Power cones in second-order cone form and dual recovery SIAM - PowerPoint PPT Presentation

Power cones in second-order cone form and dual recovery SIAM Conference on Optimization 2017 Henrik A. Friberg www.mosek.com

What is a power cone? Defined by parameter vector, α ∈ R k + , spanning: • Quadratic cone: P n = { ( x , z ) | x 1 1 ≥ � z � 2 } 1

What is a power cone? Defined by parameter vector, α ∈ R k + , spanning: • Quadratic cone: P n = { ( x , z ) | x 1 1 ≥ � z � 2 } 1 • Rotated quadratic cone in the non-self-dualized form: P n = { ( x , z ) | x 1 1 x 1 2 ≥ � z � 2 2 } 1 , 1

What is a power cone? Defined by parameter vector, α ∈ R k + , spanning: • Quadratic cone: P n = { ( x , z ) | x 1 1 ≥ � z � 2 } 1 • Rotated quadratic cone in the non-self-dualized form: P n = { ( x , z ) | x 1 1 x 1 2 ≥ � z � 2 2 } 1 , 1 • Geometric mean: P n = { ( x , z ) | x 1 1 x 1 2 · · · x 1 k ≥ � z � k 2 } 1 , 1 ,..., 1

What is a power cone? Defined by parameter vector, α ∈ R k + , spanning: • Quadratic cone: P n = { ( x , z ) | x 1 1 ≥ � z � 2 } 1 • Rotated quadratic cone in the non-self-dualized form: P n = { ( x , z ) | x 1 1 x 1 2 ≥ � z � 2 2 } 1 , 1 • Geometric mean: P n = { ( x , z ) | x 1 1 x 1 2 · · · x 1 k ≥ � z � k 2 } 1 , 1 ,..., 1 • Weighted geometric mean: 2 · · · x α k ≥ � z � α 1 + α 2 + ... + α k α 1 ,α 2 ,...,α k = { ( x , z ) | x α 1 1 x α 2 P n } 2 k

What is a power cone? Defined by parameter vector, α ∈ R k + , spanning: • Quadratic cone: P n = { ( x , z ) | x 1 1 ≥ � z � 2 } 1 • Rotated quadratic cone in the non-self-dualized form: P n = { ( x , z ) | x 1 1 x 1 2 ≥ � z � 2 2 } 1 , 1 • Geometric mean: P n = { ( x , z ) | x 1 1 x 1 2 · · · x 1 k ≥ � z � k 2 } 1 , 1 ,..., 1 • Weighted geometric mean: 2 · · · x α k ≥ � z � α 1 + α 2 + ... + α k α 1 ,α 2 ,...,α k = { ( x , z ) | x α 1 1 x α 2 P n } 2 k The power cone can be given for any α ∈ R k + as + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } , by convention of 0 0 = 1.

What is a power cone? The power cone can be given for any α ∈ R k + as + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } , by convention of 0 0 = 1.

What is a power cone? The power cone can be given for any α ∈ R k + as + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } , by convention of 0 0 = 1. Common restrictions • � k 1 α j = e T α = 1. Full generality by scale invariance P n α = P n λα for λ > 0, but only useful in barrier function to my knowledge.

What is a power cone? The power cone can be given for any α ∈ R k + as + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } , by convention of 0 0 = 1. Common restrictions • � k 1 α j = e T α = 1. Full generality by scale invariance P n α = P n λα for λ > 0, but only useful in barrier function to my knowledge. • α ∈ R k ++ . Full generality by P n (0 ,α ) = R + × P n α . When are zeros useful? • Powers, s ≥ | x | p , for any p ≥ 1: 1 p − 1 s 1 ≥ | x | p (1 , s , x ) ∈ P 3 ⇐ ⇒ ( p − 1) , 1 • p-norms, t ≥ � x � p , for any p ≥ 1: t ≥ e T s , ( t , s j , x j ) ∈ P 3 and ( p − 1) , 1 ∀ j

What is a power cone? The dual power cone was be obtained on α ⊆ R k ++ by (Chares 2009, Theorem 4.3.1) as: (e T α ) − 1 diag( α ) α ) ∗ = M P n � � 0 ( P n α , for M = ≻ 0 , 0 I n − k expanding to: α ) ∗ = { ( x , z ) ∈ R k + × R n − k | α − α x α ≥ (e T α ) − e T α � z � e T α ( P n 2 } , which is easily shown valid on all of α ⊆ R k + .

What is a power cone? The dual power cone was be obtained on α ⊆ R k ++ by (Chares 2009, Theorem 4.3.1) as: (e T α ) − 1 diag( α ) α ) ∗ = M P n � � 0 ( P n α , for M = ≻ 0 , 0 I n − k expanding to: α ) ∗ = { ( x , z ) ∈ R k + × R n − k | α − α x α ≥ (e T α ) − e T α � z � e T α ( P n 2 } , which is easily shown valid on all of α ⊆ R k + . Note self-duality of M 1 / 2 P n α in general (the self-dualized variant).

