Planning Maintenance Policies using MTBF: How Much is it Costing?
 

Planning Maintenance Policies using MTBF: How Much is it Costing?

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In an earlier article, we looked at how MTBF alone can be misleading when selecting an item for use in a design. In this ...

In an earlier article, we looked at how MTBF alone can be misleading when selecting an item for use in a design. In this
article, we’ll take a look at how the MTBF metric falls short as an input to maintenance planning. There were three items in
the referenced article, Item D, Item E, and Item F that we’ll consider. We already know each of their failure times are Weibulldistributed with the following parameters and MTBF.

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    Planning Maintenance Policies using MTBF: How Much is it Costing? Planning Maintenance Policies using MTBF: How Much is it Costing? Document Transcript

    • 1 Planning Maintenance Policies Using MTBF, How Much is it Costing? Andrew Rowland, CRE I. I NTRODUCTION In an earlier article, we looked at how MTBF alone can be misleading when selecting an item for use in a design. In this article, we’ll take a look at how the MTBF metric falls short as an input to maintenance planning. There were three items in the referenced article, Item D, Item E, and Item F that we’ll consider. We already know each of their failure times are Weibull distributed with the following parameters and MTBF. TABLE I E STIMATED PARAMETERS FOR I TEM D, I TEM E, AND I TEM F Item Item D Item E Item F Eta 101.42 107.73 100.84 Beta 0.478 1.000 4.524 MTBF 220.7 107.7 92.0 For all of the examples, R [1] was used to perform the calculations and produce the plots. II. M AINTENANCE TASKS Nowlan and Heap, in their 1978 report for the United States Department of Defense [2] defined four types of preventive maintenance tasks. These were, on-condition, scheduled rework, scheduled discard, and failure finding. Even today, maintenance tasks will fall into one of these four categories. It is the rework, discard, and failure-finding tasks whose periodicity is often based on an item’s MTBF. In this article, we will focus on the rework and discard tasks. Both of these tasks are performed at a scheduled interval, the difference being a rework task restores the existing item to a certain level of failure resistance. The discard task removes the existing item and replaces it with a new item. In both cases, the proper time to perform the task is a cost optimization problem. Mathematically, if we minimize the following function, we find the optimal periodicity to perform these types of tasks. η(T ) = Ccm (1 − R(T )) + Cpm R(T ) T R(x)dx 0 where: T is the optimum replacement periodicity. η(T ) is the average cost per unit time at time T. Ccm is the average cost of one corrective maintenance action. Cpm is the average cost of one preventive maintenance action. R(x) is the reliability function. R(T) is the value of the reliability function at time T. Once can see from the function above that the absolute values of Ccm and Cpm aren’t important. What is important is the ratio of Ccm to Cpm . For this reason, we’ll assume Cpm = 1. Typically, Cpm << Ccm . We’ll refer to this approach as the NoMTBF Method.
    • 2 III. E XAMPLES One often used method of obtaining a maintenance periodicity, especially for rework and discard tasks, is to perform the task at an interval based on a predetermined level of reliablity. Typically, when using this method, it is assumed failure times are exponentially distributed. For our examples, we’ll use the actual distribution with the parameters in Table I as well as assuming exponentially distributed failure times. We’ll refer to this as the Lx Method. Our policy will be to rework or replace when R(T) = 0.9 in both situations. Another, simpler approach is to perform the task at some percentage of the MTBF. We often hear the justification for this method being rework or replace ”right before it fails.” We’ll refer to this as the % MTBF Method. For this method, our policy will be to rework or replace at 80% of the MTBF. We will consider Ccm /Cpm ratios of 10, 20, 30, 40, 50, 60, 70, 80, 90, and 100 for each of the three methods. We’ll compare the average cost per hour of the maintenance period established by each of the three methods. Finally, to minimize computing time, let’s assume our hypothetical items will be retired after 1000 hours. A. Item D From Table I, we see that Item D has a β < 1. In other words, Item D is dominated by early life failure