Your car needs assistance

[Rui Assis]

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 a 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 500 € for each PM and 100 € 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

[Rui Assis]

Let me put the problem described in my previous post in a slightly different way.

A component degrades according to a Weibull with a location parameter 0, a shape parameter 3 and a scale parameter 5,000 hours. I want to maintain it by adopting the constant interval replacement policy (CIRP).

I estimate that the mean cost and the mean time to recover will be: 1,000 € and 5 hours for each PM action and 4,000 € and 30 hours for each CM action, respectively.

Two questions:

a) What will the hourly cost be, if I decide to replace the component every 2,500 hours (calendar time)? And the availability?

What I know for sure is that a few failures will occur before the PM time span ends (the probability of failure until 2,500 is 0,117). How many CM actions will take place in average during the interval of 2,500 hours calendar time?

b) On the other hand, if I want to find an optimum, which should the PM time interval be, in order the hourly cost is a minimum? And the availability is a maximum?

I know that, if I chose the replacement at a predetermined age (running time) policy, the answers to the above questions will be: a) 0.5579 €/hour and 99.67% b) approximately 2,800 hours at a cost of 0,5529 €/hour and 2,400 hours at 99.67% availability.

Thanks in advance.

[Rui Assis]

Correction to my first post dated Saturday, July 02, 2005 - 01:39 am:

...If you pay an average of 100 € for each PM and 500 € for each CM, how much do you spend every 10.000 Km?...

The figures 100 and 500 change places. Sorry for the mistake.

[tarik]

I can’t answer your question about expected cost and number of failures and PMs without knowing what the conversion between km and hours. What average speed are you assuming? Is it 6.365 km/hour ? If that’s the case then if I run the simulation until 1571.09h (10,000 km) then Availability=99.94% and the Number of Failures=0.0336.

The problem you have can be solved easily in BlockSim if you specify the reliability model, duration of CM repair, cost of CM repair, duration of PM repair, and cost of PM repair.

As for the optimum maintenance policy, refer to http://www.weibull.com/SystemRelWeb/preventive_maintenance.htm.
This can also be easily be calculated in BlockSim using the Optimum Replacement feature, which you can access by right clicking on a block (which already has a failure distribution model), choosing Optimum Replacement and entering the cost of PM and CM

[Rui Assis]

I am referring to the “Elapsed Time Policy” or “Periodic (Block) Replacement Policy” or still “Constant Interval Replacement Policy” and not to the “Optimum Age Replacement Policy” which is the only one the recommended link http://www.weibull.com/SystemRelWeb/preventive_maintenance.htm and BlockSim addresses.

I would rather focus on my second post and forget the first one – there is actually no connection, although they raise the same question.

My problem is to compute a solution analytically. I have been looking through a few text books and didn’t find a straight answer. I know how to model the case using simulation in Excel and I have great confidence in the results (an average of 0.122 CM actions per 2,500 hours and (1,000 + 0.122 x 4,000) / 2,500 = 0.5952 €/hour approximately). This cost conforms to what is commonly known: the Constant Interval Replacement Policy results in higher costs than the Optimum Age Replacement Policy but it is far easier to put into practice.

The problem is that I just can’t get the same results when I try an analytical solution. And this I would definitively prefer.

Any help will be appreciated.

[tarik]

You can derive the maintenance policy you are interested in by looking at the expected number of failures that the simulation produces. What is the difference between the different policies you mentioned? The link I provided you is the only replacement method that BlockSim has. I still think that the Optimum Replacement Time is what your post suggests. It can be used to optimize cost or availability (by converting uptime to cost). It can’t be used to optimize both cost and availability simultaneously because the two criteria are different (cost is in dollars the availability is in hours). But if you can relate availability to dollars then you can perform that type of optimization.

I still need to know what’s the conversion between km and hours to be able to answer your question.

As for the computation of expected number of failures when preventive/corrective maintenances are possible, the only approach that BlockSim offers is the simulation option.

[Rui Assis]

Thanks Tarik,

Under the constant interval replacement policy, replacements are performed at predetermined times regardless of the age of the units being replaced (the usual with automobiles). In addition, replacements are performed upon failure of the equipment. This is the simplest preventive maintenance and replacement policy as practitioners know. Under the replacement at predetermined age policy, the units are replaced upon failure or at age tp (calculated by BlockSim) whichever occurs first. The only drawback is that the former is more expensive, although it is far easier to put into practice, as most practitioners know.

Of course everything would be just fine if I had the possibility to calculate analytically and in a simple manner the expected number of failures during the interval tp. The question is “it seems not to exist such a simple way”. I know how to do it using simulation and I already have the answer. But as I said in my previous post, I would definitively prefer an analytical solution in order to code. This is the reason why I insist.

With regard to the conversion between km and hours, please forget as your question surely comes from your assumption that there is some connection between Km (from my first post) and hours (from my second post) and there is not. They are different examples although founded on the same ground.