&\textup{if} \ \beta+\alpha\frac{K-2n-1}{2}\leq 0,\\
              \ \\
               \frac{2K-n+1}{n+1}-\frac{2K(K-n)}{n(n+1)}\cdot \frac{M}{N-\beta} &{\textup{otherwise}},
        \end{cases}\end{aligned}$$ for any $n\in\{\max\{1,K-N+1\},...,K-1\}$, where $\alpha=\lfloor \frac{N-1}{K-n} \rfloor$ and $\beta=N-\alpha (K-n)$.

The above theorem improves Theorem \[th:general\] and the state of the art in many cases. For example, when $r\in\left[\left\lceil K-1-\frac{N-1}{\left\lceil\frac{2N}{K+1}\right\rceil} \right\rceil,K-1\right)$, the converse bound (\[eq:thm4\]) given by $n=\lfloor r+1\rfloor$ is tight and we have $R^*=R_{\textup{u}}(N,K,r)$. This result can not be proved in general using the converse bounds provided in [@ghasemi15; @lim2016information; @sengupta15; @DBLP:journals/corr/WangLG16; @tian2016symmetry; @prem2015critical] (e.g., for $K=4$, $N=10$, and $M=4$, none of these bounds give $R^*\geq 1$).

We numerically compare our two converse bounds (i.e., Theorem \[th:general\] and Theorem \[th:2\]), benchmarked against the upper bound $R_{\textup{u}}(N,K,r)$ we achieved in [@yu2016exact] under three different settings (see Fig. \[fig:compare\]). In all these cases, the two converse bounds together provide a tight characterization: Theorem \[th:general\] is tight for $r\leq 1$ and $r\geq K-1$, and Theorem \[th:2\] is tight for $1\leq r\leq K-1$. The same holds true in the proof of Theorem \[th:exact5\], where the number of users is no more than 5 but the number of files is large.

[.45]{} ![Numerical comparison among the two converse bounds presented in Theorem \[th:general\] and Theorem \[th:2\], and the upper bound achieved in [@yu2016exact]. Our 