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- えみ はやしもと
- 7 years ago
- Views:
Transcription
1
2
3 2002 7
4
5 i (mean time to failure) (mean time between failures)
6 iv %
7 v substandard standard FMEA (failure Mode and Effects Analysis) FTA (Failure Test Analysis)
8 vi A 145 A A A A
9 vii A A A
10
11 R(t), (t) a prior probability of failure m > M m < M
12 x 7.3 T substandard standard SC
13 xi
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15 C C C C S
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17 1 AGREE(Advisory Group on Reliability of Electronic Equipment) Symposium on physics of failure in electronics IEC(International Electrotechnical Commission)
18
19 (Initial Failure) (Wearout failures) (Chance failures)
20 : 1. 2.
21 Reliability is the probability of a device performing its purpose adequately for the period of time intended under the operating conditions encountered 2.2 X Y P(A) = X X +Y P(B) = Y X +Y (2.1) (2.2)
22 : P(A) 0 (2.3) : P(S) = 1 S (2.4) : P ( ) A n = P(A n ) (2.5) N n = 1, 2,, N N S A m A n 0 m n F(x) = P(X x) (2.6)
23 F( ) = 0 (2.7) F( ) = 1 (2.8) 0 F(x) 1 (2.9) x 1 < x 2 F(x 1 ) F(x 2 ) (2.10) P(x 1 < X x 2 ) = F(x 2 ) F(x 1 ) (2.11) F(x + ) = F(x) (2.12) (2.7)(2.8)(2.10)(2.12) N s N f N 0 N 0 N s + N f = N 0 (2.13) N s N 0 + N f N 0 = 1 (2.14) N s N 0 = A 1, N f N 0 = A 2 (2.15) (2.3)(2.4)(2.5)
24 8 2 R(t) N 0 N s (t) R(t) = N s(t) N 0 (2.16) N s (t) N 0 R(0) = 1 (2.17) R( ) = 0 (2.18) N f (t) Q(t) = N f (t) N 0 (2.19) (2.14) R(t), Q(t) R(t) + Q(t) = 1 (2.20)
25 q(t) = dq(t) [ dr(t) ] dt dt (2.21) t Q(t) = q(t)dt (2.22) 0 (2.22) (2.7) (2.12) (2.22)
26 10 2 (2.21) Q(t + dt) Q(t) q(t) = dt = N f (t + dt) N f (t) N 0 dt (2.23) (2.21) t Q(t) = q(t)dt (2.24) 0 (2.22) t R(t) = 1 = = 0 t 0 q(t)dt q(t)dt t q(t)dt q(t)dt (2.25) t N s t +dt N s (t +dt) N s (t) N s (t + dt) t
27 : R(t), (t) t + dt λ λ = N s(t) N s (t + dt) dt N s (t) = dn s(t) dt N s (t) = dn f (t) dt N s (t) (2.26) t λ = λ 0 (2.26) dn s (t)/(n s dt) = λ 0 d (lnn s ) = λ 0 dt lnn s = λ 0 t +C t = 0 N s (t) = N 0 lnn s = λ 0 t + lnn 0 then λ 0 = 1 ln t ( Ns N 0 ) = 1 ln R(t) (2.27) t
28 12 2 (2.26) N 0 (2.16)(2.19)(2.21) λ(t) = q(t) R(t) (2.28) (2.21) (2.28) λ(t)dt = dr(t) R(t) R(t) = exp [ ] λ(t)dt (2.29) (2.30) (2.30) R(t) = exp( λt) (2.31) (2.27) (2.30) λ (mean time to failure) MTTF t N f (t) dt N f (t + dt) N f (t + dt) N f (t) t t + dt dt N f (t + dt) N f (t) dt t t t + dt t [ Nf (t + dt) N f (t) ] t
29 N 0 1 N 0 [ Nf (t + dt) N f (t) ] t d(mt T F) = [ Nf (t + dt) N f (t) ] [ t = N f (t) + dn ] f (t) dt N f (t) t dt N 0 d(mt T F) = t dn f (t) dt = tq(t)dt N 0 dt MTTF (MT T F) = tq(t)dt (2.32) 0 (2.16) (MT T F) = 0 R(t)dt = 1 N s (t)dt (2.33) N 0 (2.30) H(t) 0 R(t) = exp[ H(t)] (2.34) t H(t) = λ(t)dt (2.35) 0
30 t N s (t) t T = tn s (t)dt (2.36) (mean time between failures) MTBF (MT BF) = N s (t)dt N s (t) N s (t + dt) (2.37)
31 t N s (t) dt [N s (t) N s (t + dt)] MTBF (2.26) MTBF λ (MT BF) = 1 λ (2.38) MTBF (2.28) (2.21) (MT BF) = 1 λ = R(t) q(t) q(t) = dr(t) dt R(t) (MT BF) = 1 [ t ] 1 q(t)dt q(t) 0 (2.39) λ MTTF MTBF (2.33) (2.30) λ (MT T F) = 0 exp( λt)dt = 1 = (MT BF) (2.40) λ (2.33)(2.40) (MT BF) = R(t)dt (2.41) 0 useful life (2.30) R(t) = exp( λt)
32 16 2 m (2.41) ( R(t) = exp t ) m (2.42) t = m R = m m λ (2.42) R(t) = 1 t m (2.43) Q(t) = t m (2.44) (t) (R) (Q) m/ m/ m/1, m/10, m/100, m/1, 000, m/10, 000, : < > ,000, ,000
