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I'm currently solving some Operations Research exercises related to Integer Programming. In one of the solutions of the exercises the author uses the following formula for the objective function:

$\sum_{i=1}^{m} \sum_{j=1}^{n} C_{ij} x_{ij} $

and for the constraints it uses the following:

$\sum_{i=1}^{m} x_{ij} = d_j , (j = 1,2,.., n) $

My question is twofold:

  1. What is the meaning of the double summation in the objective function?
  2. Could we use the double summation in the constraint, instead of using this $ (j = 1,2,.., n) $ ?

I hope my question is clear enough. If not I will try to put in different terms. Thank you very much in advance.

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Thank you. I finally understood the differences. Just to make sure, basically on the second case, we are fixing j and calculating the summation just for the i's. Is that it? –  bacchus May 2 '11 at 22:15
    
yes, that's right. –  Alon Amit May 2 '11 at 22:42

3 Answers 3

up vote 3 down vote accepted
  1. It means you sum over all combinations of $i$ and $j$. If $m=3$ and $n=4$ you'll have 12 summands in the sum.

  2. No. There are $n$ constraints and that's very different from adding them all up and having just one constraint. You must have $\sum x_{i1} = d_1$ and also $\sum x_{i2} = d_2$ etc.

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Is it much to ask for references on material about summations (i.e. books, links, etc)? –  bacchus May 2 '11 at 22:17
    
@bacchus, you could try en.wikipedia.org/wiki/Summation –  Gerry Myerson May 3 '11 at 0:01
1  
@bacchus, this is covered in many textbooks but one recommendation might by "Concrete Mathematics" by Graham, Knuth and Patashnik. –  Alon Amit May 3 '11 at 0:32

The first means you have two matrices $x$ and $C$, you take the transpose of $x$ and multiply it by $C$ (meaning matrix multiplication, not componentwise or Hadamard multiplication) and finally take the trace. So

$$\sum_{i=1}^{m} \sum_{j=1}^{n} C_{ij} x_{ij} = \text{tr}(C x^{\text{T}})$$

For the second, no. You have $n$ constraints, summing them would reduce it to 1 constraint (moreover dependent on the other $n$.)

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  1. I suggest you view the summation layer by layer, ie, fixing the outer layer index, in this case, $i$, say $i = 1$, then the inner layer summation is $\displaystyle \sum^{n}_{j=1} C_{1j} x_{1j} = (C_{11},\ldots,C_{1n}) \cdot (x_{11},\ldots,x_{1n})^T$, you could view this as the first row of matrix $C = (C_{ij})$ multiplying with the first column of transpose matrix of $x = (x_{ij})$, and the result is the 1st entry on the diagonal of the product matrix $C x^T$. Now thinking $i$ runs from 1 to $m$, you will get the summation of all the entries on the diagonal, which is the trace of $C x^T$; it is easier to understand if you visualize it as matrix product $C x^T$, an $m \times n$ matrix $C$ multiplying with an $n \times m$ matrix $x^T$ : $$ C x^T = \begin{pmatrix} C_{11} & C_{12} & \cdots & C_{1n} \\ C_{21} & C_{22} & \cdots & C_{2n} \\ &&\cdots& \\ C_{m1} & C_{m2} & \cdots & C_{mn} \end{pmatrix} \cdot \begin{pmatrix} x_{11} & x_{12} & \cdots & x_{m1} \\ x_{12} & x_{22} & \cdots & x_{m2} \\ \vdots&\vdots&&\vdots \\ x_{1n} & x_{2n} & \cdots & x_{nm} \end{pmatrix} $$ when you fix an $i$, then the summation with respect to $j$ is the $i$th row multiplying with the $i$th column of above matrix multiplication.

  2. No, the constraint is essentially saying: when you fix $j$( $j$ is the column index of the matrix $x = (x_{ij})$, also the row index of the matrix $x^T = (x_{ji})$), the sum of the entries on $j$th row of the matrix $x^T$ is a given number $d_j$; visualizing it as equation system would be like $$ \left\{ \begin{eqnarray} x_{11} + x_{21} + \cdots + x_{m1} &= d_1 \\ x_{12} + x_{22} + \cdots + x_{m2} &= d_2 \\ \cdots& \\ x_{1n} + x_{2n} + \cdots + x_{mn} &= d_n \end{eqnarray} \right. $$ if you do the summation again, it will enlarge the solution space of $x$ dramatically, the optimization of the utility function may yield a different result.

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