MMP.8:5 - Archetypal Grounding

MMP.8:5.1 - Choose a preload before or after learning an external load

An ideal static arrangement has a downward load w, an adjustable upward preload a and a residual y=w-a. A later use requires abs(y)<=1/4. The load is either 1 or 2, and a can be any value from 0 to 2. The relation is a supplied illustrative mechanical model; the pattern’s work is to formulate the choice and its information.

If w=1, acceptable choices form [3/4,5/4]; if w=2, they form [7/4,2] after the actuator limit is applied. Both sets are nonempty, so each known load has a feasible choice. Their intersection is empty. No one preload chosen before distinguishing the loads can meet the requirement for both.

Suppose a reading available before adjustment reports which of the two loads is present. The policy a=w then meets the requirement. A reading received only after the preload is locked cannot support that policy at the relevant decision.

Now the tolerance is relaxed to 3/5. The acceptable intervals overlap from 7/5 to 8/5, so a=3/2 works before any reading. Additional measurement is unnecessary for this revised requirement. The change is in the required result, not the sophistication of the computation.

MMP.8:5.2 - Decide which participant gets a scarce resource

Two work requests, L and R, may need the only available resource. Exactly one needs it. Allocation succeeds when the resource goes to that request. The instruction must choose L or R deterministically from one report received before allocation.

With no distinguishing report, there are two constant instructions: always allocate to L or always allocate to R. Each fails in one circumstance. A truthful timely report permits an instruction that follows it and succeeds in both. A report arriving after allocation cannot supply that choice.

Now the timely report is noisy. The observing procedure independently chooses one of two channels with equal probability, then sends its L/R report without naming the channel. The model supplies these conditional reporting probabilities; the remaining probability in each row produces the opposite report:

Use MMP.7 to remove the unrecorded channel by summing over it. For fixed circumstance L, P(report L)=0.5*0.9+0.5*0.7=0.8. For fixed R, P(report R)=0.5*0.5+0.5*0.9=0.7. No probability for which request actually needs the resource was needed for this construction.

There are four deterministic instructions from one two-valued report:

If the requirement is success probability at least 0.65 in each admitted circumstance, following the report satisfies it. None of these instructions gives a zero-failure guarantee. The conditional laws are sufficient to make both statements while the circumstance remains unknown and fixed.

Change the question to average success in a stream of requests modeled as L with probability 0.95 and R with probability 0.05, retaining the channel procedure. Following the report gives 0.95*0.8+0.05*0.7=0.795. Always allocating to L gives 0.95; always allocating to R gives 0.05, and choosing opposite to the report gives 0.205. The instruction with greatest average success among the four is now always L. It still fails in fixed R.

Choose the performance requirement from the work’s purpose before adopting an instruction. The observing model supplies conditional probabilities; the decision about the work determines whether performance in each circumstance or an average matters.

To retain the zero-failure requirement, the work could obtain a truthful report in time or provide enough resource to serve both requests. Compare the cost of those changes with the consequences of accepting a mistaken allocation, using C.11.DUA. A changed channel, recorded channel identity or permission to randomize the instruction changes the information or choice set; formulate the revised question accordingly.

MMP.8:5.3 - Find a strategy, rather than an answer chosen with future knowledge

A program repeatedly receives a bit b and emits a bit a. A requirement asks it to emit the same bit. If receipt precedes emission, the rule a=b works on every round. If emission must precede receipt, each possible future bit has a matching answer, but no deterministic rule using only the earlier history can guarantee a match against every admitted next bit.

To see the failure, hold the earlier history fixed. The rule chooses either 0 or 1. Both next input bits are still admitted, including the opposite one. That continuation refutes the guarantee for this history. Inspecting more successful traces cannot remove it.

A changed requirement may ask for success probability under independent fair input bits. Any earlier choice then matches with probability 1/2 in one round, including a randomized earlier choice independent of the next bit. Success in every one of N such rounds has probability 2^(-N). These probabilistic claims use the new input assumption. They do not establish success against every input stream.

This is a continuing computational interaction. The construction determines how each response may depend on incoming information. The same observation-order construction identifies what a distributed team or controller would need to know before acting.