[RESOLVED] Your car needs assistance

I have a problem and need your help.

You take your car to preventive maintenance (PM) every 10.000 Km. Now and then a failure occurs and you have it fixed by undertaking some corrective maintenance (CM) action. Suppose that there is a 36% chance for this to happen in every period of 10.000 Km at 6.365 Km in average. If you pay an average of 100 € for each PM and 500 € for each CM, how much do you spend every 10.000 Km?

I tried to solve the case analytically and also built a straight forward simulation model in Excel but, to my surprise, results differ significantly.

May I have your suggestions?

Thanks

Average € per 10000km = 500 + (.36*100)

??

You didn´t take into consideration the number of possible CM actions that might take place during the span of 10.000 Km. The figure 6.365 Km has to be included some how in the calculation.

I think it's just to let you work out extra accuracy for the percentage chance of CM being required.

Too tired to explain, but:
10.000 Km - 6.365 Km = 3.635 km, which is just over 36%

So for extra accuracy: Average € per 10000km = 500 + (.3635*100)

Quote:

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Originally Posted by si_the_geek

I think it's just to let you work out extra accuracy for the percentage chance of CM being required.

Too tired to explain, but:
10.000 Km - 6.365 Km = 3.635 km, which is just over 36%

So for extra accuracy: Average € per 10000km = 500 + (.3635*100)

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Thanks for your comment but let me tell you that there is nothing about the problem that has anything to do with accuracy and the result at which you arrived is just a coincidence. I tried different parameters on the law that models the failure occurrences and I submit now a slightly different text:

You take your car to preventive maintenance (PM) every 10.000 Km. Now and then a failure occurs and you have it fixed by undertaking some corrective maintenance (CM) action. Suppose that there is a 22.1% chance for this to happen in every period of 10,000 Km at 6,496 Km in average. If you pay an average of 100 € for each PM and 500 € for each CM, how much do you spend every 10,000 Km?

You may notice now that such a coincidence just disappeared.

My feeling for the solution is as follows: (100 + 500 x 10,000/6,496 x 0,221) = 270 € but I am not sure. Besides, the simulation model (that I trust more) returns approximately 201 €. Where am I wrong?

Thanks again.

OK how about

Average € per 10000km = 100 + (.36*(10000/6365)*500)

~ €382

Thanks, but unfortunately that is not the answer.

I finally discovered from a theoretical text book on reliability engineering that the problem is very difficult to compute analytically. You have to model it as a renewal process in order to obtain the mean number of corrective actions during the 10,000 Km period. On the way, you come to an interval which is progressively narrowed by taking quite a few iterative steps. I prefer my simulation model in Excel and I will proceed in VB.

Thank you both for taking the time and the effort.