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q
q
q
η
tmax
β
q
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q
q
t0 = 0
n
q > 1
n
q < 1
q
t0 = 0
n
q > 1
n
q < 1
ηq
T ´ t0
tf q(t)dt
T ´ t0
Rq(t)dt
q
q
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q
q
β
q < 1
β
hq(t)
q
q = 0,9
q
q = 1,5
η = 1
t0 = 0
q
q < 1
q
q < 1
q
q > 1
q
q > 1
Rq(t)
hq(t)
F q,C 3(t)
F q,C 14(t)
hq,C 3(t)
hq,C 14(t)
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f 1(x1)
f 2(x2)
C 20
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C 21
C 22
C 23
C 24
W SP C 23,C 24(t)
W SP C 23,C 24
W SP 2S C 24,C 24,C 24(t)
C 24
C 1
C 1
C 1
C 1
C 2
C 2
C 2
C 2
C 2
C 2
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Rq(t)
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q
q
β
q
c ∆i
q
C 1
C 18 tlim
q
ttimatv tlimdor
q
q
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|log(
)| > u
log(
) > u
log(
) < −u
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K
n
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β
η
q
t0
[x]+ x
0
α
∆i
ηq
γ
Γ(x)
x
F
µn
n
µn
n
τ l
τ u
2F 1 (a, b; c; z )
C c
C i i
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C p
E [T ]
f (t)
t
F (x)
x
F q(t)
t
hq(t)
t
k
P (X ≤ x)
X
x
pi i
q upper q
R2
Rq(t)
t
Rmq(t)
t
S q
T
tlim
tlim
tlim
tmax
W
Z [T ]
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1
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2 Capítulo: 1 Introdução
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1.1 Justificativa 3
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4 Capítulo: 1 Introdução
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1.3 Objetivos específicos 5
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7
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8 Capítulo: 2 Fundamentos de confiabilidade
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2.1 Introdução 9
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10 Capítulo: 2 Fundamentos de confiabilidade
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2.2 Integração de componentes e sistemas 11
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12 Capítulo: 2 Fundamentos de confiabilidade
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2.2 Integração de componentes e sistemas 13
β
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14 Capítulo: 2 Fundamentos de confiabilidade
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2.2 Integração de componentes e sistemas 15
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16 Capítulo: 2 Fundamentos de confiabilidade
θ1, θ2, . . .
t
θt
θs
θt−1
t + ∆t
t
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2.2 Integração de componentes e sistemas 17
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18 Capítulo: 2 Fundamentos de confiabilidade
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2.3 Distribuição de Weibull aplicada à confiabilidade 19
σ
n
σ
dv
dS = n dv
p
dv
S
S = 1 − (1 − ds) p ,
S = 1 − (1 − n dv) p
,
p dv = v
S = 1 −
1 −
n v
p
p
.
p
dv
v
S = 1 − limpn v→∞ 1 −
n v
p pn v n v
= 1 − exp(−n v) .
n.v = 1
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20 Capítulo: 2 Fundamentos de confiabilidade
P (X ≤ x) = F (x) = 1 − exp[−ϕ (x)] ,
ϕ (x)
x = xu
F (x) = 1 − exp
−x − xu
xom
,
xu
F (x) x0
f (t) = β.(t − δ )β−1
θβ exp
−
t − δ
θ
β
(t ≥ δ ) ,
β
θ
δ
t
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2.3 Distribuição de Weibull aplicada à confiabilidade 21
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22 Capítulo: 2 Fundamentos de confiabilidade
q
S q = k
W i
pqi − 1
1 − q ,
k
pi
i
W
S 1 = −k
W i pi ln pi
q → 1
expq(x)
lnq(x)
q
q
q
x
q
q
q
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2.4 Entropia de Tsallis 23
expq(x) =
[1 + (1 − q ) x]
1
1−q , [1 + (1 − q ) x] > 0
0,
,
lnq x = x1−q − 1
1 − q (x > 0, q = 1),
x, q ∈ R
expq(lnq x) = lnq(expq x) =
x
q → 1
lnq 1 = 0 expq 0 = 1
∀ q
q
p(x) = 1
ˆ x2 p(x)dx = σ2,
σ2
p(x)
x
q
ˆ x2[ p(x)]qdx = σ2
S q
q
p(x) = C q expq(−ξx2),
C q =
ξ (1 − q )
π
1/2 Γ((5 − 3q )/2(1 − q ))
Γ((2 − q )/(1 − q ))) ,
q < 1
C q =ξ (1 − q )
π1/2 Γ(1/(1 − q ))
Γ((3 − q )/2(1 − q ))
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24 Capítulo: 2 Fundamentos de confiabilidade
1 < q < 3
Γ(ν )
q → 1
q
P (x) =
ξ
π exp(−ξx2).
