A.1. Example of a Multi-State Deterioration model with two states 199 t 0 dS¬N(s)=→SN(0) t 0 dR(s) (A.7a) S¬N(s) t 0 =→SN(0)R(s) t 0 (A.7b) S¬N(t) →S¬N(0)=→SN(0)'R(t) →R(0)( (A.7c) S¬N(t)=SN(0)'1→R(t)(+S ¬N(0) (A.7d) Where S ¬N(0) is the probability of being in the non-nominal state at t = 0. Since SN(0)+S ¬N(0) =1, we see that if SN(0) =1, from Eq. A.7 we get that S¬N(t)=1→R(t)=1→SN(t). A di!erent way to demonstrate this is by taking Eq. A.1 and directly integrate: t 0 φSN(t) SN(t) =→ t 0 ϖ(t)dt (A.8a) t 0 φS¬N(t) SN(t) = t 0 ϖ(t)dt (A.8b) Now, using Eq. A.2c in Eq. A.8a, we get: ln|SN(s)| t 0 =ln'R(t)( (A.9a) ln|SN(t)| →ln|SN(0)| =ln'R(t)( (A.9b) ln|SN(t)| =ln'R(t)(+ln|SN(0)| (A.9c) SN(t)=SN(0)R(t) (A.9d) Replacing Eq. A.9d in Eq. A.8b, we get: t 0 S¬N(t)= t 0 ϖ(t)SN(t)dt (A.10a) S¬N(t) t 0 =SN(0)→R(t) t 0 (A.10b) S¬N(t) →S¬N(0)=SN(0)→R(t)+1 (A.10c) S¬N(t)=SN(0)'1→R(t)(+S ¬N(0) (A.10d)
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