451 lines
16 KiB
Markdown
451 lines
16 KiB
Markdown
<!-- vim: set ts=2 sw=2 et tw=80: -->
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---
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title: Midterm -- Optimization Methods
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author: Claudio Maggioni
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header-includes:
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- \usepackage{amsmath}
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- \usepackage{hyperref}
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- \usepackage[utf8]{inputenc}
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- \usepackage[margin=2.5cm]{geometry}
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- \usepackage[ruled,vlined]{algorithm2e}
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- \usepackage{float}
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- \floatplacement{figure}{H}
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---
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\maketitle
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# Exercise 1
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## Point 1
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### Question (a)
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As already covered in the course, the gradient of a standard quadratic form at a
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point $x_0$ is equal to:
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$$ \nabla f(x_0) = A x_0 - b $$
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Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b
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= 0$ (according to the first necessary condition for a minimizer), we obtain:
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$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
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\lambda v $$
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### Question (b)
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The steepest descent method takes exactly one iteration to reach the exact
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minimizer $x_m$ starting from the point $x_0$. This can be proven by first
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noticing that $x_m$ is a point standing in the line that first descent direction
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would trace, which is equal to:
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$$g(\alpha) = - \alpha \cdot \nabla f(x_0) = - \alpha \lambda v$$
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For $\alpha = \frac{1}{\lambda}$, and plugging in the definition of $x_0 = x_m +
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v$, we would reach a new iterate $x_1$ equal to:
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$$x_1 = x_0 - \alpha \lambda v = x_0 - v = x_m + v - v = x_m $$
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The only question that we need to answer now is why the SD algorithm would
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indeed choose $\alpha = \frac{1}{\lambda}$. To answer this, we recall that the
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SD algorithm chooses $\alpha$ by solving a linear minimization option along the
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step direction. Since we know $x_m$ is indeed the minimizer, $f(x_m)$ would be
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obviously strictly less that any other $f(x_1 = x_0 - \alpha \lambda v)$ with
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$\alpha \neq \frac{1}{\lambda}$.
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Therefore, since $x_1 = x_m$, we have proven SD
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converges to the minimizer in one iteration.
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## Point 2
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The right answer is choice (a), since the energy norm of the error indeed always
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decreases monotonically.
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To prove that this is true, we first consider a way to express any iterate $x_k$
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in function of the minimizer $x_s$ and of the missing iterations:
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$$x_k = x_s + \sum_{i=k}^{N} \alpha_i A^i p_0$$
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This formula makes use of the fact that step directions in CG are all
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A-orthogonal with each other, so the k-th search direction $p_k$ is equal to
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$A^k p_0$, where $p_0 = -r_0$ and $r_0$ is the first residual.
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Given that definition of iterates, we're able to express the error after
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iteration $k$ $e_k$ in a similar fashion:
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$$e_k = x_k - x_s = \sum_{i=k}^{N} \alpha_i A^i p_0$$
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We then recall the definition of energy norm $\|e_k\|_A$:
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$$\|e_k\|_A = \sqrt{\langle Ae_k, e_k \rangle}$$
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We then want to show that $\|e_k\|_A = \|x_k - x_s\|_A > \|e_{k+1}\|_A$, which
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in turn is equivalent to claim that:
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$$\langle Ae_k, e_k \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$
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Knowing that the dot product is linear w.r.t. either of its arguments, we pull
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out the sum term related to the k-th step (i.e. the first term in the sum that
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makes up $e_k$) from both sides of $\langle Ae_k, e_k \rangle$,
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obtaining the following:
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$$\langle Ae_{k+1}, e_{k+1} \rangle + \langle \alpha_k A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > \langle Ae_{k+1}, e_{k+1}
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\rangle$$
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which in turn is equivalent to claim that:
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$$\langle \alpha_k A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > 0$$
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From this expression we can collect term $\alpha_k$ thanks to linearity of the
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dot-product:
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$$\alpha_k (\langle A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1}, A^k p_0 \rangle) > 0$$
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and we can further "ignore" the $\alpha_k$ term since we know that all
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$\alpha_i$s are positive by definition:
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$$\langle A^{k+1} p_0, e_k \rangle
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+ \langle Ae_{k+1}, A^k p_0 \rangle > 0$$
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Then, we convert the dot-products in their equivalent vector to vector product
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form, and we plug in the definitions of $e_k$ and $e_{k+1}$:
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$$p_0^T (A^{k+1})^T (\sum_{i=k}^{N} \alpha_i A^i p_0) +
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p_0^T (A^{k})^T (\sum_{i=k+1}^{N} \alpha_i A^i p_0) > 0$$
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We then pull out the sum to cover all terms thanks to associativity of vector
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products:
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$$\sum_{i=k}^N (p_0^T (A^{k+1})^T A^i p_0) \alpha_i+ \sum_{i=k+1}^N
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(p_0^T (A^{k})^T A^i p_0) \alpha_i > 0$$
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We then, as before, can "ignore" all $\alpha_i$ terms since we know by
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definition that
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they are all strictly positive. We then recalled that we assumed that A is
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symmetric, so $A^T = A$. In the end we have to show that these two
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inequalities are true:
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$$p_0^T A^{k+1+i} p_0 > 0 \; \forall i \in [k,N]$$
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$$p_0^T A^{k+i} p_0 > 0 \; \forall i \in [k+1,N]$$
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To show these inequalities are indeed true, we recall that A is symmetric and
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positive definite. We then consider that if a matrix A is SPD, then $A^i$ for
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any positive $i$ is also SPD[^1]. Therefore, both inequalities are trivially
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true due to the definition of positive definite matrices.