q
q =
−∞ q = 2
q = 3 q
q = 3+m
1+m
t
m
q = n−6
n−4
n > 4 q
r
n − 2
−∞ +∞
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25
q
q
q
q
q → 1
q
t
t < T
T
f (t) = β
η − t0
t − t0η − t0
β−1
exp
−
t − t0η − t0
β
,
β > 0
η > t0
t ≥ t0
´ ∞0
f (x)dx = 1
β = 1
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26 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
q
q
q
exp[− exp(x)]
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3.1 Introdução 27
expq(x) = [1 + (1 − q ) x]1
1−q
+
x, q ∈ R [a]+ [a]+ = a a > 0 [a]+ = 0 a ≤ 0
q
q → 1
exp1 x = exp x
x
q > 1
q
expq(−x) ∼ 1/xn
n = 1/(q − 1)
q
q > 1
q = 1
q
q
q = 1
q = 2
q
q
q
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28 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
q
q
t0
q
f q(t) = (2 − q ) β
η − t0 t − t0
η − t0
β−1
expq − t − t0
η − t0
β
,
β > 0
η − t0 > 0
t − t0 ≥ 0
(2 − q )
q < 2
f q(t)
q → 1
q
f 1(t)
η − t0 θ
q
f (t) = cktc−1
sc
1 +
t
s
c −k−1
(k > 0, c > 0, s > 0),
q
β = c
η = s/(k + 1)1/c
q = (k +
2)/(k+1) > 1 t0 = 0
q
q > 1
q ≤ 1
q
Rq(t) =
ˆ ∞t
f q(t)dt
=
1 − (1 − q )
t − t0η − t0
β 2−q
1−q
+
=
expq
−
t − t0η − t0
β2−q
,
q
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3.2 Taxa de falha da distribuição q-Weibull 29
ˆ expq(ax) dx =
1
(2 − q )a [expq(ax)]2−q.
(expq x)a = expq(ax) q = 1
(expq x)a = exp1−(1−q)/a(ax) ∀q.
Rq(t) = expq −(2 − q ) t − t0
η − t0
β
,
q = 1/(2 − q )
q
F q(t)
F q(t) = 1 − Rq(t).
hq(t) ≡ f q(t)
Rq(t),
hq(t) = (2 − q )β
η − t0
t − t0η − t0
β−1
1 − (1 − q )
t − t0η − t0
β−1
+
= (2 − q )β
η − t0
t − t0η − t0
β−1
expq
−
t − t0η − t0
βq−1
,
q → 1
h1(t) = β
η − t0
t − t0η − t0
β−1
.
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30 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
q
q
hq(t)
1 ≤ q < 2
0 < β < 1
q ≤ 1
β > 1
1 < q < 2
β > 1
q < 1
0 < β < 1
q = 1
β = 1
0 2 4 6 8 10t
0
0,5
1
1,5
h q
( t )
Formato U
Unimodal
Crescente
Decrescente
q = 1,5
β = 0,5
η = 1
t0 = 0
q = 0,5
β = 2
η = 7,071
tmax = 10
t0 = 0
q = 1,5
β = 2
η = 1
t0 = 0
q = 0,5
β = 0,5
η = 2,5
tmax = 10
t0 = 0
q < 1
tmax = t0 + (η − t0) (1 − q )−1/β .
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3.2 Taxa de falha da distribuição q-Weibull 31
hq(t) =
(2 − q )β (β − 1)
(η − t0)2
t − t0η − t0
β−2
1 −
1−q1−β
t−t0η−t0
β
1 − (1 − q )
t−t0η−t0
β2+
.
H q(t) =ˆ
t
0
hq(t) dt,
H 1 → ∞
t → ∞
lim
t → tmax
H q < 1(t) = ∞ limq → 1 −
tmax = ∞
< q <
β > 1
q < 1
0 < β < 1
t∗ = t0 + (η − t0)
1 − β
1 − q
1/β
,
hq(t∗) = 2 − q
η − t0
1 − β
1 − q
(β−1)/β
.
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32 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
0 2 4 6 8 10t
-0,4
-0,2
0
0,2
0,4
0,6
0,8
1
h ′ q ( t
)
Formato U
Unimodal
q
hq(t)
q = 0,5
β = 0,5 η = 2,5
t0 = 0
q = 1,5
β = 2
η = 1 t0 = 0
hq(t)
(q = 1)
h1(t) =
β (β − 1)
(η − t0)2
t − t0η − t0
β−2
.
h
1(t) < 0
0 < β < 1
h
1(t) > 0
β > 1
β = 1
q
q
q
t0
q
t0
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3.3 Comportamento da função densidade de probabilidade q -Weibull 33
q
µ
n =´ ∞0
tnf q(t) dt
q > 1
q < 1
q < 1
q
expq(−x) = 12π
Γ
2 − q 1 − q
ˆ +∞
−∞
e1+iu
(1 + iu)2−q1−q
e−(1−q)(1+iu)xdu,
q > 1
q
expq(−x) = 1
Γ
1q−1 ˆ
∞
0
u 1q−1
−1 e−u e−(q−1)xu du (q > 1, x > 0).
