I am trying to prove that $P_{FL} \subset P_{AFL}$ where \begin{align}P_{FL}&=\left\{({\bf x},{\bf y})\,\,\middle\vert\,\,\forall i,j:\sum_{j=1}^nx_{ij}=1,x_{ij}\le y_j,0\le x_{ij},y_j\le1\right\}\\P_{AFL}&=\left\{({\bf x},{\bf y})\,\,\middle\vert\,\,\forall i,j:\sum_{j=1}^nx_{ij}=1,\sum_{i=1}^mx_{ij}\le my_j,0\le x_{ij},y_j\le1\right\}.\end{align} I could get to a point that given $x_{ij} \leq y_j$ from $P_{FL}$ summing over $m$ to get $\sum\limits_{i=1}^m x_{ij} \leq my_j$. Hence, $P_{FL}$ is at least as strong as $P_{AFL}$. However, I couldn't provide a point where $P_{FL} \setminus P_{AFL}$ to prove the inclusion is strict. How can I prove it?


1 Answer 1


Without loss of generality take $m=2$. Then

$$x_i \leq y\implies\sum_{i=1}^m x_i \leq my$$

is proven by direct summation as in the OP.

On the other hand $$ \sum_{i=1}^m x_i \leq my\quad\not\!\!\!\!\implies x_i \leq y $$ is proven by carefully choosing a counterexample. I shall take $x_1 = y+\frac{\epsilon}{2}> y , x_2 = y-\epsilon$. The FL constraint is clearly violated for $i=1$, and yet the AFL constraint is satisfied:

$$ \sum_i x_i = 2y - \frac{\epsilon}{2} < 2y.$$

  • $\begingroup$ Thanks a lot Konstantin! $\endgroup$
    – cedric
    Dec 11, 2020 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.