Align many multiline equations

Question

I am struggling trying to align a set of equations of very different lengths. All equations shell be aligned to the comma between the formula and the variable declaration.

Despite the use of split, formula 3 is still too long. If I do another line break between \left( and \right), I get an error "Extra }, or forgotten right".

As soon as the splitting is done, I would like to arrange the first line of such a multiline formula left aligned, the next lines left aligned with a small skip in front, and only the last line right aligned to the comma. Is there a possibility to do this? Or any other sugesstions of presenting many of such long formulas together in one environment in a pretty way?

I found solutions for arranging one long formula in many lines OR to arrange many formulas together around a comma or something else OR to leftalign one formula like I described (mathtools' \MoveEqLeft), but I have no idea how to do all at the same time.

\documentclass[]{scrreprt}
\usepackage{amsmath}

\begin{document}
\begin{align}
    \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1} 
        \leq h_i 
        \leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k 
             + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
    & ,     i \in \mathcal{I}_{21}
    \label{eq_1} \\
%
    \begin{split}
        \sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right) 
            + w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
    &   \\
        \geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
    & ,  i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
    \end{split} 
    \label{eq_2} \\
%
    \begin{split}   
        \left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right) 
            \cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot  \ldots 
        & \\
        \ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - 
        | h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
        = 0 
        & , i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} 
    \end{split}
    \label{eq_3} \\
%
    H_{21} = H_{12} 
    &       
    \label{eq_4}
\end{align}
\end{document}

Addition:
Until now I used the old eqnarray-environment. Using that, it looks nearly the way it should:

\documentclass{scrreprt}
\begin{document}
\begin{eqnarray}
\sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1} 
\leq h_i 
\leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
& , {} &    i \in \mathcal{I}_{21}
\label{eq:sf_height_fits_levelheight_i21}\\
%
\left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right) 
\times \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \qquad \qquad \quad
&    &      \nonumber \\
\qquad \times \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - | h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon |\right)
\geq 0
%
& ,  &      i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} \qquad \quad
\label{eq:sf_I22_not_in_new_I21} \\
H_{21} = H_{12}                             
&   &       
\label{eq:sf_def_H21}
\end{eqnarray}
\end{document}

Only problem: formular 3 is too wide, and formula numbers run out of text area (and eqnarray-environment is not the first choice).

Answer 1

Here's a solution that uses aligned environments inside the align environment. It loads the \mathtools package (and its \smashoperator macro) to squeeze the space around the summation symbols. The third equation gets an additional line break and now spans three lines.

\documentclass{scrreprt}
\usepackage{mathtools}
\DeclarePairedDelimiter\abs\lvert\rvert
\begin{document}
\setcounter{chapter}{1} % just for this example

\begin{align}
&\smashoperator[r]{\sum_{k=K_1+1}^{K_{21}-1}} 
   s_{ik} \tilde{h}_{k+1} 
   \leq h_i \leq 
   \smashoperator{\sum_{k=K_1+1}^{K_{21}}} 
   s_{ik} \tilde{h}_k + 
   \smashoperator{\sum_{k=K_1+1}^{K_{21}-1}} 
   \bar{s}_{ik} \tilde{h}_{k+1},
   \quad i\in\mathcal{I}_{21} \label{eq_1} \\[2ex]
%
&\smashoperator[r]{\sum_{l=K_{21}+1}^{K_{22}}}
  \begin{aligned}[t]
  &\alpha_{lk} (s_{il} + \bar{s}_{il}) (1+\varepsilon) 
    + w_i + 
   \smashoperator{\sum_{j=I_1+1}^{I_2}} 
   \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j\\
  &\geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
    ,\quad  i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
  \end{aligned} 
    \label{eq_2} \\[2ex]
%
&\begin{aligned}   
  &\biggl( 1- \smashoperator{\sum_{j=I_1+1}^{I_2}}
          \alpha_{ji}t_{jk}\biggr)
   \times \biggl( w_i - \frac{1}{m+2} - \varepsilon -
   \abs[\Big]{w_i - \frac{1}{m+2} - \varepsilon}
   \biggr) \times \dotsb \\
  &\quad\dotsb\times\biggl( h_i + 
   \smashoperator{\sum_{l=K_1+1}^{K_2}} 
   \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - 
   \abs[\Big]{ h_i + 
   \smashoperator{\sum_{l=K_1+1}^{K_2}} 
   \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon } 
   \biggr)
= 0,\\
  &\qquad\quad i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} 
\end{aligned} \label{eq_3} \\[2ex]
%
&H_{21} = H_{12}       
\label{eq_4}
\end{align}
\end{document}

