v2010.10.26 - Convex Optimization
v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization
360 CHAPTER 4. SEMIDEFINITE PROGRAMMING4.6.0.0.5 Example. Tractable polynomial constraint.The set of all coefficients for which a multivariate polynomial were convex isgenerally difficult to determine. But the ability to handle rank constraintsmakes any nonconvex polynomial constraint transformable to a convexconstraint. All optimization problems having polynomial objective andpolynomial constraints can be reformulated as a semidefinite program witha rank-1 constraint. [285] Suppose we requireIdentify3 + 2x − xy ≤ 0 (798)⎡ ⎤ ⎡ ⎤x [ x y 1] x 2 xy xG = ⎣ y ⎦ = ⎣ xy y 2 y ⎦∈ S 3 (799)1x y 1Then nonconvex polynomial constraint (798) is equivalent to constraint settr(GA) ≤ 0G 33 = 1(G ≽ 0)rankG = 1(800)with direct correspondence to sense of trace inequality where G is assumedsymmetric (B.1.0.2) and⎡ ⎤0 − 1 12A = ⎣ − 1 0 0 ⎦∈ S 3 (801)21 0 3Then the method of convex iteration from4.4.1 is applied to implement therank constraint.4.6.0.0.6 Exercise. Binary Pythagorean theorem.Technique of Example 4.6.0.0.5 is extensible to any quadratic constraint; e.g.,x T Ax + 2b T x + c ≤ 0 , x T Ax + 2b T x + c ≥ 0 , and x T Ax + 2b T x + c = 0.Write a rank-constrained semidefinite program to solve (Figure 105){ x + y = 1x 2 + y 2 (802)= 1whose feasible set is not connected. Implement this system in cvx [167] byconvex iteration.
4.6. CARDINALITY AND RANK CONSTRAINT EXAMPLES 3614.6.0.0.7 Example. Higher-order polynomials.Consider nonconvex problem from Canadian Mathematical Olympiad 1999:findx , y , z∈Rx, y , zsubject to x 2 y + y 2 z + z 2 x = 223 3x + y + z = 1x, y , z ≥ 0(803)We wish to solve for what is known to be a tight upper bound 223 3on the constrained polynomial x 2 y + y 2 z + z 2 x by transformation to arank-constrained semidefinite program; identify⎡ ⎤⎡⎤x [x y z 1] x 2 xy zx xG = ⎢ y⎥⎣ z ⎦ = ⎢ xy y 2 yz y⎥⎣ zx yz z 2 z ⎦ ∈ S4 (804)1x y z 1⎡ ⎤⎡⎤x 2 [ x 2 y 2 z 2 x y z 1] x 4 x 2 y 2 z 2 x 2 x 3 x 2 y zx 2 x 2y 2x 2 y 2 y 4 y 2 z 2 xy 2 y 3 y 2 z y 2z 2z 2 x 2 y 2 z 2 z 4 z 2 x yz 2 z 3 z 2X =x=x 3 xy 2 z 2 x x 2 xy zx x∈ S 7⎢ y⎥⎢ x 2 y y 3 yz 2 xy y 2 yz y⎥⎣ z ⎦⎣ zx 2 y 2 z z 3 zx yz z 2 z ⎦1x 2 y 2 z 2 x y z 1(805)then apply convex iteration (4.4.1) to implement rank constraints:findA , C∈S , bbsubject to tr(XE) = 22[3 3] A bG =b T (≽ 0)1⎡ [ ] ⎤ δ(A)X = ⎣ C[bδ(A) T b ] ⎦(≽ 0)T 1(806)1 T b = 1b ≽ 0rankG = 1rankX = 1
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360 CHAPTER 4. SEMIDEFINITE PROGRAMMING4.6.0.0.5 Example. Tractable polynomial constraint.The set of all coefficients for which a multivariate polynomial were convex isgenerally difficult to determine. But the ability to handle rank constraintsmakes any nonconvex polynomial constraint transformable to a convexconstraint. All optimization problems having polynomial objective andpolynomial constraints can be reformulated as a semidefinite program witha rank-1 constraint. [285] Suppose we requireIdentify3 + 2x − xy ≤ 0 (798)⎡ ⎤ ⎡ ⎤x [ x y 1] x 2 xy xG = ⎣ y ⎦ = ⎣ xy y 2 y ⎦∈ S 3 (799)1x y 1Then nonconvex polynomial constraint (798) is equivalent to constraint settr(GA) ≤ 0G 33 = 1(G ≽ 0)rankG = 1(800)with direct correspondence to sense of trace inequality where G is assumedsymmetric (B.1.0.2) and⎡ ⎤0 − 1 12A = ⎣ − 1 0 0 ⎦∈ S 3 (801)21 0 3Then the method of convex iteration from4.4.1 is applied to implement therank constraint.4.6.0.0.6 Exercise. Binary Pythagorean theorem.Technique of Example 4.6.0.0.5 is extensible to any quadratic constraint; e.g.,x T Ax + 2b T x + c ≤ 0 , x T Ax + 2b T x + c ≥ 0 , and x T Ax + 2b T x + c = 0.Write a rank-constrained semidefinite program to solve (Figure 105){ x + y = 1x 2 + y 2 (802)= 1whose feasible set is not connected. Implement this system in cvx [167] byconvex iteration.