 Mid-Term Exam, Fall November 2, 2017
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1 Solution to EE 67 Mid-erm Exam, Fall 207 November 2, 207
2 EE 67 Solution to Mid-erm Exam - Page 2 of 2 November 2, 207 (4 points) Convex sets (a) (2 points) Consider the set { } a R k p(0) =, p(t) for t β, where p(t) = a + a 2 t + + a k t k Is this set convex? he set can be rewritten as { } a R k p(0) =, p(t) for t β { } = a R k p(0) = { } a R k p(t) = { } a R k a = t [,β t [,β { a R k a + a 2 t + + a k t k } he set is the intersection of a hyperplane and infinitely many halfspaces herefore, it is convex (b) (2 points) he polar of a set C R n is defined as C = { y R n y x for all x C } Is the polar C of a (possibly nonconvex) set C convex? he set can be rewritten as C = { y R n y x } x C he set is the intersection of halfspaces herefore, it is convex, regardless of whether C is convex or not
3 EE 67 Solution to Mid-erm Exam - Page 3 of 2 November 2, (4 points) Convex functions (a) (2 points) he logarithmic barrier for the second-order cone constraint x 2 t is f(x, t) = log ( t 2 x x ), with domf = {(x, t) R n R x 2 < t} Show that the function f(x, t) is convex in (x, t) (Hint: use the convexity of x x/t) he domain of f is a second-order cone, and hence is a convex set We can rewrite the function as f(x, t) = log [ t (t x x/t ) = log t log ( t x x/t ) Clearly, the first term log t is convex For the second term, note that x x/t is convex, and therefore t x x/t is concave Hence, log ( t x x/t ) is concave, and log ( t x x/t ) is convex In summary, the function f(x, t) is the sum of two convex functions his concludes the proof (b) (2 points) he directional derivative of a function g : R n R is f(x) = inf Show that f : R n R is convex if g is convex g(y + x) g(y) he domain of f is R n, and hence is a convex set We show the convexity of f by definition For any x, z R n and θ [0,, we have g [y + (θx + ( θ)z) g(y) f (θx + ( θ)z) = inf Using the convexity of g, we have herefore, we have = inf g [θ (y + x) + ( θ) (y + z) g(y) g [θ (y + x) + ( θ) (y + z) θg (y + x) + ( θ)g (y + z) f (θx + ( θ)z) θg (y + x) + ( θ)g (y + z) g(y) inf θ [g (y + x) g(y) + ( θ) [g (y + z) g(y) = inf = inf [ g (y + x) g(y) θ + ( θ) g (y + z) g(y) ()
4 EE 67 Solution to Mid-erm Exam - Page 4 of 2 November 2, 207 Now we need to show that g (y + x) g(y) g (y + x) g(y) inf = lim, 0 which is true if g(y+x) g(y) is nondecreasing in For any 0 < < 2, we have [( g(y + x) = g ) y + (y + 2 x) 2 2 ( ) g(y) + g(y + 2 x), 2 2 where the last inequality comes from the convexity of g he above equation implies g (y + x) g(y) g (y + 2x) g(y) 2 As a result, we have Finally, from (), we have [ f (θx + ( θ)z) inf θ g (y + x) g(y) g (y + x) g(y) inf = lim, 0 = lim 0 g (y + x) g(y) [ g (y + x) g(y) θ + ( θ) g (y + z) g(y) + ( θ) g (y + z) g(y) g (y + x) g(y) g (y + z) g(y) = θ lim + ( θ) lim 0 0 g (y + x) g(y) = θ inf = θf(x) + ( θ)f(z) + ( θ) inf g (y + z) g(y) his concludes the proof that f is convex
5 EE 67 Solution to Mid-erm Exam - Page 5 of 2 November 2, (4 points) Convex optimization problems (a) (2 points) Consider the following optimization problem: ( ) max i=,,m a i x + b i ( ) min i=,,p c i x + d i subject to F x g, with variables x R n Assume that c i x + d i > 0 for all x that satisfy F x g Give the best convex reformulation of the problem By best, we mean LP is better than SOCP, SOCP is better than SDP, and SDP is better than general convex program he objective function is the ratio of a convex function to a concave function herefore, the problem is a convex-concave fractional problem in Exercise 47 (c) of the book We will use the same approach here he convex formulation is g 0 (y, t) subject