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Optimizing Preventive Maintenance Schedules


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Below is a step by step example of how to optimize the schedule for preventive maintenance.  The example utilizes the Reliability & Maintenance Analyst softwareClick here to download a free demo version of this software.

Click here for the technical details of optimizing schedules for preventive maintenance, predictive maintenance, and for optimizing inspection schedules in Adobe Acrobat format.  For the same article in Microsoft Word for Windows 95 format click here.


Preventive maintenance is useful for preventing failures and increasing reliability. Preventive maintenance is only useful if the failure rate is increasing with age (the shape parameter is greater than 1.0). If the failure rate is constant or decreasing, preventive maintenance either has no affect on failures (the shape parameter is equal to 1.0), or increases failures (the shape parameter is greater than 1.0).

When the failure rate is increasing, the question is how often should preventive maintenance be done. There is a cost associated with preventive maintenance, and there is a cost associated with a failure. Preventive maintenance should be scheduled at a frequency that minimizes the total of these two costs.

Example 1

It is known that a sheet metal leveler fails due to a build-up of dirt. To prevent failures, the current policy is to clean the leveler after 40 million feet of metal have been leveled. Failure data for the sheet metal leveler has been collected and is contained in the computer file "LEVELER.DAT" (This file is included with the demo version). Again check in the "Censored" box indicates the leveler was cleaned without failing.

A production manager has requested a Weibull analysis to determine what corrective action, if any, should be taken. A Weibull analysis was performed using the following steps.

  1. Select File from the main menu.
  2. Select Open, and select "LEVELER.DAT".
  3. Select Parameter Estimation.
  4. Select Weibull.
  5. Select Maximum Likelihood Estimation.

The software returns an estimate of 4.595 for the shape parameter, with an 80% confidence interval (10% on each tail) of 3.744 to 5.64. The manager is instructed that the leveler data indicates "wear-out", and that the correct procedure is to conduct preventive maintenance - cleaning the leveler.

The manager states that the policy of cleaning the leveler after every 40 million feet of leveled metal was based on meeting production schedules, and wants to know if there is a procedure for determining the "correct" schedule for cleaning the leveler.

To determine the optimum schedule for cleaning the leveler, more information is needed. The cost of cleaning the leveler is $180, and the cost of a failure is $1,220. This decision must be made based on costs. After all, if the cost of cleaning the leveler was greater than the cost of failure, the best policy is to run to failure. To determine the optimum schedule for cleaning the leveler

  1. Determine the parameters of the Weibull distribution (This has already been done).
  2. Select Maintenance.
  3. Select Optimum Schedule for Preventive Maintenance.
  4. Select the MLE parameter estimates from the table.
  5. Enter "180" for the cost of a PM and "1220" for the cost of a failure.

The software will determine the optimum time between leveler cleanings; 29.28 million feet of leveled metal, and the optimum cost, $7.888 per million feet of leveled metal.

Example 2

The manager from the previous problem has found a piece of equipment that will reduce the cost of cleaning the leveler to $80. This manager realizes that if the cost of a PM is reduced, the optimum time between PMs will be reduced, and the optimum cost will decrease. The savings gained is not the number of PMs multiplied by the savings per PM.  Before investing in this cleaning equipment, the manager wants to quantify the savings.

To do this, follow the steps in the example above, but enter $80 as the cost of a PM. The optimum time between PMs is 24.05 million feet of leveled metal at an optimum cost of $4.259 per million feet of leveled metal. Investing in this equipment would net $3.629 ($7.888-$4.259) per million feet of leveled metal.


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