33 :
34 18 2 < > 10, < > 1 3 n MTBF m 3 ( R(t) = 1 t ) n m = 1 n m t (2.45) Q(t) = n m t (2.46) n MTBF (2.46) Q(t) = n m t, Q (t) = n m t (2.47) Q(t) = n n Q (t) (2.48) < > (2.45) (2.48)
35 t = m/ t = m/ (2.45) (2.48) < > 1 MTBF m MTBF (2.47) m system = m/100 t = m/100, (2.47) Q(t) = 100 ( ) m = m 100, 000 t = m/100, t = m/100,000 Q(t) = 10 ( ) m = m 100, m system = m/10 n m m/n 2.5
36 20 2 m mean number of cycles between failure failure rate per one operating cycles ( R(c) = exp( λc) = exp c ) m (2.49) λ R(c) = 1 λc (2.50)
37 (2.20) (2.31) Q(t) = 1 exp( λt) (3.1)
38 22 3 q(t) = λ exp( λt) (3.2) λ t λ t λ t +t R(t) = exp( λ t )exp( λ t ) = exp [ (λ t + λ t ) ] (3.3) < > (3.3) tq(t) (2.32) MTTF 0 tq(t)dt = (MT T F) (3.4) (MTBF)=(MTTF) (2.40)
39 MTBF (2.41) m = R(t)dt = exp( λt)dt = 1 (3.5) 0 0 λ (3.4) ( m = tq(t)dt = t dr ) dt (3.6) 0 0 dt m = [tr(t)] 0 + R(t)dt 1 ( ) R(t) = exp λdt 0 lim tr(t) t = t lim t exp ( t 0 λdt) = 0 λ 0 λ 1/m q(t) = λ exp( λt) = 1 ( m exp t ) m ( Q(t) = 1 exp t ) m ( R(t) = exp t ) m (3.7) (3.8) (3.9)
40 24 3 λ 1, λ 2 MTBF m 1, m 2 R(t) = R 1 (t) R 2 (t) = exp[ (λ 1 + λ 2 )t] [ ( 1 = exp + 1 ) ] t m 1 m 2 (3.4) 2 2 t = t +t R(t) = exp [ λ(t +t ) ] (3.10) λ(t) = at (3.11) (2.20)(2.30) ) R(t) = exp ( at2 2 ) Q(t) = 1 exp ( at2 2 (3.12) (3.13)
41 MTBF m m q(t) = Aexp [ a(t m) 2] (3.14) MTBF MTBF (3.14) σ (3.14) q(t) = 1 ] [ σ 2π exp (t m)2 2σ 2 (2.21) Q(t) = 1 σ 2π t 0 = 1 1 σ 2π exp [ t (t m)2 2σ 2 exp [ ] dt (t m)2 2σ 2 (3.15) ] dt (3.16) 1 R(t) = 1 σ exp [ 2π t 3.3 (t m)2 2σ 2 ] dt (3.17) exp( x) exp(x) = 1 (3.18)
42 26 3 n k ( ) n f (k) = p k (1 p) n k (3.19) k p m = np (3.19) n(n 1) [n (k 1)] ( m ) k ( f (k) = 1 m ) n k k! n n ( = mk 1 1 )( 1 2 ) ( 1 k 1 )[ ( 1 m ) n/m] m ( 1 m ) k k! n n n n n (3.20) n ( 1 1 n = exp n) lim n (3.20) f (k) = mk exp( m) (3.21) k! k (3.21) ) exp( m) (1 + m + m2 2! + m3 3! + = 1 (3.22) (3.18) (3.22) (3.21) k m A (3.22) exp( m) mexp( m)
43 ( m 2 /2! ) exp( m) λ t R(t) = exp( λt) (3.23) t Q 1 (t) = (λt)exp( λt) (3.24) 2 Q 2 (t) = (λt)2 2! exp( λt) (3.25) Q(t) Q = Q 1 + Q 2 + Q 3 + = 1 R = 1 exp( λt) (3.26) 3.4 (Weibull) [ ( ) t γ m ] R(t) = exp η m η = t 1/m 0 (3.27)
44 28 3 γ µ σ γ = 0 ( σ 2 (t) = η [Γ m ( µ(t) = ηγ ) m )] ) Γ 2 ( m (3.28) (3.29) Γ (3.27) m = 1 (3.27) 2 1 lnln R(T ) = m lnt lnt 0 = m(lnt ln η) (3.30) y = lnln 1 R(t) (3.31) x = lnt (3.32) a = lnt 0 = mlnη (3.33) (3.30) y = mx + a (3.34)
45 (3.32) x (3.31) y y x m a = lnt 0 = mlnη m x = 1, y = 0 y = m(x 1) x = 0 y lnt = 1 lnln1/r = 0 lnt = 0 lnln1/rt ln(t 0 ) η (3.33) m, η (3.28)(3.29)
46 30 3
47
48 P(A), P(B) A, B P(A B) = P(A) P(B) (4.1) 2. A, B AB P(A B) = P(A) + P(B) P(A) P(B) (4.2)
49 P(A B) = P(A) + P(B) (4.3) 4. A, B P(A) + P(B) = 1 (4.4) A, B A, B A, B P(A), P(B) A, B A, B N s N 0 1 N s /N 0 R + Q = 1 (4.5) (4.4)
50 34 4 R 1 R 2 t (4.1) R S (t) = R 1 (t)r 2 (t) (4.6) (4.2) Q S (t) = Q 1 (t) + Q 2 (t) Q 1 (t)q 2 (t) (4.7) (4.5) Q S (t) = [1 R 1 (t)] + [1 R 2 (t)] [1 R 1 (t)][1 R 2 (t)] = 1 R 1 (t)r 2 (2) = 1 R S (t) (4.8) (4.2) R p (t) = R 1 (t) + R 2 (t) R 1 (t)r 2 (t) (4.9) (4.2) Q p (t) = Q 1 (t)q 2 (t) (4.5) Q p (t) = [1 R 1 (t)][1 R 2 (t)] = 1 R p (t) (4.10) 4.2 R S (t) + Q S (t) = 1 (4.11)
