\begin{eqnarray}{\label{eq:9}}
l(p_{ij}^d,\lambda,\mu,\delta)=\left \{\frac{1}{\text{ln}2} \sum_{j=1}^{M}\sum_{i\in \mathcal{D}_j} \text{ln} \left (1+\frac{p_{ij}^dg_{jj}}{\sigma_N^2+p_ch_{ij}}\right )\right\}\nonumber\\- \sum_{j=1}^{M} \sum_{i\in \mathcal{D}_j}\lambda_{ij} \left(p_{ij}^d-\frac{p_{c}g_{iB}}{(2^{R_{min}}-1) h_{jB}} + \frac{\sigma_N^2}{h_{jB}}\right)\nonumber\\-\sum_{j=1}^{M}\mu_{j}\left(\sum_{i\in \mathcal{D}_j} p_{ij}^d- P_{dmax}^j\right)\nonumber\\+\sum_{j=1}^{M}\sum_{i\in \mathcal{D}_j}\delta_{ij}p_{ij}^d
\end{eqnarray}
答案1
除了不使用严重贬低 eqnarray
环境,以下解决方案使用align
环境,使用\biggl
和而不是\biggr
和来调整大括号的大小(从而产生更一致和更合适的大小),使用而不是(以获得更好的间距),并使用直立(“罗马”)字母表示“min”和“dmax”。它还取消了第一行中的花括号,因为它们似乎只会增加视觉混乱。\left
\right
\ln
\text{ln}
通过这些调整,可以使等式占据三行而不是四行。
\documentclass[twocolumn]{article}
\usepackage{amsmath} % for 'align' environment
\begin{document}
\begin{align}\label{eq:9}
l(&p_{ij}^d,\lambda,\mu,\delta)
=\frac{1}{\ln2} \sum_{j=1}^{M} \sum_{i\in\mathcal{D}_j}
\ln \biggl( 1+\frac{p_{ij}^d g^{}_{jj}}{\sigma_N^2+p_c h_{ij}} \biggr) \nonumber\\
&- \sum_{j=1}^{M} \sum_{i\in\mathcal{D}_j}\lambda_{ij}
\biggl( p_{ij}^d-\frac{p_{c}g_{iB}}{(2^{R_{\min}}-1) h_{jB}}
+ \frac{\sigma_N^2}{h_{jB}} \biggr)\nonumber\\
&-\sum_{j=1}^{M}\mu_{j} \biggl(\,\sum_{i\in\mathcal{D}_j} p_{ij}^d
- P_{\mathrm{dmax}}^j \!\biggr)%\nonumber\\
+\sum_{j=1}^{M}\sum_{i\in\mathcal{D}_j}\delta^{}_{ij}p_{ij}^d
\end{align}
\end{document}
答案2
下面是一个适合列宽的方程式代码,其中包含一些手动间距调整:
\documentclass[ a4paper, twocolumn]{article}%
\usepackage[showframe]{geometry}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{mathtools, nccmath}
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\begin{fleqn}
\begin{equation}
\label{eq:9}
\begin{aligned}[b]
& l(p_{ij}^d,\lambda,\mu,\delta )= \\
& \mkern-4mu -\!\biggl\{\frac{1}{\ln 2} \sum_{j=1}^{M}\sum_{i\in \mathcal{D}_j} \ln\biggl(1\!+\!\frac{p_{ij}^dg_{jj}}{\sigma_N^2+p_ch_{ij}}\biggr) \biggr\}
\\&\mkern-4mu -\!\sum_{j=1}^{M} \sum_{i\in\mathcal{D}_j}\mkern-5mu \mathrlap{\lambda_{ij}\biggl(p_{ij}^d\!-\!\frac{p_{c}g_{iB}}{(2^{R_{\min}}\!-\!1) h_{jB}}\!+\!\frac{\sigma_N^2}{h_{jB}}\!\biggr)}\\%
& \mkern-4mu -\!\sum_{j=1}^{M}\mu_{j}\biggl(\sum_{i\in \mathcal{D}_j}\! p_{ij}^d\!-\! P_{d\max}^j\!\biggr)\!+\!\sum_{j=1}^{M}\!\sum_{i\in \mathcal{D}_j}\!\delta_{ij}p_{ij}^d
\end{aligned}
\end{equation}
