Financial Impact of Reliability on Supplier Selection and Maintenance Planning
Reliability is very important not only because it is related to safety, but also because it has significant financial impacts. For many applications, reliability problems usually will not cause catastrophic failures. So it may not seem crucial and may be ignored. However, many reliability issues, although they may seem to be very trivial, do cause the customer to be unsatisfied. A low customer rating will affect sales and could result in a long-term financial crisis. If you look around, you will see that this has happened and is happening in many industries. Although cheap products can win the market temporarily, if the quality and reliability do not meet customer expectations, the obtained market share will be lost.
In this article, we will use a simple example to illustrate how reliability knowledge, if applied correctly, can help decision makers make the right choices. In this example, reliability metrics are used to examine the financial impact on supplier selection and maintenance planning.
An elevator company is going to make 5,000 new elevators. Twenty buttons are installed in each elevator. Two button manufacturers are the potential suppliers. Supplier A is asking for $1.00 for each button while supplier B is asking for only $.50 each. The elevator company provides a 4 year warranty for its elevators. Each on-site repair costs $250.00. The warranty cost is the major factor in the button choice. In order to obtain the failure information for the buttons, both suppliers are required to conduct reliability tests and the failure data are provided in Table 1.
The Weibull distribution is used as the life distribution. From the data in Table 1, the model parameters are estimated and given in Table 2. This table also shows the mean time to failure (MTTF) of each button. It is 2.71E+06 for supplier A and 2.32E+06 for supplier B.
For comparison, Figure 1 shows the probability plots for these two suppliers. If each button is assumed to experience 500 cycles per day, the total number of cycles will be 730,000 by the end of a 4 year warranty period.
Given the information on the button life, the elevator company wants to know what the total cost will be for each button supplier. The total cost consists of two parts. The first part is the initial button cost at installation. The second part is the warranty cost.
When one button fails, all 20 buttons in the elevator will be replaced. Based on this replacement policy, each elevator is considered to be a system with 20 buttons competing with each other to fail. The elevator’s reliability is given by:
If we define , then the life distribution for the button failure in an elevator is a Weibull distribution with shape parameter of β and scale parameter of η'. Table 3 gives the model parameters at the elevator level.
Warranty Cost Calculation
There are several different methods that we could use to calculate the warranty costs that can be expected with each supplier’s buttons. In this article, we will examine three:
We will discuss them one by one.
MTTF is the most well-known term in reliability engineering. Many people still prefer to use it. For this elevator case, a reliability engineer decided to use the MTTF to predict the warranty costs.
The engineer assumes that all the buttons in an elevator have the same usage rate, which is 500 cycles per day. The total number of cycles by the end of a 4 year warranty is about 730,000. By dividing this number by the elevator MTTF, the expected number of failures for one elevator during the warranty period can be obtained. Using this strategy, the expected total cost for each supplier is calculated and given in Table 4.
The results in Table 4 show that the winner is supplier A. Its estimated total cost is $610,159. However, using MTTF as a sole metric can be flawed and misleading. Historically, the use of MTTF dates back to the widespread use of the exponential distribution in the early days of quantitative reliability during the 1950s and early 1960s. The exponential distribution has just one parameter (the MTTF or its reciprocal, the failure rate), which is constant. But, in reality, few products and components have a constant failure rate. In this button example, the failure distributions for both suppliers are Weibull distributions with shape parameters greater than 1, indicating that the failure rate increases with time. Therefore, simply using MTTF to estimate the number of failures and calculate the warranty cost is not appropriate. So, instead of using MTTF, solution 2 will use the probability of failure from the Weibull distribution to calculate the expected number of failures.
For supplier A, the probability of failure by the end of warranty for one elevator is:
For supplier B, the probability of failure by the end of warranty for one elevator is:
The expected number of failures is the product of the total number of elevators and the above values. Table 5 gives the expected total cost for each supplier.
From the results in Table 5, we can see that the winner again is supplier A. The estimated total cost is $101,197. Comparing Tables 4 and 5, it can be seen that the estimated total cost for the same supplier is very different. So we have determined that simply using MTTF to calculate the expected number of failures is not correct. Using MTTF implies that the beta parameter in the Weibull distribution is equal to 1. However, from the estimated beta values in Table 2, it can be seen that they are not even close to 1.
Is solution 2 the right method, then? It depends. From the calculation, we can see that solution 2 assumes that, for the same elevator, the warranty covers only the first failure. If there are more failures after the first replacement, the costs are not considered in the calculation. Obviously, this is not the warranty policy that the elevator company will offer. Rather, the warranty will cover all the costs during the 4 year warranty period, no matter how many failures occur. If the buttons have high reliability, the calculation in solution 2 will still work because the chance is very small that an elevator has more failures after the first replacement. However, if the reliability is low, as in the case of supplier B, the chance is high that more than one failure will be observed for one elevator. Therefore, all the failures should be considered in the warranty cost calculation. In solution 3, a more realistic method is given, which considers all the failures during the warranty period.
Theory on Renewal Process
Before calculating the estimated cost using solution 3, we need to discuss a little bit about the theory of a renewal process. When there is a failure, all the buttons in the elevator are replaced. The failure process starts over. Therefore, we can say that it is a renewal process with the replacement time as the renewal point. So the expected number of failures (renewals) at time t is given by :
The first term in Eqn. (1) represents the case when there are no failures before time t; the second term represents the case when the first failure occurs at time x (0<x<t). Eqn. (1) can be simplified as:
From Eqn. (2), it can be seen that if we assume r(t - x) is 0, or in other words, if the chance to have more failures after the first one is very low and can be treated as zero, then r(t) = F(t). With this assumption, Eqn. (2) becomes the same equation used in solution 2. If this assumption does not hold, a general formula for calculating r(t) is needed. In order to calculate the formula, we can take the Laplace transform to both sides of Eqn. (2).
