= Solution
At this point $N_{\rm Th}/N_{\rm Th0}=0.95$, so $a\Phi=-\ln0.95$ and $e^{-b\Phi}=0.95^{b/a}=0.345671$. The solution from part (c) gives
$$
\frac{N_{\rm U}}{N_{\rm Th0}}=0.05(0.345671)+\frac{3.08}{64.20-3.10}(0.95-0.345671)=0.0477473.
$$
Therefore
$$
\boxed{CR=\frac{3.08(0.95)}{64.20(0.0477473)}=0.9545.}
$$
The fissile population has declined by about $4.51\%$ from its initial value, while the conversion ratio has decreased slightly from $0.9595$. The replacement shortfall persists over this interval. Maintaining a specified long-term fuel inventory would require replenishment or a different balance of reactions; the prescribed-flux equations alone do not establish a constant-power operating state. Intermediate decay delays and the changing <neutron> spectrum were excluded by the original lumped model.
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