Formulating a coefficient from a table

Question

We have a table of data as follows $$\begin{array}{c|c|c|c|c} & b \leqslant a & a < b \leqslant 2a & 2a < b \leqslant 3a & 3a < b \leqslant 4a \\ \hline i=1 & e & 2d & 2d & 2d \\ \hline i=2 & - & e & d & d \\ \hline i=3 & - & - & e & \frac{2}{3}d \\ \hline i=4 & - & - & - & e \\ \hline \end{array}$$

The data inside every cell of this table corresponds to a parameter of the problem which we show by $c_i$.

How can we formulate $c_i$ without adding another index to $c$?

There seems to be a connection between the value of $i$ and the interval to which $b$ belongs (the coefficient of $a$) but I don't know how to model it.

EDIT

I think I should have mentioned that $a, b, e, d$ and $c_i$ are all parameters of the problem.

Answer 1

$z_{j} \in \{0,1\}$ equals 1 represents variable $b$ belongs to interval $j$. You can linearize interval relation as follows: $$ A_{j-1} \cdot z_{j} + \epsilon \cdot z_{j} \le b_{j} \le A_{j} \cdot z_{j} \quad \forall j \in {1,2,3,4} $$ Here, this leads to 4 binary variables $z_{j}$ and values of ${A_0}$,${A_1}$,${A_2}$,${A_3}$,${A_4}$ are $0$, $a$, $2a$, $3a$ and $4a$ respectively (assuming 0 to be lower bound of variable $b$). $\epsilon$ is introduced on LHS to handle strict less than condition.

variable $b$ is to be defined as $$ b = \sum_{j} b_{j} $$ and only one of the intervals has to be selected $$ \sum_{j} z_{j} = 1 $$ To achieve above linearization, you might consider using SOS Type-1 relation as well. If a is also a variable, you might have to write some Big-M constraints to linearize the same. finally, $c_{i}$ can be defined as follows: $$ c_{i} = \sum_{j\mid j\ge i} W_{ij} z_{j} $$ where $W_{ij}$ represents tabular parameters corresponding to row $i$ and column $j$