$C_t$ and $S_i$ are two decision variables.
(1) $C_t$=$C_{(t-1)}$+$D_i$ if t=$S_i$, $\forall${i,t}
(2) $C_t$=$C_{(t-1)}$, if $t\neq S_i$, $\forall${i,t}
How can I present the above two conditional equations as a linear constraint? Appreciated!
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The question is unclear. – Mateo Jul 08 '22 at 17:27
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I'm a little confused about the quantification here. What if $t = S_3$ but $t \neq S_2$? Then from $(1)$ we get $C_t = C_{t-1} + D_3$, but from (2) we get $C_t = C_{t-1}$. Maybe you mean something like (1) $\forall i, \forall t, ,t = S_i \implies C_t=C_{(t-1)}+D_i$ and (2) $\forall t \left( (\forall i,, t \neq S_i) \implies C_t=C_{(t-1)}\right)$? – Jair Taylor Jul 08 '22 at 17:32
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also (2) implies if $S_i = S_j$ then $D_i = D_j$, is that right? – Jair Taylor Jul 08 '22 at 20:12
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$t={1,2,3,...,T}$, and $S_i$ can be a set of times between 1 and T. If $t=S_{i}$ at any point of time, then $C_t=C_{(t-1)}+D_i$. On the other hand, If $t\neq S_{i}$ at any point of time, then $C_t=C_{(t-1)}$. These two conditions can be written as: $$C_t= \begin{cases} \ C_{(t-1)}+D_i\hspace{1cm}\text{if } t=S_{i} \hspace{1cm}\forall{i,t}\ \ C_{(t-1)}\hspace{1cm} \text{if } t\neq S_{i} \hspace{1cm} \forall{i,t} \end{cases}$$ – SMU Jul 12 '22 at 00:20