q < 1
µn = ηnΓ
1 +
n
β
Γ3−2q1−q
(1 − q )n/βΓ
3−2q1−q
+ nβ
, t0 = 0,
µn =
n j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
Γ( 3−2q
1−q )
(1−q)jβ Γ( 3−2q
1−q + j
β )
, t0 = 0,
q > 1
µn = ηnΓ
1 + n
β
Γ
2−q
q−1 − n
β
(q − 1)n/βΓ2−qq−1
, t0 = 0,
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34 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
µn =
n j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
Γ( 2−q
q−1− j
β )
(q−1)jβ Γ( 2−q
q−1)
t0 = 0,
1 < q < q upper
q upper = 1 + β/(n + β )
q → 1
µ
n = ηnΓ
1 + nβ
, t0 = 0
µ
n =n
j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
, t0 = 0
q upper
limβ→0 q upper = 1
limβ→∞
q upper = 2
limn→∞ q upper = 1
q
q > 1
q < 1
q
q upper
n
β
q
q
q < 2
µ
0 = 1
µn =n
k=0
n
k
(−1)n−kµ
k (µ1)
n−k, t0 = 0,
t0 = 0
µn =n
j=0
n j
(t0 − µ1)n− j (η − t0) j Γ
1 + jβ Γ
(
2−q
q−1
− j
β )(q−1)
jβ Γ( 2−q
q−1)
, 1 < q < 1 + β
β+n ,
µn =n
j=0
n j
(t0 − µ
1)n− j (η − t0) j Γ
1 + jβ
Γ( 3−2q
1−q )
(1−q)jβ Γ( 3−2q
1−q + jβ )
, q < 1.
q
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3.3 Comportamento da função densidade de probabilidade q -Weibull 35
= t0 + (η − t0)
ln 1
2−q0.5
q − 2
1β
,
= t0 + (η − t0)
β − 1
β + (β − 1)(1 − q )
1β
, β > 1.
η
η
q = 1
t = η
q = 1
F q(η)
q
q
1 −(1 − q )
t−t0η−t0
β
≤ 0
t = t
=
t0 + (η − t0) (1 − q )−1β
0 0,5 1 1,5 2
t
0
0,2
0,4
0,6
0,8
1
F q
( t )
q = 1
0,5
0
-1
q = 1,5 β = 0,5
η = 1
q
β = 0,5
η = 1
t0 = 0
(q =
1)
F 1(η) ≈ 0,632
F q>1(η) < F 1(η)
F q<1(η) > F 1(η)
q
q < 1
F 1(t)
t → ∞
F 0(η)
= 1
ηq
F q(ηq) = 1 − e−1 ≈ 63,2%, ∀q, ∀β
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36 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
ηq = t0 + (η − t0)− lnq e 1q−2
1β
.
q
q
lnq x ≡ x1−q − 1
1 − q , x > 0,
ln1 x = ln x
lnq(1/x) = −xq−1 lnq x
ηq = t0 + (η − t0)
e(2−q)(1−q) lnq e2−q1/β
.
ηq
limq→−∞ ηq = t0 limq→1 ηq = η
limq→2 ηq =
∞, ∀β
tmax
q
q < 1
β < 1
tmax
β
q
0 < q < 1
tmax
β
q < 0 tmax
β
β = 1
tmax = t0 + (η − t0) / (1 − q )
β → 0 tmax → ∞
0 < q < 1
tmax = t0
q < 0
0 0,2 0,4 0,6 0,8 1
β
0
1
2
3
t m a x
/ η
0,01
0,10,2
q = 0,3
-0,1-1 -5
0
tmax t0 = 0 η tmax
η
β
q < 1
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3.3 Comportamento da função densidade de probabilidade q -Weibull 37
q
tmax → ∞
q → 1−
q → −∞
tmax tmax → 0
q = 0
β = 0
limβ→0 limq→0(tmax/η) = 1
limq→0+ limβ→0(tmax/η) = ∞ limq→0− limβ→0(tmax/η) = 0
β
expq(0) = 1 ∀q
t η
t0 = 0
hq(t) ≈ (2 − q )(β/η)(t/η)β−1
β
10-7
10-6
10-5
10-4
10-3
10-2
10-1
100
101
102
t
10-2
10-1
100
101
102
103
104
h q
( t )
β = 0,1
β = 0,9
0 20 40 60 80 100t
0
0,2
0,4
0,6
0,8
h q
( t )
q = 0,9
tmax
= 100
β = 0,1
β = 0,9
β
hq(t)
q = 0,9
tmax = 100
t0 = 0
η = tmax(1 − q )1/β
η = 10−8
β = 0,1
η = 7,7426
β = 0,9
β − 1
β > 1
hq<1(0) = 0
q = 0,9
β
β
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38 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
0 20 40 60 80 100t
0
0,2
0,4
0,6
0,8
1
h q
( t
)
β = 1
β=1,2β=2
10-2
10-1
100
101
10210
-3
10-2
10-1
100
