{"id":1000,"date":"2011-04-20T14:04:53","date_gmt":"2011-04-20T19:04:53","guid":{"rendered":"http:\/\/www.livingreliability.com\/en\/?p=1000"},"modified":"2025-11-06T05:58:46","modified_gmt":"2025-11-06T10:58:46","slug":"real-meaning-of-the-six-rcm-curves","status":"publish","type":"post","link":"https:\/\/www.livingreliability.com\/en\/posts\/real-meaning-of-the-six-rcm-curves\/","title":{"rendered":"Real meaning of the six RCM curves"},"content":{"rendered":"

You may be wondering about the famous “RCM curves” shown below in Figure 1. They are the Age-Reliability behavior curves which graph the Conditional Probability of failure against age. A common question is the following.<\/em><\/p>\n

\n

\u201cFor the 139 items covered in the Nowlan & Heap (UAL) study, for example, do these curves represent the real<\/span> failure behavior of the item? Or do they represent the \u201ceffective\u201d net age-reliability relationship, given that most of these items were subjected to some form of age-based maintenance?\u201d<\/p>\n<\/blockquote>\n

\"sixRcmCurvesFigure1\"<\/a><\/p>\n

Figure 1 The six RCM failure behavior curves[1<\/a>]<\/sup><\/span><\/address>\n
\u00a0<\/address>\n

This is an excellent question. Many maintenance and reliability engineers are not completely sure of the answer nor of its underlying explanation.<\/em><\/p>\n

Introduction<\/h2>\n

When one carries out so called \u201cactuarial analysis\u201d (more simply referred to as \u201creliability analysis\u201d) to draw these curves, it is, indeed, to discover the true<\/span> failure behavior of an item or failure mode[2<\/a>]<\/sup>, regardless<\/em> of the maintenance plan currently in force. How is this possible? Are we not basing our analysis on life data from equipment that has been maintained according to the maintenance strategy and schedule currently in force?<\/p>\n

The explanation of this paradox lies in the actual observational data. At the time of maintenance (say, of the preventive renewal of a component) we will have made certain observations about the state of the maintained items. It is unlikely that we will have observed that all of the renewed items were on the verge of failure. Rather (unless the Preventive Maintenance (PM) \u00a0plan is extremely conservative)\u00a0some items will have failed prior<\/em> to the moment of PM. That is, they will have failed in service. On the other hand, some of the renewed items will not have been in a failed state at the moment of PM. On the contrary, they would have been in excellent condition. Finally, indeed, some items will have well been \u201con their last legs\u201d and just about to fail functionally. In those cases one might consider the PM as having been \u201cwell timed\u201d.[3<\/a>]<\/sup><\/p>\n

An astute manager, made aware of the above observations, would undoubtedly ask \u201cwhat is the optimal<\/em> moment at which to conduct maintenance\u201d. What PM schedule (termed a policy) will return the greatest benefit, i.e. give us the best overall availability, from a fleet perspective, over the long term? Our manager recognizes that being too conservative would be wasteful causing us to renew many perfectly good components. At the same time, he would not want to us be too liberal by selecting too long a maintenance interval. That would result in an excessive number of failures in service, increasing costs, and lowering overall reliability and availability.<\/p>\n

Let us assume the true failure behavior were as shown below in the graph of Figure 2, which is\u00a0 Pattern B of the six failure behavior curves of Figure 1.<\/p>\n

\"Figure<\/a><\/p>\n

Figure 2 Conditional probability of failure curve for an item that ages \u2013 Failure Pattern B<\/span><\/address>\n
\u00a0<\/address>\n

Figure 2 plots the hazard rate or Conditional Probability of failure[4<\/a>]<\/sup> against the working age of an item. We would like to be able to claim (to our manager) that our maintenance plan has been optimized<\/em>. That is to say, that we carry out maintenance at the working age represented by the useful life. The useful life would be the best time to do PM. Such a strategy would prevent most failures. Yet it would avoid renewal of very many items that are in perfect health. That optimum strategy would issue a PM action at a time of maximum benefit to the organization.<\/p>\n

Obviously then, the graph of Figure 2 should represent the inherent<\/em> designed-in age-reliability relationship of the item if we intend to use it to determine the useful life and, hence, the optimum age based maintenance policy. That brings us to the question, \u201cHow to draw this curve, given data from real world, maintained<\/em> equipment?\u201d<\/p>\n