Power cones in MOSEK?

Power cones in MOSEK? Absolutely ∗∗∗ ! 1 Convert α to rationals. Best rational approximations to π : 3 1 , 13 4 , 16 5 , 19 6 , 22 7 , 179 57 , 201 64 , 223 71 , 245 78 , 267 85 , 289 92 , 311 99 , 333 106 , 355 113 , 52163 16604 , ... λα with λ = lcm( denominators ) 2 Use P n α = P n to make α integer. gcd( numerators ) 3 Construct tower of variables (Ben-tal and Nemirovski 2001); here x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 ≥ ω 8 1 .

Power cones in MOSEK? Absolutely ∗∗∗ ! 1 Convert α to rationals. Best rational approximations to π : 3 1 , 13 4 , 16 5 , 19 6 , 22 7 , 179 57 , 201 64 , 223 71 , 245 78 , 267 85 , 289 92 , 311 99 , 333 106 , 355 113 , 52163 16604 , ... λα with λ = lcm( denominators ) 2 Use P n α = P n to make α integer. gcd( numerators ) 3 Construct tower of variables (Ben-tal and Nemirovski 2001); here x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 ≥ ω 8 1 . Non-unique, e.g. permute x

Power cones in MOSEK? Absolutely ∗∗∗ ! 1 Convert α to rationals. Best rational approximations to π : 3 1 , 13 4 , 16 5 , 19 6 , 22 7 , 179 57 , 201 64 , 223 71 , 245 78 , 267 85 , 289 92 , 311 99 , 333 106 , 355 113 , 52163 16604 , ... λα with λ = lcm( denominators ) 2 Use P n α = P n to make α integer. gcd( numerators ) 3 Construct tower of variables (Ben-tal and Nemirovski 2001); here x 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 ≥ ω 8 1 . Distinct, e.g., consider x 1 = x 2

Complication summary • Implementation: cumbersome and error-prone • Tower constructions: suboptimal • Dual information: where?

Complication summary • Implementation: cumbersome and error-prone • Tower constructions: suboptimal • Dual information: where? Same three complications decomposing P k +1 ( α 1 ,...,α k ) into k − 1 power cones of the form P 3 ( α 1 ,α 2 ) . See Chares (2009). Reason? Barrier parameter increases. Linear outer approximation is stronger. Hessian matrix is approximated with less effort in quasi-newton methods, e.g., using BFGS updates.

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling.

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling. � � ( α − β, e T β ) , ( β ) 2 Split α − → for any β ≤ α . Split rule x α ≥ � z � e T α x α − β x β ≥ � z � e T α ⇔ 2 , 2 x α − β u e T β ≥ � z � e T α x β ≥ u e T β ⇔ 2 ,

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling. � � ( α − β, e T β ) , ( β ) 2 Split α − → for any β ≤ α . Split rule x α ≥ � z � e T α x α − β x β ≥ � z � e T α ⇔ 2 , 2 x α − β u e T β ≥ � z � e T α x β ≥ u e T β ⇔ 2 , simple base

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling. � � ( α − β, e T β ) , ( β ) 2 Split α − → for any β ≤ α . 3 Expand α − → { ( α, β ) , 1 } for any β ∈ R + . Expansion rule x α ≥ � z � e T α x α ≥ u e T α ≥ � z � e T α ⇔ 2 , 2 x α ≥ u e T α , ⇔ u ≥ � z � 2 , x α u β ≥ u e T α + β , ⇔ u ≥ � z � 2 ,

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling. � � ( α − β, e T β ) , ( β ) 2 Split α − → for any β ≤ α . 3 Expand α − → { ( α, β ) , 1 } for any β ∈ R + . Expansion rule x α ≥ � z � e T α x α ≥ u e T α ≥ � z � e T α ⇔ 2 , 2 x α ≥ u e T α , ⇔ u ≥ � z � 2 , x α u β ≥ u e T α + β , ⇔ u ≥ � z � 2 , simple base

Let’s play Tower Tycoon Rules of the game Start with any power cone defined by α ∈ R k + : + × R n − k | x α ≥ � z � e T α P n α = { ( x , z ) ∈ R k 2 } . Rules: 1 α is invariant to permutation, zeros and positive scaling. � � ( α − β, e T β ) , ( β ) 2 Split α − → for any β ≤ α . 3 Expand α − → { ( α, β ) , 1 } for any β ∈ R + . 4 Expand α − → { ( α, β ) } for any β ∈ R + (on simple base). Expansion rule x α ≥ � z � e T α x α ≥ u e T α ≥ � z � e T α ⇔ 2 , 2 x α ≥ u e T α , ⇔ u ≥ � z � 2 , x α u β ≥ u e T α + β , ⇔ u ≥ � z � 2 ,