q = 0,9
β = 1
1,2
2
q
q = 0,9
η
tmax = 100
t0 = 0
η = 10; 14,68; 31,62
β = 1; 1,2; 2
β − 1
β > 1
hq>1(t)
q = 1,5
0 2 4 6 8 10 12 14 16 18 20t
0
0,5
1
1,5
2
h q
( t )
β = 3
β = 2
β = 1,5
0 0,5 1t
0
0,5
1
h q
( t ) β = 1,5
β = 2
β = 3
q = 1,5
q
q = 1,5
η = 1
t0 = 0
hq(t)
1 < q < 2 β > 1
hq(t)
β
1 < β < 2
hq(t)
β > 2
hq(t)
β = 2
t/η 1
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3.3 Comportamento da função densidade de probabilidade q -Weibull 39
t∗
tmax
t0 = 0
β
t∗
tmax
= (1 − β )1/β,
t∗/tmax β → 1
e−1 ≈ 0,367879
β → 0
q
q < 1
q
β = 0,5 t0 = 0
η
η = 1
tmax = ∞
q
tmax
tmax
β = 0,5
tmax = 100
η
q
q
limq→1− hq(t∗) → ∞
hq(t∗)
limβ→1 limq→−∞ hq(t∗) = 1/tmax
10-4 10-3 10-2 10-1 100 101 102 103
t10
-2
10-1
100
101
102
103
104
h q
( t )
β = 0,5η = 1
q = -100,50,80,9
q = 1
q
q < 1
β = 0,5
η = 1 t0 = 0
t = ∞
q = 1
q
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40 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
0 20 40 60 80 100t
0
0,2
0,4
0,6
0,8
1
h q
( t
)
0,5
0,9
0,95
q=0,97
0-1
β = 0,5tmax
= 100
q
q < 1
β = 0,5
tmax = 100
t0 = 0
η
η = 0,09
0,25
1
25
100
400
q = 0,97 0,95
0,9
0,5
0
−1
q
hq(t∗)
hq(t∗) → ∞
q → 1
q → −∞
β = 0,5
tmax = 100
hq(t∗) = 0,02
q
1 < q < 2
β > 1
1 < q < 2
0 < β < 1
q
0 10 20 30 40t
0
0,5
1
1,5
2
h q ( t )
1,7
1,5
1,2
1,9
1 1,5 2q
10-210
-1
100
101
102
h q
( t * )
β = 2
q
β = 2
η = 1
t0 = 0
q > 1
q
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3.3 Comportamento da função densidade de probabilidade q -Weibull 41
0 0,5 1 1,5 2t
0
1
2
3
h q
( t )q = 1,2
q = 1,5q = 1,9
q = 1,0
β = 0,5
η = 1
q
β = 0,5
η =
1
t0 = 0
q > 1
hq(t)
1 < q < 2
0 < β < 1
tmax
1 < q < 2 β > 1
τ u ≡ t
t∗ =
t
η
1 − q
1 − β
(β−1)/β
,
u
q < 1
0 < β < 1
τ l ≡ t
tmax=
t
η(1 − q )1/β,
l
γ (τ ) ≡ hq(t)/hq(t∗),
τ
τ u
τ l
hq(t)
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42 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
γ (τ u
) = βτ β−1
u
1 + (β − 1)τ βu ,
β > 1
0 < τ u < ∞
0 ≤ γ (τ u) ≤ 1 γ (0) = γ (∞) = 0
γ (τ l) = (1 − β )(1−β)/β βτ β−1l
1 − τ βl,
0 < β < 1
0 < τ l < 1
γ (τ l) ≥ 1
q η
β
τ
q
γ
γ q
γ (τ u) γ (τ l)
0 1 2 3 4 5 6τ
u
0
0,2
0,4
0,6
0,8
1
γ ( τ
u )
β = 1,5
2
3
γ (τ u) ≡ hq(t)/hq(t∗)
β > 1
τ u
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3.4 Conclusões 43
0 0,2 0,4 0,6 0,8 1τ
l
0
1
2
3
4
γ
( τ l
)
β = 0,99
0,9
0,5
0,01
γ (τ l) ≡ hq(t)/hq(t∗)
0 < β < 1
τ l
β → 1
γ (τ l)
τ ∗
β → 1
1/e
β → 0
q
η = 1
β = 0,5
q
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44 Capítulo: 3 Análise da função taxa de falha pelo modelo q -Weibull
q
q
β
0 < β < 1 β = 1 β > 1
q < 1
q = 1
1 < q < 2
h1(t)
β
β < 1
β = 1
β > 1
q
q
q
q
q
q
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45
q
q
q
q
q
q
q
q
q
q
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46Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
q
q
q
q
q
q
F q(t) = 1 − exp 1
2−q
−(2 − q )
t − t0
θ
β
,
β > 0
t > t0
q < 2 θ > 0
θ = η − t0
β = 1
q
F q(t) = 1 − exp 1
2−q
−(2 − q )
t − t0
θ
.
