I need to finish a paper but i need to fix the equations formatting of a linear programming model. I am using the environment eqnarray and i almost have it formatted the way my advisor asked. This is a sample of i got 1. I need to align the right side of the equations to the left while aligning the left side to the right (the left side is OK). Does anyone know if there is a way to make this modification with the eqnarray environment?
This is part of the latex code that generated this image:
\begin{eqnarray}
\mbox{s. t.} \qquad \sum_{j \in V_{+1} | i \neq j} x_{ij} \quad = & 1 & \forall\, i \in C \label{eq2} \\
\sum_{j \in V_{+1} | i \neq j} x_{ij} \quad \leq & 1 & \forall\, i \in \{R \cup S\} \label{eq3} \\
\sum_{i \in V_{+1} | i \neq j} x_{ji} \quad = & \sum_{i \in V_{0} | i \neq j} x_{ij} & \forall\, j \in V \label{eq4} \\
\sum_{i \in \{C \cup R\}} y_{i} \quad \leq & O & \forall i \in R \label{eq27} \\
\sum_{j \in V_{0} | i \neq j} x_{ji} \quad \leq & M \cdot y_{i} & \forall i \in R \label{eq7} \\
\sum_{j \in V_{0} | h \neq j} x_{jh} \quad \leq & M \cdot y_{i} & \forall i \in R, h \in SK_{i} \label{eq8} \\
t_{j} \quad \geq & t_{i} + ( td_{ij} + s_{i} ) \cdot x_{ij} - M \cdot (1 - x_{ij}) & \forall i \in C_{0}, j \in V_{1} | i \neq j \label{eq9} \\
t_{j} \quad \geq & t_{i} + td_{ij} \cdot x_{ij} + ct \cdot x_{ij} - M \cdot (1 - x_{ij}) & \forall i \in \{R \cup SK_{h \in R}\}, j \in V_{1} | i \neq j \label{eq11}
\begin{eqnarray}
Thank you.
eqnarray environment will not compatible for multi column align for multiple equation. So i am using alignat enviornment.
\begin{alignat}{3}
\mbox{s. t.} \qquad \sum_{j \in V_{+1} | i \neq j} x_{ij} &= 1 &&\quad \forall\, i \in C\label{eq2} \\
\sum_{j \in V_{+1} | i \neq j} x_{ij} &\leq 1 &&\quad \forall\, i \in \{R \cup S\} \label{eq3} \\
\sum_{i \in V_{+1} | i \neq j} x_{ji} & =\sum_{i \in V_{0} | i \neq j} x_{ij} &&\quad \forall\, j \in V \label{eq4} \\
\sum_{i \in \{C \cup R\}} y_{i} &\leq O &&\quad \forall i \in R \label{eq27} \\
\sum_{j \in V_{0} | i \neq j} x_{ji} & \leq M \cdot y_{i} &&\quad \forall i \in R \label{eq7} \\
\sum_{j \in V_{0} | h \neq j} x_{jh} &\leq M \cdot y_{i} &&\quad \forall i \in R, h \in SK_{i} \label{eq8} \\
t_{j} &\geq t_{i} + ( td_{ij} + s_{i} ) \cdot x_{ij}\nonumber\\
&\quad{} - M \cdot (1 - x_{ij}) &&\quad \forall i \in C_{0}, j \in V_{1} | i \neq j \label{eq9} \\
t_{j} &\geq t_{i} + td_{ij} \cdot x_{ij} + ct \cdot x_{ij}\nonumber\\
&\quad{}- M \cdot (1 - x_{ij}) &&\quad \forall i \in \{R \cup SK_{h \in R}\}, j \in V_{1} | i \neq j \label{eq11}
\end{alignat}
I think you got this!
Related
I am trying to include different equation paragraphs in my LaTex document. Is it somehow possible to visually separate the paragraphs (e.g. by horizontal lines) and give them different captions?
I don't want the captions to emerge in the table of content, just underneath the equations.
