v2009.01.01 - Convex Optimization
v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization
274 CHAPTER 4. SEMIDEFINITE PROGRAMMING 4.3.3.0.1 Example. Aδ(X) = b . This academic example demonstrates that a solution found by rank reduction can certainly have rank less than Barvinok’s upper bound (245): Assume a given vector b∈ R m belongs to the conic hull of columns of a given matrix A∈ R m×n ; ⎡ ⎤ ⎡ ⎤ A = ⎢ ⎣ −1 1 8 1 1 −3 2 8 1 2 −9 4 8 1 4 1 3 1 9 Consider the convex optimization problem ⎥ ⎦ , b = ⎢ ⎣ 1 1 2 1 4 ⎥ ⎦ (658) minimize trX X∈ S 5 subject to X ≽ 0 Aδ(X) = b (659) that minimizes the 1-norm of the main diagonal; id est, problem (659) is the same as minimize X∈ S 5 ‖δ(X)‖ 1 subject to X ≽ 0 Aδ(X) = b (660) that finds a solution to Aδ(X)=b. Rank-3 solution X ⋆ = δ(x M ) is optimal, where (confer (618)) ⎡ 2 ⎤ 128 0 x M = 5 ⎢ 128 ⎥ (661) ⎣ 0 ⎦ 90 128 Yet upper bound (245) predicts existence of at most a (⌊√ ⌋ ) 8m + 1 − 1 rank- = 2 2 (662) feasible solution from m = 3 equality constraints. To find a lower rank ρ optimal solution to (659) (barring combinatorics), we invoke Procedure 4.3.1.0.1:
4.3. RANK REDUCTION 275 Initialize: C =I , ρ=3, A j ∆ = δ(A(j, :)) , j =1, 2, 3, X ⋆ = δ(x M ) , m=3, n=5. { Iteration i=1: ⎡ Step 1: R 1 = ⎢ ⎣ √ 2 128 0 0 0 0 0 0 √ 5 0 128 0 0 √ 0 0 0 90 128 ⎤ . ⎥ ⎦ find Z 1 ∈ S 3 subject to 〈Z 1 , R T 1A j R 1 〉 = 0, j =1, 2, 3 (663) A nonzero randomly selected matrix Z 1 having 0 main diagonal is feasible and yields a nonzero perturbation matrix. Choose, arbitrarily, Z 1 = 11 T − I ∈ S 3 (664) then (rounding) ⎡ B 1 = ⎢ ⎣ Step 2: t ⋆ 1= 1 because λ(Z 1 )=[−1 −1 ⎡ X ⋆ ← δ(x M ) + B 1 = ⎢ ⎣ 0 0 0.0247 0 0.1048 0 0 0 0 0 0.0247 0 0 0 0.1657 0 0 0 0 0 0.1048 0 0.1657 0 0 2 ] T . So, ⎤ ⎥ ⎦ 2 0 0.0247 0 0.1048 128 0 0 0 0 0 0.0247 0 5 128 0 0.1657 0 0 0 0 0 0.1048 0 0.1657 0 90 128 ⎤ (665) ⎥ ⎦ (666) has rank ρ ←1 and produces the same optimal objective value. }
- Page 223 and 224: 3.1. CONVEX FUNCTION 223 Setting th
- Page 225 and 226: 3.1. CONVEX FUNCTION 225 Similarly,
- Page 227 and 228: 3.1. CONVEX FUNCTION 227 For vector
- Page 229 and 230: 3.1. CONVEX FUNCTION 229 This means
- Page 231 and 232: 3.1. CONVEX FUNCTION 231 f(Y ) −
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- Page 239 and 240: 3.3. QUASICONVEX 239 exponential al
- Page 241 and 242: 3.3. QUASICONVEX 241 Unlike convex
- Page 243 and 244: 3.4. SALIENT PROPERTIES 243 6. (af
- Page 245 and 246: Chapter 4 Semidefinite programming
- Page 247 and 248: 4.1. CONIC PROBLEM 247 where K is a
- Page 249 and 250: 4.1. CONIC PROBLEM 249 4.1.1.2 Redu
- Page 251 and 252: 4.1. CONIC PROBLEM 251 In any SDP f
- Page 253 and 254: 4.1. CONIC PROBLEM 253 Proposition
- Page 255 and 256: 4.2. FRAMEWORK 255 sets are closed
- Page 257 and 258: 4.2. FRAMEWORK 257 4.2.1.1.3 Exampl
- Page 259 and 260: 4.2. FRAMEWORK 259 4.2.2 Duals The
- Page 261 and 262: 4.2. FRAMEWORK 261 When equality is
- Page 263 and 264: 4.2. FRAMEWORK 263 The pseudoinvers
- Page 265 and 266: 4.2. FRAMEWORK 265 For the data giv
- Page 267 and 268: 4.2. FRAMEWORK 267 minimizes an aff
- Page 269 and 270: 4.3. RANK REDUCTION 269 whose rank
- Page 271 and 272: 4.3. RANK REDUCTION 271 and where m
- Page 273: 4.3. RANK REDUCTION 273 4.3.3 Optim
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- Page 305 and 306: 4.5. CONSTRAINING CARDINALITY 305 W
- Page 307 and 308: 4.5. CONSTRAINING CARDINALITY 307 t
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4.3. RANK REDUCTION 275<br />
Initialize:<br />
C =I , ρ=3, A j ∆ = δ(A(j, :)) , j =1, 2, 3, X ⋆ = δ(x M<br />
) , m=3, n=5.<br />
{<br />
Iteration i=1:<br />
⎡<br />
Step 1: R 1 =<br />
⎢<br />
⎣<br />
√<br />
2<br />
128<br />
0 0<br />
0 0 0<br />
0<br />
√<br />
5<br />
0<br />
128<br />
0 0<br />
√<br />
0<br />
0 0<br />
90<br />
128<br />
⎤<br />
.<br />
⎥<br />
⎦<br />
find Z 1 ∈ S 3<br />
subject to 〈Z 1 , R T 1A j R 1 〉 = 0, j =1, 2, 3<br />
(663)<br />
A nonzero randomly selected matrix Z 1 having 0 main diagonal<br />
is feasible and yields a nonzero perturbation matrix. Choose,<br />
arbitrarily,<br />
Z 1 = 11 T − I ∈ S 3 (664)<br />
then (rounding)<br />
⎡<br />
B 1 =<br />
⎢<br />
⎣<br />
Step 2: t ⋆ 1= 1 because λ(Z 1 )=[−1 −1<br />
⎡<br />
X ⋆ ← δ(x M<br />
) + B 1 =<br />
⎢<br />
⎣<br />
0 0 0.0247 0 0.1048<br />
0 0 0 0 0<br />
0.0247 0 0 0 0.1657<br />
0 0 0 0 0<br />
0.1048 0 0.1657 0 0<br />
2 ] T . So,<br />
⎤<br />
⎥<br />
⎦<br />
2<br />
0 0.0247 0 0.1048<br />
128<br />
0 0 0 0 0<br />
0.0247 0<br />
5<br />
128<br />
0 0.1657<br />
0 0 0 0 0<br />
0.1048 0 0.1657 0<br />
90<br />
128<br />
⎤<br />
(665)<br />
⎥<br />
⎦ (666)<br />
has rank ρ ←1 and produces the same optimal objective value.<br />
}