You can not nest longtable
: just use longtable
for the outer table, and tabular
for the inner ones.
\documentclass[a4paper]{article}
\usepackage{longtable}
\begin{document}
\begin{longtable} {|c|c|c|c|}
\hline
\# & Actividades & Objetivos &
\begin{tabular}[c]{@{}c@{}}Duracion\\ Estimada\end{tabular} \\ \hline
1 & \begin{tabular}[c]{@{}c@{}}Redaccion del documento de \\ Trabajo de Grado\end{tabular} & \multicolumn{1}{l|}{} & 12 semanas \\ \hline
2 & \begin{tabular}[c]{@{}c@{}}Redacci\'on\\ del documento de\\ Trabajo Especial de Grado\end{tabular} & Todos & 23 semanas \\ \hline
3 & \begin{tabular}[c]{@{}c@{}}Realizar un estudio de los requisitos \\ existentes en los horarios del\\ Departamento de computaci\'on\end{tabular} & 1.1 & 2 semanas \\ \hline
\multicolumn{1}{|l|}{4} & \begin{tabular}[c]{@{}c@{}}Construir\\ la representaci\'on \\ adecuada de un horario de clases \\ del departamento\\ de computaci\'on\end{tabular} & 1.1 & 2 semanas \\ \hline
5 & \begin{tabular}[c]{@{}c@{}}Definir\\ las estructuras necesarias \\ para guardar la informaci\'on de \\ todos los\\ elementos involucrados\\ en la asignaci\'on de horarios\end{tabular} & 1.1 y 1.2 & 2 semanas \\ \hline
6 & \begin{tabular}[c]{@{}c@{}}Codificar\\ ambas metaheur\'isticas \\ usando un lenguaje de \\ programaci\'on de alto\\ nivel adecuado\end{tabular} & \begin{tabular}[c]{@{}c@{}}1.2, 1.3 y\\ 1.4\end{tabular} & 5 semanas \\ \hline
7 & \begin{tabular}[c]{@{}c@{}}Determinar la matriz de feromonas\\ y n\'umero de hormigas ideal\end{tabular} & 1.2 y 1.4 & 2 semanas \\ \hline
8 & \begin{tabular}[c]{@{}c@{}}Definir los movimientos, \\ condiciones tab\'u, \\ criterios de aspiraci\'on mediante\\ la aplicaci\'on de pruebas para \\ optimizar el c\'alculo de la \\ soluci\'on final\end{tabular} & 1.3 y 1.4 & 2 semanas \\ \hline
9 & \begin{tabular}[c]{@{}c@{}}Establecer el nivel de hibridizaci\'on \\ entre ambas metaheur\'isticas\end{tabular} & \begin{tabular}[c]{@{}c@{}}1.2, 1.3 \\ y 1.4\end{tabular} & 2 semanas \\ \hline
10 & \begin{tabular}[c]{@{}c@{}}Crear\\ distintos casos de prueba\\ basados en modelos de entradas de datos\\ sintetizadospara la asignaci\'on de horarios\end{tabular} & 2.1 & 2 semanas \\ \hline
11 & \begin{tabular}[c]{@{}c@{}}Realizar pruebas finales\\ que permitan precisar \\ el comportamiento y eficiencia\\ del programa frente a los \\ distintos casos de prueba establecidos\end{tabular} & 2.1 & 2 semanas \\ \hline
12 & \begin{tabular}[c]{@{}c@{}}Realizar el an\'alisis\\ de los resultados usando\\ las m\'etricas definidas\end{tabular} & 2.2 & 2 semanas \\ \hline
\end{longtable}
\end{document}
The problem is all the nested \begin{tabular}{c}
inside your outer table, tabular
are by default vertically centred which is what you show in your image,
You could use \begin{tabular}[t]{c}
so the inner tabular are top aligned (or better, \begin{tabular}[t]{@{}c@{}}
so they don't add horizontal padding, but just removing all the inner tabular would be preferable.
I fixed several other errors in the example, undefined commands and using math italic for operators such as Exp.
\documentclass[a4paper, 8pt]{extarticle}
\usepackage{booktabs}
\usepackage[margin=0.5cm,landscape]{geometry}
\usepackage{mathtools}
\usepackage{amsmath,amsfonts}
\newcommand{\cW}{\mathcal{W}}
\newcommand{\Prob}[1]{P\left[#1\right]}
\newcommand{\Exp}[1]{\operatorname{Exp}\left[#1\right]}
\newcommand{\Var}[1]{\operatorname{Var}\left[#1\right]}
\newcommand{\riid}{\stackrel{\text{i.i.d}}{\sim}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\dset}[2]{\left\{ #1 \ \middle| \ #2 \right\}}
\DeclareMathOperator\Bin{\mathrm{Bin}}
\newcommand\N{\mathbb{N}}
\begin{document}
\begin{tabular}{lp{4cm}cccccc}
\toprule
\textbf{Verteilung $\mu_X$}&
\textbf{Beschrieb von $X$} &
\textbf{$X=$} &
\textbf{$\cW(X)=$} &
\begin{tabular}[t]{@{}c@{}}
\textbf{Gewichtsfunktion}\\
\textbf{$p_X(k)=\Prob{X=k}=$}
\end{tabular} &
\begin{tabular}[t]{@{}c@{}}
\textbf{Verteilungsfunktion}\\
\textbf{$F_X(t)=\Prob{X\leq t}=$}
\end{tabular} &
\textbf{$\Exp{X}=$} &
\textbf{$\Var{X}=$}
\\
\midrule
\vdots
\\
\hline
\begin{tabular}[t]{c}
Binomial\\
$\displaystyle X\sim \Bin(n,p)$
\end{tabular}
&
Anzahl der Erfolge $X$ bei $n$ unabhängigen 0-1-Experimenten mit
Erfolgsparameter $p$.
