\mu^{-2}$, since it contains one fewer power of $\Phi_{ij}$. This means that the non-perturbative piece can come to dominate, even though we know that it would be negligible in the $N_f=4$, $N_c=3$ case. As a result, we were forced to introduce a constant, $\Lambda_{\mbox{non-pert.}}$, multiplying the non-perturbative piece and find the largest stable value, before we could be certain that the non-perturbative effects were under control. More specifically $$\begin{aligned}
W_{non-pert} \rightarrow   W_{non-pert} = \Lambda_{\mbox{non-pert.}} N h^{N_f/N}\bigl(\Lambda_m^{-(N_f-3N)}\det(\Phi)\bigr)^{1/N}\end{aligned}$$ in what follows.

Supersymmetry Breakdown
=======================

Numerical Results {#sec:Numerics}
-----------------

$     \begin{array}[b]{|c|ccccccc|}
       \hline
       N_f  & 3 & 3 & 3 & 3 & 3 & 3 & 3 \\
       N  & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
       \mu   & ~~1.41~10^{-7} & ~~1.41~10^{-7} & ~~1.41~10^{-7} & ~~1.41~10^{-7}  & ~~1.41~10^{-7} & ~~1.41~10^{-7} & ~~1.41~10^{-7}  \\
       \Lambda_{\mbox{non-pert.}}& 0 & ~~2.00~10^{-6} & ~~2.00~10^{-4}  & ~~2.00~10^{-2} & ~~4.00~10^{-2} & ~~4.40~10^{-2} & ~~4.60~10^{-2}  \\
       \hline
       m_{3/2} & ~~1.52~10^{-14}  & ~~1.52~10^{-14} & ~~1.52~10^{-14} & ~~1.52~10^{-14} & ~~1.52~10^{-14} & ~~1.52~10^{-14} & ~~1.52~10^{-14} \\
       \<\phi_0\> & ~~1.41~10^{-7~}   & ~~1.41~10^{-7~} & ~~1.41~10^{-7~} & ~~1.41~10^{-7~} & ~~1.41~10^{-7~} & ~~1.41~10^{-7~} & ~~1.41~10^{-7~} \\
       \<\Phi_0\> & - 6.33~10^{-12} & -6.33~10^{-12} & -6.33~10^{-12} & -7.95~10^{-12} & -1.57~10^{-11} & -2.10~10^{-11} & 