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[^1]: source: [Wikipedia - Definite Matrix $\to$ Properties $\to$
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Multiplication](
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https://en.wikipedia.org/wiki/Definite_matrix#Multiplication)
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Thanks to this we have indeed proven that the delta $\|e_k\|_A - \|e_{k+1}\|_A$
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is indeed positive and thus as $i$ increases the energy norm of the error
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monotonically decreases.
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# Question 2
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## Point 1
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### (a) For which kind of minimization problems can the trust region method be
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used? What are the assumptions on the objective function?
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**TBD**
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### (b) Write down the quadratic model around a current iterate xk and explain the meaning of each term.
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$$m(p) = f + g^T p + \frac12 p^T B p \;\; \text{ s.t. } \|p\| < \Delta$$
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Here's an explaination of the meaning of each term:
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- $\Delta$ is the trust region radius, i.e. an upper bound on the step's norm
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(length);
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- $f$ is the energy function value at the current iterate, i.e. $f(x_k)$;
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- $p$ is the trust region step, the solution of $\arg\min_p m(p)$ with $\|p\| <
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\Delta$ is the optimal step to take;
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- $g$ is the gradient at the current iterate $x_k$, i.e. $\nabla f(x_k)$;
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- $B$ is the hessian at the current iterate $x_k$, i.e. $\nabla^2 f(x_k)$.
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### (c) What is the role of the trust region radius?
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The role of the trust region radius is to put an upper bound on the step length
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in order to avoid "overly ambitious" steps, i.e. steps where the the step length
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is considerably long and the quadratic model of the objective is low-quality
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(i.e. the quadratic model differs by a predetermined approximation threshold
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from the real objective).
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In layman's terms, the trust region radius makes the method switch more gradient
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based or more quadratic based steps w.r.t. the confidence in the quadratic
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approximation.
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### (d) Explain Cauchy point, sufficient decrease and Dogleg method, and the connection between them.
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**TBD**
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**sufficient decrease TBD**
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The Cauchy point provides sufficient decrease, but makes the trust region method
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essentially like linear method.
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The dogleg method allows for mixing purely linear iterations and purely quadratic
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ones. The dogleg method picks along its "dog leg shaped" path function made out
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of a gradient component and a component directed towards a purely Newton step
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picking the furthest point that is still inside the trust region radius.
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Dogleg uses cauchy point if the trust region does not allow for a proper dogleg
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step since it is too slow.
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Cauchy provides linear convergence and dogleg superlinear.
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### (e) Write down the trust region ratio and explain its meaning.
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$$\rho_k = \frac{f(x_k) - f(x_k + p_k)}{m_k(0) - m_k(p_k)}$$
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The trust region ratio measures the quality of the quadratic model built around
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the current iterate $x_k$, by measuring the ratio between the energy difference
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between the old and the new iterate according to the real energy function and
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according to the quadratic model around $x_k$.
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The ratio is used to test the adequacy of the current trust region radius. For
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an inaccurate quadratic model, the predicted energy decrease would be
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considerably higher than the effective one and thus the ratio would be low. When
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the ratio is lower than a predetermined threshold ($\frac14$ is the one chosen
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by Nocedal) the trust region radius is divided by 4. Instead, a very accurate
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quadratic model would result in little difference with the real energy function
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and thus the ratio would be close to $1$. If the trust region radius is higher
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than a certain predetermined threshold ($\frac34$ is the one chosen by Nocedal),
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then the trust region radius is doubled in order to allow for longer steps,
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since the model quality is good.
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### (f) Does the energy decrease monotonically when Trust Region method is employed? Justify your answer.