\u201cReal world\u201d implies that our calculations must take into account \u201csuspended\u201d[5<\/a>]<\/sup> data \u2013 data reflecting the actual preventive maintenance programs in place. The most popular and effective statistical method to discover the age reliability relationship is based on the Weibull model for life data. This is an empirical model discovered in the early 1950s by Walodi Weibull[6<\/a>]<\/sup>, who presented the following probability distribution before an esteemed membership (who reacted with skepticism at first) of the then reliability engineering society.<\/p>\n

Weibull Distribution \u2013 three of its forms<\/h2>\n

\"Figure<\/a><\/p>\n

\u00a0Figure 3 Three forms of the Weibull relationship<\/address>\n
\u00a0<\/address>\n
    \n
  1. Probability density \"f\left ( t\right )=\frac{\beta }{\eta }\left ( \frac{t}{\eta } \right )e^{-\left ( \frac{t}{\eta } \right )^\beta }\"\u00a0(Eqn. 1)<\/li>\n
  2. Cumulative distribution \"F\left ( t \right )=1-e^{-\left (\frac{t}{\eta } \right )^{\beta }}\"\u00a0 \u00a0 (Eqn. 2a), and
    \nReliability (aka Survival Probability) \"R(t)=1-F(t)=e^{-\left ( \frac{t}{\eta } \right )^{\beta }}\"\u00a0(Eqn. 2b)<\/li>\n
  3. Failure Rate \"h\left ( t \right )=\frac{\beta }{\eta }\left (\frac{t}{\eta } \right )^{\beta-1 }\"\u00a0(Eqn. 3)<\/li>\n<\/ol>\n

    Where:
    \n\u03b2<\/em><\/strong> (beta) is the \u201cshape\u201d parameter,
    \n\u03b7<\/strong><\/em> (eta) is the \u201cscale\u201d parameter, and
    \nt<\/strong><\/em> is the working age of the item or failure mode being modeled.
    \nF(t)<\/em><\/strong> is the Cumulative Distribution Function (CDF). It is the probability of failure in the period prior to age t<\/em>.
    \nR(t) <\/em><\/strong> the Reliability or Survival Function is the probability of the item surviving<\/em> to age\u00a0t<\/em>. It is the complement of F(t)\u00a0
    \nh(t)<\/strong><\/em><\/strong><\/em>\u00a0is the Hazard or Failure Rate function and
    \nf(t)<\/em><\/strong>\u00a0<\/strong><\/em>is the Probability Density function.\u00a0<\/strong><\/em><\/p>\n

    One can convert algebraically from one to another of the above three forms of the Weibull relationship. For example, h(t)=f(t)\/R(t).[7<\/a>]<\/sup><\/p>\n

    Then we need only determine (estimate from historical data) values for the parameters \u03b2\u00a0and \u03b7\u00a0\u00a0in order to plot the age reliability relationships of, for example, Figure 1. These graphs help us understand the age based failure behavior of items and failure modes of interest. Weibull developed a graphical method for estimating the values of the parameters \u03b2\u00a0and \u03b7\u00a0\u00a0from a set of historical failure data. Today we do not need to use Weibull\u2019s graphical<\/em> estimation methods. Computerized numerical algorithms based on the Weibull model will estimate the values of \u03b2\u00a0and \u03b7 and plot the required graphs.<\/p>\n

    Example<\/h2>\n

    For example, assume we have the following identical[8<\/a>]<\/sup> items, A, B, C, D, and E and the ages at which they failed.<\/p>\n\n\n\n\n\n\n\n\n\n\n
    Table 1:<\/th>\n<\/tr>\n<\/tbody>\n
    Item<\/td>\nFailure age<\/td>\nOrder[9<\/a>]<\/sup><\/td>\n<\/tr>\n
    A<\/td>\n67 weeks<\/td>\n1<\/td>\n<\/tr>\n
    B<\/td>\n120 weeks<\/td>\n2<\/td>\n<\/tr>\n
    C<\/td>\n130 weeks<\/td>\n3<\/td>\n<\/tr>\n
    D<\/td>\n220 weeks<\/td>\n4<\/td>\n<\/tr>\n
    E<\/td>\n290 weeks<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The Weibull method for finding the parameters \u03b2\u00a0and \u03b7 in Equation 2, \"F\left ( t \right )=1-e^{-\left (\frac{t}{\eta } \right )^{\beta }}\"\u00a0starts with a sample of life cycles whose values of F(t) at each of the failure ages 67, 120, 130, 220 and 290 weeks are known. That is, the Weibull method requires that we start by inserting\u00a0 reasonable values for the fraction of the population failing prior to the time t of each observation. That fraction will approximate the cumulative failure probability F(t)) at each of the five failure ages t.<\/p>\n

