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\end{alignat} 给出未定义的控制序列错误

tag-icon • tag-icon errors undefined alignat
\end{alignat} 给出未定义的控制序列错误

编译此代码时，它返回一个包含编译方程的 pdf。但是，此块下面的所有命令都不再执行。对于\end{alignat}，它返回未定义的控制序列错误。为什么会发生这种情况？提前致谢！

\begin{subequations} 
\allowdisplaybreaks[1]
\renewcommand{\theequation}{\theparentequation.\arabic{equation}}
\begin{alignat}{3}
    & \begin{dcases}
        \sum_{a=1}^{dm}\dfrac{(df_a)}{2}\ \text{als dm even} \\
        \sum_{a=1}^{dm}(df_a)\ \text{als dm oneven}
    \end{dcases}\\
    \text{met}& \nonumber\\
    &M = alt + 1 \nonumber\\
    & \epsilon \text{ is de precisie van de oplossing} \nonumber\\
    \text{onder voorwaarde dat}& \nonumber\\
    &C_j = \sum_{i=1}^{alt}(c_{ij}*i)\ &&\forall j=1,..,alt\\
    & \sum_{i=1}^{alt}(c_{ij}) = 1\ &&\forall j=1,..,alt \\
    & \sum_{j=1}^{alt}(c_{ij}) = 1\ &&\forall i=1,..,alt \\
    &T_{a+1} > T_a\ &&\forall a=1,..,dm-1 \\
    & \sum_{a=1}^{dm}(t_{ab}) = 1\ &&\forall b=1,..,dm \\
    & \sum_{b=1}^{dm}(t_{ab}) = 1\ &&\forall a=1,..,dm \\
    &T_b = \sum_{a=1}^{dm}(Tlinear_{ab})\ &&\forall b=1,..,dm \\
    &Tlinear_{ab} \leq t_{ab}*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Tlinear_{ab} \geq - t_{ab}*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Tlinear_{ab} \leq fi_a + (1-t_{ab})*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Tlinear_{ab} \geq fi_a - (1-t_{ab})*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &C_j \geq C_i + M*(gc_{ij} - 1)\ &&\forall i=1,..,alt-1; j=i+1,..,alt \\
    &C_i \geq C_j - M*gc_{ij}\ &&\forall i=1,..,alt-1; j=i+1,..,alt \\
    &fi_{a} = 1+ \dfrac{2\sum\limits_{j=i+1}^{alt}\sum\limits_{i=1}^{alt-1} (2gc_{ij}-1)*g_{aij}}{alt^2-alt} &&\forall a=1,..,dm \\
    &W_b = \sum_{a=1}^{dm}(Wlinear_{ab})\ &&\forall b=1,..,dm \\
    &Wlinear_{ab} \leq t_{ab}*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Wlinear_{ab} \geq -t_{ab}*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Wlinear_{ab} \leq w_a + (1-t_{ab})*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &Wlinear_{ab} \geq w_a - (1-t_{ab})*M\ &&\forall a=1,..,dm; b=1,..,dm \\
    &CW_{a} = \sum_{b=1}^{a}{W_b}\ &&\forall a=1,..,dm \\
    &CW_{a} - M*d_a \leq 0.5-\epsilon \ &&\forall a=1,..,dm \\
    &CW_{a} + (1-d_a)*M \geq 0.5\ &&\forall a=1,..,dm \\
    &df_a \leq med_a*M\ &&\forall a=1,..,dm \\
    &df_a \geq -med_a*M\ &&\forall a=1,..,dm \\
    &df_a \leq T_a + (1-med_{a})*M\ &&\forall a=1,..,dm \\
    &df_{a} \geq T_{a} - (1-med_{a})*M\ &&\forall a=1,..,dm \\
    & \sum_{a=1}^{dm}(a*med_{a}) = dm + 1 - \sum_{a=1}^{dm}d_{a}\\
    & \sum_{a=1}^{dm} med_{a} = 1
\end{alignat}
\end{subequations}%

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