TITLE: Six Octagons on a Surface with $\chi=-2$
QUESTION [5 upvotes]: Given a cubic bipartite graph $G$, living happily on an orientable surface with Euler characteristic $\chi$. Euler's formula then reads:
$$
F+V=E+\chi , 
$$
where $F=\sum f_k$, the number of $k$-gons. Now, let's restrict $G$ to $f_6$ hexagons and $f_8$ octagons only. Since its $3$-regularity it follows that:
$$
0\cdot f_6 -2 \cdot f_8=6\chi
$$
all along similar lines as given here...

How does the simplest non-trivial ($\chi=-2 \rightarrow f_8=6$) example look like?

Feel free to add as countably many hexagons, in case, you need them...

REPLY [4 votes]: If you cover the double torus with only hexagons and octagons such that edges and vertices form a 3-regular graph, then you need exactly 6 octagons. 
$$
f_6+f_8+(6f_6+8f_8)/3=(6f_6+8f_8)/2 +  \chi
$$
Since $\chi=-2$, we have $f_8=6$. 
From Euler's formula there is no restriction for the number of hexagons, but maybe there are geometric restrictions. The picture shows a covering with 6 octagons and no hexagon.