    Need for a better CDF estimate<\/h2>\n

    We cannot simply assume that the value of the Cumulative Distribution Function\u00a0F(t)<\/em> is the percentage failed at time 120 weeks, i.e. 2\/5, because that would imply that the cumulative failure probability (CDF) at 290 weeks, is 100%. Such a small sample size does not justify such a blanket, definitive statement.<\/p>\n

    To explain this more clearly, let\u2019s extend the use of the simple fraction to the absurd, by considering a sample size of 1. We would not expect the age of this\u00a0single failure to represent the age by which 100% of items in the sample\u2019s underlying population would fail. It would certainly be more realistic to regard this single failure age as representing the age by which 50%\u00a0of the underlying population would fail.<\/p>\n

    Therefore we need a better way of estimating the CDF, particularly for small samples of lifetimes, in order to apply it to the numerical Weibull solution for plotting the age-reliability relationship. The most popular approach to estimating the CDF from failure data is known as the median rank.[10<\/a>]<\/sup>\u00a0\u00a0A formula, known as \u201cBenard\u2019s probability estimator\u201d, provides an estimate for median rank for small populations, and is given in Equation 4.<\/p>\n

    Benard’s probability estimator<\/h3>\n

    \"CDF \hspace{1}estimate=\frac{i-0.3}{N+0.4}\" (Eqn. 4)<\/p>\n

    Where:
    \ni=the sequential order number of the failure; and
    \nN=the size of the sample (number of life cycles)
    \nCDF estimate=the Cumulative Distribution Function estimate, or Median Rank<\/p>\n

    Benard\u2019s formula, applied to our hypothetical single failure, gives (1-0.3)\/(1+0.4)=50%, which is intuitively reasonable.<\/p>\n

    We obtain, either from the median ranks table or from Benard\u2019s approximation, the respective cumulative probabilities of failure (i.e. the CDFs). They are .13, .31, .5, .69, and .87. Of course, when we use reliability analysis software we do not, ourselves, need to look up the median ranks in tables or use Benard\u2019s formula. A computer program applies median ranks automatically[11<\/a>]<\/sup> to each observation. The algorithm then applies a numerical (regression) technique known as the \u201cMethod of Least Squares Estimate\u201d in order to estimate the values of \u03b2\u00a0\u00a0and \u03b7 (depicted by their \u201chatted\u201d symbols a\u0302 and b\u0302 \u00a0) in the following equations:<\/p>\n

    Parameter estimation using regression<\/h2>\n

    The following equations for estimating the Weibull parameters are explained at Weibull.com[12<\/a>]<\/sup> .<\/p>\n

    \"\hat{a}=\frac{\displaystyle\sum\limits_{i=1}^r y_{i}}{r}-\hat{b}\frac{\displaystyle\sum\limits_{i=1}^r x_{i}}{r}\" (Eqn. 5)\u00a0\"$\hat{b}=\frac{\displaystyle\sum\limits_{i=1}^{r}x_{i}y_{i}-\frac{\displaystyle\sum\limits_{i=1}^{r}x_{i}\displaystyle\sum\limits_{i=1}^{r}y_{i}}{r}}{\displaystyle\sum\limits_{i=1}^{r}x_{i}^{2}-\frac{\left(\displaystyle\sum\limits_{i=1}^{r}x_{i} \right)^{2}}{r}}{}$\" (Eqn. 6)<\/p>\n

    \"x_{i}=ln\left[T_{i} \right]\" and \"y_{i}=ln\left[-ln\left[1-F(T_{i}) \right] \right]\" (Eqns. 7)<\/p>\n

    Where:
    \n^ denotes that the value is an estimate.
    \nEach F(Ti<\/sub>)<\/em> is obtained from the Median Ranks or Benard’s Formula.
    \nN=number of observations.
    \nr(\u2264N)=the number of failures.
    \ni is the sequential order number of each failure when all failures are listed in ascending order of failure age.<\/p>\n

    Once the computer algorithm has estimated a and b from Eqns. 5, 6, and 7,\u00a0 estimates for \u03b2 and \u03b7 can be found easily from a\u0302 = -\u03b2ln[\u03b7] and b\u0302 \u00a0= \u03b2. <\/em>Now any of the curves of equations 1, 2, or 3 may be plotted and examined to understand the item’s age based failure behavior.[13<\/a>]<\/sup><\/p>\n

