Fiabilidade e Controlo de Qualidade

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1 Fiabilidade e Controlo de Qualidade LMAC o. Teste o. Semestre 004/05 Duração: hora e 45 minutos 4/Maio/05 (Sab.) 0h. O esquema abaixo descreve as etapas de uma deslocação em transportes públicos de uma zona rural do sul da Inglaterra ao aeroporto de Heathrow. Há a possibilidade de ocorrência de avaria em cada uma das etapas da deslocação, contribuindo assim para que um passageiro perca o seu vôo. Considere que os transportes B, D, E e F (A e C, resp.) possuem probabilidade de avaria q (q, resp.) e funcionam de modo independente. (a) Identifique as fiabilidades dos transportes, os caminhos mínimos, os cortes mínimos e a função estrutura. Reliabilities of the components/transports p i P (breakdown of transport i), i A,..., F p B p D P E p F q p A p C q Minimal path sets P {A, B, E} P {A, D, F } P 3 {A, B, C, F } p 3 minimal path sets (.5) Minimal cut sets K {A} K {B, D} K 3 {B, F } K 4 {C, D, E} K 5 {E, F } q 5 minimal cut sets Structure function By considering X i Bernoulli(p i ), i A,..., F and applying result (.3) or (.4), we can conlude that the structure function of this system equals: (.3) p φ(x) j i P j X i

2 ( X A X B X E )( X A X D X F )( X A X B X C X F ) p ( X i ) j i K j (.4) [ ( X A )] [ ( X B )( X D )] [ ( X B )( X F )] [ ( X C )( X D )( X E )] [ ( X E )( X F )]. (b) Calcule um par de limites inferior e superior o mais estritos possível para a fiabilidade desta deslocação a Heathrow. Bounds for the reliability r(p) Since the components form a coherent system and operate in an independent fashion, we can apply Theorem.68, use the fact that p A p C q and p B p D P E p F q and succesively get: Lower bound r(p) q j ( p i ) i K j [ ( p A )] [ ( p B )( p D )] [ ( p B )( p F )] [ ( p C )( p D )( p E )] [ ( p E )( p F )] ( q) ( q ) ( q ) ( qq ) ( q ) ( q) ( q ) 3 ( qq ) (.0) Upper bound r(p) p j p i i P j ( p A p B p E )( p A p D p F )( p A p B p C p F ) [ ( q)( q) ] [ ( q)( q) ] [ ( q) ( q) ].. Um veículo pesado possui eixos, com 6 pneus cada (3 de cada lado). Este veículo opera quando os dois eixos estão a funcionar. Um eixo fica inoperacional se mais de um dos 3 pneus de cada lado estiver furado. (a) Obtenha um limite inferior para a fiabilidade do veículo, assumindo que os pneus do veículo funcionam de forma associada (positiva) e que a fiabilidade de cada pneu é de 97.5%. (3.0) System The vehicle described in the text is a system with the following characteristics: Each side of the axis corresponds to a out of 3 system because each axis is non operational if more than one of the 3 tires is non operational, that is, the system is operational if at least of the 3 tires is operational; each axis is nothing but two out of 3 sub-systems put in series; the vehicle only operates when both axes are operational, therefore the vehicle corresponds to out of 3 sub-systems put in series.

3 Reliability of the components of each out of 3 sub-system p i p 0.975, i,, 3 Lower bound for the reliability of each out of 3 sub-system Since the 3 components form a coherent sub-system and operate in a positively associated fashion, we can apply Theorem.70, namely result (.4), after having identified the minimal path sets. Minimal path sets P {, } P {, 3} P 3 {, 3} p 3 minimal path sets Lower bound for the reliability r out of 3 (p) (.4) max p i j,...,p i P j p i p max j,...,p p#p j #P j, j p p Lower bound for the reliability of the vehicle Now, we can capitalize on the previous bound for the reliability of each of the four sub-systems and apply Theorem.65 to provide the requested lower bound or the reliability of the vehicle: r vehicle (p) 4 r out of 3, i (p) i (.34) (b) Considere agora que os tempos até que os pneus fiquem furados (em milhares de Km) são com distribuição Weibull com parâmetro de escala δ e de forma α. Obtenha a função de fiabilidade da vida do veículo para uma deslocação de 400 Km. (.5) Individual durations of the components of each out of 3 sub-system T i Weibull(δ, α ), i,, 3 F i (t) F (t) e t, t 0 Duration of each out of 3 sub-system S j duration of the j th out of 3sub-system, j,..., 4 (.6) n3, k S i T (n k+) T () Reliability function of S j For j,..., 4, R Sj (t) (.8) F Binomial(3,F (t))(3 ) ( ) 3 ( e t) m [ ( e t)] 3 m m m0 3

