Completion of a life cycle cost (LCC) analysis requires expenditures for each
year of the project for calculating net present value (NPV). This requires estimating when things will
fail and the cost of the failures. A
Weibull database is helpful for selecting:
• shape factor beta (b tells about the failure mode) and
• characteristic life eta (h tells about the durability).
The Weibull statistics tell how components fail. Failure costs follow the failures. You need failure costs for the appropriate year to produce NPV details. You can also download an Excel worksheet for computing NPV from this website.
Failures in early years are highly priced in NPV terms. Failures in later years are lower priced because of discounted values. All the following examples use a discount rate = 12%, project time span of 20 years, and a 38% tax provision for calculating the NPVs. Out-of-pocket expenditures refer to the first 20 years which is a typical project life.
Figure 1 shows the rhythms of replacement. The replacement rhythms make waves of cost over the years for this wear out failure mode. Expenditures deferred to later years have strong, favorable, effects on NPV because of the discount factors; whereas expenditures early in the project life are included at almost current values which are very costly. Of course the costs are driven by the failures, and Table 1 shows the details.
You can find the annual costs to include in life cycle cost NPV spreadsheets
by use of simple
Notice in Table 1 the first failure takes ~13 years to accumulate which is equal to 91.8% of the Weibull CDF. The second and third failures each take ~9 years which is not surprising since the calculated mean time to failure is 8.998 years for b = 3.5 and h = 10 years. The cumulative failures are shown in Table 1 with red numbers for failures near whole numbers.
Figure 2’s replacement rhythms are not obvious because of the smaller beta value. Table 2 shows the failure costs.
Notice in Table 2 the first failure takes 12 years to accumulate which is 75% of the Weibull CDF. The second and third failures each take ~9 years which is not surprising since the calculated mean time to failure is 8.89 years for b = 1.8 and h = 10 years. The cumulative failures are shown in Table 2 with red numbers for failures near whole numbers.
Figure 3 shows the costs to be used each year for a chance failure mode with b = 1. Frequently this is a simplification decision (i.e., b = 1) on many life cycle cost sheets (unfortunately this decision takes major hits against NPV early in the life of the project).
When b = 1 and h = 10 years, this describes the exponential type, random failure and the mean time to failure is equal to eta (i.e., 10 years). As you can see from the costs curve we have 1/10 chance for failure each year, i.e., (1/10)*$10,000 = $1000 annual exposure. Detailed results of the simulation, regarding failures is shown in Table 3.
It’s easy to see that failures are occurring every 10 years in the simplified model of chance failures which is equal to 63.2% of the Weibull CDF. The simulated numbers for cost are ±0.2% of the expected value of $1000.
Figure 4 shows the costs to be used each year for the Weibull statistic with b = 0.7. These infant mortality failure modes (i.e., b < 1) incur high costs in the early years where expenditures carry a high value. Thus the NPV is worse than other conditions in Figure 1-3.
The early high cost failures from infant mortality have major cost impact on NPV. The failure details and cost numbers are shown in Table 4 with the first failure accumulating in 8+ years.
The first failure occurs in 8 years which is 57.5% of the Weibull CDF, the second failure 12 years later, and the third failure is also 12 years later.
These four examples show:
1. how to find failure costs for each project year with different failure modes
2. how early failure costs are killers for NPV
3. the need for Weibull failure statistics to be able to find failures and costs
4. why it’s important to know how equipment will live and die to get correct cost profiles by year
In summary, here is a comparison of these examples for repair when broken:
Figure 1’s results are the least negative and thus the preferred case. The least preferred case is shown in Figure 4 and Table 4.
You can find other LCC resources on this website:
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