Enter the cost of repairs or the number of repairs for each row of data. If you do not provide this column, then Minitab assumes a cost of 1 for all failures.

If you have interval data and you selected the Number of systems option on the main dialog box, you must enter a Cost/Frequency column to indicate the number of failures or cost of failures for the systems within each interval.

- Repair cost/number of repairs for systems
- Enter a column that contains the cost of repairs or the number of repairs for each row of data. The column must contain numeric values that are greater than or equal to 0. If each row represents a single, retired system, the value in the column must be 0. The first cost column is paired with the first data column, the second cost column is paired with the second data column, and so on.

In the following worksheet, the Start column and End column indicate the interval within which each failure (or retirement) occurred. The System column identifies the system for each failure time. The Frequency column (optional) indicates the total number of failures (or retirements) within each interval. For example, 2 failures occurred between 0 and 1 hour for system 1. 4 failures occurred between 1 and 5 hours for system 1. After 9 hours, no more observations were recorded for system 1.

C1 | C2 | C3 | C4 |
---|---|---|---|

Start | End | System | Frequency |

0 | 1 | 1 | 2 |

1 | 5 | 1 | 4 |

9 | * | 1 | 0 |

0 | 4 | 2 | 2 |

5 | 7 | 2 | 4 |

10 | * | 2 | 0 |

8 | 9 | 3 | 4 |

9 | 8 | 3 | 2 |

11 | * | 3 | 0 |