\documentclass{article}
\usepackage[utf8]{inputenc}
\begin{document}
\section{Mathematical Formulation}
\begin{equation}
\operatorname{Max} F=\sum_{u,v \in V \mid u<v} \left(d_{uv} + d_{vu}\right) x_{uv}
\end{equation}
such that:
\begin{equation}
\sum_{l \in L_{u v}} y_{l} \geq x_{u v} \hspace{4em} u, v \in V \mid u<v
\end{equation}
\begin{equation}
\sum_{l \in \tilde L_{i j}} y_{l} \geq 1 \hspace{5em} \ \lbrack i,j \rbrack \in E
\end{equation}
%Next paragraph here
\begin{equation}
\operatorname{Min} F=\sum_{u \in V} \sum_{l \in L} \sum_{(i, j) \in \vec {E_{l}}} t_{l i j} x_{u l i j}
\end{equation}
\begin{equation}
\sum_{l \in L} \sum_{(i, v) \in \vec{E}_{l}} x_{u l i v}-\sum_{l \in L} \sum_{(v, j) \in \vec{E}_{l}} x_{u l v j}=d_{u v} \hspace{4em} u,v \in V \mid u \ne v
\end{equation}
\end{document}
The caption should look similar to the one for my GAMS code:
You could create a custom command like this:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\newcounter{foo}
\newcommand{\mycaption}[1]{
{
\medskip
\centering
\setcounter{foo}{\theequation}
\addtocounter{foo}{1}
Equation \thefoo: #1
\vskip-0.5\baselineskip
\noindent\hrulefill
}
}
\begin{document}
\section{Mathematical Formulation}
\begin{equation}
\operatorname{Max} F=\sum_{u,v \in V \mid u<v} \left(d_{uv} + d_{vu}\right) x_{uv}
\end{equation}
such that:
\begin{equation}
\sum_{l \in L_{u v}} y_{l} \geq x_{u v} \hspace{4em} u, v \in V \mid u<v
\end{equation}
\begin{equation}
\sum_{l \in \tilde L_{i j}} y_{l} \geq 1 \hspace{5em} \ \lbrack i,j \rbrack \in E
\end{equation}
%Next paragraph here
\mycaption{some text}
\begin{equation}
\operatorname{Min} F=\sum_{u \in V} \sum_{l \in L} \sum_{(i, j) \in \vec {E_{l}}} t_{l i j} x_{u l i j}
\end{equation}
\begin{equation}
\sum_{l \in L} \sum_{(i, v) \in \vec{E}_{l}} x_{u l i v}-\sum_{l \in L} \sum_{(v, j) \in \vec{E}_{l}} x_{u l v j}=d_{u v} \hspace{4em} u,v \in V \mid u \ne v
\end{equation}
\end{document}
Or if you prefer a separate counter:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\newcounter{foo}
\newcommand{\mycaption}[1]{
{
\medskip
\centering
% \setcounter{foo}{\theequation}
\addtocounter{foo}{1}
Model \thefoo: #1
\vskip-0.5\baselineskip
\noindent\hrulefill
}
}
\begin{document}
\section{Mathematical Formulation}
\begin{equation}
\operatorname{Max} F=\sum_{u,v \in V \mid u<v} \left(d_{uv} + d_{vu}\right) x_{uv}
\end{equation}
such that:
\begin{equation}
\sum_{l \in L_{u v}} y_{l} \geq x_{u v} \hspace{4em} u, v \in V \mid u<v
\end{equation}
\begin{equation}
\sum_{l \in \tilde L_{i j}} y_{l} \geq 1 \hspace{5em} \ \lbrack i,j \rbrack \in E
\end{equation}
%Next paragraph here
\mycaption{some text}
\begin{equation*}
\operatorname{Min} F=\sum_{u \in V} \sum_{l \in L} \sum_{(i, j) \in \vec {E_{l}}} t_{l i j} x_{u l i j}
\end{equation*}
\begin{equation}
\sum_{l \in L} \sum_{(i, v) \in \vec{E}_{l}} x_{u l i v}-\sum_{l \in L} \sum_{(v, j) \in \vec{E}_{l}} x_{u l v j}=d_{u v} \hspace{4em} u,v \in V \mid u \ne v
\end{equation}
\end{document}
So, I've been trying to create a flashcard of a (rather simple) mathematical proof I did on my own. Now I want to align the equals signs, and it seems to work up to 11 terms, but not more. This doesn't make any sense in my opinion and my mistake is probably somewhere else.