\end{fleqn}
\lipsum[2]
\end{document}
如果您接受方程的该部分为中等大小(〜80%\displaystyle
),\medmath
则来自的命令nccmath
允许使用更简单的代码,无需手动调整:
\begin{fleqn}
\begin{equation}
\label{eq:9}
\begin{aligned}[b]
& l(p_{ij}^d,\lambda,\mu,\delta )= \\
& \medmath{\biggl\{\frac{1}{\ln 2} \sum_{j=1}^{M}\sum_{i \in \mathcal{D}_j} \ln\biggl (1+\frac{p_{ij}^dg_{jj}}{\sigma_N^2+p_ch_{ij}}\biggr ) \biggr\}} \\[-1ex]
&\medmath{-\sum_{j=1}^{M} \sum_{i \in \mathcal{D}_j}\lambda_{ij} \biggl(p_{ij}^d-\frac{p_{c}g_{iB}}{(2^{R_{\min}}\!-\!1) h_{jB}} + \frac{\sigma_N^2}{h_{jB}}\biggr)} \\[-1ex]%
&\medmath{-\sum_{j=1}^{M}\mu_{j}\biggl(\sum_{i \in \mathcal{D}_j} p_{ij}^d - P_{d\max}^j\biggr) +\sum_{j=1}^{M}\sum_{i \in \mathcal{D}_j}\delta_{ij}p_{ij}^d}
\end{aligned}
\end{equation}
\end{fleqn}
答案3
使用\MoveEqLeft
from的解决方案mathtools
:
\documentclass[twocolumn]{article}
\usepackage{mathtools} % for '\MoveEqLeft' com.
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\begin{align}\label{eq:9}
\MoveEqLeft
l(p_{ij}^d,\lambda,\mu,\delta) \notag\\
& = \frac{1}{\ln 2} \sum_{j=1}^{M}\sum_{i\in\mathcal{D}_j}
\ln \biggl( 1+\frac{p_{ij}^d g_{jj}}{\sigma_N^2+p_c h_{ij}} \biggr) \notag\\
& - \sum_{j=1}^{M} \sum_{i\in\mathcal{D}_j}\lambda_{ij}
\biggl( p_{ij}^d-\frac{p_{c}g_{iB}}{(2^{R_{\min}}-1) h_{jB}}
+ \frac{\sigma_N^2}{h_{jB}} \biggr)\notag\\
& - \sum_{j=1}^{M}\mu_{j} \biggl(\,\sum_{i\in\mathcal{D}_j} p_{ij}^d
- P_{d\max}^j \biggr)\notag\\
& + \sum_{j=1}^{M}\sum_{i\in\mathcal{D}_j}\delta_{ij}p_{ij}^d
\end{align}
\lipsum[2]
\end{document}
附录:
使用包\medmath
中的nccmat
这个公式可以分为四行:
\documentclass[twocolumn]{article}
\usepackage{mathtools} % for '\MoveEqLeft' command
\usepackage{nccmath} % for medium size of equation
\usepackage{lipsum}
\begin{document}
\lipsum[1]
\begin{equation}
\medmath{\begin{aligned}[b]\label{eq:9}
\MoveEqLeft
l(p_{ij}^d,\lambda,\mu,\delta) \\
& = \frac{1}{\ln 2} \sum_{j=1}^{M}\sum_{i\in\mathcal{D}_j}
\ln \biggl( 1+\frac{p_{ij}^d g_{jj}}{\sigma_N^2+p_c h_{ij}} \biggr) \\
& - \sum_{j=1}^{M} \sum_{i\in\mathcal{D}_j}\lambda_{ij}
\biggl( p_{ij}^d-\frac{p_{c}g_{iB}}{(2^{R_{\min}}-1) h_{jB}}
+ \frac{\sigma_N^2}{h_{jB}} \biggr) \\
& - \sum_{j=1}^{M}\mu_{j} \biggl(\,\sum_{i\in\mathcal{D}_j} p_{ij}^d
- P_{d\max}^j \biggr) + \sum_{j=1}^{M}\sum_{i\in\mathcal{D}_j}\delta_{ij}p_{ij}^d
\end{aligned}
}
\end{equation}
\lipsum[2]
\end{document}