# Optimum Overhaul Example

Jump to navigation
Jump to search

*This example appears in the Reliability Growth and Repairable System Analysis Reference*.

Field data have been collected for a system that begins its wearout phase at time zero. 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 the cost of an overhaul is four times more expensive than a repair. The table below presents the data. Do the following:

- Estimate the parameters of the Power Law model.
- Determine the optimum overhaul interval.
- If , would it be cost-effective to implement an overhaul policy?

Field Data
| ||

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 |

**Solution**

- The next figure shows the estimated Power Law parameters.
- The QCP can be used to calculate the optimum overhaul interval, as shown next.
- Since the systems are wearing out and it would be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out.