Here is the "code":
[latex\]\begin{align} \sigma \left(X\right) &= \sqrt{\sum_{k=0}^n \left(k - µ\right)^2 \cdot P\left(X=k\right)} \\
&= \sqrt{\sum_{k=0}^n \left(k^2 - 2kµ + µ^2\right) \cdot P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n \left(k^2 \cdot P\left(X=k\right) - 2kµ \cdot P\left(X=k\right) + µ^2\cdot P\left(X=k\right)\right)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P\left(X=k\right) + \sum_{k=0}^n - 2kµ \cdot P\left(X=k\right) + \sum_{k=0}^n µ^2\cdot P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P\left(X=k\right) + - 2µ \cdot \sum_{k=0}^n k \cdot P\left(X=k\right) + µ^2\cdot \sum_{k=0}^n P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n \left( k^2 \cdot \binom{n}{k} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - 2n^2 p^2 + n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left( k^2 \cdot \binom{n}{k} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\left(k-1+1\right) \cdot \frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} + \frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \right) \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) + \sum_{k=0}^n \left(\frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) + np - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(n\cdot\left(n-1\right)\cdot\frac{(n-2)!}{\left(k-2\right)! \cdot \left(n-k\right)!} \cdot p^2 \cdot p^{k-2} \cdot \left(1-p\right)^{(n-2)-(k-2)} \right) + np - n^2 p^2} \\
&=\sqrt{n\cdot\left(n-1\right)\cdot p^2 \cdot \sum_{k=2}^n \left( \binom{n-2}{k-2} \cdot p^{k-2} \cdot \left(1-p\right)^{(n-2)-(k-2)} \right) + np - n^2 p^2} \\
&=\sqrt{n\cdot\left(n-1\right)\cdot p^2 \cdot \sum_{k=0}^{n-2} \left( \binom{n-2}{k} \cdot p^{k} \cdot \left(1-p\right)^{(n-2)-k} \right) + np - n^2 p^2} \\
&=\sqrt{n\cdot\left(n-1\right)\cdot p^2 \cdot \left(p + 1-p\right)^{n-2} + np - n^2 p^2} \\
&=\sqrt{n\cdot\left(n-1\right)\cdot p^2 + np - n^2 p^2} \\
&=\sqrt{\left(n^2 - n\right)\cdot p^2 + np - n^2 p^2} \\
&=\sqrt{n^2 p^2 - n p^2 + np - n^2 p^2} \\
&=\sqrt{- n p^2 + np} \\
&=\sqrt{n\cdot p\cdot\left(1-p\right)}\end{align} \[/latex]
[latex] q.e.d. [/latex]
Here is the same code without \left and \right to make it more readable:
[latex\]\begin{align} \sigma (X) &= \sqrt{\sum_{k=0}^n (k - µ)^2 \cdot P(X=k)} \\
&= \sqrt{\sum_{k=0}^n (k^2 - 2kµ + µ^2) \cdot P(X=k)} \\
&=\sqrt{\sum_{k=0}^n (k^2 \cdot P(X=k) - 2kµ \cdot P(X=k) + µ^2\cdot P(X=k))} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P(X=k) + \sum_{k=0}^n - 2kµ \cdot P(X=k) + \sum_{k=0}^n µ^2\cdot P(X=k)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P(X=k) + - 2µ \cdot \sum_{k=0}^n k \cdot P(X=k) + µ^2\cdot \sum_{k=0}^n P(X=k)} \\
&=\sqrt{\sum_{k=0}^n ( k^2 \cdot \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} ) - 2n^2 p^2 + n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ( k^2 \cdot \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ((k-1+1) \cdot \frac{n!}{(k-1)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ((\frac{n!}{(k-2)! \cdot (n-k)!} + \frac{n!}{(k-1)! \cdot (n-k)!} ) \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n (\frac{n!}{(k-2)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) + \sum_{k=0}^n (\frac{n!}{(k-1)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n (\frac{n!}{(k-2)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) + np - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n (n\cdot(n-1)\cdot\frac{(n-2)!}{(k-2)! \cdot (n-k)!} \cdot p^2 \cdot p^{k-2} \cdot (1-p)^{(n-2)-(k-2)} ) + np - n^2 p^2} \\
