 Solution

The diagram has all possible undirected edges except $AB$: it is a complete graph on five vertices with one edge removed. Therefore $A$ and $B$ each have three neighbours, while $C,D,E$ each have four.

The neighbours of $A$ and $B$ form a triangle, so their clustering coefficients equal one. Each of $C,D,E$ has five of the six possible neighbour–neighbour edges, giving $5/6$. The nodal average shortest path length is ${\ell }_i=\frac{1}{4}{\sum }_{j\ne i}d(i,j)$. All graph distances equal one except $d(A,B)=d(B,A)=2$, henceAveraging the nodal quantities givesThe ordered-pair distance sum is $22$, and the graph diameter is two.

If the intended unnormalized “small-worldness” is the ratio of clustering to path length, the computed quantities giveThat ratio has no universal threshold for identifying a small-world network. The standard reference-normalized small-worldness, $\sigma =(\bar{C}/C_{\mathrm{ref}})/(L/L_{\mathrm{ref}})$, cannot receive a numerical value when no reference is specified and the question forbids using a network null model.^[ref]

The graph has high clustering and short paths in the descriptive sense. With nine of ten possible edges, however, those properties follow directly from its near completeness; they do not by themselves establish small-world network structure relative to a reference. This distinguishes the metrics that can be computed directly from the stronger classification that requires a stated convention.