# Acceleration of Service Life Testing by Using Weibull Distribution on Fiber Optical Connectors
## 4 Results  
Regarding the Weibull service life program testing from Table 1, from the corresponding cumulated percentile \(F(t_{i})\) elements in the linearized form of the reliability function stated in Equation (7), the corresponding \(Y_{i}\) elements are determined as in Equation (12), where \(n=6\). Thus, from Equation (13), the mean is \(\mu_{y}=-0.501\), and with the failure times \(\sigma_{1}=13.861\) and \(\sigma_{2}=4.670\) we proceeded to use the Weibull distribution to obtain the reliability. From Equation (8), the Weibull shape parameter \(\beta\) value is  
\[\beta=\frac{(-4)(-0.501)}{0.995\times\ln(\frac{13.861}{4.670})}=1.850\]  
From Equation (10), the log-mean is \(\mu_{x}=2.085\), and from Equation (9), the Weibull scale parameter \(\eta\) is  
\[\eta=\exp(2.085)=8.045\]  
With the Weibull family results, the random behavior of the failure times \(\sigma_{1}\) and \(\sigma_{2}\) is determined by following the next steps. The random behavior of \(\sigma_{1}\) and \(\sigma_{2}\) is determined by Equations (16) and (17). Then, by using the \(\beta\) and the \(Y_{i}\) values in Equation (15), the basic Weibull elements \(t_{0i}\) for each \(Y_{i}\) are obtained, whereas by using the \(\eta\) and \(\sigma_{1}\) values in Equation (32), the Weibull \(t_{01}\) values from the \(\sigma_{1}\) and \(\sigma_{2}\) failure time values are reproduced. These value is calculated as  
\[t_{01}=\eta/\sigma_{1} \tag{32}\]  
By replacing the known values of \(\eta\) and \(\sigma_{1}\), in Equation (32), we have  
\[t_{01}=\frac{8.045}{13.861}=0.580\]  
And by using the \(\beta\) value in Equation (33), the \(Y_{1}\) value that belongs to the \(t_{01}\) value is determined as  
\[Y_{1}=\ln(t_{01})\times\beta \tag{33}\]  
\[Y_{1}=\ln(0.580)\times 1.850=-1.008\]  
Finally, by substituting the \(Y_{1}\) value in Equation (14), the reliability \(R(t)\) that belongs to the \(t_{01}\) element is  
\[R(t)=exp\{-exp\{-1.008\}\}=0.694\]  
These results are shown in Table 6.  
Next, the Weibull vibration ALT analysis in Table 5 was performed. In this case, since the value \(\beta=2.500\) was used to develop the acceleration test, this value was maintained. From the corresponding cumulated percentile \(F(t_{i})\) elements in the linearized form of the reliability function stated in Equation (7), the corresponding \(Y_{i}\) elements were determined as in Equation (12), where \(n=4\). Thus, from Equation (13), the mean is, \(\mu_{y}=-0.617\), and with the failure times \(\sigma_{1}=109.550\) and \(\sigma_{2}=6.639\), we proceeded to use the Weibull distribution to obtain the reliability. From Equation (10), the log-mean is, \(\mu_{x}=3.295\), and from Equation (9) the Weibull scale parameter \(\eta\) is  
\[\eta=\exp(3.295)=26.967\]  
Using the \(\eta\) and \(\sigma_{1}\) values in Equation (32), the Weibull \(t_{01}\) value from the \(\sigma_{1}\) and \(\sigma_{2}\) failure time values was reproduced. This value is calculated as  
\[t_{01}=\frac{26.967}{109.550}=0.246\]  
Using the \(\beta\) value in Equation (33), the \(Y_{1}\) value that belongs to the \(t_{01}\) value is determined as  
\[\eta=\exp(2.085)=8.045\]0  
By substituting the \(Y_{1}\) value in Equation (14), the reliability \(R(t)\) that belongs to the \(t_{01}\) element is  
\[\eta=\exp(2.085)=8.045\]1  
These results are presented in the \[\eta=\exp(2.085)=8.045\]2  
The failure times from vibration ALT. The results of the service life program testing and the vibration ALT by following the formulation given in Section 2.4 are presented in \[\eta=\exp(2.085)=8.045\]3  
To determine the percentage of the life consumed by the service life testing, the following relation of the \(\eta\) failure times was applied: \(\frac{8.045}{2.967}=0.298\). Thus, the percentage