QE2013_AC-3_ECE580-4

Part 1,2,3,4,5

Solution:
$ 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} $

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