q < 2
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4.2 Distribuições de tempos de vida 47
q < 2
β > 0
β > 0
q = 1
q
β = 1
q = 1
β = 1 q = 1
q
q
q
h1
hq(t)
1 < q < 2
0 < β < 1
q < 1
β > 1
1 < q < 2 β > 1
q < 1
0 < β < 1
q = 1
β = 1
q = 1
q
q
q
β = 1
q = 1
β = 1
λ = 1
θ.
q q
y = βx + b
y = ln
− ln 1
2−q[1 − F q(t)]
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48Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
x = ln(t − t0) b = −β ln
θ
(2−q)1β
F i = i − 0,3
n + 0,4,
n
i
1
n
n
i
P (X ≥ n) = 0,5
p
X ∼ B(n, p)
q
ti
xi = ln(ti − t0)
yi = ln
− ln 1
2−q(1 − F i)
.
F i
R2
R2 = 1 −
ni=1
[yi− yi]2ni=1
[yi−yi]2
,
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4.2 Distribuições de tempos de vida 49
β > 0
θ > 0
t0 < tminq < 2,
yi
ln
− ln 1
2−q[1 − F q(ti)]
yi
yi
n tmin
R2 ≤ 1
R2
q = 1
q
=
F q (ti) − F i
2/n
= n lnRSS
n
+ 2K,
n
xi
yi RSS
K
c = n ln
RSS
n
+ 2K +
2K (K + 1)
n − K − 1 .
c c ∆i
∆i = ci −
[ c] ,
[
c] c
∆i = 0
∆i
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50Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
q
q
β < 1
q < 1
β > 1
R2
q
q
q
q
β θ
η
t0 q R2
×10−3
q
×10−3
q
q
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4.3 Aplicação em equipamentos de poços de petróleo 51
q-Weibull
Weibull
101
102
103
101
100
t dias
R q
q-Weibull
Weibull
0 1 2 3 4 5 6 7 8 9
2,4
2,6
2,8
3
3,2
t 102dias
h q
1 0
3 d
i a s
1
q
≤ t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤
< t ≤ < t ≤
< t ≤
< t ≤
q
q
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52Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
β θ
η
t0 q R2
×10−4
q
×10−5
q
q
q-Weibull
Weibull
102
103
103
102
101
t dias
R q
q-WeibullWeibull
5 10 15
3
4
5
6
7
t 102
dias
h q
1 0
3 d i a s
1
q
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4.3 Aplicação em equipamentos de poços de petróleo 53
R2
q
β > 1 q > 1
β > 1
q
β θ
η
t0 q R2
×10−4
q ×10−4
q
Weibull
q-Weibull
100 101 102
103
102
101
100
t dias
R q -Weibull
Weibull
0 1 2 3 4 5 6 7
4
6
8
10
12
t 102
dias
h q
1 0
3 d i a s
1
q
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54Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
q
q −
q
1250
q
q
y
ln(t − t0)
10,5 11 11,5 12 12,5
0,5
1
1,5
2
2,5
ln t t 0
l n
l n 1
F t
Β 1,00 q 1,00
Θ 32 081 mint0 57 165 min
R² 0,9652
a
9,5 10 10,5 11 11,5 12
0,5
1
1,5
2
ln t t 0
l n
l n 1
F t
Β 0,59 q 1,00
Θ 7 610 mint0 13 942 min
R² 0,9736
b
10,5 11 11,5 12 12,5
0,5
1
1,5
2
2,5
ln t t 0
l n
l n q '
1
F t
Β 1,00 q 1,10
Θ 19 421 min
t0 38 240 min
R² 0,9706
c
7 8 9 10 11 12
0
0,1
0,2
0,3
0,4
0,5
0,6
ln t t 0
l n
l n
q '
1
F t
Β 0,13 q 0,01Θ 216 890 min
t0 2 261 min
R² 0,9970
d
q
q
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4.4 Aplicação em estação de solda robotizada 55
t = 40 000
t = 110 000
t0
R2 = 0.9652
β = 1
q = 1
0 < β < 1
β > 1