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**TBD**
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## Point 2
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The trust region algorithm is the following:
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\begin{algorithm}[H]
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\SetAlgoLined
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Given $\hat{\Delta} > 0, \Delta_0 \in (0,\hat{\Delta})$,
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and $\eta \in [0, \frac14)$\;
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\For{$k = 0, 1, 2, \ldots$}{%
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Obtain $p_k$ by using Cauchy or Dogleg method\;
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$\rho_k \gets \frac{f(x_k) - f(x_k + p_k)}{m_k(0) - m_k(p_k)}$\;
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\uIf{$\rho_k < \frac14$}{%
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$\Delta_{k+1} \gets \frac14 \Delta_k$\;
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}\Else{%
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\uIf{$\rho_k > \frac34$ and $\|\rho_k\| = \Delta_k$}{%
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$\Delta_{k+1} \gets \min(2\Delta_k, \hat{\Delta})$\;
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}
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\Else{%
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$\Delta_{k+1} \gets \Delta_k$\;
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}}
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\uIf{$\rho_k > \eta$}{%
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$x_{k+1} \gets x_k + p_k$\;
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}
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\Else{
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$x_{k+1} \gets x_k$\;
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}
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}
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\caption{Trust region method}
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\end{algorithm}
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The Cauchy point algorithm is the following:
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\begin{algorithm}[H]
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\SetAlgoLined
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Input $B$ (quadratic term), $g$ (linear term), $\Delta_k$\;
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\uIf{$g^T B g \geq 0$}{%
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$\tau \gets 1$\;
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}\Else{%
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$\tau \gets \min(\frac{\|g\|^3}{\Delta_k \cdot g^T B g}, 1)$\;
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}
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$p_k \gets -\tau \cdot \frac{\Delta_k}{\|g\|^2 \cdot g}$\;
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\Return{$p_k$}
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\caption{Cauchy point}
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\end{algorithm}
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Finally, the Dogleg method algorithm is the following:
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\begin{algorithm}[H]
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\SetAlgoLined
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Input $B$ (quadratic term), $g$ (linear term), $\Delta_k$\;
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$p_N \gets - B^{-1} g$\;
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\uIf{$\|p_N\| < \Delta_k$}{%
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$p_k \gets p_N$\;
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}\Else{%
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$p_u = - \frac{g^T g}{g^T B g} g$\;
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\uIf{$\|p_u\| > \Delta_k$}{%
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compute $p_k$ with Cauchy point algorithm\;
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}\Else{%
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solve for $\tau$ the equality $\|p_u + \tau * (p_N - p_u)\|^2 =
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\Delta_k^2$\;
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$p_k \gets p_u + \tau \cdot (p_N - p_u)$\;
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}
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}
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\caption{Dogleg method}
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\end{algorithm}
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## Point 3
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The trust region, dogleg and Cauchy point algorithms were implemented
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respectively in the files `trust_region.m`, `dogleg.m`, and `cauchy.m`.
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## Point 4
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### Taylor expansion
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The Taylor expansion up the second order of the function is the following:
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$$f(x_0, w) = f(x_0) + \langle\begin{bmatrix}48x^3 - 16xy + 2x - 2\\2y - 8x^2
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\end{bmatrix}, w\rangle + \frac12 \langle\begin{bmatrix}144x^2 -16y + 2 - 16 &
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-16 \\ -16 & 2 \end{bmatrix}w, w\rangle$$
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### Minimization
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The code used to minimize the function can be found in the MATLAB script
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`main.m` under section 2.4. The resulting minimizer (found in 10 iterations) is:
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$$x_m = \begin{bmatrix}1\\4\end{bmatrix}$$
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### Energy landscape
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The following figure shows a `surf` plot of the objective function overlayed
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with the iterates used to reach the minimizer:
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The code used to generate such plot can be found in the MATLAB script `main.m`
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under section 2.4c.
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## Point 5
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### Minimization
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The code used to minimize the function can be found in the MATLAB script
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`main.m` under section 2.5. The resulting minimizer (found in 25 iterations) is:
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$$x_m = \begin{bmatrix}1\\5\end{bmatrix}$$
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### Energy landscape
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The following figure shows a `surf` plot of the objective function overlayed
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with the iterates used to reach the minimizer:
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The code used to generate such plot can be found in the MATLAB script `main.m`
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under section 2.5b.
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### Gradient norms
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The following figure shows the logarithm of the norm of the gradient w.r.t.
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iterations:
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The code used to generate such plot can be found in the MATLAB script `main.m`
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under section 2.5c.
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Comparing the behaviour shown above with the figures obtained in the previous
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assignment for the Newton method with backtracking and the gradient descent with
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backtracking, we notice that the trust-region method really behaves like a
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compromise between the two methods. First of all, we notice that TR converges in
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25 iterations, almost double of the number of iterations of regular NM +
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backtracking. The actual behaviour of the curve is somewhat similar to the
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Netwon gradient norms curve w.r.t. to the presence of spikes, which however are
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less evident in the Trust region curve (probably due to Trust region method
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alternating quadratic steps with linear or almost linear steps while iterating).
