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Let $i,j \in N$ with $i \neq j$, and let $t \in T$.

The sum is written as : $$\sum_{j \in N \setminus\{i\} }(x_{i,j,t} + x_{j,i,t}) = 2 \qquad \forall i \in N, t \in T$$

Would this be the same as writing:

$$\sum_{t \in T}\sum_{i \in N} \sum_{j \in N \setminus\{i\}} (x_{i,j,t} + x_{j,i,t}) = 2 $$

If not, is there a way to express the first sum without specifying $\;\forall i \in N, t \in T$ and only using sums?

Markus Scheuer
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mathplzfun
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1 Answers1

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We consider \begin{align*} \sum_{j \in N \setminus\{i\} }(x_{i,j,t} + x_{j,i,t}) &= 2 \qquad\qquad \forall i \in N, t \in T\tag{1}\\ \sum_{t \in T}\sum_{i \in N} \sum_{j \in N \setminus\{i\}} (x_{i,j,t} + x_{j,i,t}) &= 2\tag{2} \end{align*}

  • In (1) we have a system of equations. For each pair $(i,t)\in N\times T$ we consider the equation \begin{align*} \sum_{j \in N \setminus\{i\} }(x_{i,j,t} + x_{j,i,t})=2 \end{align*}

  • In (2) we have a single equation: \begin{align*} \sum_{t \in T}\sum_{i \in N} \sum_{j \in N \setminus\{i\}} (x_{i,j,t} + x_{j,i,t}) = 2 \end{align*} where we take the sum in (1) and sum it up over all $t\in T$ and $i\in n$.

Note if $N$ and $T$ are finite, we can sum up all equations of (1) and get \begin{align*} \color{blue}{\sum_{t \in T}\sum_{i \in N}\sum_{j \in N \setminus\{i\} }(x_{i,j,t} + x_{j,i,t})=2|N| |T|} \end{align*}

Markus Scheuer
  • 108,315