&=\sqrt{n\cdot(n-1)\cdot p^2 \cdot \sum_{k=2}^n ( \binom{n-2}{k-2} \cdot p^{k-2} \cdot (1-p)^{(n-2)-(k-2)} ) + np - n^2 p^2} \\
&=\sqrt{n\cdot(n-1)\cdot p^2 \cdot \sum_{k=0}^{n-2} ( \binom{n-2}{k} \cdot p^{k} \cdot (1-p)^{(n-2)-k} ) + np - n^2 p^2} \\
&=\sqrt{n\cdot(n-1)\cdot p^2 \cdot (p + 1-p)^{n-2} + np - n^2 p^2} \\
&=\sqrt{n\cdot(n-1)\cdot p^2 + np - n^2 p^2} \\
&=\sqrt{(n^2 - n)\cdot p^2 + np - n^2 p^2} \\
&=\sqrt{n^2 p^2 - n p^2 + np - n^2 p^2} \\
&=\sqrt{- n p^2 + np} \\
&=\sqrt{n\cdot p\cdot(1-p)}\end{align} \[/latex]
[latex] q.e.d. [/latex]
I don't know why ctrl + k (in StackOverflow) formats it in different ways, but if always used it this way in my code: [latex][/latex]
However, it shows just a blank line and under it "q.e.d." (code with \right and \left):
However, if I just write the first 11 lines, it works:
[latex\]\begin{align} \sigma \left(X\right) &= \sqrt{\sum_{k=0}^n \left(k - µ\right)^2 \cdot P\left(X=k\right)} \\
&= \sqrt{\sum_{k=0}^n \left(k^2 - 2kµ + µ^2\right) \cdot P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n \left(k^2 \cdot P\left(X=k\right) - 2kµ \cdot P\left(X=k\right) + µ^2\cdot P\left(X=k\right)\right)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P\left(X=k\right) + \sum_{k=0}^n - 2kµ \cdot P\left(X=k\right) + \sum_{k=0}^n µ^2\cdot P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P\left(X=k\right) + - 2µ \cdot \sum_{k=0}^n k \cdot P\left(X=k\right) + µ^2\cdot \sum_{k=0}^n P\left(X=k\right)} \\
&=\sqrt{\sum_{k=0}^n \left( k^2 \cdot \binom{n}{k} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - 2n^2 p^2 + n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left( k^2 \cdot \binom{n}{k} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\left(k-1+1\right) \cdot \frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} + \frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \right) \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) + \sum_{k=0}^n \left(\frac{n!}{\left(k-1\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n \left(\frac{n!}{\left(k-2\right)! \cdot \left(n-k\right)!} \cdot p^k \cdot \left(1-p\right)^{n-k} \right) + np - n^2 p^2} \end{align} \[/latex]
Here is the same code without \left and \right to make it more readable:
[latex\]\begin{align} \sigma (X) &= \sqrt{\sum_{k=0}^n (k - µ)^2 \cdot P(X=k)} \\
&= \sqrt{\sum_{k=0}^n (k^2 - 2kµ + µ^2) \cdot P(X=k)} \\
&=\sqrt{\sum_{k=0}^n (k^2 \cdot P(X=k) - 2kµ \cdot P(X=k) + µ^2\cdot P(X=k))} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P(X=k) + \sum_{k=0}^n - 2kµ \cdot P(X=k) + \sum_{k=0}^n µ^2\cdot P(X=k)} \\
&=\sqrt{\sum_{k=0}^n k^2 \cdot P(X=k) + - 2µ \cdot \sum_{k=0}^n k \cdot P(X=k) + µ^2\cdot \sum_{k=0}^n P(X=k)} \\