51 R S (t), Q S (t) R p (t) + Q p (t) = 1 (4.12) R p (t), Q p (t) R S (t) = exp[ (λ 1 + λ 2 )t] (4.13) Q S (t) = 1 exp[ (λ 1 + λ 2 )t] (4.14) R p (t) = exp( λ 1 t) + exp( λ 2 t) exp[ (λ 1 + λ 2 )t] (4.15) Q p (t) = [1 exp( λ 1 t)][1 exp( λ 2 t)] (4.16) R S = R 1 R 2 R n (4.17) R i Q p = Q 1 Q 2 Q n (4.18)
52 36 4 < > λ t = λ d = λ r = λ c = : λ i = 4λ t + 10λ d + 20λ r + 10λ c = R S (t) = exp( t) 10 R S (10) = exp( ) = , m = 1/λ = 10,000 [ ] 10,000 10,000 e 1 = [%] 63 [%] 1) 1)
53 λ 1, λ 2 (4.15)(4.16) R p2 (t) = exp( λ 1 t) + exp( λ 2 t) exp[ (λ 1 + λ 2 )t] (4.19) Q p2 (t) = [1 exp( λ 1 t)][1 exp( λ 2 t)] (4.20) (3.5) m p2 = R p (t)dt = (4.21) 0 λ 1 λ 2 λ 1 + λ 2 λ 1, λ 2, λ 3 R p3 (t) = exp( λ 1 t) + exp( λ 2 t) + exp( λ 3 t) exp[ (λ 1 + λ 2 )t] exp[ (λ 2 + λ 3 )t] exp[ (λ 3 + λ 1 )t] + exp[ (λ 1 + λ 2 + λ 3 )t] (4.22) Q p3 (t) = [1 exp( λ 1 t)][1 exp( λ 2 t)][1 exp( λ 3 t)] (4.23) MTBF m p3 = 0 R p (t)dt = λ 1 λ 2 λ 3 λ 1 + λ 2 λ 2 + λ 3 λ 2 + λ 3 λ 3 + λ 1 1 (4.24) λ 1 + λ 2 + λ 3
54 R 1 = exp( λ 1 t), R 2 = exp( λ 2 t), R 3 = exp( λ 3 t) (4.25) Q 1 = 1 exp( λ 1 t), Q 2 = 1 exp( λ 2 t), Q 3 = 1 exp( λ 3 t) (4.26) (4.19)(4.20) R p2 + Q p2 = 1 = R 1 + R 2 R 1 R 2 + Q 1 Q 2 = R 1 (R 2 + Q 2 ) + R 2 (R 1 + Q 1 ) R 1 R 2 + Q 1 Q 2 = R 1 R 2 + R 1 Q 2 + R 2 Q 1 + Q 1 Q 2 = (R 1 + Q 1 )(R 2 + Q 2 ) (4.27) (4.22)(4.23) 1 = (R 1 + Q 1 )(R 2 + Q 2 )(R 3 + Q 3 ) (4.28) n 1 = (R 1 + Q 1 )(R 2 + Q 2 )(R 3 + Q 3 ) (R n + Q n ) (4.29) R p2 = 1 Q p2 = 1 [1 exp( λt)] 2 = 2 exp( λt) exp( 2λt) (4.30) Q p2 = Q 1 Q 2 = Q 2 = [1 exp( λt)] 2 (4.31)
55 m p2 = 1 λ 1 2λ = 3 2λ (4.32) R p3 = 1 Q p3 = 3exp( λt) 3exp( 2λt) + exp( 3λt) (4.33) Q p3 = Q 3 = [1 exp( λt)] 3 (4.34) MTBF m p3 = 3 λ + 3 2λ + 1 3λ = 29 6λ (4.35) n (4.29) (R + Q) n = R n + nr n 1 Q + n(n 1) R n 2 Q Q n = 1 (4.36) 2! (4.36) (R + Q) 3 = R 3 + 3R 2 Q + 3RQ 2 + Q 3 = 1 (4.37) Q 3 Q p = Q 3 (4.38) R p = R 3 + 3R 2 Q + 3RQ 2 (4.39)
56 40 4 < > Q p = 3RQ 2 + Q 3 (4.40) R p = R 3 + 3R 2 Q (4.41) λ = R = exp( 0.1) = Q = 1 R = = Q p = Q 3 = ( ) 3 = R p = 1 Q p = = m p = 3 λ + 3 2λ + 1 3λ = = < >
57 : λ i λ = λ i = R = exp( ) = R p = R 3 + 3R 2 Q = 3R 2 2R 3 =
58 : % n+1
59 ] exp( λt) [1 + λt + (λt)2 + (λt)3 + = 1 (4.42) 2! 3! exp( λt) exp( λt) (λt) 1 exp( λt)(λt) 2 /2! R 2 (t) = exp( λt) + exp( λt) (λt) (4.43) Q 2 (t) = 1 R(t) = exp( λt) (λt)2 2! + exp( λt) (λt)3 3! + (4.44) R 3 (t) = exp( λt) + exp( λt) (λt) + exp( λt) (λt)2 2! Q 3 (t) = 1 R(t) = exp( λt) (λt)3 3! + exp( λt) (λt)4 4! (4.45) (4.46) MTBF 2 m 2 = R 2 dt = 1 0 λ + λ λ 2 = 2 λ (4.47) 3 m 3 = 1 λ + 1 λ + 1 λ = 3 λ (4.48)
60 44 4 ] R n (t) = exp( λt) [1 + (λt) + (λt)2 + + (λt)n 1 (4.49) 2! (n 1)! [ ] (λt) n Q n (t) = exp( λt) + (λt)n+1 (4.50) n! (n + 1)! < > m n = n λ (4.51) λ = 0.01 R b = exp( λt) + exp( λt) (λt) = = R P = 1 ( ) 2 = (2.21) q = dr(t) dt (4.52) q(t) = λ exp( λt) (4.53) λ 1, λ 2 2 λ 1 λ 2 t 1
61 t 2 = t t 1 2 q 1 (t 1 ) = λ 1 exp( λ 1 t 1 ) (4.54) 2 q 2 (t 2 ) = λ 2 exp[ λ 2 (t t 1 )] (4.55) (2.21) dr = q(t)dt (4.54) dr 1 = q 1 (t 1 )dt 1 (4.55) dr 2 = q 2 (t 2 )dt 2 dr dr = dr 1 dr 2 = q 1 (t 1 )dt 1 q 2 (t 2 )dt 2 dr(t) = q 1 (t 1 )q 2 (t t 1 )dt 1 (dt dt 1 ) = q 1 (t 1 )q 2 (t t 1 )dt 1 dt q(t) = dr(t) = q 1 (t 1 )q 2 (t t 1 )dt 1 (4.56) dt t 1 2 t q(t) = q 1 (t 1 )q 2 (t t 1 )dt 1 (4.57) 0 q 1, q 2 q(t) 5 2