4 e 3t + 3 ( e t) e t t Duration of the vehicle S vehicle min j,...,4 S j Reliability function of S vehicle (.3) 4 R Svehicle (t) R Sj (t) j [ e 3t + 3 ( e t) e t] 4 t Um sistema em ponte possui componentes, com durações (em horas) e função taxa de falha comum t, 0 < t.5 λ(t) 3 t,.5 < t t, t >. (a) Qual o comportamento monótono da função taxa de falha das durações das componentes e da duração da estrutura? Justifique. Individual durations and common hazard rate function T i, i,..., 5, r.v. with hazard rate function λ(t). (3.0) Stochastic aging character of T i λ(t) is not a monotone (or monotonic) function. Thus, T i IHR or T i DHR. However, t Λ(t) t λ(u) du t 0 0, t < 0 t t 0 u du, 0 t.5 t u du + t.5 (3 u) du,.5 < t t u du +.5 (3 u) du + t (u ) du, t > 0, t < 0 t t, ) 0 t.5.5 t + (3u u,.5 < t [ (t t) ( ) ], t >... 0, t < 0 t t, 0 t t t,.5 < t t t +.75, t > d t Λ(t) dt, 0 t.5.5 t,.5 < t.75 t, t >. 4

5 Please note that: > 0;.5 t.5 is a decreasing function, thus, the minimum value is this function in (.5, ] is > 0;.75 is an increasing function, thus, the minimum value is this function in (, + ] is t > 0. As a consequence t Λ(t) is an increasing monotone function for t 0, thus, T i IHRA, i,..., 5. Duration of the bridge system T Stochastic ageing character of T According to Table 3., the IHRA property of the components of a system is preserved after the formation of a coherent system, like the bridge system. Thus, T IHRA. (b) Estabeleça um limite inferior para a duração esperada da vida da estrutura, sabendo que a duração esperada comum das componentes é de aproximadamente 5.6 horas. (.0) Lower bound for E(T ) We are dealing with a coherent system characterized as follows: the 5 components have durations T i IHRA, i,..., 5 and, thus, according to Proposition 3.36, T i NBUE, i,..., 5; the expected value of the duration of each of the 5 components is equal to µ µ i E(T i ) 5.6. the minimal path sets are P {, 4} P {, 5} P 3 {, 3, 5} P 4 {, 3, 4} p 5 minimal path sets. Now, we can apply Theorem 3.65, and conclude that µ E(T ) max µ j,...,p i i P j ) } µ i µ max j,...,p { (#Pj µ µ min j,...,p {#P j } µ.8. 5

6 4. Um veículo de teste foi sujeito a uma rodagem de 5000Km, tendo-se registado os Kms percorridos entre falhas eléctricas graves sucessivas: 63, 5, 466, 85, 88, 34, 47, 3463, 486, 07 Km. (a) Averigue a adequação do modelo exponencial a este conjunto de dados considerando um nível de significância de 5%. Life test Since the test was scheduled to end after exactly t Km and the exercise mentions just one vehicle submitted to this life test, we are dealing with a Type I/item censored testing with replacement. (.0) R.v. T (i) time (in Km) of the i th serious electrical failure of the military vehicle Z i T (i) T (i ) run time (in Km) between the i th and (i ) th serious electrical failure of the military vehicle Z i Z, i IN Censored data n r 0 electrical failures during the life test (z,..., z r ) 63, 5, 466, 85, 88, 34, 47, 3463, 486, 07) (t (),..., t (r) ) (63, 4, 480, 605, 7393, 8707, 979, 64, 38, 445) Cumulative total time in test According to Definition 5.7, the cumulative total time in test is given by: t n t Km Hypotheses H 0 : Z Exponential(λ) H : Z Weibull(λ, α), α Significance level α 0 5% Test statistic (Bartlett s test) [ ( (5.9) r ri ) Z i B r + r+ ln r r 6r a H0 χ (r ) Rejection region of H ( 0 ) W 0, F (α χ 0 /) (r ) r0, α (0,.700) (9.0, + ) ] r ln (Z i ) i Decision The observed value of [ the test statistic is ( r ri ) z i b r + r+ ln ] r ln (z i ) r r 6r i ( ) F ( α χ 0 /), + (r ) 6

7 0 [ ( ) ln ] (0,.700) (9.0, + ), therefore we should not reject H 0 for any significance level α 5%. (b) Após ter enunciado as hipóteses de trabalho que entender convenientes, obtenha uma estimativa centrada da função de fiabilidade aos 00 Km e um intervalo de confiança a 95% para a kilometragem associada a 0% das falhas. (3.0) Distribution assumption T i Exponential(λ), i,..., 6. This is fairly reasonable since we did not reject H 0 in (a). Unknown parameter R T (t) e λ t Unbiased estimate of R T (t) According to Table 5.4, the UMVUE of R T (t) is, for t 00 < t 5000 and r > 0, equal to ( ) r R T (t) t t ( ) Confidence interval for λ According to Table 5.6 of the lecture notes, CI ( α) 00% (λ) [λ L ; λ U ] F χ (α/) F (r) χ ( α/) (r) ; t t a) F χ CI 95% (λ) (0.05) F (0) 5000 ; χ (0.975) (0) 5000 [ ; ] [0.0009; ]. Another unknown parameter FT (0.) λ ln( 0.), which is a decreasing function of λ > 0. Confidence interval for FT (0.) ( ) [ ] CI 95% FT (0.) ln( 0.); ln( 0.) λ U λ L [ ; ]. 7