The Laplace transform for r(t) is:
The inverse Laplace transform of Eqn. (3) is the expression for r(t). It seems simple. Unfortunately, for commonly used lifetime distributions such as the Weibull and lognormal distributions, the closed form inverse Laplace transform does not exist. It has a closed form only in some special cases, such as the exponential distribution (a special case of Weibull distribution with a shape parameter equal to 1). We will use the exponential distribution to illustrate how to get the closed form of r(t).
The pdf of an exponential distribution is given by :
and the cumulative distribution function (cdf) is:
So Eqn. (3) becomes:
And then take the inverse:
For the button example, since β is not equal to 1, no closed form expression for r(t) can be found. So simulation becomes an option. In solution 3, we will use simulation to estimate r(t), the expected number of failures for one elevator, and then use it to calculate the warranty cost.
For supplier A, given the high reliability of its buttons, solution 3 is expected to give a result close to solution 2. However, for supplier B, because of its low reliability, the costs calculated using solutions 2 and 3 should be quite different.
BlockSim, a simulation software package from ReliaSoft, is used to estimate the expected number of failures for one elevator in the warranty period. Table 6 gives the results.
Table 6 shows that the expected number of failures for one elevator with buttons from supplier A is 0.0009 and, with supplier B, it is 0.745. Using these numbers, Table 7 gives the estimated total cost for each supplier.
As we expected, because of the high reliability in the warranty period for supplier A, solutions 2 and 3 have similar results. For supplier B, its low reliability will cause more failures, so the results from solutions 2 and 3 are quite different. Theoretically, solution 3 should give a higher warranty cost than the results from solution 2 for both suppliers. However, because of the randomness of simulation, only very few failures were generated in the warranty period for supplier A. Therefore, the result from solution 3 for supplier A is slightly lower than the value from solution 2.
From the calculations described above, the elevator company decided to choose supplier A to provide the buttons for the new design. The expected total cost is about $101,125.
Optimal Maintenance Interval
After choosing supplier A, the elevator company also wants the manufacturer to recommend an optimal maintenance interval for replacing all the buttons. For a planned maintenance, the cost is $80.00 per incident. For an unscheduled corrective maintenance due to failure, the cost is $250.00 per incident. Based on this information, supplier A needs to find a maintenance interval that can minimize the cost per cycle.
Assume that the maintenance interval is TP. A replacement occurs either when there is a failure or when the elevator has operated for TP cycles. So the expected number of cycles between two adjacent replacements is given by:
The first term represents the case when the replacement occurs at the scheduled interval TP. The second term represents the case when the replacement occurs at the first failure time x (0<x<TP).
The expected cost per unit time is given by :
For supplier A, we know that the distribution for a panel of 20 buttons is a Weibull distribution with β = 7.23 and η' = 1,909,055. Using these parameters, Eqn. (5) can be written as:
To get the optimal TP, we can take the derivative of TP in Eqn. (6) and set it to 0.
The derivative is a nonlinear equation as given by:
Or if we simplify the equation, it becomes:
γ(a.x) is the lower incomplete gamma function:
We solve Eqn. (7) or (8) to get the optimal time, TP = 1.336 x 106. With this maintenance interval, the expected cost per unit time is 6.979 x 10-5.
Solving Eqn. (7) or (8) is not easy; mathematical software packages are required. Fortunately, ReliaSoft’s BlockSim software can automatically calculate the optimal replacement time if the failure distribution and the costs are provided. Another ReliaSoft product, Weibull++, also has a simple function that can be used to calculate the expected cost per unit time. Using this function, we can get the costs for a list of cycles and visually identify the optimal maintenance interval. Table 8 gives these costs.
From the plot of cost per unit time vs. cycles shown in Figure 2, the optimal maintenance interval, which is about 1.3 x 106, can be easily identified. This result matches the exact analytical result given in Eqns. (7) and (8) very well.
Since the usage of the buttons is 182,500 cycles per year, the optimal maintenance interval is estimated to be about 7 years.
In this article, we used a example to illustrate how to use reliability knowledge to make decisions on component choice and maintenance planning. Since, for most products, the reliability is relatively high during the warranty period, the simple and straightforward method of calculating costs based on probability of failure (solution 2) can be applied to estimate the warranty cost. However, if the warranty period is long or if the reliability is low, this approach will underestimate the cost. In those cases, a simulation approach (solution 3) should be used instead. From this article, it can be seen that if reliability knowledge is used correctly, it will help decision makers implement the right choices.
 Kececioglu, D., Maintainability, Availability, and Operational Readiness Engineering, Prentice-Hall, 1995.
 ReliaSoft Corporation, Life Data Analysis Reference, ReliaSoft Publishing, 2005.
 Ross, S. M., Introduction to Probability Models, 7th Edition, Academic Press, 2000.
More Example Examples
As with many other features that appear in the Reliability Edge, this article uses a simple example to illustrate how particular reliability techniques can be applied to address the types of challenges that practitioners face on a daily basis. If you are looking for more examples of this type, you will find them interspersed throughout ReliaSoft's training seminars (http://www.ReliaSoft.com/seminars/), in the sample files that are shipped with ReliaSoft's software, in the step-by-step software training guides (http://www.ReliaSoft.com/guides/), in the monthly HotWire eMagazine (http://www.weibull.com/hotwire/) and also throughout the free reliability engineering resource website (http://www.weibull.com).
[Editor's Note: In the printed edition of Volume 9, Issue 2, there was a typographical error Table 4 on page 3. The Warranty Costs for Supplier B are calculated by multiplying 5751.677 Failures by $250 (cost per repair). This has been corrected in this web version of the article.]