62 46 4 t q(t) = λ 1 λ 2 exp( λ 1 t 1 )exp[ λ 2 (t t 1 )]dt 1 = 0 λ 1 λ 2 λ 2 λ 1 [exp( λ 1 t) exp( λ 2 t)] (4.58) MTBF (2.25)(2.41) R b (t) = m p = t q(t)dt = exp( λ 1 t) + λ 1 [exp( λ 1 t) exp( λ 2 t)] (4.59) λ 2 λ 1 0 R b (t)dt = 1 λ λ 2 = m 1 + m 2 (4.60) (4.59)(4.60) λ 1 λ, λ 2 λ + x x (4.43)(4.47) q 1 (t 1 ) = λ 1 exp( λ 1 t 1 ) q 2 (t 2 ) = λ 2 exp( λ 2 t 2 ) (4.61) q 3 (t 3 ) = λ 3 exp( λ 3 t 3 ) t 2 = t t 1 0 t 1 t 2 (4.62) t 3 = t t 2 0 t 2 t 3 (4.63) q(t) = λ 1 λ 2 λ 3 t t 2 =0 t2 t 1 =0 exp( λ 1 t 1 )exp[ λ 2 (t 2 t 1 )]exp[ λ 2 (t t 2 )]dt 1 dt 2 (4.64)
63 (2.25) R b = λ 2 λ 3 exp( λ 1 t) (λ 2 λ 1 )(λ 3 λ 1 ) + λ 1λ 3 exp( λ 2 t) (λ 1 λ 2 )(λ 3 λ 2 ) + λ 1λ 2 exp( λ 3 t) (λ 1 λ 3 )(λ 2 λ 3 ) n (4.65) R b = λ 2 λ 3 λ n exp( λ 1 t) (λ 2 λ 1 )(λ 3 λ 1 ) (λ n λ 1 ) λ 1 λ 3 λ n exp( λ 2 t) + (λ 1 λ 2 )(λ 3 λ 2 ) (λ n λ 2 ) + λ 1 λ 2 λ n 1 exp( λ n t) + (λ 1 λ n )(λ 2 λ n ) (λ n 1 λ n ) (4.66) MTBF m p = 1 λ λ λ n (4.67) [%] 100 [%] R ss 1 R b (t) = exp( λt) + R ss exp( λt) λt (4.68) 2 R b (t) = exp( λ 1 t) + R ss λ 1 λ 2 λ 1 [exp( λ 1 t) exp( λ 2 t)] (4.69)
64 48 4 λ 1 λ 2 λ 3 2 R b (t) = exp( λ 1 t) + λ 1 λ 1 + λ 3 λ 2 {exp( λ 2 t) exp[ (λ 1 + λ 3 )t]} (4.70) R b (t) = exp( λt) + R ss λ 1 λ 1 + λ 3 λ 2 {exp( λ 2 t) exp[ (λ 1 + λ 3 )t]} (4.71) < > λ 1 = λ 2 = R ss = : t = R b = exp( t) [exp( t) exp( 0.001t)] = = ,000 2 R = 1 [1 exp( t)] 2 =
65 MTBF m b = = 0 0 R b dt { } λ 1 exp( λ 1 t) + exp( λt) [exp( λ 1 t) exp( λ 2 t)] dt λ 2 λ 1 ( 1 λ 1 + λ 1 ) (4.72) λ 2 + λ = 1 λ 1 + λ 1 λ 2 λ 1 λ MTBF , λ = ln = MTBF N λ n N (n + 1) λ Nλ MTBF 1/(Nλ) n n MTBF 1/(Nλ) n + 1 (n +1)/Nλ
66 50 4 ( R S (t) = exp Nλ ) n + 1 t (4.73) < > 30 λ = R = exp( ) = (4.73) ( ) R = exp 10 =
67 A C B C A A B B A-A C-A C-B B-B 5.2 A B C A B 4
68 : 5.2: R = [1 (1 R A )(1 R B )(1 R C )][1 (1 R A )(1 R B )] P(A) = P(A B i )P(B i ) + P(A B j )P(B j ) (5.1)
69 P(system f ailure i f component X is good) P(X is good) + P(system f ailure i f component is bad) P(X is bad) (5.2) Q S R x X Q x (5.2) Q S = Q S (i f X is good) R x + Q S (i f X is bad) Q x (5.3) R S = 1 Q S (5.4) 5.1 X C Q S = Q S (i f C is good) R C + Q S (i f C is bad) Q C (5.5) C A, B A, B C Q S (i f C is good) = (1 R A )(1 R B ) (5.6) C A A, B B C Q S (i f C is bad) = (1 R A R A )(1 R B R B ) (5.7) Q S = (1 R A )(1 R B ) R C + (1 R A R A )(1 R B R B ) Q C (5.8)
70 : 5.2 (5.3) 5.3 R 1, R 2 1 X R 2 Q S = (1 R 1 ) R (1 R 2 ) (5.9) X 2 X X X 1 (1 R 2 ) R 1 + Q 1 (5.9) (1 R 1 ) R (1 R 2 ) = R 2 R 1 R R 2 = 1 R 1 R 2 (5.10)
71 : Q S = 0 R 2 + (1 R 1 )(1 R 2 ) (5.11) 0 R (5.11) R 1, R 2, R 3, R 4
72 :
73 Q S = Q S (i f 1 is good) R 1 + Q S (i f 1 is bad) Q 1 (5.12) Q S (i f 1 is good) Q S (i f 1 is bad) 2 Q S (i f 1 is bad) = Q S (i f 2 is good) R 2 + Q S (i f 2 is bad) Q 2 (5.13) 1 (5.13) Q S (i f 1 is bad) = Q S (i f 2 is bad) Q 2 (5.14) 1 (5.13) Q S = Q S (i f 2 is bad) Q 2 Q 1 (5.15) Q S = Q 1 Q 2 Q 3 Q 4 (5.16) Q S = Q S (i f 1 is good) R 1 + Q S (i f 1 is bad) Q 1 (5.17) 1 1
74 Q S = 1 R 1 R 2 R 3 R 1 R 2 R 4 R 1 R 3 R 4 R 2 R 3 R 4 + 3R 1 R 2 R 3 R 4 (5.20) main : 2005/4/26(22:28) 58 5 (5.17) Q S (i f 1 is good) = Q S (i f 2 is good) R 2 + Q S (i f 2 is bad) Q 2 = (1 R 3 )(1 R 4 ) R 2 + (1 R 3 R 4 ) Q 2 (5.18) Q S (i f 1 is bad) Q S (i f 1 is bad) = Q S (i f 2 is good) R 2 + Q S (i f 2 is bad) Q 2 = (1 R 3 R 4 )R Q 2 (5.19) (R 1 + Q 1 )(R 2 + Q 2 )(R 3 + Q 3 )(R 4 + Q 4 ) = 1
75