to g i (y, t) 0, i =,, m h(y, t), ( ) where g 0 is the perspective of f 0 (x) = max i=,,m a ( ) i x + b i, h is the perspective of h(x) = min i=,,p c i x + d i, and gi is the perspective of f i (x) = fi x g i, where fi is the ith row of matrix F and g i is the ith element of vector g Next, we compute the above functions explicitly he objective function is g 0 : R n R ++ R, which can be calculated as ( ) ( g 0 (y, t) = tf 0 (y/t) = t max a i (y/t) + b i = max a i y + b i t ) i=,,m i=,,m Similarly, we can compute h : R n R ++ R as ( ) ( h(y, t) = th(y/t) = t min c i (y/t) + d i = min c i y + d i t ) i=,,p i=,,p Finally, we have g i : R n R ++ R as g i (y, t) = tf i (y/t) = t (f i (y/t) g i ) = f i y g i t herefore, the original problem can be reformulated as ( a i y + b i t ) which is equivalent to the LP subject to max i=,,m fi y g i t 0, i =,, m ( min c i y + d i t ), i=,,p subject to z a i y + b i t z, i =,, m F y tg 0 c i y + d i t, i =,, p
6 EE 67 Solution to Mid-erm Exam - Page 6 of 2 November 2, 207 (b) (2 points) Consider the following complex least l -norm problem: x subject to Ax = b, with complex variables x C n and complex parameters A C m n and b C m Note that x = max x i i=,,n Formulate the complex least l -norm problem as a problem with real parameters and real variables Moreover, give the best convex reformulation of the new problem By best, we mean LP is better than SOCP, SOCP is better than SDP, and SDP is better than general convex program (Hint: use z = (Re(x), Im(x)) R 2n as the variable) We first write the objective function and constraints in real variables z = (Re(x), Im(x)) R 2n he objective function is x = max x ( i = max Re(xi ) 2 + Im(x i ) 2) /2 i=,,n i=,,n Each term ( Re(x i ) 2 + Im(x i ) 2) /2 can be written compactly as ( Re(xi ) 2 + Im(x i ) 2) /2 = [ Re(xi ) Im(x i ) where C i R 2 2n is a matrix such that [ Re(xi ) Im(x i ) 2 = C i 2 = C i [ Re(x) Im(x) [ Re(x) Im(x) 2 = C i z 2 2, In particular, the matrix C i has zero elements except (C i ),i = and (C i ) 2,n+i = Using j =, the linear equality can be rewritten as Ax = [Re(A) + jim(a) [Re(x) + jim(x) = [Re(A)Re(x) Im(A)Im(x) + j [Re(A)Im(x) + Im(A)Re(x) = [ Re(A) Im(A) [ Re(x) + j [ Im(A) Re(A) [ Re(x) Im(x) Im(x) Noting that b = Re(b) + jim(b), the linear equality can be written compactly as [ [ [ Re(A) Im(A) Re(x) Re(b) = Im(A) Re(A) Im(x) Im(b) }{{}}{{} Ã Ãz = b herefore, the original problem in real variables can be written as b 2 subject to max i=,,n C iz 2 2 Ãz = b,
7 EE 67 Solution to Mid-erm Exam - Page 7 of 2 November 2, 207 which is equivalent to the following SOCP subject to t C i z 2 t, i =,, n Ãz = b
8 EE 67 Solution to Mid-erm Exam - Page 8 of 2 November 2, (4 points) Dual problems Consider the problem of projecting a point a R n on the unit ball in l -norm: 2 x a 2 2 subject to x, with variables x R n Derive the dual problem (Hint: use the additional variable y = x a) With the additional variable y = x a, the problem is equivalent to 2 y 2 2 subject to x, with optimization variable x, y R n he Lagrangian is he dual function is y = x a, L(x, y, λ, ν) = 2 y λ ( x ) + ν (y x + a) g(λ, ν) = inf x,y he optimization over x is = inf x inf x [ 2 y λ ( x ) + ν (y x + a) ( λ x ν x ) ( ) + inf y 2 y y + ν y λ a ν ( λ x ν x ) = inf x n (λ x i ν i x i ) If λ < ν i, we will let x i, which makes the objective value unbounded below If λ ν i, the objective value satisfies i= λ x i ν i x i λ x i ν i x i = (λ ν i ) x i 0, where the inequality holds with equality when x i = 0 herefore, when λ ν i, the optimal x i = 0, and the objective value is 0 he optimization over y is minimization of a convex quadratic function herefore, the optimal y = ν, and the objective value is 2 ν ν Combining the above two parts, the dual function is { g(λ, ν) = 2 ν ν a ν λ if ν λ otherwise After making the implicit constraints explicit, the dual problem is maximize subject to ν λ, with optimization variable λ R and ν R n 2 ν ν a ν λ λ 0,
