How Often to Change a Part That Wears Out Slowly

How to read this sheet

Check the answer with the two columns rather than trusting it. The last hour costs you a figure, and the average hour at the right interval costs a figure, and at the right interval those two are the same number. If the last hour costs more than the average, the part has been left on too long; if it costs less, the change is being spread over too few hours. That comparison can be made at the machine with nothing but the sheet from last month, and it is the whole method. The number everybody will want to skip. How fast the running cost climbs with age is not written down anywhere and it decides everything. It is also the easiest thing on the sheet to estimate, because everybody in the department knows what a tired part does: take the difference between a fresh one and one at the end of its present life, in waste or seconds or ends down or speed, divide by the hours between them, and enter that. An estimate that is wrong by a third moves the interval by about a sixth, because the answer goes as a square root - which is the same reason it is worth doing at all and the reason not to agonise over it. Why early is the dearer mistake. Changing too soon pays the whole cost of a change over too few hours and gets nothing for it; leaving it too long pays deterioration that builds gradually. So the cost curve is steep on the early side and gentle on the late one, and the seeded sheet finds being thirty per cent early costs about twice being thirty per cent late. Maintenance departments err early to be safe. It is the more expensive direction to be safe in. What a quick change is worth, honestly. Most of it is the standing hours it saves on the changes already being made, which is the case everybody already makes. There is a second part nobody counts: a shorter stop makes changing cheaper, which makes changing more often correct, which means the plant runs on fresher parts all the time. On the seeded sheet that second part is only a few per cent of the total, and the sheet reports both figures rather than letting the smaller one be sold as the larger. Where the stop dominates the cost of a change it will be worth more, and the sheet will say so. What is not modelled. Deterioration is taken as flat and then linear, which is what most wearing parts do and is wrong for anything that degrades suddenly at the end - if the part has a cliff rather than a slope, this is not the sheet for it. Changes are assumed to happen when planned, where in practice they wait for a shift change or a lot change, and that scheduling reality will pull the interval towards whatever the natural stopping points are. And nothing here covers the part that fails outright, which is a reliability problem with a different shape and a different answer.