template_HCMUT
Auteur:
Chìu Tuấn Bình
Last Updated:
il y a 7 ans
License:
Creative Commons CC BY 4.0
Résumé:
Template for HCMUT Report
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
Template for HCMUT Report
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[a4paper]{article}
\usepackage{vntex}
%\usepackage[english,vietnam]{babel}
%\usepackage[utf8]{inputenc}
%\usepackage[utf8]{inputenc}
%\usepackage[francais]{babel}
\usepackage{a4wide,amssymb,epsfig,latexsym,multicol,array,hhline,fancyhdr}
\usepackage{booktabs}
\usepackage{amsmath}
\usepackage{lastpage}
\usepackage[lined,boxed,commentsnumbered]{algorithm2e}
\usepackage{enumerate}
\usepackage{color}
\usepackage{graphicx} % Standard graphics package
\usepackage{array}
\usepackage{tabularx, caption}
\usepackage{multirow}
\usepackage[framemethod=tikz]{mdframed}% For highlighting paragraph backgrounds
\usepackage{multicol}
\usepackage{rotating}
\usepackage{graphics}
\usepackage{geometry}
\usepackage{setspace}
\usepackage{epsfig}
\usepackage{tikz}
\usepackage{listings}
\usetikzlibrary{arrows,snakes,backgrounds}
\usepackage{hyperref}
\hypersetup{urlcolor=blue,linkcolor=black,citecolor=black,colorlinks=true}
%\usepackage{pstcol} % PSTricks with the standard color package
\newtheorem{theorem}{{\bf Định lý}}
\newtheorem{property}{{\bf Tính chất}}
\newtheorem{proposition}{{\bf Mệnh đề}}
\newtheorem{corollary}[proposition]{{\bf Hệ quả}}
\newtheorem{lemma}[proposition]{{\bf Bổ đề}}
\everymath{\color{blue}}
%\usepackage{fancyhdr}
\setlength{\headheight}{40pt}
\pagestyle{fancy}
\fancyhead{} % clear all header fields
\fancyhead[L]{
\begin{tabular}{rl}
\begin{picture}(25,15)(0,0)
\put(0,-8){\includegraphics[width=8mm, height=8mm]{hcmut.png}}
%\put(0,-8){\epsfig{width=10mm,figure=hcmut.eps}}
\end{picture}&
%\includegraphics[width=8mm, height=8mm]{hcmut.png} & %
\begin{tabular}{l}
\textbf{\bf \ttfamily Trường Đại Học Bách Khoa Tp.Hồ Chí Minh}\\
\textbf{\bf \ttfamily Khoa Khoa Học và Kỹ Thuật Máy Tính}
\end{tabular}
\end{tabular}
}
\fancyhead[R]{
\begin{tabular}{l}
\tiny \bf \\
\tiny \bf
\end{tabular} }
\fancyfoot{} % clear all footer fields
\fancyfoot[L]{\scriptsize \ttfamily Bài tập lớn môn Mô Hình Hoá Toán Học - Niên khóa 2016-2017}
\fancyfoot[R]{\scriptsize \ttfamily Trang {\thepage}/\pageref{LastPage}}
\renewcommand{\headrulewidth}{0.3pt}
\renewcommand{\footrulewidth}{0.3pt}
%%%
\setcounter{secnumdepth}{4}
\setcounter{tocdepth}{3}
\makeatletter
\newcounter {subsubsubsection}[subsubsection]
\renewcommand\thesubsubsubsection{\thesubsubsection .\@alph\c@subsubsubsection}
\newcommand\subsubsubsection{\@startsection{subsubsubsection}{4}{\z@}%
{-3.25ex\@plus -1ex \@minus -.2ex}%
{1.5ex \@plus .2ex}%
{\normalfont\normalsize\bfseries}}
\newcommand*\l@subsubsubsection{\@dottedtocline{3}{10.0em}{4.1em}}
\newcommand*{\subsubsubsectionmark}[1]{}
\makeatother
\definecolor{dkgreen}{rgb}{0,0.6,0}
\definecolor{gray}{rgb}{0.5,0.5,0.5}
\definecolor{mauve}{rgb}{0.58,0,0.82}
\lstset{frame=tb,
language=Matlab,
aboveskip=3mm,
belowskip=3mm,
showstringspaces=false,
columns=flexible,
basicstyle={\small\ttfamily},
numbers=none,
numberstyle=\tiny\color{gray},
keywordstyle=\color{blue},
commentstyle=\color{dkgreen},
stringstyle=\color{mauve},
breaklines=true,
breakatwhitespace=true,
tabsize=3,
numbers=left,
