I am currently stuck on the following exercise:
Let $G$ be a simple, undirected, connected and bipartite graph with weights $c: E(G) \to \mathbb{N}_{>0}$. Let $$ \mathcal{I} = \{F \subseteq E(G) \mid F \text{ is a matching and } G - F \text{ remains connected} \}.$$ For which (smallest) number $k$ is the Matroid greedy algorithm an $1/k$ approximation-algorithm for the problem of finding a maximum weight set $F \in \mathcal{I}$. Prove that the choice of $k$ is tight.
Here is what I've got so far:
A reasonable choice would be $k=3$: Let $V(G) = A \dot \cup B$ be the bipartition. Define the following independence systems: $$ \mathcal{I}_1 := \{F \subseteq E(G) \mid |\delta_G (v) \cap F| \leq 1 \text{ for all } v \in A \} \\ \mathcal{I}_2 := \{F \subseteq E(G) \mid |\delta_G (v) \cap F| \leq 1 \text{ for all } v \in B \} \\ \mathcal{I}_3 := \{F \subseteq E(G) \mid G - F \text{ remains connected} \} \quad$$
Now, $\mathcal{I} = \mathcal{I}_1 \cap \mathcal{I}_2 \cap \mathcal{I}_3 $ and these three are actually matroids as $\mathcal{I}_3$ is the dual matroid of the graphic matroid. Therefore $$ \frac{|B_\min|}{ |B_\max|} \geq \frac{1}{3}$$ which is a lower bound for the approximation ratio, where $B_\max$ is a base of $\mathcal{I}$ with maximum cardinality and $B_\min$ respectively.
Therefore, $k = 3$ is a legitimate choice and it remains to show that it is indeed tight. This is the problem that I'm having. We are looking for a graph where there are two bases $B_1, B_2$ of $\mathcal{I}$ with $$ \frac{|B_1|}{|B_2| } = \frac{1}{3}.$$
I tried some subgraphs of $K_{3,3}$, but they don't seem to work out. That is why I am wondering whether my choice of $k$ is actually tight or one could improve it. If not, what is a graph where there are two such bases? Any help is appreciated.