PSD.12:5.1 - An explicit reversal boundary for the pump comparison

Use the illustrative F and M slice from PSD.11: F costs 8 incremental budget units and loses 2 service hours with normal access or 3 with road loss; M costs 5 and loses 1 or 9 hours. N remains the baseline and S still has an unqualified whole-arrangement result. The following calculation tests only F against M; it does not close the wider candidate comparison.

For an illustrative analytical question, suppose a declared value model minimizes expected service-loss hours plus lambda times incremental budget units. Here lambda is an explicitly elicited value conversion, in service-hour-equivalent value per budget unit; it is not a measurement conversion or an unspoken public preference. Let p be the road-loss probability if a qualified probability model supports one.

Under those assumptions:

• F’s value loss is 2 + p + 8*lambda.
• M’s value loss is 1 + 8*p + 5*lambda.
• F has lower value loss precisely when 7*p > 1 + 3*lambda; equality is a tie.

For lambda = 0.5, the reversal is at p = 5/14, approximately 0.357. At p = 0.2, F and M give 6.2 and 5.1 respectively, so M is better under this model. At p = 0.6, they give 6.6 and 8.3, so F is better. These are invented sensitivity settings, not a forecast.

The source comparison supplied no probability. Therefore the valid return is the conditional boundary, not a claim that either setting is likely. If a probability model cannot be qualified, retain the scenario comparison instead of assigning equal chances.

A different declared test asks whether modeled service loss is at most 4 hours in both stated access conditions. F satisfies that illustrative test; M fails it under road loss. This is a different robustness criterion, not a hidden replacement for the value model. The four-hour cut is an example, not a domain standard. F’s result covers only those two modeled conditions.

The account returns the reversal boundary, the limited threshold result, the unresolved probability and value premises, and S’s candidate gap. Reachable assistance, wider property consequences, and protected conditions still require their direct results. None of the calculations authorizes a pump investment.

PSD.12:5.1.1 - A separate constructed delay-response question

Keep the preceding probability and value-reversal question separate. For this new illustration, fix a thirty-day service window and one access interruption. A stipulated model supplies the service-loss hours for F and M at three access delays; lower service loss is preferred. Each arrangement and all other modeled conditions stay fixed while delay varies.

With x = 1 day and h = 1 day, M’s central response is 3 hours, its endpoint mean is (1 + 9)/2 = 5 hours, and D = 2 hours. F’s central response and endpoint mean are both 2.5 hours, so its D = 0. F has no midpoint gap at these three points; this does not establish a linear response between them.

Under the separately declared four-hour service-loss criterion, M’s two-day response of 9 hours fails. F satisfies that criterion at the three stated points only. Neither endpoint mean is an expected real loss without a probability basis, and a central response below four hours does not settle the endpoint test.

The useful return is the finite response difference, M’s failure condition and the remaining model and coverage limits. F’s additional investment and the feasibility of any alternative access protection still need their whole-configuration comparison. If the stipulated model lacks support for real use, retain the calculation as conditional or illustrative; do not report an established pump-performance result. No additional observations or investment are authorized by it.