MMP.17:5 - Archetypal Grounding

MMP.17:5.1 - Refine an interpolation only until it settles the comparison

A repeated calculation needs the response (r(x)) of a costly reference model for (0\leq x\leq1). The present question is whether (r(0.4)\leq0.3). Available analysis of the reference model gives (|r’’(x)|\leq2). Its accurate case values are (r(0)=0) and (r(1)=1).

The first surrogate is the line (\widehat r(x)=x). For linear interpolation on an interval of width (h), the curvature bound gives an error at most (2h^2/8). With (h=1), the response at (0.4) is therefore enclosed by (0.4\pm0.25). This interval crosses (0.3); the replacement has not answered the question.

Query the reference model at (x=0.5), obtaining (r(0.5)=0.25), and use the two half-intervals. In the first half, (\widehat r(x)=0.5x), so (\widehat r(0.4)=0.2). The bound is now (2(0.5)^2/8=0.0625), giving

[ r(0.4)\in[0.1375,,0.2625]. ]

The upper endpoint is below (0.3). One added case and a specified interpolation rule settle the comparison under the curvature premise. The three values alone would not justify the bound.

Now the receiving limit changes to (0.18). The same enclosure crosses that limit. For this one query, the practitioner returns to the reference model at (0.4), which gives (0.16), and can answer the stricter comparison. If many similar queries are expected, further subdivision may instead be worthwhile. There is no need to improve the surrogate over the entire interval to finish the single question.

These calculations establish agreement with the reference model under its stated smoothness and case-accuracy conditions. Whether that model’s response represents the subject phenomenon remains a separate modeling question.

MMP.17:5.2 - Correct a cheap allocation model, then change its regime

A network model returns delivered quantities (q_1(d)) and (q_2(d)) through two channels for demand (2\leq d\leq4). In the normal regime all demand is served, so (q_1+q_2=d). A cheap approximation splits it equally: (L(d)=(d/2,d/2)).

The source supplies (q(2)=(1.2,0.8)) and (q(4)=(3,1)). The first-channel discrepancies from (L) are (0.2) and (1). Linear interpolation of that discrepancy gives

[ \widehat d_1(d)=0.4d-0.6,\qquad \widehat q_1(d)=0.9d-0.6,\qquad \widehat q_2(d)=d-\widehat q_1(d). ]

At demand (3), the replacement returns ((2.1,0.9)). It preserves total delivery and nonnegativity throughout the declared interval. Those properties follow from the construction; agreement between the cases has not yet been established.

The receiving question is whether the first delivery stays at or below (2.25) at demand (3). The provisional value (2.1) is only (0.15) below the limit, and there is no supported error bound for this interpolation. A source query at (3) returns ((2.4,0.6)), ruling out the proposed limit. Add its discrepancy (0.9) and interpolate separately over ([2,3]) and ([3,4]). The new surrogate matches all three cases and preserves the balance. Further-input accuracy remains an empirical question or requires a separate bound.

Now channel 2 becomes unavailable. The normal-regime fit does not describe this input. Suppose the changed source account states that channel 1 serves all demand up to (3) units and any excess remains unserved. The needed output now includes unserved demand (u):

[ q_1=\min(d,3),\qquad q_2=0,\qquad u=d-q_1. ]

At demand (4), the result is ((3,0,1)). The balance has become (q_1+q_2+u=d). This branch follows from the newly supplied operating rule, not from extrapolating the normal-regime cases. Its formula is already cheap enough to use; training another replacement adds no benefit here. The caller can retain the normal-regime surrogate for its supported questions and use this formula for the stated unavailable-channel regime.

MMP.17:5.3 - Reuse stochastic cases when the requested response changes

A stochastic loss model returns either (0) or (5). Its source structure states that the probability (p(x)) of loss (5) is affine for (0\leq x\leq1); the two endpoint probabilities are unknown. All 2,000 runs in the construction are independent: 1,000 runs give 100 losses of (5) at (x=0), and another 1,000 give 300 at (x=1).

Fit the endpoint proportions and interpolate:

[ \widehat p(x)=0.1+0.2x,\qquad \widehat\mu(x)=5\widehat p(x)=0.5+x. ]

The affine premise comes from the source structure, not from the two observed proportions. The mean surrogate is evaluable without rerunning the stochastic source.

At (x=0.5), the estimated mean is (1). A simple uncertainty calculation illustrates what must accompany that value. Each endpoint proportion has variance at most (1/(4{,}000)). Independence gives variance at most (1/(8{,}000)) for their average, an unbiased estimator of (p(0.5)) under the affine premise. Chebyshev’s inequality therefore gives coverage of at least 95% for a half-width of (0.05) around that average. The resulting probability interval is ([0.15,0.25]), and the corresponding mean interval is ([0.75,1.25]). For a receiver using this 95% confidence procedure, the upper endpoint supports the mean-at-most-(1.4) comparison; it is not a deterministic bound. MMP.13 permits sharper uncertainty calculations when the receiving use needs them.

The receiver now asks whether (\Pr(Y_{0.5}>4)\leq0.18). The mean alone cannot answer: a constant loss of (1) has the same mean but a different tail. Here the retained two-point support supplies the relation (\Pr(Y_x>4)=p(x)=\mu(x)/5). Reuse the same cases and the probability surrogate; no new fit is needed. The interval ([0.15,0.25]) crosses (0.18), so this uncertainty result leaves the new comparison unresolved.

The interval concerns the fixed input (0.5) under independent runs, the stated support and the affine probability law. It is not a simultaneous guarantee for all inputs, and it does not cover error in those source premises. A sharper inference, a useful bound, more runs or a qualified unresolved answer are different possible continuations; choose among them for the receiving question.