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Let $x\in \mathbb{R}^{n}$, $Y\in \mathbb{R}^{mxn}$.

We can then define:

$row_{i}(Y)=$ $i^{th}$ row of $Y$
$column_{i}(Y)=$ $i^{th}$ column of $Y$
$x_{i}=i^{th}$ element of $x$
$sum(x)=$ sum of the element of $x$
$card(x)=$ number of non-zero element of $x$.

My Goal is to devise a method to determine if it is possible to construct a matrix $k\in \mathbb{R}_{\geq 0}^{mxn}$ given some $a\in \mathbb{R}^{n}$, $b, c\in\mathbb{R}^{m}$ for some $m, n \in \mathbb{N}_{>0}$ where $sum(a)=sum(b)$

s.t. for all $ i, j \in \mathbb{N}_{\geq0}$ with $0<i<m$ and $0<j<n$:

$\bullet$ $sum(row_{i}(K))=b_{i}$
$\bullet$ $sum(column_{j}(K))=a_{j}$
$\bullet$ $card(row_{i}(K))=c_{i}$

Any suggestions?

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  • $\begingroup$ In your two final statements under Goal (the only interesting, the rest is redundant really), don't you mean rows in one of them, and columns in the other, and awj instead of awi? $\endgroup$ – Johan Löfberg Dec 13 '13 at 15:08
  • $\begingroup$ You are right, I changed the question to make it more straightforward. $\endgroup$ – Angel Eyes Dec 13 '13 at 22:10
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So you are essentially asking how you can constrain the cardinality of a vector. The answer below is based on a standard approach called big-M modelling.

Let $x$ be a vector for which you want exactly $c$ non-zeros. Introduce a new binary vector $y$ which indicates if an element is zero or non-zero. Let $M$ be an upper bound on the possible value of an element in $x$ at optimality (from domain knowledge)

If $y_i=0$ then $x_i$ should be $0$:

$x \leq My$

This also ensures $y_i=1$ when $x_i$ is non-zero.

If $y_i=1$ then $x_i$ non-zero. This is a bit tricky, since you have to define what non-zero is in the numerical precision you are working with:

$ x \geq 0.00001y$

It also ensures $y_i=0$ if $x_i=0$

Exactly $c$ elements are non-zero

$\sum y = c$

All remaining constraints are standard linear programming constraints, and the logic above is applied on every row of $K$.

The code below implements an example in the MATLAB Toolbox YALMIP (disclaimer:developed by me)

m = 5;
n = 10;
% create a known solution
K0 = sprand(m,n,.25)
a = sum(K0,2);
b = sum(K0,1);
c = sum(K0>0,2);

% Define decision variables
K = sdpvar(m,n,'full')
Y = binvar(m,n,'full');
M = max(max(a),max(b));

Model = [sum(K,1)==b, sum(K,2)==a, K>=0];
Model = [Model, K <= M*Y, K >= 0.00001*Y,sum(Y,2)==c];

% Let us find a solution with smallest largest element
Objective = max(max(K));
solvesdp(Model,Objective);
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