&
\begin{tabular}[t]{@{}c@{}}
$\displaystyle \sum_{i=1}^{n} I_{A_i} =\sum_{i=1}^{n}Y_i$ \\
Sind die ZV $Y_1,\ldots,Y_n\riid Be(p)$, \\
so ist $X:=Y_1+\cdots+Y_n\sim Bin(n,p)$.
\end{tabular}
&
$\displaystyle \set{0,1,\ldots,n}$
&
$\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$
&
$\displaystyle \sum_{k=0}^t \binom{n}{k}p^k(1-p)^{n-k}$
&
$\displaystyle np$
&
$np(1-p)$
\\
\hline
\begin{tabular}[t]{c}
Geometrisch\\
$\displaystyle X\sim Geom(p)$ \\
\\
``gedächtnislos''
\end{tabular} &
Wartezeit $X$ auf den ersten Erfolg bei einer unendlichen Folge von
0-1-Experimenten mit Erfolgsparameter $p$. Spezialfall von $NB(1,p)$ &
\begin{tabular}[t]{@{}c@{}}
$\displaystyle\inf\dset{i\in\N}{A_i\text{ tritt ein}}$\\
$\displaystyle{}=\inf\dset{n\in\N}{Y_i=1}$
\end{tabular} &
$\displaystyle\set{1,2,3,\ldots}$ &
$\displaystyle(1-p)^{k-1}p$ &
$\displaystyle1-(1-p)^t$ &
$\displaystyle \frac{1}{p}$ &
%???$\displaystyle \frac{1-p}{p^2}$ &
\\\hline
\vdots
\\\bottomrule
\end{tabular}
\end{document}
Or perhaps, despite the question title, you suggest in comments you really wanted centre alignment:
\documentclass[a4paper, 8pt]{extarticle}
\usepackage{booktabs,array}
\usepackage[margin=0.5cm,landscape]{geometry}
\usepackage{mathtools}
\usepackage{amsmath,amsfonts}
\newcommand{\cW}{\mathcal{W}}
\newcommand{\Prob}[1]{P\left[#1\right]}
\newcommand{\Exp}[1]{\operatorname{Exp}\left[#1\right]}
\newcommand{\Var}[1]{\operatorname{Var}\left[#1\right]}
\newcommand{\riid}{\stackrel{\text{i.i.d}}{\sim}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\dset}[2]{\left\{ #1 \ \middle| \ #2 \right\}}
\DeclareMathOperator\Bin{\mathrm{Bin}}
\newcommand\N{\mathbb{N}}
\begin{document}
\begin{tabular}{lm{4cm}cccccc}
\toprule
\textbf{Verteilung $\mu_X$}&
\textbf{Beschrieb von $X$} &
\textbf{$X=$} &
\textbf{$\cW(X)=$} &
\begin{tabular}[c]{@{}c@{}}
\textbf{Gewichtsfunktion}\\
\textbf{$p_X(k)=\Prob{X=k}=$}
\end{tabular} &
\begin{tabular}[c]{@{}c@{}}
\textbf{Verteilungsfunktion}\\
\textbf{$F_X(t)=\Prob{X\leq t}=$}
\end{tabular} &
\textbf{$\Exp{X}=$} &
\textbf{$\Var{X}=$}
\\
\midrule
\vdots
\\
\hline
\begin{tabular}[c]{c}
Binomial\\
$\displaystyle X\sim \Bin(n,p)$
\end{tabular}
&
Anzahl der Erfolge $X$ bei $n$ unabhängigen 0-1-Experimenten mit
Erfolgsparameter $p$.
&
\begin{tabular}[c]{@{}c@{}}
$\displaystyle \sum_{i=1}^{n} I_{A_i} =\sum_{i=1}^{n}Y_i$ \\
Sind die ZV $Y_1,\ldots,Y_n\riid Be(p)$, \\
so ist $X:=Y_1+\cdots+Y_n\sim Bin(n,p)$.
\end{tabular}
&
$\displaystyle \set{0,1,\ldots,n}$
&
$\displaystyle \binom{n}{k}p^k(1-p)^{n-k}$
&
$\displaystyle \sum_{k=0}^t \binom{n}{k}p^k(1-p)^{n-k}$
&
$\displaystyle np$
&
$np(1-p)$
\\
\hline
\begin{tabular}[c]{c}
Geometrisch\\
$\displaystyle X\sim Geom(p)$ \\
\\
``gedächtnislos''
\end{tabular} &
Wartezeit $X$ auf den ersten Erfolg bei einer unendlichen Folge von
0-1-Experimenten mit Erfolgsparameter $p$. Spezialfall von $NB(1,p)$ &
\begin{tabular}[c]{@{}c@{}}
$\displaystyle\inf\dset{i\in\N}{A_i\text{ tritt ein}}$\\
$\displaystyle{}=\inf\dset{n\in\N}{Y_i=1}$
\end{tabular} &
$\displaystyle\set{1,2,3,\ldots}$ &
$\displaystyle(1-p)^{k-1}p$ &
$\displaystyle1-(1-p)^t$ &
$\displaystyle \frac{1}{p}$ &
%???$\displaystyle \frac{1-p}{p^2}$ &
\\\hline
\vdots
\\\bottomrule
\end{tabular}
\end{document}
Best Answer
Elaborating a little on Harish Kumar's answer, you'll probably want the
longtable
to the left.To do that, don't use the optional argument for
longtable
but issueThis should give you what you want:
Output
or, if you prefer perfect alignment (thanks Mico for the comment)
Output