Reliability HotWire

Issue 64, June 2006

Reliability Basics
Economical Life Model for Repairable Systems

One consideration in reducing the cost of maintaining repairable systems is to establish an overhaul policy that will minimize the total life cost of the system. However, an overhaul policy only makes sense if the systems suffer from wearout problems. In other words, the β parameter of the Crow-AMSAA model should be greater than 1. Implementing an overhaul policy if wearout is not present is not economically beneficial, because an overhauled unit would have the same or a higher probability of failing compared to a unit that was not overhauled. If we assume that there is a point at which it is cheaper to overhaul a system than to continue repairs, what is the overhaul time that will minimize the total life cycle cost while considering repair cost and the cost of overhaul?

This article presents a methodology, implemented in RGA 6, for estimating the optimum overhaul time based on the Economical Life Model for Repairable Systems.

Denote C1 as the average repair cost (unscheduled), C2 as the replacement or overhaul cost and C3  as the average cost of scheduled maintenance. Scheduled maintenance is performed for every S miles or time interval. In addition, let N1 be the number of failures in [0, t] and let N2 be the number of replacements in [0, t]. Suppose that replacement or overhaul occurs at times T, 2T, 3T. The problem is to select the optimum overhaul time T = T0 so as to minimize the long term average system cost (unscheduled maintenance, replacement cost and scheduled maintenance). Since β > 1, the average system cost is minimized when the system is overhauled (or replaced) at time T0, such that the instantaneous maintenance cost equals the average system cost.

Total system cost between overhaul or replacement is:

MATH Eqn. 1

So the average system cost is:

MATH  Eqn. 2

The instantaneous maintenance cost at time T is equal to:

MATH Eqn. 3

The following equation holds at optimum overhaul time T0:

MATH Eqn. 4

Therefore:

MATH  Eqn. 5

When there is no scheduled maintenance, Eqn. 4 becomes:

MATH Eqn. 6

The optimum overhaul time, T0, is the same as Eqn. 5, so for periodic maintenance scheduled every S miles, the replacement or overhaul time is the same as for the unscheduled and replacement or overhaul cost model.

In RGA 6, the Economical Life Model can be applied for repairable systems or fleet analysis, which are two types of analysis of fielded systems.

Example

A sample of field data has been collected for a fleet of systems which suffer from wearout problems. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that an overhaul is four times more expensive than a repair. The data set is presented in the next table.
 

System 1

System 2

System 3

1006.3

722.7

619.1

2261.2

1950.9

1519.1

2367

3259.6

2956.6

2615.5

4733.9

3114.8

2848.1

5105.1

3657.9

4073

5624.1

4268.9

5708.1

5806.3

6690.2

6464.1

5855.6

6803.1

6519.7

6325.2

7323.9

6799.1

6999.4

7501.4

7342.9

7084.4

7641.2

7736

7105.9

7851.6

8246.1

7290.9

8147.6

 

7614.2

8221.9

 

8332.1

9560.5

 

8368.5

9575.4

 

8947.9

 

 

9012.3

 

 

9135.9

 

 

9147.5

 

 

9601

 

The data set is modeled using the Crow-AMSAA (NHPP) model and an increment length of 6000miles. The estimated parameters are shown in the next figure.

Since β > 1, the systems are wearing out.

The QCP can be used to calculate the optimum overhaul interval, as shown next.

Note that in the QCP, you can enter either the actual repair and overhaul costs or a factor that describes the overhaul cost in comparison to the repair cost (as in this example).

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