&=\sqrt{\sum_{k=0}^n ( k^2 \cdot \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} ) - 2n^2 p^2 + n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ( k^2 \cdot \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ((k-1+1) \cdot \frac{n!}{(k-1)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n ((\frac{n!}{(k-2)! \cdot (n-k)!} + \frac{n!}{(k-1)! \cdot (n-k)!} ) \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n (\frac{n!}{(k-2)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) + \sum_{k=0}^n (\frac{n!}{(k-1)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) - n^2 p^2} \\
&=\sqrt{\sum_{k=0}^n (\frac{n!}{(k-2)! \cdot (n-k)!} \cdot p^k \cdot (1-p)^{n-k} ) + np - n^2 p^2} \end{align} \[/latex]
This is the output (code with \right and \left, without I just seems to work with 10 lines instead of 11):
I hope you can help me. However, I am a beginner in LaTeX, so I need a rather detailed explanation.
I am trying to typeset the following equation in the align-evoirement
\begin{align}
t_2' &= t_2 + \frac{L/C} \\
t_1' &= t_1 + \frac{L + v\Delta t\cos\theta}{c} \\
t_2' - t_1' &= (t_2-t_1) + \frac{L-L-v\Delta t\cos\theta}{C} \\
\Delta t' &= \Delta t - \frac{v\Delta t \cos \theta}{c} \\
\Delta t' &= \Delta t \left(1-\frac{v\cos\theta}{c}\right) \\
\frac{1}{\nu} &= \frac{1}{\nu_0 \sqrt{1-\frac{v^2}{c^2}}}~\left(1-\frac{v\cos \theta {c}\right) \\
\nu &= \nu_0\frac{ \sqrt{1-\dfrac{v^2}{c^2}}}{1-\dfrac{v\cos \theta }{c}}
\end{align}
But when I tried this I got the following message:
[45]
! Missing } inserted.
<inserted text>
}
l.938 \end{align}
?
I copied the equations in Matcha (without the &), where it was perfectly working... I tried some things, but those did not seem to work...
Does anybody know what I did wrong?
At two occasions you write \frac{..} without the mandatary second argument. You must write \frac{...}{...} instead.
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
t_2' &= t_2 + \frac{L}{C} \\
t_1' &= t_1 + \frac{L + v\Delta t\cos\theta}{c} \\
t_2' - t_1' &= (t_2-t_1) + \frac{L-L-v\Delta t\cos\theta}{C} \\
\Delta t' &= \Delta t - \frac{v\Delta t \cos \theta}{c} \\
\Delta t' &= \Delta t \left(1-\frac{v\cos\theta}{c}\right) \\
\frac{1}{\nu} &= \frac{1}{\nu_0 \sqrt{1-\frac{v^2}{c^2}}}~\left(1-\frac{v\cos \theta}{c}\right) \\
\nu &= \nu_0\frac{ \sqrt{1-\dfrac{v^2}{c^2}}}{1-\dfrac{v\cos \theta }{c}}
\end{align}
\end{document}
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I am a beginner in latex. I have the following piece of latex code. The code is working fine but I wish all the equality operators and the "if kflag=n" of each equation be aligned and written in one equation box with one equation counting number. How can it be done?
\begin{equation} %kflag=0
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=0
\end{equation}
\begin{equation} %kflag=1
\left \{
\begin{array}{rl}
T =& \alpha\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& \frac{\alpha\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=1
\end{equation}
\begin{equation} %kflag=2
\left \{
\begin{array}{rl}
T =& (1-D)\sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}+T_{max}-\\
&(1-D)\sigma_{max,0}\times e\times\frac{\Delta_{max}}{\delta_n}+\\
&10\times \sigma_{max,0}\times e^{1-\frac{\Delta_n}{\delta_n}}\times \frac{\Delta_n}{\delta_n}\\
K =& 11\times\frac{(1-D)\sigma_{max,0}}{\delta_n}\times e^{1-\frac{\Delta_n}{\delta_n}}\times (1-\frac{\Delta_n}{\delta_n})\\
\end{array}
\right.
