 Monotonicity of the multiplier equation for capped tolerances

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]