β = 1
R2
R2 = 0.9736
β < 1
q
q
β = 1
q = 1
q
1 < q < 2
q < 1
q
Rq(t)
q
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56Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
q-exponential
exponencial
Weibull
q-Weibull
104 105
103
102
101
100
t min
R
Β Θ min t 0 min q R2
exponencial 1,00 32081 57165 1,00 0,9652
Weibull 0,59 7610 13942 1,00 0,9736
q exponencial 1,00 19421 38240 1,10 0,9706
q Weibull 0,13 2 16 894 2 261 0,01 0,9970
q
q
1250
q = 1
t ≤ 40000
β < 1
40000
< t ≤ 110 000
β ≈ 1
t > 110000
β > 1
40 000
110 000
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4.4 Aplicação em estação de solda robotizada 57
8 8,5 9 9,5 10 10,5
0
0,25
0,5
0,751
1,25
1,5
1,75
ln t t 0
l n
l n 1
F t
Β 0,65
Θ 3 124 min
t0 2 080 minR² 0,9847
a
7 8 9 10 11
4
3
2
1
0
1
ln t t 0
l n
l n 1
F t
Β 1,04
Θ 30 747 min
t0 39 785 min
R² 0,9796
b
9 9,5 10 10,5 11
2
1,5
1
0,50
0,5
ln t t 0
l n
l n 1
F t
Β 1,74
Θ 45 502 min
t0 98 964 min
R² 0,9716
c
≤ 40000
40000 <
≤ 110 000
110000
β < 1
β ≈ 1)
β > 1
q
q
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58Capítulo: 4 Análise comparativa das distribuições generalizadas q -Weibull e q -exponencial aplicadas à
engenharia de confiabilidade
q-exponencial
-WeibullWeibull
exponencial
0 5 10 15
2
3
4
5
6
t 104min
h q
1 0 5 m i n 1
Β Θ min t 0 min q R2
exponencial 1,00 32081 57 165 1,00 0,9652
Weibull 0,59 7610 13 942 1,00 0,9736q exponencial 1,00 19421 38 240 1,10 0,9706
q Weibull 0,13 216 890 2 261 0,01 0,9970
q
q
c
∆i q
∆i
c ∆i
c ∆i
q
q
q
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4.5 Conclusões 59
q
q
q
q = 1
q
q
q
q
q
q
q
q
q
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61
q
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62 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
q
q
q
ν
i
ν = {β i, ηi, t0i , q i} .
f q,i(t) =
βi(2−qi)
(ηi−t0i) t−t0i
ηi−t0iβi−1
expqi
− t−t0i
ηi−t0iβi
, t t0i
0, t < t0i ,
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5.2 Portas das Árvores de Falha Dinâmicas 63
F q,i(t) =
1 − exp 1
2−qi− (2 − q i) t−t0i
ηi−t0iβi , t t0
i
0, t < t0i ,
F q,i(t) = 1 −
1 + (q i − 1)
[t − t0i ]+ηi − t0i
βi2−qi
1−qi
+
q i = 1,
Rq,i(t) =
exp 1
2−qi
− (2 − q i)
t−t0iηi−t0i
βi
, t t0i
1, t < t0i,
Rq,i(t) =
1 + (q i − 1)
[t − t0i]+ηi − t0i
βi 2−qi
1−qi
+
q i = 1.
hq,i
hq,i(t) = f q,i(t)
Rq,i(t).
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64 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
X
q
F −1
q (•)
X
U
q
12−qi [1 − F q,i(t)] =
− (2 − q i)
t − t0iηi − t0i
βi,
1βi
1
2−qi
[1 − F q,i(t)]
q i − 2
1
β i=
t − t0iηi − t0i
.
t = t0i + (ηi − t0i)
1
2−qi
[1 − F q,i(t)]
q i − 2
1
β i.
U
F q,i(t)
U ∼ U (0, 1) .
U
1 − U
X = F −1 (U ) = t0i + (ηi − t0i)
1
2−qi
U
q i − 2
1
β i,
q
X = F −1 (U ) = t0i + (ηi − t0i)
U
1−qi2−qi −1qi−1
1
β i , q = 1
(
U )1
β i , q = 1
.