    Suspensions<\/h2>\n

    Order<\/h3>\n

    Now that we have discussed, in general terms, the Weibull method for plotting the age-reliability relationship (using historical failure data), we turn our attention to the problem of \u201csuspensions\u201d, which is the main subject of this article.[14<\/a>]<\/sup> Let\u2019s begin with the sample of lifetimes in Table 2.:<\/p>\n\n\n\n\n\n\n\n\n
    Table 2<\/th>\n<\/tr>\n
    Item<\/td>\nFailure age<\/td>\nOrder<\/td>\n<\/tr>\n
    A<\/td>\n84 weeks<\/td>\n1<\/td>\n<\/tr>\n
    B<\/td>\n91 weeks<\/td>\n2<\/td>\n<\/tr>\n
    C<\/td>\n122 weeks<\/td>\n3<\/td>\n<\/tr>\n
    D<\/td>\n274 weeks<\/td>\n4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The items are arranged in sequence according to their survival ages. Let N be the total number of observations, in this case, 4, and let \u201ci\u201d be the order (either 1, 2, 3, or 4) of a given failure observation. We can easily apply Benard\u2019s formula to estimate the CDF at each of the four observations. Then we may proceed according to the methods discussed earlier, to determine the Weibull parameters and thus the age-reliability relationship.<\/p>\n

    Order with suspensions<\/h3>\n

    But what if item B, for example, did not fail, but was renewed preventively, as directed, say, by the PM system? In such a case what would be the order “i” (of failure of items C and D) to be applied in Benard’s formula? We no longer know the orders of failure of items C and D because we do not know exactly when (beyond 91 weeks) item B would<\/em> have failed (had it not been preemptively renewed).<\/p>\n\n\n\n\n\n\n\n\n
    Table 3<\/th>\n<\/tr>\n
    Item<\/td>\nFailure (F#) or Suspension (S#)<\/td>\nFailure or suspension age<\/td>\nOrder<\/td>\n<\/tr>\n
    A<\/td>\nF1<\/td>\n84 weeks<\/td>\n1<\/td>\n<\/tr>\n
    B<\/td>\nS1<\/td>\n91 weeks<\/td>\n<\/td>\n<\/tr>\n
    C<\/td>\nF2<\/td>\n122 weeks<\/td>\n?<\/td>\n<\/tr>\n
    D<\/td>\nF3<\/td>\n274 weeks<\/td>\n?<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    The table shows that the first failure was at 84 weeks. Then at 91 weeks an item was taken out of service for reasons other than failure. (that is, its life cycle was \u201csuspended\u201d). Two more failures occurred at 122 and 274 weeks. But the actual order of the failures of C, and D are unknown since they would depend on when B would have\u00a0failed. Therefore their orders are indicated by “?”.<\/p>\n

    Using average orders<\/h3>\n

    Suspended data is handled by assigning an average\u00a0order number<\/em> to each failure. The “average” is calculated by considering all the possible sequences as follows:<\/p>\n

    Had the suspended part been allowed to fail there are three possible scenarios (depending on when Item B might<\/em> have failed had it not been suspended). In each of these scenarios the hypothetical failure of Item B is denoted by \u201cS1 ->F\u201d in Table 4.<\/p>\n\n\n\n\n\n\n\n
    Table 4<\/th>\n<\/tr>\n
    Possible Scenarios<\/strong><\/td>\ni=1<\/strong><\/td>\ni=2<\/strong><\/td>\ni=3<\/strong><\/td>\ni=4<\/strong><\/td>\n<\/tr>\n
    Possibility 1<\/strong><\/td>\nF1<\/td>\nS1-> F<\/td>\nF2<\/td>\nF3<\/td>\n<\/tr>\n
    Possibility 2<\/strong><\/td>\nF1<\/td>\nF2<\/td>\nS1->F<\/td>\nF3<\/td>\n<\/tr>\n
    Possibility 3<\/strong><\/td>\nF1<\/td>\nF2<\/td>\nF3<\/td>\nS1->F<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Where:
    \nF1 represents the first failed unit, Item A
    \nF2 represents the second failed unit, Item C
    \nF3 represents the third failed unit, Item D, and
    \nS1 represents the first (and only) suspended unit, Item B, in this sample.<\/p>\n

    The first observed failure (F1) will always be in the first position and have order i=1. However, for the second failure (F2) there are two out of three scenarios in which it can be in position 2 (order number of i=2), and one way that it can be in position 3 (order number of i=3). Thus the average order for the second failure is:<\/p>\n

