-
Notifications
You must be signed in to change notification settings - Fork 2
/
report.tex
338 lines (288 loc) · 11.9 KB
/
report.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
\documentclass[12pt]{article}
% Any percent sign marks a comment to the end of the line
% Every latex document starts with a documentclass declaration like this
% The option dvips allows for graphics, 12pt is the font size, and article
% is the style
\usepackage[pdftex]{graphicx}
\usepackage{amsfonts}
\usepackage{amsmath}
\DeclareMathOperator*{\max_bottom}{max}
\usepackage{url}
\usepackage{hyperref}
\usepackage{caption}
\usepackage{subcaption}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{adjustbox}
\usepackage{listings}
\usepackage{amsmath}
\lstset{language=python}
\hypersetup{
colorlinks=true,
linkcolor=blue,
filecolor=magenta,
urlcolor=cyan,
pdftitle={Sharelatex Example},
bookmarks=true,
pdfpagemode=FullScreen,
}
\usepackage{graphicx}
\graphicspath{ {./images/} }
% These are additional packages for "pdflatex", graphics, and to include
% hyperlinks inside a document.
\setlength{\oddsidemargin}{0.5cm}
\setlength{\evensidemargin}{0.5cm}
\setlength{\topmargin}{-1.6cm}
\setlength{\leftmargin}{0.5cm}
\setlength{\rightmargin}{0.5cm}
\setlength{\textheight}{24.00cm}
\setlength{\textwidth}{15.00cm}
\parindent 0pt
\parskip 5pt
\pagestyle{plain}
% These force using more of the margins that is the default style
\newcommand{\namelistlabel}[1]{\mbox{#1}\hfil}
\newenvironment{namelist}[1]{%1
\begin{list}{}
{
\let\makelabel\namelistlabel
\settowidth{\labelwidth}{#1}
\setlength{\leftmargin}{1.1\labelwidth}
}
}{%1
\end{list}}
\begin{document}
\title{\Huge Introduction to Optimization - HW 4}
\author{
\textbf{Uri Kirstein}\\
311137095 \\ sukirstn@campus.technion.ac.il
\\ \\
\textbf{Pavel Rastopchin}\\
321082026 \\ pavelr@campus.technion.ac.il
\\ \\
}
\maketitle
\newpage
\section{Task 1, Q1}
\subsection{Problem}
\[\min_x x^T M x + c^T x\]
\[s.t: Ax - b = 0\]
\[i.e: h(x)= Ax - b = 0\]
\[M \succ 0\]
\[i.e: M^T=M\]
\subsection{Solution}
\paragraph{Writing Lagrangian}
\[L(x, \mu)=f(x) + \mu ^T h(x) = x^T M x + c^T x + \mu ^T A x - \mu ^ T b\]
\paragraph{Finding Gradient of Lagrangian}
\[dL(x, \mu)= dx^T Mx + x^T Mdx + c^T dx + \mu ^T Adx \]
\[(dx^T Mx)\ is \ a \ scalar, \ thus: (dx^T Mx) = (dx^T Mx)^T \ thus:\]
\[dL(x, \mu) = x^T M^T dx + x^T M dx + c^T dx + \mu ^T Adx \]
\[dL(x, \mu)= (x^T M^T + x^T M + c^T + \mu ^T A)dx \]
\[dL(x, \mu)= (x^T M^T + x^T M + c^T + \mu ^T A)dx \]
\[g^T = \nabla_x L^T = (x^T M^T + x^T M + c^T + \mu ^T A)\]
\[\nabla_x L = (x^T M^T + x^T M + c^T + \mu ^T A)^T\]
\[\nabla_x L = Mx + M^T x + c + A^T \mu\]
\[M \ is \ symetric, \ thus: Mx + M^T x\ thus:\]
\[\nabla_x L = 2Mx + c + A^T \mu\]
\paragraph{KKT}
\[\nabla_x L = 2Mx + c + A^T \mu = 0 \]
\paragraph{Writing x}
\[2Mx + c + A^T \mu = 0 \]
\[2Mx = - c - A^T \mu\]
\[x = - \frac{1}{2} M^{-1}c - \frac{1}{2} M^{-1}A^T \mu\]
\paragraph{Substitute x into constraint}
\[recall\ that: Ax - b = 0\ thus:\]
\[A(- \frac{1}{2} M^{-1}c - \frac{1}{2} M^{-1}A^T \mu) - b = 0 \]
\[- \frac{1}{2} AM^{-1}c - \frac{1}{2} AM^{-1}A^T \mu - b = 0 \]
\paragraph{Writing $\mu$}
\[\frac{1}{2} AM^{-1}A^T \mu = - \frac{1}{2} AM^{-1}c - b \]
\[AM^{-1}A^T \mu = - AM^{-1}c - 2b \]