76 DFR(decreasing failure rate) 2. CFR(constant failure rate) 3. IFR(increasing failure rate) :
77 q(t ) = 1 ] [ σ 2π exp (T M)2 2σ 2 (6.1) M σ T M MTBF MTBF MTBF MTBF 6.2 MTBF MTBF 63 [%] 37 [%]
78 :
79 T t 6.2 a priori probability of failure t 1 t 2 T 1 T 2 P t2 t 1 = P T2 T 1 = t2 t 1 q(t)dt (6.2) T2 T 1 q(t )dt (6.3) t t Q(t) = q(t)dt (6.4) (6.1) [%] 2.14[%] T = 0 T = M 3σ [%] T = 0 T = M + 3σ [%] 2.14 [%] 2.14 [%] [%]
80 :
81 :
82 : a prior probability of failure
83 T 2 T 1 T 2 T 1 a prior probability of failure T = 0 T 1 T 1 F T2 T 1 = P T 2 T 1 R(T 1 ) = Q(T 2) Q(T 1 ) R(T 1 ) (6.5) T 1 T 2 a posterori probability of failure < > 2.14 [%] 2.14 [%] R T2 T 1 (t) = 1 F T2 T 1 (6.6) F M 3σ, M 2σ = / = R M 3σ, M 2σ = 1 F = F M+3σ, M+2σ = / = R M+3σ, M+2σ = 1 F = ϕ(t ) = σq(t ) (6.7)
84 68 6 (6.1) (6.1) (6.7) ϕ(t ) (6.1) r(t ) = ϕ(t ) R W (T ) (6.8) λ W (T ) = q(t ) R W (T ) = ϕ(t ) σr W (T ) = r(t ) σ (6.9) 6.6 z R W (T ) (6.9) (M<3σ) T = 0 q(t ) = 0 q(t ) = 1 ] [ σ 2π exp (logt M)2 2σ 2 (6.10)
85 :
86 70 6 M logt σ logt T = 0 T T Q(T ) = q(t )dt (6.11) 0 1 T 1 Q(T ) = 1 q(t )dt (6.12) R(T ) = T T q(t )dt (6.13) T W T W Q(t) (wearout failure) W < > T W = M 3σ T = T W = M 3σ
87 Q W = T W = M 4σ Q W = T W = M 5σ Q W = ,000 10,000 < > 1 T W = M 3σ 2 q w = T W = M 2σ Q W = T T T +t a prior probability of failure T +t λ exp( λt)dt = exp( λt) exp[ λ(t +t)] T exp( λt) [ λ(t +t)] F(t) = exp( λt) = 1 exp( λt) (6.14) T = 0 T = t
88 72 6 F(t) = Q(t) = 1 exp( λt) T t T T t t Q(t) = Q C (t) + F W (t) Q C (t)f W (t) (6.15) C W Q C (t) = 1 exp( λt) T F W (t) (6.5)(6.13) F W (t) = 1 σ T +t 2π T 1 σ 2π exp [ (T M) 2 /2σ 2] dt T exp[ (T M)2 /2σ 2 ]dt (6.16) T t R W (t) = 1 F W (t) R(t) = exp( λt) exp( λt) F W (t) = exp( λt) R W (t) (6.17) (6.16) R W (t) = 1 σ 2π 1 σ 2π T +t exp[ (T M) 2 /2σ 2] dt T exp[ (T M)2 /2σ 2 ]dt (6.18)
89 T T + t T +t, T t (6.1) (6.13) R W (t) = R W (T +t) R W (T ) (6.19) t (6.19) R(t) = exp( λt) R W (T +t) R W (T ) (6.20) < > R S (t) = exp( λ i t) R W i (T i +t) R Wi (T i ) (6.21) [R Wi (T i +t)/r Wi (T i )] T W t T Wi R W i (T i ) (6.22) R Wi (T Wi ) R Wi (T i ) R Wi (T Wi ) T Wi t T Wi (6.20) (2.28) λ C λ W
90 74 6 λ = λ C + λ W (6.23) (6.20) R(t) = exp( λ C t) R W (T +t) R W (T ) ( T +t ) = exp λdt T (6.24) 6.6 λ W = r/σ (6.24) λ = λ C +λ W T T +t T +t T λdt = T +t T = λ C t + (λ C + λ W )dt T +t T λ W dt t λ W = 1 2 [λ W T + λ W (T +t)] (2.28) λ W = 1 [ ] q(t ) q(t +t) + 2 R(T ) R(T +t) (6.25) T +t λ W dt = λ W t (6.26) T R(t) = exp[ (λ C + λ Wm )t] (6.27)
91 7 7.1 T = 0 (6.20) t = T R(T ) = exp( λt) R W (0 + T ) R W (0) = exp( λt)r W (T ) (7.1) 1 R W (0) = 1 (2.21) (6.1) R W (T ) = 1 σ 2π ] (T M)2 exp [ 2σ 2 dt (7.2)
92 76 7 T M exp( λt) R W (T ) m 7.1: m > M 7.2: m < M (7.1) T = 0 T 0 R(t) = exp( λt) R W (T 0 +t) R W (T 0 ) (7.3) 7.3 t = 0 (7.3) 1 (T =0) 7.1 M 9 4 T 0 = 0.4M 7.3
93 : T 0 t = T 2 M 3.5σ M 3σ (6.23) 7.4 n = 1000 m = 1,000,000, M = 10,000, σ = 2,000 1,
94 : m = m/1000 = 1000 [ ] 1,000 < > , : 5,400 [ ] 9,000 [ ] 9,970 [ ]
95 : 2 2M = 14,400 σ 2 = 2σ 1 = 1,200 10,000 [ ] M=21,600 λ r = 1/M = λ S = N λ r λ r wearout replacement rate T = nm (n = M/3σ 1 ) n = 7200/1800 = 4 T = 4M = 28, < >
96 : M 6σ 1 = 3,600 [ ] 10,000 [ ] 3,600 7,200 7,200 10, ,000 [ ] 10,000 [ ] [ ] ,200 [ ] 100,000 [ ] λ S = 1 (7.4) M i M i R S (t) = exp( λ S t) = exp ( 1 ) t M i (7.5)