mechanisms. Knowing this, we would intuitively know that a rework or discard task is not warranted. However, if the only information we have is the MTBF, our intuition wouldn’t have the chance to guide us. Finding the optimal maintenance period for Item D using the function above, we find that T = 1000; in other words, Item D should be “run to failure.” This should come as no surprise. Our intiution already told us this would be the case, our calculations simply confirm. Figure 1 plots the average cost per unit time functions for T = [0, 1000] hours. We see that, regardless of the Ccm /Cpm ratio, the function a minimum at 1000 hours. For a Ccm /Cpm ratio of 10, the average cost per hour is $0.055. Fig. 1. Cost per Time Functions for Item D Using the Lx Method, we find a maintenance interval of 0.92 hours. This is the time at which the reliability of Item D is 0.9. The average cost per hour, assuming a 10:1 Ccm /Cpm ratio, if Item D is maintained at this interval is $11.20, approximately 200 times the cost of “running to failure.” The Lx Method assuming exponential failure times arrives at an interval of 23.3
    • 3 hours with an average cost per hour of $0.57, more than 10 times the average cost per hour of “running to failure.” Using the % MTBF Method, we would set the maintenance period at 176.6 hours or 80% of the 220.7 hours MTBF we see in Table I. The average cost per hour for the 10:1 Ccm /Cpm ratio of this maintenance interval is $0.124, more than twice the average cost of the optimal periodicity. Table II summarizes the results for the three methods. Note there are two columns for the Lx Method; one for the Weibull assumption and the other for the exponential assumption. TABLE II T IME OF M INIMUM C OST FOR I TEM D Ccm /Cpm Ratio 10 20 30 40 50 60 70 80 90 100 NoMTBF Method Optimum Time (hours) 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 NoMTBF Method Minimum Cost ($ / hour) 0.055 0.110 0.165 0.220 0.275 0.330 0.385 0.441 0.496 0.551 Lx Method (WEI) Minimum Cost ($ / hour) 11.197 22.335 33.473 44.612 55.750 66.888 78.026 89.164 100.302 111.440 Lx Method (EXP) Minimum Cost ($ / hour) 0.569 1.135 1.700 2.266 2.832 3.398 3.963 4.529 5.095 5.661 % MTBF Method ($ / hour) 0.124 0.248 0.372 0.495 0.619 0.743 0.866 0.990 1.113 1.237 B. Item E From Table I, we see that Item E has a β = 1. In other words, Item E has exponentially distributed failure times. Again, knowing this, we would intuitively know that a rework or discard task is not warranted. Finding the optimal maintenance period for Item E using the function above, we find once again that T = 1000 and Item E should be “run to failure” as well. Figure 2 plots the average cost per unit time functions for T = [0, 1000] hours. We see that, regardless of the Ccm /Cpm ratio, the function is at a minimum at 1000 hours. For a Ccm /Cpm ratio of 10, the average cost per hour is $0.093. Using the Lx Method, we find a maintenance interval of 11.35 hours. This is the time at which the reliability of Item E is 0.9. The average cost per hour for the 10:1 Ccm /Cpm ratio if Item E is maintained at this interval is $0.93, 10 times the cost of “running to failure.” Using the % MTBF Method, we would set the maintenance period at 86.2 hours or 80% of the 107.7 hour MTBF we see in Table I. The average cost per hour for the 10:1 Ccm /Cpm ratio of this maintenance interval is $0.17, almost twice the average cost of the optimal periodicity. Table III summarizes the results for the three methods. C. Item F From Table I, we see that Item F has a β > 1. In other words, Item F is dominated by wearout failure mechanisms. Knowing this, we intuitively know that a rework or discard task may be warranted. The knowledge that Item F is dominated by wearout is not something an MTBF statistic alone can tell us. Finding the optimal maintenance period for Item F using the function above, we find that T ¡ 1000 and Item F should be reworked or replaced at some interval. We see in Table IV that the interval is dependent on the Ccm /Cpm ratio. For a Ccm /Cpm ratio of 10, the optimal maintenance interval is 47 hours and the average cost per hour is $0.027. Figure 2 plots the average cost per unit time functions for T = [0, 200] hours. Using the Lx Method, we find a maintenance interval of 61.3 hours. This is the time at which the reliability of Item F is 0.9. For a Ccm /Cpm ratio of 10, the average cost per hour if Item F is maintained every 61.3 hours is $0.17, 6 times the average cost per hour of maintaining Item F every 47 hours. Worse yet, the additional cost is driven by corrective maintenance costs. Assuming exponential failure times, the Lx Method arrives at an interval of 9.7 hours with an average cost per hour of $1.03,
