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Integer frame-count choice under a Raman saturation limit chooses the number of pump-on/off exposure pairs as well as their durations. More pairs incur more readout overhead; too few pairs can saturate before using the available wall-clock budget. Under the model of maximum-exposure constraint in paired Raman acquisition, for positive readout time $d$ and frame cap $c$, each candidate integer $n$ with $2nd<W$ has usable live timeIts unused budget is $W-2nd-T_n\ge 0$. This wall-clock slack in saturation-limited Raman frame allocation is allowed: failure to consume the entire budget does not make a shorter acquisition physically impossible.

For each candidate, allocate $T_{\mathrm{on}}+T_{\mathrm{off}}=T_n$ with each total at most $nc$. Clip the unconstrained on total $T_n\sqrt{\alpha }/(\sqrt{\alpha }+\sqrt{\beta })$ to $[\max (0,T_n-nc),\min (T_n,nc)]$, then evaluate $V_n=\alpha /T_{\mathrm{on}}+\beta /T_{\mathrm{off}}$. This solves the continuous subproblem; comparing the finitely many integer candidates solves the stated frame-count problem. Positive counts and no other costs make additional feasible live time beneficial.

In an original example, use $W=0.60\mathrm{s}$, $d=0.010\mathrm{s}$, $c=0.097\mathrm{s}$, $\alpha =2\times 10^{-6}\mathrm{s}$ and $\beta =0.5\times 10^{-6}\mathrm{s}$. The relevant candidates areFour pairs are globally optimal in this model: one or two pairs use both states at their caps and improve as $n$ increases; three is evaluated explicitly; for every $n\ge 4$ the unconstrained ratio is feasible and $V_n=4.5\times 10^{-6}/(0.60-0.02n)$ increases with $n$. At four pairs the on/off totals are about $0.3467$ and $0.1733\mathrm{s}$.

The primary Raman apparatus used 97 ms exposures to avoid saturation and 10 ms shuttered readout, motivating these stipulated example timings.^[ref] Actual frame choice can also be constrained by read noise, drift and minimum switching cadence. Such costs must be included before treating this shot-noise-only optimum as an experimental operating prescription.

The marginal value of unused time below a Raman frame cap measures how the optimum predicted Raman ratio variance changes with the allowed acquisition budget. Hold the pair count $n$, readout time $d$ and per-frame exposure cap $c$ fixed. Put $H=nc$ and $T=W-2nd>0$. The wall-clock slack in saturation-limited Raman frame allocation becomes positive once $T>2H$; additional budget then cannot increase exposure at that pair count.

For shot-noise coefficients $\alpha >\beta >0$, define $K=(\sqrt{\alpha }+\sqrt{\beta }{)}^2$. The unconstrained on fraction is $p=\sqrt{\alpha }/(\sqrt{\alpha }+\sqrt{\beta })$. It reaches its cap when $T=H/p$. The minimized variance is thereforeThis follows by using the unconstrained ratio, then holding the weaker on state at its cap while extending the off state, and finally holding both at their caps.

Away from the transition points, the positive marginal improvement per extra wall-clock second isThe slope agrees on both sides of the first transition. At the final cap it changes to zero, because new time is unused.

For an original example, choose $n=2$, $c=0.10\mathrm{s}$, $d=0.010\mathrm{s}$, $\alpha =4\times 10^{-6}\mathrm{s}$ and $\beta =10^{-6}\mathrm{s}$. The wall-clock transitions are $0.34$ and $0.44\mathrm{s}$:At the last setting, $0.05\mathrm{s}$ is slack. Optimizing the pair count as well introduces frame-count switching in Raman acquisition-budget sensitivity; the fixed-count slopes above need not describe that full integer optimization.

The primary Raman experiment capped exposures to avoid saturation and used shuttered readout between pump states.^[ref] The numerical timings here are illustrative. These marginal values concern the stated noise objective, without additional drift, read-noise or switching-cadence costs.