Optimizing Your Preventative Maintenance Plan

Dear Statman-

We would like to optimize the preventive maintenance plan for our pump valves. Right now, we don’t know how to set it up. Can you give us some advice?

Signed,

Un-pumped

Dear Un-Pumped:

You are lucky. It turns out that statisticians and reliability engineers have long ago worked out a formula for how to optimize your preventive maintenance (PM) plan. For those of you who aren’t familiar with PM, I’ll explain.

Let’s take the above example. A pump has parts (like valves, bearings, wiring, etc.) that wear out with time. As these parts wear out, they have to be replaced. One strategy would be to wait for the part to fail and then replace it. However, what if the failure causes a lot of damage to the pump? Then you don’t want to wait until it fails. You would rather replace it before it fails so that no damage will occur. On the other hand, you don’t want to replace the valve too early because it may have a lot of life left in it. A compromise is necessary. What to do?

The first thing you have to find out is how the failure times pile up. Let’s suppose you have a long history on this pump and you have dutifully noted the failure times of the valves. Some valves last 3 weeks, others last 10 weeks, etc. Clearly, they won’t all last the same amount of time. Suppose we made a plot of failure times in the form of a histogram. A histogram is basically a pile of data. Usually, most of the data will pile up in the middle, say 8 weeks. Sometimes, a valve will only last a couple of weeks and other times it might last up to 20 weeks. From the failure times, there are statistical methods we can use to determine a formula describing its shape. Let’s assume that this formula comes from the Weibull distribution. The following figure shows a histogram and the resulting fitted curve for some fictitious data. The Weibull uses two parameters: the characteristic life, a, and the shape factor, b. The fitted curve over the histogram is called the Weibull density function. It is a function of a and b and is given by

 

where t is time. The probability of a part failing on or before a particular time, F(t), is the integral of f(t) and is given by

 

The parameters a and b are estimated from the data. There are some previous Statman articles that deal with the Weibull distribution and how to find values for a and b. These can be found here. Let’s assume that we can find the values of a and b.

 

Now that we know the distribution and its parameters, we need just two more pieces of information. First, what is the cost of the valve? This is usually pretty easy to determine. I am sure your accounting department can help you there. Let’s call this cost C1. Second, we need to know the cost of not replacing the valve before failure. For example, how much damage does the failure do to the pump? How long will it take to repair? How much production is lost as a result? Anyway, you get the idea. The total cost of valve failure can be harder to quantify. Still, you might be able to get a ballpark idea. Let’s call this cost C2. We want to pick a replacement time, T, which minimizes the long-term average cost per unit time. The formula for the cost uses an integral (please don’t faint) and is given by:

 

where all the terms have been defined. It turns out that the integral in the above equation has a difficult analytical solution when using the Weibull distribution. Rather than boring you with too much math, look at the plot below. It shows a graph of the cost/time for C1=10, C2=50, a=11 and b=2. The plot has a minimum at about four weeks. So, based on this, we would minimize our long-term cost if we replaced the valve about every month. Generally, the replacement time will increase with higher C1, lower C2, higher a, and lower b.

If you have several parts to replace on the pump, you can use the same approach for each one. Assuming each part fails according to a Weibull distribution, first find values for a and b for each part. Then, use the formula to find the replacement time, T, which minimizes the cost. The assumption I am making here is that these failures are independent. For example, when the valve fails it doesn’t cause the bearings to fail any sooner (or later).

Of course, the above formula will work for other distributions too, like the lognormal. Statman has done a little investigating of both the Weibull and lognormal. I suggest you fit your data to both to see which gives the better fit before you calculate T. Many times the Weibull (or lognormal) are used blindly because that is what has been done in the past. I encourage you to question that assumption and see if you can’t find something that works better.

Thanks, Statman.