9 EE 67 Solution to Mid-erm Exam - Page 9 of 2 November 2, (4 points) Semidefinite relaxation of nonconvex problem A signal ŝ {, } n is sent through a noisy channel, and received as y = Hŝ + w, where H R m n, y R m, and w is the Gaussian noise he maximum likelihood detection of the signal can be formulated as the following nonconvex problem: y Hs 2 2 (2) subject to s 2 i =, i =,, n In the next series of questions, we explore the semidefinite relaxation (SDR) of the original nonconvex problem (2), and relationship between their dual problems (a) ( point) Show that the original problem (2) can be rewritten in homogenous form: x Lx (3) subject to x 2 i =, i =,, n, x n+ = Specify L as a function of the original problem data H and y (Hint: define x = he objective function of the original problem (2) can be written as y Hs 2 2 = (y Hs) (y Hs) = s H Hs 2y Hs+y y = [ [ s H herefore, by defining x = and L = H H y y H y y be rewritten in homogenous form: [ s [ s [ H H H y y H y y ) [ s, the original problem can x Lx (4) subject to x 2 i =, i =,, n, x n+ =
10 EE 67 Solution to Mid-erm Exam - Page 0 of 2 November 2, 207 (b) (2 points) Derive the dual problem of the problem in homogenous form (5) he problem in (5) is equivalent to x Lx (5) subject to x 2 i =, i =,, n + Note that the equality constraints are all quadratic terms of x i in this formulation, which makes it easier to derive the dual problem he Lagrangian is he dual function is n+ L(x, ν) = x Lx + ν i (x 2 i ) g(ν) = inf x = inf x he dual function is unbounded below unless herefore, we have i= [ n+ x Lx + ν i (x 2 i ) i= [ n+ x (L + diag(ν)) x ν i L + diag(ν) 0 i= { n+ g(ν) = i= ν i if L + diag(ν) 0 otherwise he dual problem is then maximize n+ ν i i= subject to L + diag(ν) 0
11 EE 67 Solution to Mid-erm Exam - Page of 2 November 2, 207 (c) ( point) Now we consider the SDR of the nonconvex problem (5) variable X = xx S n + he problem in (5) is equivalent to We define a new tr(lx) subject to diag(x) = n+, X 0, rank(x) =, where n+ is (n + )-dimensional vector of s By removing the nonconvex rank constraint, we have the following SDR: tr(lx) (6) subject to diag(x) = n+, X 0, which is a semidefinite program (SDP) Derive the dual problem of the SDR (6) and show that it is the same as the dual of the original problem derived in part (b) (In other words, the SDR is the dual of the dual of the original nonconvex problem) (Hint: the Lagrangian multiplier associate with the constraint X 0 is Z 0, and we need to use the term tr(zx) in the Lagrangian) Write ν R n+ as the Lagrangian multiplier associated with the linear constraint diag(x) = n+, and Z S n + as the Lagrangian multiplier associated with the constraint X 0 hen the Lagrangian is Observing that the Lagrangian can be rewritten as he dual function is L(X, Z) = tr(lx) + ν (diag(x) n+ ) tr(zx) ν diag(x) = tr (diag(ν)x), L(X, Z) = tr ((L + diag(ν) Z) X) n+ν g(z) = inf X [ tr ((L + diag(ν) Z) X) n+ ν If L + diag(ν) Z 0, we can make tr ((L + diag(ν) Z) X) unbounded below If L + diag(ν) Z = 0, we have tr ((L + diag(ν) Z) X) = 0 herefore, the dual function is { g(z) = n+ ν if L + diag(ν) = Z otherwise he dual problem is then maximize subject to n+ν L + diag(ν) = Z Z 0,
12 EE 67 Solution to Mid-erm Exam - Page 2 of 2 November 2, 207 which is equivalent to maximize n+ν subject to L + diag(ν) 0, which is the same as the dual problem derived in part (b)
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