## Cost Benefit Analysis of a Reliability Improvement Program

A manufacturer has reliability program in place. They have a very good understanding on the failure modes (in terms of its failure rate behavior) of a particular product. The manufacturer wants to know what is the cost benefit, specifically warranty cost reduction, of a reliability improvement program.

The pump has the following failure modes and corresponding failure distributions.

Table 1, Pump failure modes with its failure distributions

Time unit is in Day.

The manufacture provides a 200-days on-site warranty.

Labor cost to replace pump on-site = \$1,100

The cost of manufacturing and selling of the pump is \$600

The selling price of the pump is \$1,000

The cost of a pump fails within warranty time: cost of pump (\$600) + Labor cost (\$1,100) = \$1700

What is the expected profit of selling a pump?

A Reliability Block Diagram (RBD) is created using Weibull Toolbox.

System Reliability:

From Weibull Toolbox, query system unreliability at 200 days.

The probability of failure at warranty time of 200 days is 0.1233

Hence, for every pump sold, the warranty cost = 0.1233 x 1,700 = \$209.61

The expected profit of each pump = (1,000 – 600) – 209.61 = \$190.39

Reliability Importance metric

Obtain the Static Reliability Importance plot at t = 200 days.

An item with the highest Reliability Importance, (a value between 0 and 1) indicates that the change in system reliability is most sensitive to the change in reliability of this item. Therefore, to get the maximum effect of your improvement effort, you would target the item with highest Reliability Importance when possible.

It shows that Shaft Seal (SSL) has the highest Reliability Importance and its reliability at 200 days is 0.93.

The R&D department verify that reliability of SSL at 200 days can be improved to 0.993 (Weibull with beta=3, eta=800 days), but it cost an additional \$50 to make the pump.

The manufacturer seeks your advice whether to improve the product reliability or not, i.e., which option will make more profit.

Update the reliability information of SSL in the RBD, and recompute system reliability at 200 days.

System reliability at 200 days reduces to 0.06992

Therefore, the projected warranty cost for each pump = 0.06992 x 1,750 = \$122.36

The expected profit of each pump = (1,000 – 650) – 122.36 = \$227.64

This represents a 19.6% increase in profit.

Conclusion

This example uses Reliability Importance metric to identify the weakest component in a product. The decision whether to improve it, is based on cost-benefit analysis which takes various cost components and reliability information into consideration.

What if the analysis shows that there is no increase in profit? Worse still, what if your profit drops with the improvement program?

When competitions heat up, increase profit may not be the reason for such analysis. Rather, can you stay in the competition and still make a profit is what you want to know.

With the ability to perform quantitative reliability analysis, manufacturers will be acutely aware of how the warranty cost affects their bottom-line.

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