Answer 2

with combination of align and multlined environments from mathtools. i also will not use \cdot for multiplications (it is self understanding that terms in braces are multiplied):

\documentclass{scrreprt}
\usepackage{mathtools}
\DeclarePairedDelimiter\abs{\lvert}{\rvert}

\setcounter{chapter}{1} % just for this example
\begin{document}
\begin{align}
&   \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1}
        \leq h_i \leq
    \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k
    + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1},
     \qquad  i \in \mathcal{I}_{21}
    \label{eq_1}        \\[1em]
%
&   \begin{multlined}[0.7\linewidth]
    \sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk}(s_{il} + \bar{s}_{il})
    (1 + \varepsilon) + w_i    \\
    + \sum_{j=I_1+1}^{I_2} \alpha_{ji}(s_{jk} + \bar{s}_{jk}) w_j
        \geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik}),
    \qquad i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
    \end{multlined}
    \label{eq_2}        \\[1em]
%
&   \begin{multlined}[0.7\linewidth]
        \biggl( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\biggr)
            \biggl( w_i - \frac{1}{m+2} - \varepsilon -
            \abs*{w_i - \frac{1}{m+2} - \varepsilon}\biggr) \dotsm  \\
    \dotsm \biggl( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l
        - H_{12} - \varepsilon - \abs[\bigg]{h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk}
        \tilde{h}_l - H_{12} - \varepsilon }\biggr) = 0,    \\
        i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21}
    \end{multlined}
    \label{eq_3}    \\[1em]
%
&    H_{21} = H_{12}
    \label{eq_4}
\end{align}
\end{document}

Answer 3

You could try using aligned instead of split. Is this what you are after?

\documentclass[]{scrreprt}
\usepackage{amsmath}

\begin{document}
\begin{align}
    \begin{aligned}
    \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1} 
        \leq h_i 
        \leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k 
             + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
    & ,     i \in \mathcal{I}_{21}
    \end{aligned}
    \label{eq_1} \\
%
    \begin{aligned}
        \sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right) 
            + w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
    &   \\
        \geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
    & ,  i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
    \end{aligned} 
    \label{eq_2} \\
%
    \begin{aligned}   
        \left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right) 
            \cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot  \ldots 
        & \\
        \ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - 
        | h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
        = 0 
        & , i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} 
    \end{aligned}
    \label{eq_3} \\
%
    H_{21} = H_{12} 
    &       
    \label{eq_4}
\end{align}
\end{document}

Answer 4

Perhaps this is more along the lines of what you are looking for. I was able to split up the last equation in the middle of the parentheses (using $\left.$ and $\right.$), and I opted to use horizontal spacing to achieve your indentation.

\documentclass[]{scrreprt}
\usepackage{amsmath}

\begin{document}
\begin{align}
   & \sum_{k=K_1+1}^{K_{21}-1} s_{ik} \tilde{h}_{k+1} 
        \leq h_i 
        \leq \sum_{k=K_1+1}^{K_{21}} s_{ik} \tilde{h}_k 
             + \sum_{k=K_1+1}^{K_{21}-1} \bar{s}_{ik} \tilde{h}_{k+1}
    & , & i \in \mathcal{I}_{21}
    \label{eq_1} \\
%
   \begin{split}
       & \sum_{l=K_{21}+1}^{K_{22}} \alpha_{lk} (s_{il} + \bar{s}_{il}) \left( 1 + \varepsilon \right) 
            + w_i + \sum_{j=I_1+1}^{I_2} \alpha_{ji} (s_{jk} + \bar{s}_{jk}) w_j
    \\
       & \kern 2cm \geq (1 + \varepsilon)(1-s_{ik}-\bar{s}_{ik})
    \end{split} 
    & , & i \in \mathcal{I}_{22}, k \in \mathcal{K}_{22}
    \label{eq_2} \\
%
   \begin{split}   
        & \left( 1- \sum_{j=I_1+1}^{I_2}\alpha_{ji}t_{jk}\right) 
            \cdot \left( w_i - \frac{1}{m+2} - \varepsilon -|w_i - \frac{1}{m+2} - \varepsilon|\right) \cdot  \ldots 
        \\
        & \kern 2cm \ldots \cdot \left( h_i + \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon - 
        | h_i + \right.
        \\
        & \kern 3.85cm \left. \sum_{l=K_1+1}^{K_2} \alpha_{lk} \tilde{h}_l - H_{12} - \varepsilon | \right)
        = 0 
    \end{split}
    & , & i \in \mathcal{I}_{22}, k \in\mathcal{K}_{21} 
    \label{eq_3} \\
%
    & H_{21} = H_{12}       
    \label{eq_4}
\end{align}
\end{document}