Minimum cost closed walk in a graph
This problem is equivalent to TSP. Compute all pairwise shortest distances between the vertices in the given graph $G$. Then take the complete graph $K_n$ that is weighted with the shortest-distances of the original graph $G$. The TSP tour of the complete graph corresponds to the shortest min-cost closed walk. 

More precisely, a shortest tour $\pi$ in $K_n$ decodes a closed walk in $G$: just replace every edge $(u,v)$ in $\pi$ by the shortest path from $u$ to $v$. Clearly, the costs are preserved. Assume that there would be a shorter closed walk in $G$. Select select the vertices in order they appear first. This permutation implies a tour in $K_n$ (maybe even shorter than the closed walk in $G$), hence we have a contradiction.

See the figure for your example.