Can Systems be Certified Distributively? Scalable Analysis Methods - PDF document

Can Systems be Certified Distributively? Scalable Analysis Methods for Sparse Large-scale Systems Anders Rantzer LCCC — Lund Center for Control of Complex engineering systems Automatic Control LTH, Lund University, Sweden Can a global performance certificate be split into component specifications without introducing conservatism? Can Systems be Certified Distributively? Outline ○ Introduction Distributed Positive Test for Matrices • ○ Distributed Nonconservative System Verification ○ A Scalable Robustness Test What freedom do we have to modify a component without redesigning the whole system? A Matrix Decomposition Theorem Proof idea The decomposition follows immediately from the band structure of the Cholesky factors: The sparse matrix on the left is positive semi-definite if and only x x x x x x x if it can be written as a sum of positive semi-definite matrices 0 x x x x x x x x x with the structure on the right. x x x x x x x x x x x = x x x x x x x x x x x x x x x x x x x x x x 0 0 x x x x x x 0 0 0 0 x x x x x x x x x x x x x x x x x x x 0 0 x x x x x x x x x x x 0 + ... + x x x x x x x + = x x x x x 0 x x x 0 x x x x x 0 0 x x x x x x x 0 0 x x x 0 0 0 0 x x x 0 0 x x x [Martin and Wilkinson, 1965] Example Generalization Cholesky factors inherit the sparsity structure of the symmetric matrix if and only if the sparsity pattern corresponds to a The simplest decomposition is to just split each coefficient “chordal” graph. equally between the squares where it belong. This could work if the matrix is diagonally dominant: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * = + + + * * * * * * * * * * * * * * * * * * * * * 0 t 11 t 12 t 13 t 11 t 12 t 13 0 * * * * * 0 0 t 22 t 23 0 t 21 t 22 t 23 t 24 t 21 2 2 t 32 t 33 0 t 31 t 32 t 33 t 34 t 35 t 31 + ... + = 2 3 4 0 0 t 42 t 43 t 44 t 45 t 46 0 t 55 t 56 t 53 t 54 t 55 t 56 t 57 t 57 3 2 2 5 7 0 t 65 t 66 t 64 t 65 t 66 t 67 t 67 0 2 2 0 0 0 t 75 t 76 t 77 t 75 t 76 t 77 1 3 6 [Blair & Peyton, An introduction to chordal graphs and clique trees, 1992] 1

Example: Non-chordal graph Example: Chordal graphs If T is a tree, then T k is chordal for every k ≥ 1 . T 2 T A Theorem on Positive Extensions Outline A matrix with entries specified according to a chordal graph has a positive definite completion if and only if all fully specified principal minors are positive definite. [Grone, et.al, 1984] ○ Introduction ○ Distributed Positive Test for Matrices 3 2 1 * * * * Distributed Nonconservative System Verification • 2 4 2 1 * * * ○ A Scalable Robustness Test 1 2 4 1 1 * * 1 1 3 1 1 * * 1 1 5 2 1 * * 1 2 4 1 * * * 1 1 3 * * * * A System with Tridiagonal Structure A Sparse Stability Test For the sparse matrix A , let the left hand side illustrate the w 1 w 2 w n structure of ( sI − A ) ∗ ( sI − A ) . Then the matrix is stable if and x 1 x 2 x n − 1 only if the right hand side split can be done with all squares positive definite for s in the right half plane. x 2 x 3 x n 0 0 x x x x x x 0 0 0 0 x x x x x x x x x x 0 x x x x x x x x x x x 0 + ... + +   =       x x x x x 0 x x x 0 a 11 a 12 0 x 1 ( t ) ˙ x 1 ( t ) w 1 ( t ) x x x x x 0 0 x x x ...   x x x x 0 0 x x x x 2 ( t ) ˙ x 2 ( t ) w 2 ( t )       a 21 a 22 0 0   0 0 x x x 0 0 x x x        =  + .   . .   ...     . . . � �� �   . . .     a ( n − 1 ) n   ( sI − A ) ∗ ( sI − A ) x n ( t ) x n ( t ) w n ( t ) ˙ 0 a n ( n − 1 ) a nn Hence global stability can always be verified by local tests! A Sparse Stability Test A Sparse Gain Bound Find stable minimum-phase rational Φ 1 ( s ) ,... , Φ m ( s ) with Solutions to ˙ x ( t ) = Ax ( t ) + w ( t ) , x ( 0 ) = 0 satisfy � T � T 0 0 x x x x x x 0 0 0 0 � x ( t )� 2 dt ≤ γ 2 � w ( t )� 2 dt x x x x x x x x x x 0 x x x x x x x x x x x 0 + ... + + = 0 0 x x x x x 0 x x x 0 x x x x x 0 0 x x x if and only if x x x x 0 0 x x x 0 0 0 0 x x x 0 0 x x x � �� � � �� � � �� � � �� � 0 0 x x x x x x 0 0 0 0 ( sI − A ) ∗ ( sI − A ) Φ 1 ( s ) ∗ Φ 1 ( s ) Φ 2 ( s ) ∗ Φ 2 ( s ) Φ m ( s ) ∗ Φ m ( s ) x x x x x x x x x x 0 x x x x x x x x x x x 0 + ... + + = x x x x x 0 x x x 0 x x x x x 0 0 x x x for s = i ω . Then A is Hurwitz iff equality holds also for Re s > 0 . x x x x 0 0 x x x 0 0 0 0 x x x 0 0 x x x � �� � Proof idea: If A Hurwitz, the maximum modulus theorem gives γ 2 ( sI − A ) ∗ ( sI − A )− I ( sI − A ) ∗ ( sI − A ) � Φ 1 ( s ) ∗ Φ 1 ( s ) + ⋅ ⋅ ⋅ + Φ m ( s ) ∗ Φ m ( s ) Re s > 0 where the terms on the right hand side are positive definite for s in the right half plane. 2

