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# Optimum Maintenance Intervals in RCM++

Preventive maintenance can yield significant cost benefits by increasing the availability of a system and reducing the total maintenance costs. The question of how often the task should be performed, however, is important to consider. If the preventive maintenance interval is too short, then the maintenance costs associated with preventive maintenance can be too high. If, on the other hand, the interval is too long, then the costs associated with corrective maintenance can be too high. RCM++ provides calculations to help determine the optimum maintenance interval, based on the probability of occurrence of a failure event and the costs of performing different types of maintenance. In this article we will provide the theory behind this utility and give an example that illustrates its application in the software.

## Introduction

In order for preventive maintenance to be beneficial, the failure rate of the system must increase over time and the cost of the planned preventive maintenance must be less than the cost of the unplanned corrective maintenance. If both of those conditions are met, then the preventive maintenance should be performed. However, the time interval for performing preventive maintenance should be such that the total maintenance costs are minimized, as shown in Figure 1. In order to do that, the time interval that minimizes the maintenance cost function must be found. Figure 1: Cost vs. Time

The maintenance cost per unit time function is given by: where:

• R(t) is the reliability at time t.
• CP is the preventive maintenance cost per incident (planned maintenance).
• CU is the corrective maintenance cost per incident (unplanned maintenance).

The optimum replacement time interval, t, is the time that minimizes CPUT(t). This can be found by solving for t such that: or by solving for a time, t, that satisfies: For more information on the maintenance cost per unit time equation you can read Preventive Maintenance and the Cost Per Unit Time Equation from Issue 96 of the Reliability HotWire.

## Application

Imagine that we are engineers in an automotive company and we are responsible for the maintenance of the assembly line. Recently, we’ve focused our efforts on the induction hardening process for drive shafts. Figure 2 shows the FMEA for this process. Figure 2: FMEA of Induction Hardening Process

One of the controls that we have identified is a preventive maintenance task on the induction hardening machines. Using past data, we have determined that the machine's probability of failure follows a Weibull distribution with a beta of 3 and an eta of 800 hours, as shown in Figure 3. Figure 3: Probability of Failure of the Hardening Machine

Figure 4 shows the maintenance costs associated with the corrective maintenance of the machine. The typical task duration for repairing the machine is 5 hours. However, given that when the machine fails unexpectedly there is a delay for the repair crew to arrive and the spare parts to be obtained, there is a total downtime of 7.7 hours per incident. Since the cost per hour of downtime is \$1,000, this translates to a cost of downtime of \$7,700 per unexpected failure. With the other cost inputs, including the materials cost of \$200 per incident and the calculated total labor cost of \$250 (5 hours for the task multiplied by the labor rate of \$50 per hour), the total cost per corrective maintenance incident is equal to \$8,150.00. Figure 4: Corrective Maintenance Costs

Using the probability of failure of the machine and the associated corrective maintenance costs, we can run a simulation to determine the average availability of the machine for one year of operation (or 4,160 hours, given that the machine operates for 16 hours a day, 5 days a week) with a "run to failure" maintenance strategy . The average availability is 99.01% and the total operating cost is \$43,463.95, as shown in Figure 5. These figures reflect the availability and cost assuming that no preventive maintenance is performed. Figure 5: Calculated Average Availability and Total Operating Cost for Corrective Maintenance Only (no Preventive Maintenance)

As mentioned above, we have determined that preventive maintenance should be performed on the hardening machine. We now need to decide how often the preventive maintenance should be scheduled. As we’ve seen, given the corrective and preventive maintenance costs and the probability of failure, we can find a time interval that minimizes the total cost function. This task can be performed easily in RCM++.

Figure 6 shows the costs associated with the preventive maintenance. Since preventive maintenance is a planned task, the total duration of the incident is considerably lower compared to the corrective maintenance task. As a result, the downtime cost and the total cost per incident will also be lower. Figure 6: Preventive Maintenance Costs

Now that we have determined the maintenance costs, we can calculate the optimum interval for performing the preventive maintenance. Figure 7 shows that the optimum interval is found to be 468.984 hours. RCM++ gives us the option to set this as the assigned interval and use it in our calculations, set it as a proposed interval in order to keep it as a record without using it, or not use it at all. In this case, we will set it as the proposed interval. Figure 7: Calculation of Optimum Maintenance Interval

For the actual assigned interval, we choose to round the calculated figure to 470 hours. Using this maintenance interval, we can run a simulation again to calculate the average availability and total operating cost for a year of operation. As we can see in Figure 8, the average availability from implementing the preventive maintenance strategy is calculated as 99.36% and the total operating cost \$29,390.25. So we can see that by using the optimum maintenance interval to perform preventive maintenance, the availability is increased (99.36% compared to 99.01%) and the operating cost is reduced (\$29,390.25 compared to \$43,463.95). Figure 8: Calculated Average Availability and Total Operating Cost for Preventive Maintenance Strategy with Optimum Maintenance Interval 