\[ \mu = (AM^{-1}A^T)^{-1}(- AM^{-1}c - 2b)\]
\[ \mu = -(AM^{-1}A^T)^{-1}(AM^{-1}c + 2b)\]
\paragraph{Substitute $\mu$ into x}
\[recall\ that: x = - \frac{1}{2} M^{-1}c - \frac{1}{2} M^{-1}A^T \mu \ \ thus:\]
\[x = - \frac{1}{2} M^{-1}c - \frac{1}{2} M^{-1}A^T (-(AM^{-1}A^T)^{-1}(AM^{-1}c + 2b))\]
\paragraph{Final answer}
\[x = - \frac{1}{2} M^{-1}c + \frac{1}{2} M^{-1}A^T (AM^{-1}A^T)^{-1}(AM^{-1}c + 2b)\]
\newpage
\section{Task 1, Q2.1}
\subsection{Problem}
\[min_x ||x-c||_2 ^2 = min_x (x-c)^T (x-c) = x^T x - x^T c - c^T x + c^T c\]
\[h(x)= Ax -b = 0 \]
\subsection{Solution}
\paragraph{Lagrangian}
\[L(x,\mu)=f(x) + \mu ^T h(x) = x^T x - x^T c - c^T x + c^T c + \mu ^T Ax - \mu ^T b\]
\[dL = dx^T x + x^T dx - dx^T c - c^T dx + \mu ^T Adx \]
\[those \ are \ scalars: dx^T x \ \ and \ \ dx^T c \ \ thus: \]
\[dx^T x = x^T dx \ , \ dx^T c = c^T dx \ thus: \]
\[dL = x^T dx + x^T dx - c^T dx - c^T dx + \mu ^T Adx \]
\[dL = 2x^T dx - 2c^T dx + \mu ^T Adx \]
\[dL = (2x^T - 2c^T + \mu ^T A)dx \]
\[\nabla_x L = (2x^T - 2c^T + \mu ^T A)^T = 0 \]
\[\nabla_x L = 2x - 2c + A^T \mu = 0 \]
\[x= c - \frac{1}{2} A^T \mu \]
\paragraph{Substitute x into constraint}
\[A(c- \frac{1}{2} A^T \mu) = b\]
\[Ac - \frac{1}{2} AAT^ \mu = b\]
\[AA^T \mu = 2Ac - 2b\]
\[\mu = 2 (AA^T)^{-1} (Ac - b)\]
\paragraph{Substitute $\mu$ into x}
\[x = c - \frac{1}{2} A^T ( 2 ( AA^T )^{-1} ( Ac - b ) ) \]
\[x = c - A^T (AA^T)^{-1} (Ac - b)\]
\newpage
\section{Task 1, Q2.2}
\subsection{Problem}
\[min_x x^T A x + b^T x = min_x x_1 ^2 + 2 x_1 x_2 + 2 x_2 ^2 + x_1 - x_2 + x_3 \]
\[h(x)= x_1 + x_2 + x_3 - 1 = 0\]
\[g(x)= x_3 - 1 \leq 0 \]
\subsection{Is the problem is convex?}
The problem is convex if objective function is convex and all constraints are convex. If matrix $A$ is positive definite or positive semi-definite, then the objective is convex. Constraints are liner, so they are convex. Let's eigenvalues of $A$:
\[|A- \lambda I | = (1- \lambda)(2 -\lambda)(-\lambda)+ \lambda = 0 \]
\[\lambda (- \lambda ^2 + 3 \lambda - 1 = 0\]
\[\lambda_{1,2} = \frac{+3 \pm \sqrt{5}}{2} > 0 \ , \ \lambda_3 = 0 \ \ thus: \]
\[A \succeq 0 \longrightarrow the \ problem \ is \ convex \]
\subsection{Dual problem}
\paragraph{Minimization of L in x}
\[L(x, \lambda, \mu) = x_1 ^2 + 2 x_1 x_2 + 2 x_2 ^2 + x_1 - x_2 + x_3 + \lambda (x_3 - 1) + \mu (x_1 + x_2 + x_3 - 1)\]
\[\frac{\delta L}{\delta x_1} = 2x_1 + 2x_2 + 1 + \mu = 0 \]
\[\frac{\delta L}{\delta x_2} = 2x_1 + 4x_2 - 1 + \mu = 0 \]
\[\frac{\delta L}{\delta x_3} = 1 + \lambda + \mu = 0\]
Solving equations system and denoting $\mu = -1 -\lambda$ we get:
\[x_1 = \frac{1}{2} \lambda -1 \ , \ x_2 =1 \ , \ x_3 = 1 - \frac{1}{2} \lambda\]
\paragraph{Substitute x into L to get the dual problem}
\[\eta(\lambda , \mu) = \min_x L(x, \lambda, \mu) = L(x^*, \lambda, \mu) = \]
\[= (\frac{1}{2} \lambda - 1)^2 +2(\frac{1}{2} \lambda - 1)+2+(\frac{1}{2} \lambda - 1)-1+(1 - \frac{1}{2} \lambda)+ \lambda(1 - \frac{1}{2} - 1) + \mu(\frac{1}{2} \lambda - 1 + 1 + 1 - \frac{1}{2} \lambda - 1) = \]
\[= - \frac{1}{4} \lambda ^2 \longrightarrow \eta(\lambda , \mu) = - \frac{1}{4} \lambda ^2\]
\paragraph{Maximization of the dual problem}
Instead of maximizing, we can minimize negative dual problem.
