Replacement Of Items That Fail Completely

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.

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

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

Example 1

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

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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

Example 2

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

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.

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The given problem can be divided into two parts.

1. Individual replacement.
2. 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_ip_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

N₀ = Number of items in the beginning = 10,000
N₁ = Number of items during the 1^st month X probability that an item fails during 1st month of installation
= 10000 X 0.03 = 300

N₂ = 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)

=N₀P₂ + N₁P₁
=(10000 X 0.07) + (300 X 0.03) = 709

N₃ = N₀P₃ + N₁P₂ + N₂P₁
= (10000 X 0.20) + (300 X 0.07)+ (709 X 0.03) = 2042

N₄ = N₀P₄ + N₁P₃ + N₂P₂+ N₃P₁
= (10000 X 0.40) + (300 X 0.20)+ (709 X 0.07) + (2042 X 0.03) = 4171

N₅ = N₀P₅ + N₁P₄ + N₂P₃+ N₃P₂+ N₄P₁
= (10000 X 0.15) + (300 X 0.40)+ (709 X 0.20) + (2042 X 0.07) + (4171 X 0.03) = 2030

N₆ = N₀P₆ + N₁P₅ + N₂P₄+ N₃P₃ + N₄P₂ + N₅P₁
= (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.

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.