A Sparse Passivity Test Outline Suppose x = Ax + Bx + w ˙ x ( 0 ) = 0 y = Cx ○ Introduction Then ○ Distributed Positive Test for Matrices � T � γ 2 u ( t ) y ( t ) + � w ( t )� 2 � ○ Distributed Nonconservative System Verification dt ≥ 0 for all u , w , T 0 A Scalable Robustness Test • if and only if the matrix � ( sI − A ) ∗ ( sI − A ) � γ 2 C T − ( sI − A ) ∗ B B T B γ 2 C − B ∗ ( sI − A ) is positive semi-definite for Re s ≥ 0 . Passivity can be tested componentwise without conservatism! Robustness Analysis for Chained System Scalable Distributed Computations ✲ δ 2 δ 1 ✲ δ m δ m x x x 0 0 ✲ − 1 ✲ 0 0 0 0 W 1 x x x x 0 W 2 x x x x x 0 + ... + + = x x x x x 0 0 ✛ ✛ ✛ ✛ x x x x x 0 0 ✛ ✛ ✛ q q q ✛ ✛ ✛ W m ✲ ✲ ✲ q ✲ ✲ ✲ x x x x 0 0 q q − 2 0 0 0 0 x x x 0 0 Many robustness analysis problems can be reduced to proving that The matrix G X G ∗ − X is negative definite if and only if there exist ( I − ∆ ( s ) G ( s )) − 1 is stable for ∆ = diag { δ 1 ,... , δ m } with � δ i ( i ω )� ≤ 1 . y i , z i , w i such that the following are negative definite: This can be done by finding X ( ω ) = diag { x 1 ( ω ) ,..., x m ( ω )} ≻ 0 with X ( ω ) ≻ G ( i ω ) X ( ω ) G ( i ω ) ∗ where ∗  � 1  �  � 1    h 1 h 1 x 1 0 0 � x 1 0 W 1 = � 2 � 2 − x 2 + w 1 f 1 f 1 0 y 1     0 x 2     � 1 h 1 0 0 0 0 f 2 0 f 2 0 y ∗ z 1 1 f 1 � 2 h 2 0 0     ... ... ∗         h 2 h 2 w 1 y 1 0 G ( i ω ) = 0 f 2 0    x 3 y ∗   W 2 = � 3 � 3 + z 1 − w 2 − x 3 − y 2 ...     1    − y ∗ 0 0 � m − 1 h m − 1 f 3 f 3 0 − z 2   2 0 0 0 f m − 1 � m . . . Note that each x i influences at most nine elements of X − G X G ∗ . W m − 2 = ... Can Systems be Certified Distributively? Yes, distributed tests can be constructed without conservatism! Local trade-offs between accuracy and complexity! Chordal graphs are common and natural! Nonlinear versions?! 3