\[ \min_{\lambda \geq 0} -\eta(\lambda , \mu) = \min_{\lambda \geq 0} \frac{1}{4} \lambda ^2\]
\[\frac{\delta \eta}{\delta \lambda} = \frac{1}{2} \lambda = 0 \longrightarrow \lambda=0 \]
\[recall: \ \mu = -1 - \lambda \longrightarrow \mu = -1\]
\[x_1 = \frac{1}{2} \lambda -1 = 0 \ , \ x_2 =1 \ , \ x_3 = 1 - \frac{1}{2} \lambda\ = 1\]
\newpage
\section{Task 1, Q3}
\subsection{Problem}
\[min_x c^T x\]
\[s.t. \ g_1 (x) = Ax-b \leq 0 \]
\[s.t. \ g_2 (x) = -x \leq 0 \]
\paragraph{Minimization of L in x}
\[L(x,\lambda_1, \lambda_2) = c^T x + \lambda_1 ^T (Ax-b) + \lambda_2 ^T (-x)\]
\[L(x,\lambda_1, \lambda_2) = c^T x + \lambda_1 ^T Ax - \lambda_1 ^T b - \lambda_2 ^T x\]
\[dL = c^T dx + \lambda_1 ^T A dx - \lambda_2 ^ T dx \]
\[dL = (c^T + \lambda_1 ^T A - \lambda_2 ^T)dx \]
\[\nabla_x L = (c^T + \lambda_1 ^T A - \lambda_2 ^T)^T = c+A^T \lambda_1 - \lambda_2 =0\]
\[c=\lambda_2 - A^T \lambda_1\]
\paragraph{Substitute c into L to get the dual function}
\[L(x,\lambda_1, \lambda_2) = (\lambda_2 - A^T \lambda_1)^T x + \lambda_1 ^T (Ax-b) + \lambda_2 ^T (-x) = -\lambda_1 ^T b\]
\[
\eta(\lambda_1 , \lambda_2) =\left\{
\begin{array}{ll}
-\lambda_1 ^T b \ \ ; \ c=\lambda_2 - A^T \lambda_1 \\
- \inf \ \ ; \ otherwise
\end{array}
\right.
\]
\paragraph{Dual problem}
Instead of maximizing we can minimize negative dual function.
\[\min_{\lambda \geq 0} \lambda_1 ^T b\]
\[s.t. \ c=\lambda_2 - A^T \lambda_1\]
\newpage
\section{Task 2}
\subsection{Problem}
\[\min_x 2(x_1-5)^2+(x_2-1)^2\]
\[s.t \ \ x_2 \leq 1- \frac{x_1}{2} \longrightarrow g_1(x)=x_2-1
+\frac{x_1}{2} \leq 0 \]
\[x_2 \geq x_1 \longrightarrow g_2(x)=x_1-x_2 \leq 0 \]
\[x_2 \geq - x_1 \longrightarrow g_3(x)=-x_1-x_2 \leq 0 \]
\subsection{Feasible area}
\begin{figure}[h]
\centering
\includegraphics[width=0.7\textwidth]{pics/feas}
\end{figure}
Active constrains are red and blue lines. which is $g_1$ and $g_2$, thus $\lambda_3 = 0$.
\subsection{Optimal solution at the intersection of the active constraints}
\[x_1=x_2 , x_2=1-\frac{x_1}{2} \longrightarrow x_2 = \frac{2}{3}, x_1=\frac{2}{3} \]
\[f(x_1,x_2) = 37 \frac{2}{3} \]
\subsection{Calculate the Lagrange multipliers}
\[\L=2(x_1-5)^2+(x_2-1)^2 +\lambda_1(x_2-1
+\frac{x_1}{2}) +\lambda_2(x_1-x_2) \]
\[\frac{\delta L}{\delta x_1} = 4 x_1 -20 + \frac{1}{2} \lambda_1 + \lambda_2 = 0 \longrightarrow x_1^* = \frac{20-\frac{1}{2}\lambda_1-\lambda_2}{4} \]
\[\frac{\delta L}{\delta x_2} = 2x_2-2+\lambda_1-\lambda_2=0 \longrightarrow x_2^*=\frac{2-\lambda_1+\lambda_2}{2} \]
\[by \ \ KKT: \lambda_i g_i(x^*)=0 \]
\[
\left\{
\begin{array}{ll}
-9\lambda_1+6\lambda_2+40=0 \\
3\lambda_1-6\lambda_2+32=0
\end{array}
\right.
