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Use threeparttable package with endfloat

養花種魚數月亮賞星星 於 Monday, August 10, 2009 6:30 PM 發表

由於threeparttable並非浮動物件,但是他可以放在table環境中。所以若使用endfloat套件時,必須將threeparttable放置在table環境中。


\documentclass[a4paper,12pt]{article}
\usepackage{booktabs,threeparttable}
\usepackage{endfloat}

\begin{document}
\noindent \textbf{Example 1.} We first redo the same example of Chen and
Chen \cite{Chen04} while considering the time increasing demand. $%
f\left( t,p \right) =\left( 300-120p \right)e^{0.06t}$, $c_{f}=40$, $%
c_{h}=0.02 $, $c_{s}=0.5$, $\theta =0.2$, $H=12$, $r=0.02$. Besides, we
assume that the time-dependent backlogging rate is $c_{v}(t)=e^{0.01t}$, $%
\beta \left( x\right) =e^{-0.2x} $ and take $c_{l}=0.6$. By applying (\ref%
{neoq}), we obtain the estimated number of replenishments $n=6$. Then,
applying the Algorithm 1 and 2, we get $TP(5)=474.1$, $TP(6)=487.4$ and $%
TP(7)=487.2$. %$TP(8)=478.428$.
Therefore, the optimal number of replenishments is 6, and the optimal
pricing and time schedule are shown in Table 1. The behavior of inventory
system over the planning horizon and the convergence result of PSO
algorithms for optimal solution are depicted in Fig. \ref{example1} and \ref%
{iter:a}, respectively.

\begin{table}[!htpb]
\centering
\begin{threeparttable}[tbp]
\caption{Optimal pricing and time schedule for Example 1}
\begin{tabular}{rrrrrrrr}
\hline
\multicolumn{1}{c}{$i$} & \multicolumn{1}{c}{$p_i$} & \multicolumn{1}{c}{$cv (t_i)$} & \multicolumn{1}{c}{$t_i$} & \multicolumn{1}{c}{$s_i$} & \multicolumn{1}{c}{$Q_i$} & \multicolumn{1}{c}{$LS_i$~\tnote{a}} & \multicolumn{1}{c}{$LI_i$~\tnote{b}} \\ \hline
1 & 1.8630 & 1.0067 & 0.6643 & 2.2946 & 137.4 & 0.6643 & 1.6303 \\
2 & 1.8688 & 1.0296 & 2.9202 & 4.4519 & 90.8 & 0.6256 & 1.5317 \\
3 & 1.8750 & 1.0517 & 5.0443 & 6.4886 & 73.4 & 0.5924 & 1.4443 \\
4 & 1.8812 & 1.0731 & 7.0520 & 8.4184 & 67.8 & 0.5634 & 1.3664 \\
5 & 1.8876 & 1.0937 & 8.9559 & 10.2523 & 67.2 & 0.5375 & 1.2964 \\
6 & 1.8940 & 1.1137 & 10.7670 & 12.0000 & 68.9 & 0.5147 & 1.2330\\ \hline
\end{tabular}%
\begin{tablenotes}
\footnotesize
\item[a] $LS_i = t_i -s_{i-1}$
\item[b] $LI_i = s_i -t_i$
\end{tablenotes}
\end{threeparttable}
\end{table}
\end{document}

排版的結果


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