Linear equation | Finding the unknown values

Question

I have a set of equations as follow,

• $a_1\cdot x_1 + a_2\cdot x_2 + a_3\cdot x_3 + a_4\cdot x_4 + \dots + a_{11} \cdot x_{12} + a_{12}\cdot x_{12} = 6000$
• $x_1 + x_2 + x_3 + x_4 + \dots x_{11} + x_{12} = 100 $
• $\{a_1,a_2,a_3, \dots , a_{11}, a_{12}\} \leq 60$

How can I find the values for all $a_i, \text{ and } x_i$,not all values of $a_i$ can be equal to $60$ also if possible integer, else the general solution.

• How the question get affected if I change the constrain on $a_i$ to some $X$, say $0 < a_i \leq X$,
• Also if the constrain on $a_i$ is , say $a_i>0$.

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If solvable in Matlab please show the code also .

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If not solvable please mention why.

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What I tried is $$ \begin{bmatrix} a_1&a_2 & a_3 &\dots & a_{11} & a_{12} \\ 1 & 1 & 1& \dots & 1 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots\\ x_{11} \\ x_{12} \end{bmatrix} = \begin{bmatrix} 6000 \\ 100 \end{bmatrix} $$

The rank will be almost $2$ so atmost I can get $10$ free variables. but since the values for $a_i$'s are also unknown I don't know how to proceed.

Answer 1

Assume that $\exists i\in \{1, 2, ..., 12\}: a_i < 60$, then $$6000 = a_1 x_1 + ... + a_{12} x_{12} < 60x_1 + ...+ 60x_{12} = 60(x_1 + ...+x_{12}) = 6000 $$ is contradiction. So $\forall i \in \{1, 2, ..., 12\}: a_i \geq 60$, so $a_1 = a_2 = ... = a_{12} = 60$

Answer 2

So we have $$ \left\{ \matrix{ 0 \le a_{\,k} \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 0 \le x_{\,k} \le 100\quad \left| {\,1 \le k \le 12} \right. \hfill \cr x_{\,1} + x_{\,2} + \cdots + x_{\,12} = 100 \hfill \cr {\bf a} \cdot {\bf x} = 6000 \hfill \cr} \right. $$

If we put $b_k = 60 -a_k$ then $$ \left\{ \matrix{ b_{\,k} = 60 - a_{\,k} \quad \left| {\,1 \le k \le 12} \right. \hfill \cr u_{\,k} = 1\quad \left| {\,1 \le k \le 12} \right. \hfill \cr 0 \le x_{\,k} \le 100 \hfill \cr 0 \le b_{\,k} \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr {\bf u} \cdot {\bf x} = 100 \hfill \cr 6000 = {\bf a} \cdot {\bf x} = \left( {60\,{\bf u} - {\bf b}} \right) \cdot {\bf x} = \hfill \cr = 6000 - {\bf b} \cdot {\bf x}\quad \Rightarrow \quad {\bf b} \cdot {\bf x} = 0 \hfill \cr} \right. $$

and clearly, with ${\bf b}, \, {\bf x}$ having non-negative components we must have $$ \left\{ \matrix{ {\bf b} = {\bf 0}\; \Rightarrow \;{\bf a} = 60\,{\bf u} \hfill \cr 0 \le x_{\,k} \le 100 \hfill \cr {\bf u} \cdot {\bf x} = 100 \hfill \cr} \right. $$

That is, $\bf x$ are the vectors lying on the diagonal plane of a 12-D hypercube of side $[0,100]$.

There are $101^{12} \approx 10^{24}$ points inside that cube, and those on the diagonal plane are exactly the number of weak compositions of $100$ into $12$ parts , i.e. $\binom {100+12-1}{11} \approx 2.8 \cdot 10^{14}$.

So your first care should be how to manage such a huge amount of data, if you want to list all the solutions.

You can somewhat reduce the size by taking advantage of the permutation invariance.
If $(x_1,x_2, \cdots)$ is a solution, then any permutation of the twelve components will be.
So you can limit to list the solutions in order, e.g. non-decreasing.
The number in this case will be reduced to that of the partitions of $100$ into up to twelve parts (because the components may be null), which is $ \approx 1.7 \, \cdot 10^7$.