$p_{k}(M^0_{\mu})$ reduces to finding the largest capacity $c>0$ such that the vector $$v_{0}:=v_{\mu,c}^{0}=\left(\mu+1-c\,; \mu-c, c^{\times (k-1)}, 1-c\right)$$ belongs to the closure of the symplectic cone of $X_{k+1}$.

The initial step of the reduction algorithm already gives nontrivial results. Indeed, the vector $v_{0}$ is non-negative only if $c\leq 1$, which is equivalent to the fact that the Gromov width is ${\mathsf{w}}_{1}(M^0_{\mu})=1$. This obviously gives an upper bound for ${\mathsf{w}}_{k}$, and we note that this bound is stronger than the volume condition whenever $\mu\geq \frac{k}{2}$. Now, the vector $v_{0}$ is ordered only if we suppose $c\geq 1/2$. When $0<c\leq\frac{1}{2}$, the reordered vector $\hat{v}_{0}$ is $$\hat{v}_{0} =
  \left(\mu+1-c\,; \mu-c, 1-c, c^{\times (k-1)}\right),$$ which is positive and reduced (since its defect is zero). Now, we observe that $c_{\operatorname{vol}}\leq 1/2$ if and only if $k\geq8\mu$. Hence, ${\mathsf{w}}_{k}(M_{\mu}^{0}) = c_{\operatorname{vol}}$ whenever $\mu\leq \frac{k}{8}$. That implies we have full packing by $k$ balls for all $\mu\in[1, k/8]$ while, for $\mu\in(k/8, \infty)$, we have the lower and upper bounds $1/2<{\mathsf{w}}_{k}\leq\min\{1,c_{\operatorname{vol}}\}$.

Given $n\in{\mathbb{N}}$, let us write $$\lambda_{n} = \mu+1-c-nd_{0}$$ where $d_{0}=2c-1$. Using this notation, we have $$v_{0} = \left( \lambda_{0}\,;\lambda_{0}-1, c^{k-1}, (1-c) \right).$$ Given $\mu>k/8$ and $c\in(1/2,1]$, the vector $v_{0}$ is ordered, and has defect $d_{0}=2c-1>0$. Applying a Cremona 