= Integer allocation near a continuous isotope-reference optimum
An <integer allocation near a continuous isotope-reference optimum> converts a real-valued budget solution into feasible counts of sample and reference observations. For the model in <reference-versus-sample allocation in repeated isotope-ratio measurements>, minimize
$$
V=4\left(\frac{s^2}{n_s}+\frac{r^2}{n_r}+s_c^2\right),\qquad
c_sn_s+c_rn_r\le T,
$$
with positive integer counts. The shared component $s_c^2$ does not affect the choice when it is genuinely independent of the allocation. Counting new reference readings does not make a shared calibration error independent.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[Appendix A, equation A-3, covariance propagation]{ref}
For equal unit costs and a budget of $N$ readings, put $n_r=N-n_s$. With $s,r>0$, the count-dependent objective $f(n)=s^2/n+r^2/(N-n)$ has
$$
f''(n)=\frac{2s^2}{n^3}+\frac{2r^2}{(N-n)^3}>0.
$$
Its continuous minimum is $n_*=Ns/(s+r)$. Therefore an integer minimum lies among the feasible floor and ceiling of $n_*$, clipped to $1,\ldots,N-1$. Comparing those candidates is sufficient because the function decreases before its minimum and increases after it. Rounding both continuous counts independently can violate the budget.
For an original example, let $N=17$, $s=0.003$ and $r=0.001$. The continuous sample count is12.75. Counts $(n_s,n_r)=(12,5)$ give variance $3.8000\times10^{-6}$ before a common component, while $(13,4)$ give approximately $3.7692\times10^{-6}$. The latter is preferable, with log-ratio standard uncertainty about0.1941%. The nearby $(14,3)$ gives approximately $3.9048\times10^{-6}$.
Unequal costs introduce a stepped feasible boundary. One exact finite search enumerates positive $n_r$ and, for each, uses
$$
n_s=\left\lfloor\frac{T-c_rn_r}{c_s}\right\rfloor
$$
when that value is positive. For fixed $n_r$, using the largest feasible $n_s$ improves this stated objective. For an original $T=20$, $c_s=3$, $c_r=2$, $s=3$ and $r=1$ in consistent planning units, the best pair is $(5,2)$, consuming19 units. Its count-dependent expression $s^2/n_s+r^2/n_r$ is2.3; $(6,1)$ and $(4,4)$ both give2.5. An unused budget unit can be a legitimate integer remainder.
Acquisition ordering, paired exposures, minimum replicate counts or drift constraints can restrict the feasible set further. The continuous optimum remains a useful guide, but only comparisons among genuinely feasible integer designs establish the chosen discrete optimum.
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