Latest revision as of 11:21, 25 March 2015

QE2013_AC-3_ECE580-4

Part 1,2,3,4,5

Solution 1:
$ x(3) = x(2) + 2u(2) = x(1) + 2u(1) + 2u(2) = x(0) + 2u(0) + 2u(1) + 2u(2) $

$ \because x(0) = 3, x(3) = 9 $

$ \therefore \ Constraint: 2u(0) + 2u(1) + 2u(2) = 6 $

Let $ f(u) = u^2(0) + u^2(1) + u^2(2), h(u) = 2u(0) + 2u(1) + 2u(2) - 6 $ (1/2 in the objective function can be ignored for now)

Let u* be a local minimizer. Lagrange theorem says there exists a λ such that:

$ \nabla f(u^*) + \lambda \nabla h(u^*) = 0 \\ h(u^*) = 0 $

Therefore,

$ 2 u^*(0) + 2 \lambda = 0 \\ 2 u^*(1) + 2 \lambda = 0 \\ 2 u^*(2) + 2 \lambda = 0 \\ 2 u^*(0) + 2 u^*(1) + 2 u^*(2) - 6 = 0 \\ \\ \therefore u^*(0) = u^*(1) = u^*(2) = 1 $

Optimal performance index is $ \frac{3}{2} $

Solution 2:

$ J = \frac{1}{2} u^T u \\ x(k+1) = x(k) + 2u(k), x(0) = 3, x(3) = 9 \\ x(1) = x(0) + 2u(0) \\ x(2) = x(1) + 2u(1) = x(0) + 2u(0) + 2u(1) \\ x(3) = x(2) + 2u(2) = x(0) + 2u(0) + 2u(1) + 2u(2) \\ \therefore 2u(0) + 2u(1) + 2u(2) = 9 - 3 = 6 $

$ Au = 6 $ where A = [2 2 2]

$ min\ \frac{1}{2} u^T u \\ subject\ to\ Au = 6 \\ $

$ f(u) = \frac{1}{2} u^T u \\ h(u) = Au - 6 \\ l(u, \lambda) = f(u) + \lambda h(u) = \frac{1}{2} u^T u + \lambda Au - 6 \lambda \\ D_u l(u, \lambda) = u^T + \lambda A = \begin{bmatrix} u(0) + 2\lambda & u(1) + 2\lambda & u(2) + 2\lambda \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \end{bmatrix} \\ D_\lambda l(u, \lambda) = Au - 6 = 2u(0) + 2u(1) + 2u(2) - 6 = 0 \\ 2(-2 \lambda) + 2(-2 \lambda) + 2(-2 \lambda) - 6 = 12 \lambda - 6 = 0 \Rightarrow \lambda^* = -\frac{1}{2} \\ u^*(0) = u^*(1) = u^*(2) = -2\lambda = 1 \\ L(u, \lambda) = F(u) + \lambda H(u) = 1 > 0 \\ $

$ \therefore u^* = \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}^T $is a minimizer

Comment: Solution 1 and 2 are the same. Solution 1 skips a few steps with formulas while Solution 2 does not. They both have the same results.

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