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Compute Residual in Kuhn-Tucker Conditions

Syntax
[F] = ckbs_kuhn_tucker(
      mu, s, y, u, b, d, Bdia, Hdia, Hlow)

Purpose
This routine computes the residual in the Kuhn-Tucker conditions for the  \mu -relaxed affine constrained Kalman-Bucy smoother problem.

Problem
Given  \mu \in \B{R}_+ ,  H \in \B{R}^{p \times p} ,  d \in \B{R}^p ,  b \in \B{R}^r , and  B \in \B{R}^{r \times p} , the  \mu -relaxed affine constrained Kalman-Bucy smoother problem is:  \[
\begin{array}{rl}
{\rm minimize} & \frac{1}{2} y^\R{T} H y + d^\R{T} y
- \mu \sum_{i=1}^r \log(s_i)
\; {\rm w.r.t} \; y \in \B{R}^p \; , \; s \in \B{R}_+^r
\\
{\rm subject \; to} & s + b + B y  = 0
\end{array}
\]  In addition,  H is symmetric block tri-diagonal with each block of size  n \times n and  B is block diagonal with each block of size  m \times n (there is an integer  N such that  p = n * N and  r = m * N ).

Lagrangian
We use  u \in \B{R}^r to denote the Lagrange multipliers corresponding to the constraint equation. The corresponding Lagrangian is  \[
L(y, s, u)  =
\frac{1}{2} y^\R{T} H y + d^\R{T} y
- \mu \sum_{i=1}^r \log(s_i)
+ u^\R{T} (s + b + B y)
\]  The partial gradients of the Lagrangian are given by  \[
\begin{array}{rcl}
\nabla_y L(y, s, u ) & = & H y + B^\R{T} u + d  \\
\nabla_s L(y, s, u ) & = & u - \mu / s \\
\nabla_u L(y, s, u ) & = & s + b + B y \\
\end{array}
\]  where  \mu / s  is the component by component division of  \mu  by the components of the  s . Note, from the second equation, that we only need consider  u \geq 0 because  s \geq 0 .

Kuhn-Tucker Residual
We use  D(s) to denote the diagonal matrix with  s along its diagonal and  1_r to denote the vector, of length  r with all its components equal to one. The Kuhn-Tucker Residual function  F : \B{R}^{r + p + r} \rightarrow \B{R}^{r + p + r} is defined by  \[
F(s, y, u)
=
\left(
\begin{array}{c}
s + b + B y       \\
H y + B^\R{T} u + d   \\
D(s) D(u) 1_r - \mu 1_r
\end{array}
\right)
\]  The Kuhn-Tucker conditions for a solution of the  \mu -relaxed constrained affine Kalman-Bucy smoother problem is  F(s, y, u) = 0  .

mu
The argument mu is a positive scalar specifying the relaxation parameter  \mu .

s
The argument s is a column vector of length  r . All the elements of s are greater than zero.

y
The argument y is a column vector of length  p

u
The argument u is a column vector of length  r . All the elements of s are greater than zero.

b
The argument b is a column vector of length  r .

d
The argument d is a column vector of length  p

Bdia
The argument Bdia is an  m \times n \times N array. For  k = 1 , \ldots , N we define  B_k \in \B{R}^{m \times n} by  \[
      B_k = Bdia(:, :, k)
\] 

B
The matrix  B is defined by  \[
B
=
\left( \begin{array}{cccc}
B_1 & 0      & 0      &           \\
0   & B_2    & 0      & 0         \\
0   & 0      & \ddots & 0         \\
    & 0      & 0      & B_N
\end{array} \right)
\] 

Hdia
The argument Hdia is an  n \times n \times N array. For  k = 1 , \ldots , N we define  H_k \in \B{R}^{n \times n} by  \[
      H_k = Hdia(:, :, k)
\] 

Hlow
The argument Hlow is an  n \times n \times N array. For  k = 1 , \ldots , N we define  L_k \in \B{R}^{n \times n} by  \[
      L_k = Hlow(:, :, k)
\] 

H
The matrix  H is defined by  \[
H
=
\left( \begin{array}{cccc}
H_1 & L_2^\R{T} & 0         &           \\
L_2 & H_2       & L_3^\R{T} & 0         \\
0   & \ddots    & \ddots    & \ddots    \\
    & 0         & L_N       & H_N
\end{array} \right)
\] 

F
The result F is a column vector of length  r + p + r containing the value of the Kuhn-Tucker residual ; i.e.,  F(s, y, u) .

Example
The file kuhn_tucker_ok.m contains an example and test of ckbs_kuhn_tucker. It returns true if ckbs_kuhn_tucker passes the test and false otherwise.
Input File: src/ckbs_kuhn_tucker.m