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5.2 Portas das Árvores de Falha Dinâmicas 65
q
tlim tlimatv
tlimdor
α
q
C 1
C 18 tlim
β η
t0
q α tlim
C 1 ∞
C 2 ∞
C 3
C 4 ∞
C 5
C 6 ∞
C 7
C 8 ∞
C 9
C 10 ∞
C 11 ∞
C 12
∞
C 13
C 14 ∞
C 15 ∞
C 16 ∞
C 17 ∞
C 18 ∞
q
ttimatv tlimdor
β η
t0
q α tlimatv tlimdor
C 20 ∞ ∞
C 21 ∞ ∞
C 22 ∞ ∞
C 23 ∞ ∞
C 24
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66 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
C 3
C 14 C 15
C 3
tlim = 10717, 7
F q,C 3(t > tlim) = 1
C 14
1
t → ∞
t lim 10,72 103h
C 3
C 14
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
t 103h
F q ,
C 3
t ,
F q ,
C 1 4
t
F q,C 3(t)
F q,C 14(t)
tlim
tlim
tlim
t > tlim
t lim 10,72 103h
C 3C 14
4 6 8 10 12 14
0
0,5
1
1,5
2
2,5
3
3,5
t 103h
h q
1 0
2 h
1
hq,C 3(t)
hq,C 14(t)
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5.2 Portas das Árvores de Falha Dinâmicas 67
q
∩
∩
∪
∪
∩
∪
∩
∩ ∪ ∪ ∩
∩
∪ ∪ ∩ ∪
∩ ∪
∩ ∪
∩
∪
q
∩ ∩ ∪ ∪
∩ ∪ ∪ ∪ ∪
∩
∪
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68 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
ANDlist(t) =n
j=1
F q,listj (t) ,
list
n
t
C 5
C 8 ANDC 5,C 8(t) =
2 j=1
F q,listj(t)
list = {5; 8}
q
ANDlist(t) =
i=list1,2,··· ,n
1 −
1 + (q i − 1)
[t−t0i ]+ηi − t0i
βi qi−2
qi−1
+
, q i = 1
1 −
−
[t−t0i]+ηi − t0i
βi
, q i = 1
AND
list(t) =n
i=1
n
j=1
F q,listj (t) , i = j
f q,listj(t) , i = j
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5.2 Portas das Árvores de Falha Dinâmicas 69
0 − 0
0 < t ≤ 10
10 < t ≤ 20
20 < t ≤ 30
30 < t ≤ 40
40 < t ≤ 50
50 < t ≤ 60
60 < t ≤ 70
70 < t ≤ 80
80 < t ≤ 90
F (10) = 0, 347
m
q
· · ·
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70 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
q
· · ·
ANDC 4,C 7
C 4
C 7
0 2 4 6 8 10 12 14 16
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
ANDC 4,C 7(t)
F q,C 4(t)
F q,C 7(t)
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5.2 Portas das Árvores de Falha Dinâmicas 71
0 5 10 15 20 25
0
0,05
0,1
0,15
0,2
0,25
t 103h
h q
1 0
3
h
1
0 2 4 6 8 10
0
0,1
0,2
0,3
0,4
0,5
t 103h
h q
1 0
3 h
1
0 5 10 15 20 25
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 10 20 30 40 50
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
C 1
C 2 C 1
C 15 C 1
C 4
C 1 C 9
C 2 C 15
C 2 C 4
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72 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0
3 h
1
0 5 10 15 20 25 30
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0
3 h
1
C 12
C 5 C 15
C 8 C 15
C 5
C 4
C 7
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5.2 Portas das Árvores de Falha Dinâmicas 73
C 1
C 2
C 1
C 15
C 1
C 4
C 1
C 9
C 2
C 15
C 2
C 4
C 12
C 5
C 15
C 8
C 15
C 5
C 4
C 7
ORlist(t) = 1 −n
j=1
1 − F q,listj (t)
,
t
n
list
C 11
C 2 list = {1
ORC 11,C 2
q
ORlist(t) = 1 −
i=list1,2,··· ,n
1 + (q i − 1) [t−t0i ]+ηi − t0i
βi
qi−2qi−1
+
, q i = 1
−
[t−t0i ]+ηi − t0i
βi
, q i = 1
,
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74 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
OR
list(t) =
ni=1
n j=1
1 − F q,listj(t) , i = j
f q,listj(t) , i = j
,
· · ·
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5.2 Portas das Árvores de Falha Dinâmicas 75
0 1 2 3 4 5 6
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
ORC 10,C 9(t)
F q,C 10(t)
F q,C 9(t)
q
q
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76 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 5 10 15 20 25 30 35
0,2
0,4
0,6
0,8
t 103h
h q
1 0
3 h
1
0 1 2 3 4 5 6 7 8
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 10 20 30 40 50 60 70
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15
0,25
0,5
0,75
1
1,25
1,5
1,75
2
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 10 20 30 40
0
0,1
0,2
0,3
0,4
0,5
t 103h
h q
1 0
3 h
1
C 1
C 2 C 1
C 15 C 1
C 6
C 1 C 5
C 2 C 15
C 2 C 8
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5.2 Portas das Árvores de Falha Dinâmicas 77
0 5 10 15 20 25
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 1 2 3 4 5 6 7 8
0
0,25
0,5
0,75
1
1,25
1,5
1,75
2
t 103h
h q
1 0
3
h
1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
0,25
0,5
0,75
1
1,25
1,5
1,75
t 103h
h q
1 0
3 h
1
C 2
C 7 C 15
C 10 C 15
C 5
C 10
C 9