97 R S (t) = n R W i (T i +t) R Wi (T i ) (7.6) n t (6.13) R Wi (t) = R W i (T i +t) R Wi (T i ) T = i +t q i(t)dt T i q i (t)dt q i i T i i T Wi T Wi = T i +t R S (t) = R W i (T Wi ) R Wi (T Wi t) (7.7) T W (7.7) R S (t) = R W i (T W ) R Wi (T W t) (7.8)
98 82 7 (7.7) (7.8) λ S = λ Ci + 1 M i (7.9) R S (t) = exp [ ) ] (λ Ci + 1Mi t (7.10) λ r = 1/M λ = λ C + λ r (7.10) R S (t) = exp [ (λ Ci + λ r ) i t ] (7.11) λ i (7.11)
99 8 8.1 substandard components MTBF
100 84 8 N N E substandard N E N substandard MTBF m e ( Q = 1 exp t ) (8.1) m e ( Q = 1 exp N ) Et m e (8.2) 100 [%] N ( R = exp N ) Et m e (8.3) substandard < > 10 m e = R = exp( 1) = [ ] R = exp( 10) = R = exp( 10 1 ) = R = exp( ) = N E MTBF m e N E m e N E 1 m e (4.67) ( E(t) = m e ) (8.4) N E
101 E(t) = 3m e = 30 [ ] substandard substandard standard : substandard standard T max T max 1
102 : N B N B /N N N E 1 (N B /N) ( N 2 = N E N E 1 N ) B N = N EN B N (8.5)
103 (8.5) ( N 2 = N E N E 1 N ) E N = N2 E N (8.6) N 2 N 3 = N E N N 2 = N3 E N 2 (8.7) 1 (N E /N) [%] λ g N G λ e N E λ g λ e λ S = N G λ g + N E λ e (8.8) N = N G + N E (8.9) N
104 88 8 λ S = Nλ g (8.10) (8.10) (8.8) λ S = Nλ g n n G (8.11) n G n n n n G = n E (8.12) (8.12) (8.11) ( λ S = Nλ g 1 + n ) E (8.13) n G n E /n G the incremental failure rate factor n G /n repair efficiency 8.3 N E N G λ e λ g
105 : N E λ e substandard N G λ g substandard N E = N E N E + N G N T [ λ = λ g + λ 0 exp T ] E(t) (8.14) (8.15) λ 0 = (λ e λ g ) N E N (8.16) E(t) (8.4) T = 0 λ i = λ g + λ 0 = λ gn G + λ e N E N (8.17)
106 90 8 λ = λ g (8.18) 1 [ T ] R(T ) = exp λdt = exp 0 { T = exp 0 [ λ g + λ 0 E(t)exp ( T [ λ g T λ 0 E(t) + λ 0 E(t)exp )] E(t) ( T E(t) } dt )] (8.19) 8.2 (8.15) λ W [ λ L = λ g + λ 0 exp T ] + λ W (8.20) E(t) λ W = q(t ) R(T ) = ] (T M)2 [ 2σ 2 exp ] T [ exp (T M)2 dt 2σ 2 (8.21) λ W = q(t ) R(T ) = q(t ) T q(t )dt (8.22) λ L = λ g + λ 0 exp [ T ] E(t) + + q(t ) T q(t )dt (8.23)
107 L(T ) T = 0 L(T ) = exp [ T ] λ L dt 0 L(T ) T (8.24) T T +t t [ T +t ] R(t, T ) = exp λ L dt T (8.25) T R(t) = exp( λ g t) (8.26) (8.24)(8.25) L(T ) = R C (T )R e (T )R W (T ) = exp( λ C T )exp [ T exp 0 R(t, T ) = R C (T )R e (T )R W (T ) = exp( λ C t)exp [ T +t exp T { λ 0 [1 exp( αt )] α ] q(t ) T q(t )dt dt { λ 0 α } } exp( αt )[1 exp( αt)] ] q(t ) T q(t )dt dt α = 1/E(t) (8.19) n (8.27) (8.28) R S (t) = R i (t, T i ) (8.29) (8.28)
108
109 9 9.1 (off-schedule maintenance) (schedule maintenance)
110 94 9 T p m t t m = t + t + m 1 m 2 = = n t i=1 m i n i=1 λ i t (9.1) T i H o = t m 1 T 1 + t m 2 T 2 + = = n T i i=1 n i=1 m i t (λ i T i )t (9.2) T i H o T o T p t
111 T m = T p + T o (9.3) T r system utilization factor U = t T p + T o + T r +t (9.4) maximum possible system utilization factor U = = t T p + T o +t t T m +t (9.5) (9.1) (9.2) 1) 3/2λ 2) 3/2λ 1/λ 1/2λ 1) 2) (4.32)
112 96 9 τ 1 e λτ m = 0 L(t)dt (9.6) L(t) 8 (8.27) (r + q) 3 = r 3 + 3r 2 q + 3rq 2 + q 3 (9.7) T n (9.7)
113 r 3 n 3r 2 qn 3rq 2 n q 3 n n N N = n(3r 2 q + 2 3rq q 3 ) = n(3r 2 q + 6rq 2 + 3q 3 ) (9.8) h 1 h total = Nh 1 (9.9) h 2 nh 2 h over all = Nh 1 + nh 2 (9.10) repair rate h over all /nt 9.2 maximum possible system utilization factor (9.5) m T m m T m = T m m t (9.11)
