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Active-set clipping in cost-weighted baseline-tolerance allocation enforces the reference caps that an interior solution of cost-weighted allocation of baseline-response tolerances may violate. Keep the parent's stipulated logarithmic effort modelHere $B_j{\delta }_j$ is a physical bias contribution in the common units of $E$. If $E\ge {\sum }_j{B}_j{\delta }_j^{\mathrm{ref}}$, no tightening is needed. Otherwise the budget binds, and the Karush-Kuhn-Tucker conditions giveThe positive Lagrange multiplier $\lambda$ prices scarce bias budget. For a set $A$ of capped components and remaining free set $F$, stationarity givesprovided the denominator is positive and the resulting choices agree with the proposed capped/free status. The monotonicity of the multiplier equation for capped tolerances permits a scalar search; strict convexity of the objective gives a unique allocation.

For an original cap-activating version of the parent's apparatus example, let $E=10\mathrm{nV}$, $a_1=1$, $a_2=3$, $B_1=0.01\mathrm{V}$, and $B_2=100\mathrm{V}{\mathrm{s}}^{-1}$. Set reference bias contributions $B_1{\delta }_1^{\mathrm{ref}}=1\mathrm{nV}$ and $B_2{\delta }_2^{\mathrm{ref}}=20\mathrm{nV}$. The unconstrained shares $2.5$ and $7.5\mathrm{nV}$ violate the first cap. Capping it leaves $9\mathrm{nV}$ for the second component, soMerely clipping the first share and leaving the second at $7.5\mathrm{nV}$ wastes $1.5\mathrm{nV}$ of permitted bias and unnecessarily increases the chosen effort. Reallocation lowers that effort by $3\log (9/7.5)\approx 0.547$ units.

The physical motivation remains leakage from residual offset and drift in a pulsed-radiometer baseline, documented in NIST's calibration apparatus.^[ref] The logarithmic cost and the numerical caps here are illustrative assumptions, not reported apparatus constants.

Cost-weighted allocation of baseline-response tolerances chooses how tightly to enforce each baseline constraint when different tolerances require different effort. The allocation of a deterministic bias budget among baseline constraints gives the physical restriction ${\sum }_j{B}_j{\delta }_j\le E$. An optimization also needs an explicit cost model; the bias bound alone does not choose a unique allocation.

For one illustrative model, let $0<{\delta }_j\le {\delta }_j^{\mathrm{ref}}$ and minimizeTighter tolerances cost more, and the logarithm compares quantities in matching units. This is a stipulated convex effort model, not an empirical law for every numerical solver or measurement procedure.

When the budget binds and no reference upper bound is active, a constraint multiplier gives $-a_j/{\delta }_j+\lambda B_j=0$. ThereforeThe positive curvature $a_j/{\delta }_j^2$ establishes the unique interior minimum. A more expensive-to-tighten condition receives more of the allowed physical bias, while its coefficient bound converts that share to response units.

For an original two-constraint example, take $E=10\mathrm{nV}$, offset bound $B_1=10\mathrm{mV}$, drift-rate bound $B_2=100V/s$ and costs $a_1=1$, $a_2=3$. The optimum allocates $2.5$ and $7.5\mathrm{nV}$, giving ${\delta }_1=2.5\times 10^{-7}$ and ${\delta }_2=7.5\times 10^{-11}\mathrm{s}$. If their reference limits are $10^{-5}$ and $10^{-8}\mathrm{s}$, both are interior. Compared with equal physical bias shares, this optimum reduces the chosen cost by $\log (1/2)+3\log (3/2)\approx 0.523$ cost units.

Active-set clipping in cost-weighted baseline-tolerance allocation is needed if an interior solution exceeds a reference limit. If all reference limits already satisfy the bias budget, no tightening is needed in this cost model. Otherwise cap active components and redistribute the remaining budget among the others.

NIST's pulse-energy baselines motivate controlling residual offset and drift responses.^[ref] The selected tolerances control deterministic leakage within the stated coefficient bounds. Noise variance and unmodelled background shapes remain separate parts of the measurement analysis.

Monotonicity of the multiplier equation for capped tolerances makes the allocation in active-set clipping in cost-weighted baseline-tolerance allocation computable by a scalar search, even at a change of active constraints. Express each capped physical bias contribution as $c_j=B_j{\delta }_j^{\mathrm{ref}}>0$. The total allocated bias at positive Lagrange multiplier $\lambda$ isEach summand is continuous and nonincreasing. Its breakpoint is $\lambda =a_j/c_j$; below that value it is capped, and above it it is strictly decreasing. Between breakpoints,Thus $Q$ is initially constant at ${\sum }_j{c}_j$ and becomes strictly decreasing once the first component is free. Continuity persists when the derivative changes.

For a binding budget $0<E<{\sum }_j{c}_j$, there is exactly one solution of $Q(\lambda )=E$. It lies in the explicit positive bracketbecause $Q({\lambda }_{\mathrm{lo}})={\sum }_j{c}_j>E$ and $Q({\lambda }_{\mathrm{hi}})\le E$. The bisection method preserves a bracket without requiring a derivative at the root. A tolerance-budget residual bound for multiplier bisection can then relate the stopping rule to the physical bias budget instead of only to the multiplier's units.

For an original dimensionless illustration, take $(a_1,a_2,a_3)=(1,2,1)$, $(c_1,c_2,c_3)=(1,1,4)$, and $E=3$, with bias expressed in one common reference unit. The bracket is $[1/4,4/3]$. At $\lambda =1$, the contributions are $(1,1,1)$ and sum to three. This root is itself a breakpoint: the first contribution changes between capped and free there. The derivative is $-1$ just below it and $-2$ just above it, while the function remains continuous and strictly decreasing.

The argument uses the positive coefficients and logarithmic effort model specified by the parent. NIST's pulse-radiometer discussion motivates controlling offset and leakage, but does not prescribe that mathematical cost model.^[ref]