198 Appendix A. Appendix: Introduction f(t)=→R↑(t) (A.2a) ϖ(t)= f(t) R(t) (A.2b) R(t)=exp→ t 0 ϖ(s)ds (A.2c) Where t ⇒0 represents time and it is positively defined; f(t) is the probability density function; and R(t) is the reliability function. Now, by replacing these relations in Eq. A.1, we get: φSN(t) dt = R↑(t) R(t) SN(t) (A.3a) φS¬N(t) dt =→ R↑(t) R(t) SN(t) (A.3b) Let’s focus in Eq. A.3a, we solve it as follows: t 0 dSN(s) SN(s) = t 0 dR(s) ds R(s) ds = t 0 dR(s) R(s) (A.4a) ln|SN(s)| t 0 =ln|R(s)| t 0 (A.4b) ln|SN(t)| →ln|SN(0)| =ln|R(t)| →ln|R(0)| (A.4c) In Eq. A.4.c, SN(0) is known as the initial state probability, and by definition R(0)=1, then: ln|SN(t)| =ln|R(t)| +ln|SN(0)| (A.5a) SN(t)=SN(0)R(t) (A.5b) From Eq. A.5, we observe that if SN(0)=1, then SN(t)=R(t). This means that when the initial state probability in the nominal condition is 1, the probability of being in the nominal state SN(t) is equivalent to the reliability function R(t). Let’s now consider Eq. A.3b. dS¬N(t)=→ R↑(t) R(t) SN(t)dt, (A.6) Replacing Eq. A.5b in Eq. A.6, and making a change of variable, we get:
RkJQdWJsaXNoZXIy MjY0ODMw