I found the fractional covering number of the graph in the figure. And as a result, I came across the result of 5/2. Next I wanted to show its equality with fractional packing number. I found the n-fold packings. I took advantage of their multiplicity while finding the packings.
- $m(1)+m(2) \le n$
- $m(1)+m(3) \le n$
- $m(2)+m(3) \le n$
- $m(2)+m(4) \le n$
- $m(3)+m(5) \le n$
- $m(4)+m(5) \le n$
When we add them all together, the result is as follows: $2m(1)+3m(2)+3m(3)+4m(2)+5m(2)+6m(2) \le 6n$
Then : $\sum_{}^{} 2(m_i) + m(2) + m(3) \le 6n$ we reach the conclusion.
We define$\sum_{}^{} m_i = \left| Y \right|$.
We can say that $2\left| Y \right| \le 6n$.
Here comes the result $\left| Y \right| \le 3n$. I can't reach the result of $\frac{5n}{2}$.
I couldn't find the equality. If $m(2)+m(3)$ did not exist, we could reach the result of $\frac{5n}{2}$, but since it is there, there is redundancy. I think I got it right while finding the fractional covering number, it provided all the conditions.
Don't just add the constraints. Instead, add the first three and twice the last one, yielding $2|Y| \le 5n$, as desired. These "magic constant multipliers" $(1,1,1,0,0,2)$ arise from linear programming dual variables.