114 98 9 m, T m t, T m maximum possible system utilization factor U m system availability A A = m m + T m (9.12) m, T m t, T m maximum possible system utilization factor U m 100 [%] A + B = 1 complementary probability B B system unavailability T m B = m + T m (9.13) B 8760 [ ] T m T m (9.12) (9.13) A = 1 T m 8760 T m B = 8760 (9.14) (9.15) T m 9.3 system dependability D
115 off-schedule maintenance T o (9.12)(9.13) D = 1 T o 8760 (9.16) (9.2) t off-schedule maintenance T o (9.2) (9.2) H o t = n i=1 (λ i T i ) (9.17) off-schedule H o = T o r r = T o /t r = T o /t 0ff-schedule D = = = t T o +t T o /t r (9.18) D = (λ i T i ) (9.19)
116
117 FMEA (failure Mode and Effects Analysis)
118 102 10
119 10.1. FMEA (failure Mode and Effects Analysis) 103 (10.1) 10.4 C S = (C 1 C 2 C 3 C 4 C 5 ) 1/5 (10.1) C S C 1 C 2 C 3 C 4 C 5 C C C
120 : C : C 2
121 10.1. FMEA (failure Mode and Effects Analysis) : C 3 C S 7 C S < 10 4 C S < 7 2 C S < 4 C < : C S
122 FTA (Failure Test Analysis) (Fault Tree diagram)
123 10.2. FTA (Failure Test Analysis) Q A, Q B 1 (1 Q A )(1 Q B ) = Q A + Q B Q A Q B = Q A + Q B (10.2) (10.2) OR AND OR AND
124 :
125 10.2. FTA (Failure Test Analysis) : (X 1 + X 2 )(X 1 + X 3 ) = X 1 + X 1 X 3 + X 1 X 2 + X 2 X 3 = X 1 + X 1 X 2 + X 2 X 3 = X 1 + X 2 X 3 (10.3) 10.3: 10.4:
126 Z 8115
127
128
129
130 114 11
131 : 1) C A B C A 11.2: man 2. machine 1)
132 method
133 [%] SC
134 :
135 :
136 :
137 :
138 main : 2005/4/26(22:28) 122 第 11 章 品質保証体系 図 11.5: 信頼性試験試験結果
139 : (a) (b) (c) (d) (e) (f) (g) 2. (a) (b) (c) 3. (a) (b) (c)
140 : 2SC1815
141 SC1815 1
142 V CBO 60 [V ] V CEO 50 [V ] V EBO 5 [V ] I C 150 [ma] I B 50 [ma] P C 400 [mw] T j 125 [ ] T stg [ ] 12.1: N 1 N 1 N 2 N 0
143 N 0 + N 0 N 1 + N 0 N 1 N 2 + N 0 N 1 N2 2 [ + = N N1 (1 + N 2 + N2 2 + N2 3 ( ) + )] 1 = N N 1 1 N 2 = N N 1 N 2 1 N 2 (12.1) N 2 1 I = I s 1 + N 1 N 2 1 N 2 (12.2) : M M = 1 1 αdx (12.3)
144 M = V B n (V /V B ) n (12.4) [V /cm] Ψ 0 Ψ 0 = kt ( ) q ln NA N D n 2 i (12.5)
145 V (BR)CBO >V (BR)CEO >V (BR)CES >V (BR)CEX V B I CBO V (BR)CBO β µa β 12.2
146 : V (BR)CBO V (BR)CEO V (BR)CES V (BR)CER V (BR)CEX
147 T = R P (12.6) T R [ /W] P < > J.Fourie q x = ka x dt dx q x k [m /w] A x x (12.7) x (12.6) (12.7) (12.6)
148 : 12.4
149 : P R i R o T j T c T a R T = R i + R o (12.8) R T = R i + R o(r s + R c + R f ) R o + R s + R c + R f (12.9) R o R T = R i + R s + R c + R f (12.10)
150 : R s R c R f
151 [mv/deg] : 12.7 r t r t = t ( 1 R j + 1 t ) 1 r t (t +t 1 ) r t (t) + r t (t 1 ) t 0 t 12.8 R s + R c 12.2
152 : 12.8:
153 θ C + θ S [deg/w] TO ( µ) TO 66 ( µ) ( µ) TO H1A TO 220AB ( µ) TO TO 3P BIA ( µ) CIA ( µ) TO 3P(L) F1A ( µ) TO 3P(H) D2A ( µ) TO 3P(H) E2A 12.2: ( K = Aexp φ ) kt K (12.11)
154 : A φ k T a τ = a/k A = a/a (12.11) ( ) φ τ 1 = A exp (12.12) kt (12.12) ( ) ( ) φ φ τ 1 = A exp, τ 2 = A exp kt 1 kt 1 (12.13) lnτ 1 lna lnτ 2 lna = T 2 (12.14) T 1 [ ( φ 1 1 )] (12.15) k T 2 T 1 α = τ 2 τ 1 = exp
155 α [ev ] SiO Si SiO : ln(τ) = A + B α ln(s) (12.16) T A, B S 12.4
156 :
157
158
159
160
161 A A.1 1. (a) (b) 2. (a) (b)
162 146 A (c) (d) (e) (f) 3. (a) (b) (c) (d) X 1, X 2,, X n m = E(X 1 ) = E(X 2 ) = = E(X n ) (A.1) σ = σ(x 1 ) = σ(x 2 ) = σ(x n ) (A.2)
163 A E(aX + by ) = ae(x) + be(y ) (A.3) V (ax + by ) = a 2 V (X) + b 2 V (Y ) (A.4) X 1, X 2,, X n X = X 1 + X 2 + X n n (A.5) S 2 = [ (X 1 X) 2 + (X 2 X) 2 + (X n X) 2] (A.6)