\quad \text{if} \quad kflag=2
\end{equation}
\begin{equation} %kflag=3
\left \{
\begin{array}{rl}
T =& T_{max}+K\times(\Delta_n-\Delta_{max})\\
K =& \frac{(1-D)\sigma_{max,0}*e}{\delta_n}\\
\end{array}
\right.
\quad \text{if} \quad kflag=3
\end{equation}
Right now the equations look like this
Here's an option using a multitude of nested structures - equation for the numbering, aligned for the horizontal alignment of structures and dcases (or cases) for the left-braced content.
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\newcommand{\Ddn}{\frac{\Delta_n}{\delta_n}}
\newcommand{\smz}{\sigma_{\mathrm{max}, 0}}
\begin{equation}
\begin{aligned}
&\begin{dcases}
\phantom{K}\mathllap{T} = (1 - D) \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 0 \\ % k-flag = 0
&\begin{dcases}
\phantom{K}\mathllap{T} = \alpha \smz \times e^{1 - \Ddn} \times \Ddn \\
K = \frac{\alpha \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 1 \\ % k-flag = 1
&\begin{dcases}
\phantom{K}\mathllap{T} = \begin{aligned}[t]
&(1 - D) \smz \times e^{1 - \Ddn} \times \Ddn + T_{\mathrm{max}} \\
&{} - (1 - D) \smz \times e \times \frac{\Delta_{\mathrm{max}}}{\delta_n} \\
&{} + 10 \times \smz \times e^{1 - \Ddn} \times \Ddn
\end{aligned} \\
K = 11 \times \frac{(1 - D) \smz}{\delta_n} \times e^{1 - \Ddn} \times \bigl( 1 - \Ddn \bigr)
\end{dcases} & \text{if $k$-flag} = 2 \\ % k-flag = 2
&\begin{dcases}
\phantom{K}\mathllap{T} = T_{\mathrm{max}} + K \times (\Delta_n - \Delta_{\mathrm{max}}) \\
K = \frac{(1 - D) \smz \times e}{\delta_n}
\end{dcases} & \text{if $k$-flag} = 3
\end{aligned}
\end{equation}
\end{document}
I am working on my thesis (a report type file) in which I have several consecutive equations associated to some mathematical model, the thing is that those are too separate to each other and looks not pretty good, so I would like to know how can customize this. I am using a report type file and I have been around though to many pages without finding a good answer to this.
