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