    • 4 Fig. 2. Cost per Time Functions for Item E TABLE III T IME OF M INIMUM C OST FOR I TEM E Ccm /Cpm Ratio 10 20 30 40 50 60 70 80 90 100 NoMTBF Method Optimum Time (hours) 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 NoMTBF Method Minimum Cost ($ / hour) 0.093 0.186 0.278 0.371 0.464 0.557 0.650 0.743 0.835 0.928 Lx Method (WEI) Minimum Cost ($ / hour) 0.928 1.856 2.784 3.713 4.641 5.569 6.497 7.425 8.353 9.282 Lx Method (EXP) Minimum Cost ($ / hour) 0.928 1.856 2.784 3.713 4.641 5.569 6.497 7.425 8.353 9.282 % MTBF Method ($ / hour) 0.169 0.337 0.506 0.674 0.843 1.011 1.180 1.348 1.517 1.686 over 38 times the average cost per hour of the optimal interval. Using the % MTBF Method, we would set the maintenance period at 73.6 hours or 80% of the 92.0 hour MTBF we see in Table I. For a Ccm /Cpm ratio of 10, the average cost per hour of this maintenance interval is $0.14, five times the average cost of the optimal periodicity. Table IV summarizes the results for the three methods. IV. C ONCLUSION We have taken a look at three items with different reliability characteristics. One dominated by early lfie failure mechanisms, one by wearout mechanisms, and one whose failure times are exponentially distributed. For each item, we’ve calculated a maintenance interval and the average cost per hour for various Ccm /Cpm ratios using three different methods. Hopefully we have seen that a simple MTBF statistic alone is not enough to calculate cost optimized maintenance intervals. In fact, we’ve seen that a MTBF statistic alone is not enough to determine if an item is even a candidate for preventive maintenance.
    • 5 Fig. 3. Cost per Time Functions for Item F TABLE IV T IME OF M INIMUM C OST FOR I TEM F Ccm /Cpm Ratio 10 20 30 40 50 60 70 80 90 100 NoMTBF Method Optimum Time (hours) 47 40 36 34 32 31 30 29 28 28 NoMTBF Method Minimum Cost ($ / hour) 0.027 0.032 0.035 0.038 0.040 0.041 0.043 0.044 0.045 0.046 Lx Method (WEI) Minimum Cost ($ / hour) 0.166 0.332 0.498 0.665 0.831 0.997 1.163 1.329 1.495 1.662 Lx Method (EXP) Minimum Cost ($ / hour) 1.031 2.062 3.093 4.124 5.155 6.186 7.218 8.249 9.280 10.311 % MTBF Method ($ / hour) 0.142 0.283 0.425 0.566 0.708 0.850 0.991 1.133 1.274 1.416 Having an understanding of the reliability characteristics of an item is critical to developing appropriate maintenance strategies. The MTBF alone failes to provide the necessary understanding. In the case of Item D, we make the system less reliable as a result of preventive replacement because we are replacing an item dominated by early life failure mechanisms. Because Item E has exponentially distributed failure times, preventive maintenance will do nothing to improve the reliability. Preventive maintenance on Item E is simply an exercise in spare parts usage. Only Item F possess the reliability characteristics to make preventive maintenance applicable. We saw that with only the MTBF available to us, the methods to calculate maintenance intervals result in average hourly costs significantly in excess of maintenance intervals calculated with a fuller understanding of the reliability characteristics. How much is your organization spending because MTBF has led you to sub-optimal maintenance intervals? MTBF is ingrained in the reliability community as well as throughout most companies. It is unlikely that we will ever see the end of MTBF. Ultimately it comes down to us, as reliability engineers, to understand the limitations of MTBF and educate those around us to its shortcomings. If the reliability community gets in lock-step, we can be the tugboats that change the ships heading.
    • 6 R EFERENCES [1] R Development Core Team, R: A Language and Environment for Statistical Computing. Vienna, Austria: R Foundation for Statistical Computing, 2010. [2] F. Stanley Nowlan and Howard F. Heap, AD-A066-579 Reliability-Centered Maintenance. Washington, D.C.: Office of Assistant Secretary of Defense, 1978. This paper is c 2013, Andrew Rowland. Andrew Rowland is a Reliability Consultant. He previously worked as a Reliability and Safety Engineer in the aerospace, defense, and civil nuclear industries. Mr. Rowland received a BSEE in 1999 and a MS in Statistics in 2006. He is an American Society for Quality Certified Reliability Engineer, a member of the IEEE Reliability Society, and the American Statistical Association. He may be contacted by email at andrew.rowland@reliaqual.com.