The code is the following:
\begin{equation}
\sum_{k \in K} \sum_{i \in V_{1}}v_{ijkg} \leq d_{jg} \hspace*{0.3 cm} \forall j \in V_{2}, \hspace{0.1 cm} \forall g \in G
\end{equation}
\begin{equation}
\sum_{i \in V_{1}}\sum_{j \in V_{2}} \sum_{g \in G} v_{ijkg} \leq Q_{k} \hspace*{0.3 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{j \in V_{2}} \sum_{k \in K} v_{ijkg} \leq W_{ig}, \hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall g \in G
\end{equation}
\begin{equation}
d_{jg}y_{jk} \geq v_{ijkg}, \hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall j \in V_{2}, \hspace*{0.1 cm} \forall k \in K, \hspace*{0.1 cm} \forall g \in G
\end{equation}
\begin{equation}
v_{ijkg} \leq Q_{k} \ast z_{i}, \hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall j \in V_{2}, \hspace*{0.1 cm} \forall k \in K, \hspace*{0.1 cm} \forall g \in G
\end{equation}
\begin{equation}
v_{ijkg} \leq Q_{k} \ast F_{ik}, \hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall j \in V_{2}, \hspace*{0.1 cm} \forall k \in K, \hspace*{0.1 cm} \forall g \in G
\end{equation}
\begin{equation}
\sum_{g \in G}v_{ijkg} \leq M \ast \sum_{s \in V} x_{sjk}, \hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall j \in V_{2}, \hspace*{0.1 cm} k \in K
\end{equation}
\begin{equation}
\sum_{s \in V,s \neq h} x_{shk} - \sum_{s \in V,s \neq h} x_{hsk}=0,\hspace*{0.3 cm} \forall h \in V, \hspace{0.1 cm} \forall k \in K
\end{equation}
\begin{equation}
u_{jkg}-u_{lkg}+Q_{k}x_{jlk} \leq Q_{k}-\sum_{i \in V_{1}}v_{ijkg}, \hspace*{0.3 cm} \forall j,l \in V_{2},j \neq l, \hspace*{0.1 cm} \forall g \in G, \hspace*{0.1 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{j \in V_{2}}x_{ijk} + \sum_{j \in V_{2}} x_{jik} \leq 2,\hspace*{0.3 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{i \in V_{1}} \sum_{j \in V_{2}} x_{ijk} \leq 1, \hspace*{0.3 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{s \in V,s \neq j}x_{sjk}=y_{jk}, \hspace*{0.3 cm} \forall j \in V_{2}, \hspace*{0.1 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{k \in K}F_{ik} \leq I \ast z_{i}, \hspace*{0.3 cm} \forall i \in V_{1}
\end{equation}
\begin{equation}
\sum_{j \in V_{2}} x_{ijk} = F_{ik}, \hspace*{0.3 cm} \forall i \in V_{1} \hspace*{0.1 cm} \forall k \in K
\end{equation}
\begin{equation}
\sum_{s \in V} \sum_{p \in V} T^{c}_{sp} x_{spk} \leq SRT, \hspace*{0.3 cm} \forall k \in K_{1}
\end{equation}
\begin{equation}
\sum_{s \in V} \sum_{p \in V} T^{h}_{sp} x_{spk} \leq SRT, \hspace*{0.3 cm} \forall k \in K_{2}
\end{equation}
\begin{equation}
\begin{split}
x_{spk} \in \{0,1\} \hspace*{0.1 cm} \forall s,p \in V,s \neq p, & \hspace*{0.1 cm} \forall k \in K \hspace*{0.2 cm} \\ y_{ik} \in \{0,1\}, \hspace*{0.1 cm} i \in V_{1} \hspace*{0.1 cm} \forall k \in K & \\ v_{ijkg} \in \mathbb{Z}^{+0}, \hspace*{0.1 cm} \forall i \in V_{1}, \hspace*{0.1 cm} \forall j \in V_{2}, \hspace*{0.1 cm} \forall & k \in K, \hspace*{0.1 cm} \forall g \in G \\ z_{i} \in \{0,1\}, \hspace*{0.1 cm} \forall i \in V_{1}& \\ F_{ik} \in \{0,1\}, \hspace*{0.1 cm} i \in V_{1} \hspace*{0.1 cm} \forall & ,k \in K \\ u_{jkg} \in \mathbb{Z}^{+}, \hspace*{0.1 cm} j \in V_{2} \hspace*{0.1 cm} \forall k \in & K, \hspace*{0.1 cm} \forall g \in G
\end{split}
\end{equation}
Here is a picture of some part of the output. The equations list take two entirely sheets.
any ideas? thanks in advance.
You could temporarily reduce the space above/below of the equations:
{
\addtolength{\belowdisplayskip}{-0.5ex}
\addtolength{\belowdisplayshortskip}{-0.5ex}
\addtolength{\abovedisplayskip}{-0.5ex}
\addtolength{\abovedisplayshortskip}{-0.5ex}
\begin{equation}
....
\end{equation}
\begin{equation}
....
\end{equation}
}