164 148 A 7. (a) (b) 8. (a) (b) A.2 X 1, X 2,, X n (A.5) ( ) X1, X 2,, X n E( X) = E n = 1 n (X 1, X 2,, X n ) = m (A.7)
165 A ( ) X1, X 2,, X n V ( X) = V = 1 n n 2V(X 1, X 2,, X n ) = 1 n 2 (σ 2 + σ 2 + σ 2 ) = σ 2 2. ( ) X1, X 2,, X n V ( X) = V n = 1 n 2V(X 1, X 2,, X n ) n (A.8) = σ 2 n N n N 1 (A.9) E(S 2 ) = { 1 [ E (X1 X) 2 + (X 2 X) (X n X) 2]} n = 1 { n E [(X 1 m) X m)] 2 + [(X 2 m) X m)] [(X n m) X m)] 2} = 1 n E [ (X 1 m) 2 + (X 2 m) 2 + (X n m) 2] {[ ] } 1 E n (X 1 + X 2 + X n ) ( X m) + E [ ( X m) 2] = 1 n (σ 2 + σ 2 + σ 2 ) σ 2 n = n 1 n σ 2 (A.10) (n 1)/n 3. 1
166 150 A 1 m σ 2 n X n ( ) N m, σ 2 n 0 1 W N = Y N Y N = N i=1 (X i X i ) = 1 N σ Y N N i=1 σ X 2 NσX i X) i i=1(x W N Φ WN (ω) = E [exp( jωw N )] { [ ]} N jω = E exp NσX i X) i=1(x = [ { [ ]}] jω N E exp NσX (X i X) { [ ]} jω E exp = NσX (X i X) [ E 1 jω ( ) jω (Xi X) (X i X) R ] N = 1 ω2 NσX NσX 2 N 2N + E [R N] N ln[φ WN (ω)] = N ln [1 ω2 2N + E [R ] N] N ln[φ WN (ω)] = ω2 2 + E [R N] N [ ω 2 2 2N E [R ] 2 N] + N
167 A N 1 A.3 ) lim φ WN(ω) = exp ( ω2 N 2 (A.11), A.3.1,, X s 2, m σ 2 m = X = 1 n (x 1 + x x n ) (A.12) σ 2 = s 2 = 1 n 1 n i=1 (x i X) 2 (A.13) < > R.A.Fisher x 1, x 2,, x n n F = f (x 1 ) f (x 2 ) f )x n ) (A.14) x 1, x 2,, x n θ F theta theta theta F θ = 0 (A.15)
168 152 A, ln θ = 0 (A.16) A.3.2 θ Θ 1 Θ 2 γ Θ 1, Θ 2 P(Θ 1 θ Θ 2 ) = γ γ 0.90, 0.95, 0.99, 0.999,,,,, ( ) 2 X 1, X 2, X n m 1, m 2, m n σ 1, σ 2, σ n m = M 1 + m m n (A.17) σ = σ1 2 + σ σ 2 (A.18) 3 X 1, X 2, X n m σ X = 1 n (X 1 + X 2 + X n ) (A.19) m σ 2 /n
169 A X 1, X 2, X n m σ T = n X m Σ (A.20) n 1 t Y = (n 1) Σ2 σ 2 (A.21) n 1 Σ 2 = 1 n 1 n i=1 (X i X) 2 X = 1 n (X 1 + X 2 + X n ) 1. γ 2. γ c 3. X 4. k = cn/ n X k m X + k 1. γ 2. n 1 t, c F(c) = 1 (1 + γ) 2 3. X σ 2 4. k = cσ/ n X k m X + k 1. γ
170 154 A 2. n 1 t, c 1, c 2 F(c 1 ) = 1 2 (1 γ), F(c 2) = 1 (1 + γ) 2 3. σ 2,(n 1)σ 2 4. k 1 = (n 1)σ 2 /c 1, k 2 = (n 1)σ 2 /c 2 k 2 Σ 2 k 1 (A.11) P( Z < k) = P( X m σ n ) < k (A.22) (A.22) k = k 0 P P 0 1 P 0 (A.22) ( ) σ σ P X k 0 n < m < X + k 0 n = P 0 (A.23) σ X k 0 n σ < m < X +k 0 n P 0 (.A.10) P 0
171 A A.3.3 B(n, p) E(X) = np, V (X) = np(1 p) (A.24) Z = X np np(1 p) (A.25) P 0 Z < k 0 ) X np P( < k 0 = P 0 (A.26) np(1 p) p = X n (A.27) (A.26)(A.27) ( p(1 p) P p k 0 n ) p(1 p) < p < p + k 0 = P 0 (A.28) n p = p (A.28) ( ) p(1 p) p(1 p) P p k 0 < p < p + k 0 = P 0 (A.29) n n
172 156 A P 0 p k 0 < p < p+k 0 p(1 p) n p(1 p) n A.4 < > σ n X Z = X m σ n N(0, 1) 0.05 P( Z 1.96) = 0.05 X m 1.96 σ n
173 A < > (A.24. A.25)
174
175 IEEE VA QC VA QC VA QC VA QC
176
177 [1] Igor Basovsky Reliability Theory and Practice Prentice-Hall Inc. [2] [3] Peyton Z. Peebles, Jr. [4] Erwin Kreyszig
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24.15章.微分方程式
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基礎から学ぶトラヒック理論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.
基礎から学ぶトラヒック理論 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/085221 このサンプルページの内容は, 初版 1 刷発行時のものです. i +α 3 1 2 4 5 1 2 ii 3 4 5 6 7 8 9 9.3 2014 6 iii 1 1 2 5 2.1 5 2.2 7
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地域総合研究第40巻第1号
* abstract This paper attempts to show a method to estimate joint distribution for income and age with copula function. Further, we estimate the joint distribution from National Survey of Family Income
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