stepnumber=1,
numbersep=1pt,
firstnumber=1,
numberfirstline=true
}
\begin{document}
\begin{titlepage}
\begin{center}
ĐẠI HỌC QUỐC GIA THÀNH PHỐ HỒ CHÍ MINH \\
TRƯỜNG ĐẠI HỌC BÁCH KHOA \\
KHOA KHOA HỌC - KỸ THUẬT MÁY TÍNH
\end{center}
\vspace{1cm}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=3cm]{hcmut.png}
\end{center}
\end{figure}
\vspace{1cm}
\begin{center}
\begin{tabular}{c}
\multicolumn{1}{l}{\textbf{{\Large MÔ HÌNH HOÁ TOÁN HỌC}}}\\
~~\\
\hline
\\
\multicolumn{1}{l}{\textbf{{\Large Xử lý tối ưu bài toán}}}\\
\\
\textbf{{\Huge Hospitals \& Residents trên Matlab}}\\
\\
\hline
\end{tabular}
\end{center}
\vspace{3cm}
\begin{table}[h]
\begin{tabular}{rrl}
\hspace{5 cm} & GVHD: &Lê Hồng Trang\\
& SV: & Chìu Tuấn Bình - 1510221\\
& & Mai Đức Tú - 1513924 \\
& & Phồng Quang Tuấn - 1513865\\
& & Lê Duy Hiển - 1511057 \\
& & Nguyễn Đỗ Đức Anh - 1510062\\
\end{tabular}
\end{table}
\begin{center}
{\footnotesize TP. HỒ CHÍ MINH, THÁNG 4/2017}
\end{center}
\end{titlepage}
\thispagestyle{empty}
\newpage
\tableofcontents
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{mdframed}[hidealllines=true,backgroundcolor=magenta!10]
\begin{lstlisting}
% ------------------------------- %
% XOA MAN HINH VA CAC BIEN %
% ------------------------------- %
clear
clc
% ------------------------------- %
% NHAP DU LIEU BAI TOAN %
% ------------------------------- %
n = ...; % So nguoi dan
m = ...; % So benh vien
% Ma tran D bieu dien thu tu uu tien cua benh vien doi voi benh nhan
% ung voi tung hang
D = [...];
% Ma tran B bieu dien thu tu uu tien cua benh nhan doi voi benh vien
% ung voi tung cot
B = [...];
% Ma tran c bieu dien suc chua cua tung benh vien
c = [...];
% Ma tran a bieu dien moi benh nhan chi duoc chon lua mot benh vien
a = ones(n,1);
% ------------------------------- %
% GIAI BAI TOAN BANG SOLVER MOSEK %
% ------------------------------- %
cvx_begin
cvx_solver mosek
% Bien x(i,j) chi nhan gia tri 0 hoac 1
% ung voi su ghep goi benh nhan r_i voi benh vien h_j
variable x(n,m) binary
% Toi da tong cac bien x(i,j)
% tuc la cang nhieu cap duoc ghep doi cang tot
maximize( 0 )
subject to
% Tong cac hang trong cung mot cot (so benh nhan duoc chon)
% nho hon hoac bang suc chua cua benh vien
sum(x,1) <= c;
% Tong cac cot trong cung mot cot (so benh vien duoc chon)
% nho hon hoac bang 1
sum(x,2) <= a;
% Bao dam khong co cac cap chan
for u = 1:n
for v = 1:m
%Tinh so hang dau tien trong ham dieu kien on dinh
t1 = 0;
for j = 1:m
t1 = t1 + lt(D(u,j),D(u,v)) * x(u,j);
end
%Tinh so hang thu hai trong ham dieu kien on dinh
t2 = 0;
for i = 1:n
t2 = t2 + lt(B(i,v),B(u,v)) * x(i,v) / c(v);
end
%Xac lap ham dieu kien on dinh
t1 + t2 + x(u,v) >= 1;
%Ham dam bao cac cap (r_u,h_v) duoc xet nam trong A, neu
%cap do khong nam trong A thi x_uv = 0
if D(u,v) == m+n+1 || B(u,v) == m+n+1
(eq(D(u,v),m+n+1) + eq(B(u,v),m+n+1)) * x(u,v) == 0;
end
end
end
cvx_end
% ------------------------------- %
% HIEN THI KET QUA RA MAN HINH %
% ------------------------------- %
D
B
c
x % Cac cap duoc ghep doi
\end{lstlisting}
\end{mdframed}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{thebibliography}{80}
\bibitem{CVX}
CVX Introduction
``\textbf{link: http://cvxr.com/cvx/doc/intro.html/}'',
\textit{What is CVX}, lần truy cập cuối: 15/04/2017.
\end{thebibliography}
\end{document}