## 3 Results  
In this section, a decision model application was presented through which results were shown to check the consistency of the model proposed for the selection of inspection intervals. This application was based on a confidential case study in a sanitary ceramics company. Although the figures were not the real data, they were conveniently changed to represent a realistic and consistent context.  
In this company, a component that comprised part of the natural gas burning system of a tunnel oven was subjected to erosion and fatigue and failed according to a Weibull distribution with two parameters \(\alpha=2\) and \(\beta=8000\) h. This component had accumulated 3000 h of running time. The equipment worked round the clock (365 days per year \(\times\) 24 h a day = 8760 h per year) and was presently scheduled to be overhauled by 9000 h. The nominal production rate was 350 units/hour, and the contribution margin was EUR 100/unit. Based on historical data, as symptoms were aggravated, the throughput had to be lowered and the cumulated production loss was estimated over time from new as being described by a third order polynomial \(d(\%)=-4.8983+0.0032M_{n}-6.10^{-7}{M_{n}}^{2}+7.10^{-11}{M_{n}}^{3}\). If a potential failure is noticed in due time, the repair will take place immediately and its cost is estimated to be EUR 35,000. If a functional failure arises, the equipment will have to stop unexpectedly costing EUR 100,000 due to repair and lost production mainly. Inspections will be performed for EUR 4000 each. The interest tax rate currently used in the company is 25% per year. Periods P-F and M-F are approximately 500 and 50 h, respectively. The component has already accumulated 2.000 h of work (\(H=2000\)).  
If the company sets a minimum reliability between inspections of 0.9, what will the hourly cost of this course of action be?  
The limit time interval during which an inspection must take place after point P has been noticed is (P-M) = (P-F) \(-\) (M-F) = 500 \(-\) 50 = 450 h. If inspections are accomplished at moments \(M\), each STW will therefore extend 450 h backwards in time. The interest tax rate must be adjusted for the hour as time is measured in hours in this example.  
\(j=(1+\text{ITR})^{1/m}-1=(1+0.25)^{1/8760}-1=0.002547\%\).hour.  
Expected costs until 9000 h from now will have to be computed. Excel enabled all the calculations in a tabular form. Table 1 and Figure 3 show the results. In Table 1, one can see that 17 inspections were necessary to accomplish a minimum reliability of 0.9 between every two inspections until 9000 h from the present moment. From the ninth inspection onwards, inspections were performed with a periodicity of (P-M) = 450 h because only STW remained. The expected costs obtained were: Failure: EUR 15,931; Inspections: EUR 30,881; Repair: EUR 25,404; Degraded production: EUR 4238, the sum of which yielded a total of EUR 76,454 over almost 11,000 h.  
A failure inside a UTW takes place at 665 h on average (9). A failure inside a STW takes place every 4594 h on average (10). When no failures occur, the expected life from until 9000 h from now is 0.1674 \(\times\) 9000 = 1841 h (11). The total expected life is therefore (1841 + 665 + 4594) = 7100 h (12).  
Finally, the expected hourly cost can be found (13):  
\[\overline{C_{h}}=(76454).\frac{(0.00002547).(1+0.00002547)^{7101}}{(1+0.00002 547)^{7101}-1}=\text{EUR 11.77}/\text{h}\]