Wall-clock slack in saturation-limited Raman frame allocation ($S_n=\max (0,W-2n(d+c))$)

Wall-clock slack in saturation-limited Raman frame allocation is time that remains unused when a fixed number of detector frames has already reached its permitted exposure caps. With $n$ pump-on/off pairs, per-frame cap $c$ and readout time $d$ per frame,assuming $W>2nd$. The integer frame-count choice under a Raman saturation limit compares these possibilities over different pair counts. Slack at fixed $n$ does not imply that the continuous exposure allocation was optimized incorrectly.

If $S_n>0$, both states can use their full live totals $nc$, and the shot-noise relative variance is $V_n=(\alpha +\beta )/(nc)$. Increasing only the wall-clock allowance cannot improve it: the marginal value of unused time below a Raman frame cap is zero in this capped regime. Another frame pair changes both the overhead and the allowable division of live time.

For an original example, set $W=0.50\mathrm{s}$, $n=2$, $c=0.10\mathrm{s}$ and $d=0.020\mathrm{s}$. The live time is $0.40\mathrm{s}$, readout uses $0.08\mathrm{s}$, and slack is $0.02\mathrm{s}$. With $\alpha =2\times 10^{-6}\mathrm{s}$ and $\beta =0.5\times 10^{-6}\mathrm{s}$, equal capped totals $0.20\mathrm{s}$ give $V_2=12.5\times 10^{-6}$.

Three pairs increase readout time to $0.12\mathrm{s}$, leaving only $0.38\mathrm{s}$ live time. Yet the larger per-state cap permits the preferred ratio two to one: $T_{\mathrm{on}}\approx 0.2533\mathrm{s}$ and $T_{\mathrm{off}}\approx 0.1267\mathrm{s}$. Their variance is $V_3=4.5\times 10^{-6}/0.38\approx 11.8421\times 10^{-6}$, lower despite less total live time. Relieving the weak state's cap outweighs the extra readout cost in this chosen model.

The primary Raman apparatus limited exposures to avoid detector saturation and alternated pump state during shuttered readout.^[ref] The timings above are separate illustrative choices. Read noise, sample drift and switching constraints can change the comparison; unused time and useful precision are different optimization quantities.

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.

Frame-count switching in Raman acquisition-budget sensitivity occurs when a different integer number of exposure pairs becomes optimal as the available time changes. For each feasible pair count $n$, let $V_n(W)$ be the minimum predicted ratio variance after optimizing its live exposures. The complete design isThe marginal value of unused time below a Raman frame cap gives slopes for a fixed $n$. At a branch switch, those slopes need not be the derivative of the lower envelope on both sides.

A simple original example uses equal shot-noise coefficients $\alpha =\beta =a$, readout duration $d$ per frame and cap $c$ per exposure. Equal on/off totals are optimal, givingTake $a=10^{-6}\mathrm{s}$, $c=0.10\mathrm{s}$ and $d=0.010\mathrm{s}$. Two pairs saturate their $0.40\mathrm{s}$ live-time capacity once $W\ge 0.44\mathrm{s}$. Three pairs have live time $W-0.06\mathrm{s}$ near this budget. Their variances agree atJust below this value the two-pair variance is the constant $10\times 10^{-6}$. Just above it the three-pair variance is $4\times 10^{-6}/(W-0.06\mathrm{s})$ and begins decreasing. One pair has lower exposure capacity, while four or more pairs incur more readout and have less useful live time near the switch, so these two branches determine the global optimum there.

The one-sided derivatives at an optimal Raman frame-count switch are zero on the left and $-25\times 10^{-6}{\mathrm{s}}^{-1}$ on the right. The optimal variance is continuous at this switch, but has a kink. At $W=0.45\mathrm{s}$ two pairs give $10\times 10^{-6}$; at $0.47\mathrm{s}$ three pairs give about $9.7561\times 10^{-6}$. Extra time becomes useful again after the discrete design changes, despite the previous fixed-count plateau.

The primary Raman experiment's saturation-limited exposures and shuttered on/off readout motivate these competing resources.^[ref] The example uses chosen rounded timings. Read noise, drift and switching requirements can move the branch crossings and must be included when choosing an actual acquisition protocol.