    \"\frac{3 \times1+2 \times 2}{3}=2.33\"<\/p>\n

    For the third failure there are two ways it can be in position 4 (order number of i=4) and one way it can be in position 3 (order number of i = 3).<\/p>\n

    \"\frac{4 \times 2+3 \times 1}{3}=3.67\"<\/p>\n

    We will use these average position values to calculate the median rank from Benard’s formula\u201d (Equation 4), as in column 5 of Table 5, for use within Equation 7.<\/p>\n

    New increment<\/h3>\n\n\n\n\n\n\n\n\n
    Table 5<\/th>\n<\/tr>\n
    Item<\/strong><\/td>\nFailure or Suspension<\/strong><\/td>\nFailure or suspension age<\/strong><\/td>\nOrder, i<\/strong><\/td>\nMedian rank<\/strong><\/td>\n<\/tr>\n
    A<\/td>\nF<\/td>\n84 weeks<\/td>\n1<\/td>\n0.16<\/td>\n<\/tr>\n
    B<\/td>\nS<\/td>\n91 weeks<\/td>\n?<\/td>\n<\/td>\n<\/tr>\n
    C<\/td>\nF<\/td>\n122 weeks<\/td>\n2.33<\/td>\n0.46<\/td>\n<\/tr>\n
    D<\/td>\nF<\/td>\n274 weeks<\/td>\n3.67<\/td>\n0.77<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Obviously, finding all the sequences for a mixture of several suspensions and failures and then calculating the average order numbers is a time-consuming process. Fortunately a formula is available for calculating the order numbers. The formula produces what is termed a \u201cnew increment\u201d. The new Increment I is given by:<\/p>\n

    \"I=\frac{\left(n+1 \right)-\left(previous \;order \; number \right)}{1+number \;of\; items\; following\; the\; suspended\; set}\"<\/p>\n

    Where
    \nn=the total number of items in the sample;
    \nI=the increment<\/p>\n

    So for failures at 122 weeks and 274 weeks the Increment will be:<\/p>\n

    \"I=\frac{\left(4+1 \right)-\left(1 \right)}{1+2}=1.33\"<\/p>\n

    The Increment will remain 1.33 until the next suspension, where it will be recalculated and used for the subsequent failures.\u00a0\u00a0This process is repeated for each set of failures following each set of suspensions. The order number of each failure is obtained by adding the \u201cincrement\u201d of that failure to the order number of the previous failure. For example referring to Table 5, acknowledge that the Order of Item C (2.33) is indeed 1+1.33, and the Order of Item D (3.67) is indeed \u00a0approximately 2.33+1.33.<\/p>\n

    Now that we have the adjusted orders of the failures, we can provide the F(t)’<\/em>s[15<\/a>]<\/sup> required by Eqn. 7 for estimating the Weibull parameters using Regression.<\/p>\n

    Software<\/h3>\n
    \"\"<\/a>
    Figure 4 Analysis selection in SimuMatic<\/figcaption><\/figure>\n

    Although the foregoing seems rather complicated, the availability of powerful reliability software packages renders the job quite easy. For example, Reliasoft\u2019s tool, SimuMatic, requires the user merely to enter or import the data and select the desired calculation method from the dialog (see Figure 4 ).<\/p>\n

    The Figure implies that there are several alternative calculation methods that can be used to estimate the age-reliability relationship. We have briefly described one of them, the Method of Least Squares Estimation with median ranking. Another popular method is known as Maximum Likelihood Estimation (MLE)[16<\/a>]<\/sup>. This article has attempted to provide the reader with some insight into the famous six RCM failure patterns (shown in the figure in the Introduction to this section), and, into the question, \u201cHow to know the true<\/em> age-reliability relationship, from real world samples containing suspended lifetimes?\u201d. The next part discusses the combining of the age-reliability relationship with business factors in order to establish an optimal preventive maintenance interval.<\/p>\n

    Optimizing TBM – cost model<\/h2>\n

    Density function<\/h3>\n

    Having understood the foregoing method for plotting the age reliability relationship, a practical question comes to mind. We may ask, quite legitimately, how knowledge of the age-reliability relationship can assist in optimizing the maintenance decision process? The following addresses this question.<\/p>\n

    Two useful ways to express the Weibull age-reliability relationship are as a:<\/p>\n

      \n
    1. Hazard function\u00a0 \"h(t)=\frac{\beta}{\eta}\left(\frac{t}{\eta} \right)^{\beta-1}\" [17<\/a>]<\/sup>\n