\]
\[\lambda_1=12 \ , \ \lambda_2=11\frac{1}{3} \]
\subsection{Optimal value of objective function}
\[
\left\{
\begin{array}{ll}
x_1^* = \frac{20-\frac{1}{2}\lambda_1-\lambda_2}{4} \longrightarrow x_1^* = \frac{2}{3}\\
x_2^*=\frac{2-\lambda_1+\lambda_2}{2} \longrightarrow x_2^* = \frac{2}{3}
\end{array}
\right.
\]
\[f(x_1^*,x_2^*)=37 \frac{2}{3} \]
\subsection{Dual problem}
\[\eta(\lambda_1, \lambda_2) = 2(-\frac{1}{8}\lambda_1-\frac{1}{4}\lambda_2)^2 + (-\frac{1}{2}\lambda_1+\frac{1}{2}\lambda_2)^2 + \lambda_1 (-\frac{9}{16}\lambda_1 +\frac{3}{8}\lambda_2+\frac{5}{2})+\lambda_2 (4+\frac{3}{8}\lambda_1-\frac{3}{4}\lambda_2) \]
\[\min_{\lambda \geq 0} -\eta(\lambda_1, \lambda_2) \]
\subsection{Substitute the obtained in Q.3 Lagrange multipliers}
Is the dual optimum achieved at that point? Yes.
\[\eta(\lambda_1^*, \lambda_2^*) = 37 \frac{2}{3} \]
\newpage
\section{Wet part}
\subsection{Penalty method}
\subparagraph{Lagrangian} In order to implement our code properly we decided to write all expressions we will need during the work.
\[L_{p\mu}(x) = f(x) + \mu_i \varphi_{p }(g_{i(x)}) + ... \]
\[\nabla_x L_{p\mu}(x) = \nabla_x f(x) + \mu_i \varphi_{p }^{'} (g_{i(x)}) \nabla_x g_{i(x)} + ... \]
\[\nabla_x^2 L_{p\mu}(x) = \nabla_x^2 f(x) + \mu_i \varphi_{p}^{''} (g_{i(x)}) \nabla_x g_{i(x)}\nabla_x^T g_{i(x)} + \mu_i \varphi_{p}^{'} (g_{i(x)}) \nabla_x^2 g_{i(x)} + ... = \]
\[\nabla_x^2 L_{p\mu}(x) = \nabla_x^2 f(x) + \mu_i \varphi_{p}^{''} (g_{i(x)}) \nabla_x g_{i(x)}\nabla_x^T g_{i(x)} + 0 + ... \]
\subparagraph{Penalty Function}
\[
\varphi_{p}(g_{(x)}) =\left\{
\begin{array}{ll}
\frac{1}{p}(\frac{p^2 g_{(x)}^2}{2} + pg_{(x)}) \ ; \ pg_{(x)} \geq -\frac{1}{2} \\
\frac{1}{p} [-\frac{1}{4} \log(-2pg_{(x)})-\frac{3}{8}] \ ; \ pg_{(x)} -\frac{1}{2}
\end{array}
\right.
\]
\[
\varphi_{p}^{'}(g_{(x)}) =\left\{
\begin{array}{ll}
(pg_{(x)}+1) \ ; \ pg_{(x)} \geq -\frac{1}{2} \\
-\frac{1}{4pg_{(x)}}\ ; \ pg_{(x)} < -\frac{1}{2}
\end{array}
\right.
\]
\[
\varphi_{p}^{''}(g_{(x)}) =\left\{
\begin{array}{ll}
p \ ; \ pg_{(x)} \geq -\frac{1}{2} \\
\frac{1}{4pg_{(x)}^2}\ ; \ pg_{(x)} < -\frac{1}{2}
\end{array}
\right.
\]
\subsection{Results}
We got same results as in analytic solution:
\begin{verbatim}
The solution is:
[[0.66666667]
[0.66666667]]
Lagrange multipliers are:
[array([12.00000029]), array([11.33333356]), array([2.82832242e-22])]
\end{verbatim}
\newpage
\subsection{Plots}
\begin{figure}[h!]
\centering
\includegraphics[width=.4\textwidth]{pics/Augmented_Lagrangian_gradient_fig}
\includegraphics[width=.4\textwidth]{pics/Convergence_rate_fig}
\includegraphics[width=.4\textwidth]{pics/Maximal_constraint_violation_fig}
\includegraphics[width=.4\textwidth]{pics/Optimal_point_and_optimal_multipliers_Distance_fig}
\end{figure}
\end{document}