One-sided derivatives at an optimal Raman frame-count switch describe the change in the best acquisition variance when several integer designs tie. Let $V_{\ast }(W)={\min }_n{V}_n(W)$ as in frame-count switching in Raman acquisition-budget sensitivity, and let $A$ be the active designs with $V_n(W_{\ast })=V_{\ast }(W_{\ast })$. Suppose the relevant branches are differentiable near the tie and nonactive designs have a positive value gap.

Writing their slopes as $s_n=V_n^{\prime }(W_{\ast })$ givesIndeed, $V_n(W_{\ast }+h)=V_{\ast }(W_{\ast })+s_{n}h+o(h)$. Positive $h$ selects the smallest slope; negative $h$ selects the largest because multiplication reverses their ordering. A single derivative exists only when the selected left and right slopes agree. If an active branch itself has a kink, use its appropriate one-sided slope on each side.

For an original balanced-noise example, take $\alpha =\beta =a=10^{-6}\mathrm{s}$, exposure cap $c=0.080\mathrm{s}$ and per-frame readout $d=0.015\mathrm{s}$. Near the three-pair/four-pair switch,The branches tie at $W_{\ast }=6c+8d=0.60\mathrm{s}$, with variance about $8.3333\times 10^{-6}$. The three-pair branch has slope zero; the four-pair branch has slope $-4a/(6c{)}^2\approx -17.3611\times 10^{-6}{\mathrm{s}}^{-1}$. Thus these are the left and right optimum slopes. Smaller pair counts have less capacity and larger counts have more readout near this tie, so they do not change the local optimum.

Joint time-and-readout perturbations at a Raman design tie require following the direction in both resource coordinates; the scalar slopes above hold when only $W$ changes. They also quantify a local sensitivity, not a universal prescription to change exposure count whenever a small budget fluctuation occurs.

The primary stimulated-Raman apparatus limited exposures for saturation and switched pump state during shuttered readout.^[ref] The rounded timings here are a separate mathematical example. Read noise, drift or switching costs can change the competing branches and their derivatives.

Joint time-and-readout perturbations at a Raman design tie describe the directional response of the best acquisition variance when both available wall time and per-frame readout cost change. A tie generally has different derivatives on opposite sides, so there need not be one gradient for the optimum. The active branch derivatives must be combined along the actual perturbation direction.

Use the original balanced-noise model in one-sided derivatives at an optimal Raman frame-count switch. With $a=10^{-6}\mathrm{s}$, per-exposure cap $c=0.080\mathrm{s}$ and readout $d$, the locally competing branches areThey tie along $W=6c+8d$. At $(W_{\ast },d_{\ast })=(0.60,0.015)\mathrm{s}$, define $K=4a/(6c{)}^2\approx 17.3611\times 10^{-6}{\mathrm{s}}^{-1}$. The three-pair branch is locally constant, while the four-pair gradient is $(-K,8K)$.

For a path $W=W_{\ast }+ϵu_W$, $d=d_{\ast }+ϵu_d$ with $ϵ\downarrow 0$, the optimum has directional derivativeThe sign of the net live-time change $u_W-8u_d$ selects the improving branch. This conclusion assumes the branches remain feasible and nonactive designs have a positive value gap nearby.

For original finite examples, adding8ms of budget while adding1ms to each readout leaves $W-8d$ exactly unchanged, so these two branches remain tied. Adding10ms while adding1ms per readout increases that live time by2ms: the four-pair variance becomes $4\times 10^{-6}/0.482\approx 8.2988\times 10^{-6}$, below the three-pair value $8.3333\times 10^{-6}$. The first-order improvement is $-K(0.002)\approx -3.4722\times 10^{-8}$. Adding only6ms with the same readout increase instead makes the four-pair branch worse; the three-pair optimum remains unchanged locally.

The resource distinction has a physical motivation. The primary stimulated-Raman experiment used alternating pump states and shuttered detector readout; its exposure and readout intervals were separate parts of the acquisition sequence.^[ref] The rounded timings and noise parameter above are a mathematical planning example, not that instrument's measured values. Changing exposure caps, state-switching costs or noise terms requires the corresponding branch model to be recomputed.