In some situations, failure of a certain item occurs all of a sudden, instead of gradual deterioration (e.g., failure of light bulbs, tubes, etc.).

The failure of the item may result in complete breakdown of the system. The breakdown implies loss of production, idle inventory, idle labour, etc. Therefore, an organization must prepare itself against these **failures.**

*“Failure to prepare is preparing to fail.” – Ben Franklin*

Thus, to avoid the possibility of a complete breakdown, it is desirable to formulate a suitable replacement policy. The following two courses can be followed in such situations.

**Individual replacement policy.**Under this policy, an item may be replaced immediately after its failure.**Group replacement policy**. Under this policy, the items are replaced in group after a certain period, say t, irrespective of the fact that items have failed or not. If any item fails before its preventive replacement is due, then individual replacement policy is used.

In situations where the items fail completely, the formulation of replacement policy depends upon the probability of failure. Mortality tables or Life testing techniques may be used to obtain a probability distribution of the failure of items in a system.

Mortality Tables

M(t) = Number of items surviving at time t

M(t – 1) = Number of items surviving at time (t – 1)

N = Total number of items in the system

The probability of failure of items during the interval t and (t – 1) is given by

M(t – 1) – M(t) |

The conditional probability that any item survived upto age (t – 1) and will fail in the next period is given by

M(t – 1) – M(t) |

### Example 1

Following mortality rates have been observed for certain type of light bulbs.

Time (weeks) |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

Number of bulbs still operating |
100 | 94 | 82 | 58 | 40 | 28 | 19 | 13 | 7 | 3 | 0 |

Calculate the probability of failure.

Solution.

Here, t is the time (weeks) and M (t) is the number of bulbs still operating. The probability of failure can be calculated as shown in the following table.

Table

Time(t) |
M (t) |
Probability of failurep _{i} = [ M (t – 1)- M (t) ] / N |

0 | 100 | —- |

1 | 94 | (100 – 94)/100 = 0.06 |

2 | 82 | (94 – 82)/100 = 0.12 |

3 | 58 | (82 – 58)/100 = 0.24 |

4 | 40 | (58 – 40)/100 = 0.18 |

5 | 28 | (40 – 28)/100 = 0.12 |

6 | 19 | (28 – 19)/100 = 0.09 |

7 | 13 | (19 – 13)/100 = 0.06 |

8 | 7 | (13 – 7)/100 = 0.06 |

9 | 3 | (7 – 3)/100 = 0.04 |

10 | 0 | (3 – 0)/100 = 0.03 |

### Example 2

Following mortality rates have been observed for a certain type of electronic component.

Month |
0 | 1 | 2 | 3 | 4 | 5 | 6 |

% surviving at the end of the month |
100 | 97 | 90 | 70 | 30 | 15 | 0 |

There are 10000 items in operation. It costs Re 1 to replace an individual item and 35 paise per item, if all items are replaced simultaneously. It is decided to replace all items at fixed intervals & to continue replacing individual items as and when they fail. At what intervals should all items be replaced? Assume that the items failing during a month are replaced at the end of the month.

Solution.

Month |
% surviving at the end of the month |
Probability of failurep _{i} |

0 | 100 | —- |

1 | 97 | (100 – 97)/100 = 0.03 |

2 | 90 | (97 – 90)/100 = 0.07 |

3 | 70 | (90 – 70)/100 = 0.20 |

4 | 30 | (70 – 30)/100 = 0.40 |

5 | 15 | (30 – 15)/100 = 0.15 |

6 | 0 | (15 – 0)/100 = 0.15 |

The given problem can be divided into two parts.

- Individual replacement.
- Group replacement.

Case I

It should be noted that no item survives for more than 6 months. Thus, an item which has survived for 5 months is sure to fail during sixth month.

The expected life of each item is given by

= Σ x_{i}p_{i}, where x_{i} is the month and p_{i} is the corresponding probability of failure.

= (1 X 0.03) + (2 X 0.07) + (3 X 0.20) + (4 X 0.40) + (5 X 0.15) + (6 X 0.15)

= 4.02 months.

**∴** Average number of replacement every month = N/(average expected life) = 10000/4.02 = 2487.5

= 2488 items (approx.).

Here average cost of monthly individual replacement policy = 2488 X 1 = Rs. 2488/-, (the cost being Re 1/- per item).

Case II

Let N_{i} denote the number of items replaced at the end of ith month.

Calculating values for N_{i}

N_{0} = Number of items in the beginning = 10,000

N_{1} = Number of items during the 1^{st} month X probability that an item fails during 1st month of installation

= 10000 X 0.03 = 300

N_{2} = Number of items replaced by the end of second month

=(Number of items in beginning X probability that these items will fail in 2^{nd} month) + (Number of items replaced in first month X probability that these items will fail during second month)

=N_{0}P_{2} + N_{1}P_{1}=(10000 X 0.07) + (300 X 0.03) = 709

N_{3} = N_{0}P_{3} + N_{1}P_{2} + N_{2}P_{1}= (10000 X 0.20) + (300 X 0.07)+ (709 X 0.03) = 2042

N_{4} = N_{0}P_{4} + N_{1}P_{3} + N_{2}P_{2}+ N_{3}P_{1}= (10000 X 0.40) + (300 X 0.20)+ (709 X 0.07) + (2042 X 0.03) = 4171

N_{5} = N_{0}P_{5} + N_{1}P_{4} + N_{2}P_{3}+ N_{3}P_{2}+ N_{4}P_{1}= (10000 X 0.15) + (300 X 0.40)+ (709 X 0.20) + (2042 X 0.07) + (4171 X 0.03) = 2030

N_{6} = N_{0}P_{6} + N_{1}P_{5} + N_{2}P_{4}+ N_{3}P_{3} + N_{4}P_{2 }+ N_{5}P_{1}= (10000 X 0.15) + (300 X 0.15)+ (709 X 0.40) + (2042 X 0.20) + (4171 X 0.07) + (2030 X 0.03) = 2590.

From the above calculations, it is observed that N_{i} increases upto fourth month and then decreases. It can also be seen that N_{i} will later tend to increase and the value of N_{i} will oscillate till the system acquires a steady state.

The optimum replacement cycle under group replacement is given in the following table.

End ofmonth |
Total no. ofitems failed |
Cumulative no. of failure |
Cost of replacement after failure (Re 1/ item) |
Cost of all replacement (Rs. 0.35/ item) |
Total cost(Rs.) |
Average cost per month(Rs.) |

1 | 300 | 300 | 300 | 3500 | 3800 | 3800 |

2 | 709 | 1009 | 1009 | 3500 | 4509 | 2254.50 |

3 | 2042 | 3051 | 3051 | 3500 | 6551 | 2183.66 |

4 | 4171 | 7222 | 7222 | 3500 | 10722 | 2680.50 |

5 | 2030 | 9252 | 9252 | 3500 | 12752 | 2550.40 |

6 | 2590 | 11842 | 11842 | 3500 | 15342 | 2557.00 |

The above table shows that the average cost during the third month is minimum. Thus, it would be economical to replace all the items every three months.

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