This presentation demonstrates how Life Data Analysis is used to compare the reliability of a component from two suppliers (the concept can also apply to comparing different designs). It also briefly describes how to compute Mean life (and other reliability metrices) when dataset contains suspension data.

Your regular Supplier A produces an auto part with which your company plans to build 1 million vehicles. The part (Part A) costs US$10.

A new supplier (Supplier B) offers US$5 for the part (Part B) with the same functional specifications.

You are the engineer assisting to make the purchasing decision. You want to make sure Part B has a better or at least equal mean life (or Mean-Time-To-Failure, MTTF) over that of your current Supplier A.

Each supplier is required to conduct a life test which must have at least 10 failures. Followings are the reliability test results.

Supplier A tested 10 units and all failed by 180 days.

Supplier B tested 20 units and after 180 days, 10 units were still working.

The mean life from Supplier A can be estimated by (Sum(TTF)/N)= 133.4 days.

This method is easy for the Part A, but how to calculate the mean life of Part B (10 units still operating after 180 days)?

First, we fit the two datasets to an appropriate distribution (Life Data Analysis). We choose Weibull distribution here (the choice of distribution is beyond the scope of this presentation). Figure 1 shows the Life Data Analysis for Supplier B dataset.

Now, we have the distribution parameter (beta=1.32, and Eta=229.7 days) for Part B. We can now calculate the mean, m

This is not a trivial calculation. We use the Weibull Toolbox Calculator to perform this computation.

The mean life for Part B is 211.2 days

The same analysis is repeated for Supplier A dataset... The mean life for Part A is 133.4 days.

The mean lives of Part A and Part B are 133.4 and 211.2 days respectively.

These mean values are indicated in **reliability vs time** and **PDF** plots below.

So, the auto part from Supplier B is more reliable and is cheaper than that of Supplier A. You decide to go with Supplier B.

If you stay with Supplier A, it would have costed US$10 x 1 million = US$ 10 million.

With Supplier B, it costs US$ 5 million. Not only that you save US$5 million, you enjoy longer mean life!

Now, consider that you provide a warranty of 60 days to your customers. Assuming a failure of your vehicle due to this part will cost you an average of US$200 (including the cost of the part) to fix it within warranty time.

Using the LDA Calculator, the probability of failure of Part B at t=60 days:

The probability of failure for Part B at t=60 days is 0.1541

For a million vehicles that you produce, 154,100 unit would have failed by the warranty time. The potential warranty claim is US$30.82 million!

If you stay with your regular supplier (Supplier A),

The probability of failure for Part A at t=60 days is 0.007081

For a million vehicles that you produce, an estimation of 7,081 of them would have failed by the warranty time. The potential warranty claim is US$1.416 million.

If you stay with your current supplier, your initial cost is US$10 million, and the warranty cost is US$1.416 million.

If you switch to Supplier B, your initial cost is US$5 million, and the warranty cost is US$30.82 million.

With the above information, are you going to change supplier?

Conclusion

Using Mean Life or MTTF (Mean-Time-To-Failure) as a sole reliability matrix to make decision is not a good idea!

In this case study, we performed a cost-benefit analysis using Life Data Analysis to assess the financial impact due to **cost of failure**,
**warranty time** and the **component failure rate behavior**.

There are many possible scenarios: What if the component from Supplier B is cheaper and its reliability at warranty time is slightly lower than that of Supplier A? What if you are the supplier B, what are you going to do?

For a large manufacturing organisation, does it make good business sense to have reliability engineers with strong quantitative analysis knowledge?

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