Align forall to the right hand side

alignequations

I have a few equations, and I would like to align the forall statements of these equations to the right-hand side as below:

However, when I add a new equation (eq19), I have no idea why I cannot get a result as above. Would you please help me to resolve this problem using a similar notation as above (since I do not want to change all equations in my main file)?

\documentclass{article}

\usepackage{amsmath}

\begin{document}

\begin{flalign}
  & o_{cs} \leq mO 
  & \forall c \in \mathcal{C}; s \in \mathcal{S} \label{eq18}\\
%   & a_{jcs} - a_{jbs} \geq 
%     (\sum_{i \in \mathcal{T} \cup \alpha} 
%      x_{ijcs} + 
%      \sum_{i \in \mathcal{T} \cup \alpha} 
%      x_{ijbs} - 2)\Omega \nonumber \\
%   & 
%   & \forall j \in \mathcal{T}; b, c \in \mathcal{C}; s \in \mathcal{S}; NC_{js} > 1 \label{eq19}\\
  & \sum_{i \in \mathcal{T} \cup \alpha}
    \sum_{j \in \mathcal{T}}
    x_{ijcs} \: T_{js} \leq (ES_s-BS_s) \: {fR_c} 
  & \forall c \in \mathcal{C}; s \in \mathcal{S} \label{eq19a}
\end{flalign}
\end{document}

Best Answer

The problem is that your longest left aligned equation + longest right aligned equation + number will be wider than your total available width.

Here, the equations will fit:

But as soon as you add the \in or more to the right hand equation, the total width will be overfull and the flalign environment won't be able to section it up into two good boxes and thereby can't have the equations and numbering on the same line:

Thereby you either have to use another environment than flalign or widen the text width using the geometry package.

\usepackage{geometry}

Which will adjust the widths for your case to work, but the widths of the page can still be set manually with the geometry package.

\documentclass{article}
\usepackage{geometry}
\usepackage{amsmath}

\begin{document}

\begin{flalign}
  % Line 1
  & o_{cs} \leq mO 
  & \forall c \in \mathcal{C}; s \in \mathcal{S} \label{eq18}\\
  % Line 2
  & a_{jcs} - a_{jbs} \geq 
    (\sum_{i \in \mathcal{T} \cup \alpha} 
     x_{ijcs} + 
     \sum_{i \in \mathcal{T} \cup \alpha} 
     x_{ijbs} - 2)\Omega
  & \forall j \in \mathcal{T}; b, c \in \mathcal{C}; s \in \mathcal{S}; NC_{js} > 1 \label{eq19}\\
  % Line 3
  & \sum_{i \in \mathcal{T} \cup \alpha}
    \sum_{j \in \mathcal{T}}
    x_{ijcs} \: T_{js} \leq (ES_s-BS_s) \: {fR_c} 
  & \forall c \in \mathcal{C}; s \in \mathcal{S} \label{eq19a}
\end{flalign}
\end{document}

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I would definitely go the second way with $\displaystyle: I find that it looks strange to have something centered after a bullet. To get the vertical alignment correct, just add [t] after \begin{aligned}, where "t" means "top". (See also the top of page 8 of the User's guide for the amsmath Package.)

[Tex/LaTex] Aligning the right hand side of multiline equations

\documentclass[preview,border=12pt]{standalone}
\usepackage[a4paper,margin=1cm]{geometry}
\usepackage{mathtools}
\begin{document}
\abovedisplayskip=0pt\relax% don't use this line in your production (egreg does not like this)
\begin{equation}
\begin{split}
\frac{\partial P(y_{i}\succ0)} {\partial z_{ij}}
&=
\!
\begin{multlined}[t][10cm]
\tfrac{a(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}a^{2}(1-\tau^{2})^{2}\left(z_{1i}\prime\gamma_{1j}\right)^{2}}
\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}
-a\tau \left(z_{1i}\prime\gamma_{1j}\right)\right)
\gamma_{1j}\\
+\tfrac {b(1-\tau^{2})^{1/2}}{(2 \pi)^{1/2}} e^{-\tfrac{1}{2}b^{2j}(1-\tau^{2})^{2}\left(\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)^{2}}\\
\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}\\
&=
\!
\begin{multlined}[t][10cm]
\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(b\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-a\tau z_{1i}\prime\gamma_{1j}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(a\left(z_{1i}\prime\gamma_{1j}\right)-b\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}\\
&=
\!
\begin{multlined}[t][10cm]
\phi\left(z_{1i}\prime\gamma_{1}\right)\Phi\left(\tfrac{\tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}-\tau z_{1i}\prime\gamma_{1}}{(1-\tau^{2})^{1/2}}\right)\gamma_{1j}\\
+\phi\left(\tfrac{z_{2i}\prime\gamma_{2}}{\sigma}\right)\Phi\left(\tfrac{z_{1i}\prime\gamma_{1j}-\tau \tfrac{z_{2i}\prime\gamma_{2j}}{\sigma}}{(1-\tau^{2})^{1/2}}\right)\tfrac{\gamma_{2j}}{\tau}
\end{multlined}
\end{split}
\end{equation}
\end{document}

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Best Answer

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