3

first my code for creating the table:

\documentclass{report}
\usepackage{booktabs}
\usepackage{multirow}

\begin{document}

\begin{table}[!ht]
\begin{tabular}{lccc} 
 \textbf{Cases} & \textbf{First Order Condition} & \multicolumn{2}{c}{\textbf{Debt Level Chosen}} \\[5pt] \toprule

 \multirow{2}{*}{$b(1+r)\leq \tilde{b}$} & \multirow{2}{*}{$u'(1+b)=p-\frac{p b}{2 \epsilon_H}$} & $b^{G}$ & with $\tilde{b}>b^{G}$  \\[3pt]

 & & $\tilde{b}$ & with $\tilde{b}\leq b^{G}$ \\ [5pt] \midrule

$b(1+r)>\tilde{b}$ & $u'(1+b)=p$  & $b^{NG}$ & with $\tilde{b}<b^{NG}$  \\[5pt] \bottomrule
\end{tabular}
\caption{caption here}
\end{table}
\end{document}

For spacing I used the ["any number"pt] command after breaking the line. It is defining the space beneath that specific row.

Is it possible to define the space above a row?

Placing the [pt] command after for instance \toprule, increases the thickness of the line. I did not use \renewcommand{\arraystretch}{1.5} as I do not want the same row spacing for all table rows in my paper.

3

Since you're using booktabs, you can certainly define the space before and after a rule, and between rows with the \addlinespace command:

\documentclass[12pt,a4paper,notitlepage]{report}
\usepackage{booktabs}
\usepackage{multirow}

\begin{document}
\begin{table}[!ht]
\centering
\begin{tabular}{lccc} 
\textbf{Cases} & \textbf{First Order Condition} & \multicolumn{2}{c}{\textbf{Debt Level Chosen}} \\ \toprule
\multirow{2}{*}{$b(1+r)\leq \tilde{b}$} & \multirow{2}{*}{$u'(1+b)=p-\frac{p b}{2 \epsilon_H}$}
  & $b^{G}$     & with $\tilde{b}>b^{G}$ \\
& & $\tilde{b}$ & with $\tilde{b}\leq b^{G}$ \\ \midrule
$b(1+r)>\tilde{b}$ & $u'(1+b)=p$  & $b^{NG}$ & with $\tilde{b}<b^{NG}$ \\ \bottomrule
\end{tabular}
\caption{Original}
\end{table}

\begin{table}[!ht]
\centering
\begin{tabular}{lccc} 
\textbf{Cases} & \textbf{First Order Condition} & \multicolumn{2}{c}{\textbf{Debt Level Chosen}} \\ \toprule
\multirow{2}{*}{$b(1+r)\leq \tilde{b}$} & \multirow{2}{*}{$u'(1+b)=p-\frac{p b}{2 \epsilon_H}$}
  & $b^{G}$     & with $\tilde{b}>b^{G}$ \\ \addlinespace[3em]
& & $\tilde{b}$ & with $\tilde{b}\leq b^{G}$ \\ \addlinespace[2em] \midrule \addlinespace[1em]
$b(1+r)>\tilde{b}$ & $u'(1+b)=p$  & $b^{NG}$ & with $\tilde{b}<b^{NG}$ \\ \bottomrule
\end{tabular}
\caption{Modified, 3em after first row, 2em before midrule, 1em after midrule}
\end{table}
\end{document}

enter image description here

  • Hello Mike Renfro, thank you a lot for your suggestion. It is exactly what I was looking for. Problem solved! – Claude Mar 21 '16 at 10:21

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