 Optimal allocation of detector repetitions across calibration blocks

Optimal allocation of detector repetitions across calibration blocks chooses block sizes using the dependence and acquisition-cost model, not just the desired number of stored traces. For fixed total $N$, number of independent blocks $K$, common success probability, and positive within-block correlation $\rho$, effective repetition count across independent calibration blocks givesMaximizing it is equivalent to minimizing ${\sum }_j{n}_j^2$. If two integer block sizes differ by at least two, transferring one repetition from the larger to the smaller decreases that sum by $2(n_i-n_j-1)>0$. Therefore the optimal integer sizes differ by at most one.

For an original fixed-count example, $N=23$ and $K=5$ give balanced sizes $(5,5,5,4,4)$. At $\rho =0.1$, their effective count is $529/31.4\approx 16.85$. Sizes $(11,3,3,3,3)$ give only $529/36.4\approx 14.53$ under the same assumptions.

More independent blocks can help, but preparing one may consume time. In a continuous planning model let each block cost setup time $h>0$, each repetition cost $\tau >0$, and each block contain $n$ repetitions. With total time $T$, the effective count becomesFor $0<\rho <1$, differentiation gives the interior choice $n_{\ast }=\sqrt{h(1-\rho )/(\tau \rho )}$. Integer block counts, minimum sizes, and hardware restrictions must subsequently be imposed.

For chosen $h=9\mathrm{s}$, $\tau =1\mathrm{s}$, $\rho =0.1$ and $T=180\mathrm{s}$, the continuous optimum $n_{\ast }=9$ is feasible with ten blocks and gives $N_{\mathrm{eff}}=50$. Single-repetition blocks give 18, while two blocks of 81 repetitions also give 18. These are model-based comparisons, not measured apparatus timings.

Unequal block noise or different drift correlations require unequal-noise allocation across calibration blocks. Independence between prepared blocks must also be justified by the reference and environmental model; NIST explicitly retains shared-input covariance.^[ref]