MNT.2:5.1 - Construct an age rule from a failure model

In a separate constructed comparison, a replaceable component lasts 6 or 10 operating-time units, each with probability one half. Replacements meet the assumptions in section 4.2. The three available candidates are replacement at age 4, at age 8, or at failure. All are admissible for this use. Planned replacement costs 2 units and corrective replacement costs 10, including the case’s failure consequences.

At age 4 every cycle ends with planned replacement: cost per operating-time unit is 2/4 = 0.5. At age 8, half the cycles fail at 6 and half reach planned replacement at 8: expected cost is 0.5*10 + 0.5*2 = 6, expected duration is 0.5*6 + 0.5*8 = 7, and the rate is 6/7, approximately 0.857. At failure the rate is 10/[0.5*6 + 0.5*10] = 1.25. Age 4 is preferred among these three candidates on the declared cost criterion; this comparison has not searched all possible ages.

Now change only corrective cost to 3. The rates become 0.5, 2.5/7, approximately 0.357, and 3/8 = 0.375. Age 8 becomes preferred. The failure model and task family remain unchanged; the consequence change reopens the interval choice.