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Finally, we notice that TR is the only method to have neighbouring iterations
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having the exact same norm: this is probably due to some proposed iterations
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steps not being validated by the acceptance criteria, which makes the method mot
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move for some iterations.
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# Exercise 3
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We first show that the lemma holds for $\tau \in [0,1]$. Since
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$$\|\tilde{p}(\tau)\| = \|\tau p^U\| = \tau \|p^U\| \text{ for } \tau \in [0,1]$$
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Then the norm of the step $\tilde{p}$ clearly increases monotonically as $\tau$
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increases. For the second criterion, we compute the quadratic model for a
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generic $\tau \in [0,1]$:
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$$m(\tilde{p}(\tau)) = f - \frac{\tau^2 \|g\|^2}{g^TBg} - \frac12 \frac{\tau^2
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\|g\|^2}{(g^TBg)^2} g^TBg = f - \frac12 \frac{\tau^2 \|g\|^2}{g^TBg}$$
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$g^T B g > 0$ since we assume $B$ is positive definite, therefore the entire
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term in function
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of $\tau^2$ is negative and thus the model for an increasing $\tau \in [0,1]$
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decreases monotonically (to be precise quadratically).
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Now we show that the monotonicity claims hold also for $\tau \in [1,2]$. We
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define a function $h(\alpha)$ (where $\alpha = \tau - 1$) with same monotonicity
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as $\|\tilde{p}(\tau)\|$ and we show that this function monotonically increases:
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$$h(\alpha) = \frac12 \|\tilde{p}(1 - \alpha)\|^2 = \frac12 \|p^U + \alpha(p^B -
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p^U)\|^2 = \frac12 \|p^U\|^2 + \frac12 \alpha^2 \|p^B - p^U\|^2 + \alpha (p^U)^T
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(p^B - p^U)$$
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We now take the derivative of $h(\alpha)$ and we show it is always positive,
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i.e. that $h(\alpha)$ has always positive gradient and thus that is it
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monotonically increasing w.r.t. $\alpha$:
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$$h'(\alpha) = \alpha \|p^B - p^U\|^2 + (p^U)^T (p^B - p^U) \geq (p^U)^T (p^B - p^U)
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= \frac{g^Tg}{g^TBg}g^T\left(- \frac{g^Tg}{g^TBg}g + B^{-1}g\right) =$$$$= \|g\|^2
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\frac{g^TB^{-1}g}{g^TBg}\left(1 - \frac{\|g\|^2}{(g^TBg)(g^TB^{-1}g)}\right) $$
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Since we know $B$ is symmetric and positive definite, then $B^{-1}$ is as well.
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Therefore, we know that the term outside of the parenthesis is always positive
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or 0. Therefore, we now only need to show that:
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$$\frac{\|g\|^2}{(g^TBg)(g^TB^{-1}g)} \leq 1 \Leftrightarrow \|g\|^2 \leq
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(g^TBg)(g^TB^{-1}g)$$
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since both factors in the denominator are positive for what we shown before.
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We now define a inner product space $\forall a, b \in R^N, \; {\langle a,
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b\rangle}_B = a^T B b$. We now prove that this is indeed a linear product space
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by proving all properties of such space:
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- **Linearity w.r.t. the first argument:**
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${\langle x, y \rangle}_B + {\langle z,
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y \rangle}_B = x^TBy + z^TBy = (x + z)^TBy = {\langle (x + z), y \rangle}_B$;
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- **Symmetry:**
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${\langle x, y \rangle}_B = x^T B y = (x^T B y)^T = y^TB^Tx$, and
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since $B$ is symmetric, $y^TB^Tx = y^TBx = {\langle y,x \rangle}_B$;
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- **Positive definiteness:**
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${\langle x, x \rangle_B} = x^T B x > 0$ is true since B is positive definite.
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Since ${\langle x, y \rangle}_B$ is indeed a linear product space, then:
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$${\langle g, B^{-1} g \rangle}_B \leq {\langle g, g \rangle}_B, {\langle B^{-1}
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g, B^{-1} g \rangle}_B$$
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holds according to the Cauchy-Schwartz inequality. Now, if we expand each inner
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product we obtain:
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$$g^T B B^{-1} g \leq (g^T B g) (g^T (B^{-1})^T B B^{-1} g)$$
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Which, since $B$ is symmetric, in turn is equivalent to writing:
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$$g^T g \leq (g^TBg) (g^T B^{-1} g)$$
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which is what we needed to show to prove that the first monotonicity constraint
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holds for $\tau \in [1,2]$.
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**TBD**
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