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EXAM QUESTIONS AND SOLUTIONS FOR OPTIMISATION
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B1 Optimization – SolutionsA. Zisserman, Michaelmas Term 2018
1. The Rosenbrock function is
f (x, y) = 100(y − x2 )2 + (1 − x)2
(a) Compute the gradient and Hessian of f (x, y).
(b) Show that that f (x, y) has zero gradient at the point (1, 1).
(c) By considering the Hessian matrix at (x, y) = (1, 1), show that this point is a mini-
mum.
(a) Gradient and Hessian
400x3 − 400xy + 2x − 2 1200x2 − 400y + 2 −400x
∇f = H=
200(y − x2 ) −400x 200
(b) gradient at the point (1, 1)
400x3 − 400xy + 2x − 2
0
∇f = 2 =
200(y − x ) 0
(b) Hessian at the point (1, 1)
1200x2 − 400y + 2 −400x
802 −400
H= =
−400x 200 −400 200
Examine eigenvalues:
• det is positive, so eigenvalues have same sign (thus not saddle point)
• trace is positive, so eigenvalues are positive
• Thus a minimum
• λ1 = 1001.6006, λ2 = 0.39936077
1
, 2. In Newton type minimization schemes the update step is of the form
δx = −H−1 g
where g = ∇f . By considering g.δx compare convergence of:
(a) Newton, to
(b) Gauss Newton
for a general function f (x) (i.e. where H may not be positive definite).
A note on positive definite matrices
An n × n symmetric matrix M is positive definite if
• x> Mx > 0 for all non-zero vectors x
• All the eigen-values of M are positive
In each case consider df = g.δx. This should be negative for convergence.
(a) Newton
g.δx = −g> H−1 g
Can be positive if H not positive definite.
(b) Gauss Newton
g.δx = −g> (2J> J)−1 g
Non positive, since J> J is positive definite.
2
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