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78 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
C 1
C 2
C 1
C 15
C 1
C 6
C 1
C 5
C 2
C 15
C 2
C 8
C 2
C 7
C 15
C 10
C 15
C 5
C 10
C 9
n
k
n
k
KofN list, k(t) =
=2m−1x=0
m j=1
x j + (1 − 2x j ) F q,listj (t)
,
m j=1
x j ≤ (m − k)
0,m
j=1
x j > (m − k) ,
KofN
list, k(t) =
=2m−1
x=0
mi=1
m
j=1
x j + (1 − 2x j) ×
×F q,listj (t)
, i = j
(1 − 2x j ) f q,listj (t) , i = j
,
m j=1
x j ≤ (m − k)
0,m
j=1
x j > (m − k) ,
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5.2 Portas das Árvores de Falha Dinâmicas 79
list
m
k
t
x j j
x
j = 1
list = {2, 5, 7}
k = 2
m = 3
x
0
23 − 1 = 7
x = 6
x1 = 1
x2 = 1
x3 = 0
610 = 1102 C 2
C 5 C 7
k
k
k
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80 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
· · ·
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
KofN {C 1,C 10,C 13},2(t)
F q,C 1(t)
F q,C 10(t)
F q,C 13(t)
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5.2 Portas das Árvores de Falha Dinâmicas 81
0 2 4 6 8 10
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 10 20 30 40
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15 17,5
0
1
2
3
4
5
t 103h
h q
1 0
4 h
1
0 2 4 6 8 10 1 2
0
1
2
3
4
5
t 103h
h q
1 0
3 h
1
C 1
C 2 C 15
C 1 C 2
C 4 C 1
C 11 C 5
C 1 C 12
C 5 C 1
C 2 C 5
C 1 C 15
C 10
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82 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2 4 6 8 10 1 2
0
1
2
3
4
t 103h
h q
1 0
3
h
1
0 2 4 6 8 10 12
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
2
4
6
8
10
12
14
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
1
2
3
4
5
6
7
8
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
1
2
3
4
5
6
t 103h
h q
1 0
3 h
1
C 1
C 15 C 9
C 1 C 4
C 9 C 17
C 15 C 18
C 11 C 15
C 13 C 2
C 10 C 13
C 15 C 10
C 13
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5.2 Portas das Árvores de Falha Dinâmicas 83
C 1
C 2 C 15
C 1
C 2 C 4
C 1
C 11 C 5
C 1
C 12 C 5
C 1
C 2 C 5
C 1
C 15 C 10
C 1
C 15
C 9
C 1
C 4 C 9
C 17
C 15
C 18
C 11
C 15 C 13
C 2
C 10 C 13
C 15
C 10 C 13
(T )
pd ≤ 1
pd = 1
F T (t) =
tˆ
0
f T (t1)dt1,
f T (t1)
t1
t
F q,i(t)
F C
q,i(t) = Rq,i(t)
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84 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
pd
PDepi,T (t) = F T (t)F q,i(t) + F C T (t)F q,i(t) + F T (t)F C
q,i(t) pd
= F q,i(t)
F T (t) + F C T (t)
+ F T (t)F C
q,i(t) pd
= F q,i(t) + (1 − F q,i(t)) F T (t) pd,
PDep
i,T (t) = f q,i(t) + (1 − F q,i(t))
F T (t) pd + (1 − F q,i(t)) f T (t) pd
= f q,i(t) − f q,i(t)F T (t) pd + (1 − F q,i(t)) f T (t) pd
= f q,i(t) (1 − F T (t) pd) + (1 − F q,i(t)) f T (t) pd.
T
n
pd
pd
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5.2 Portas das Árvores de Falha Dinâmicas 85
pd
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86 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
PDepC 14,C 10(t)
pd = 0, 9
F q,C 14(t)
F q,C 10(t)
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5.2 Portas das Árvores de Falha Dinâmicas 87
0 5 10 15 20 25 30
0,2
0,4
0,6
0,8
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
0,5
1
1,5
2
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10
0
0,5
1
1,5
2
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,25
0,5
0,75
1
1,25
1,5
1,75
2
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,25
0,5
0,75
1
1,25
1,5
1,75
2
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12
0
0,25
0,5
0,75
1
1,25
1,5
1,75
t 103h
h q
1 0
3 h
1
C 1
C 2 C 1
C 15 C 1
C 10 C 1
C 5 C 2
C 14
C 11
C 10
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88 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2,5 5 7,5 10 12,5 15 17,5
0,4
0,5
0,6
0,7
0,8
0,9
1
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 2,5 5 7,5 10 12,5 15
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0
3 h
1
C 12
C 5 C 14
C 10 C 15
C 5
C 10
C 9
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5.2 Portas das Árvores de Falha Dinâmicas 89
C 1
C 2
C 1
C 15
C 1
C 10
C 1
C 5
C 2
C 14
C 11
C 10
C 12
C 5
C 14
C 10
C 15
C 5
C 10
C 9
X 1
X 2 x1
x2
f 1 (x1) f 2 (x2)
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90 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
f 1(x1)
f 2(x2)
X 1
x2
P (X 1 < x2) =
x1=x2ˆ
x1=0
f 1(x1)dx1,
X 1 < X 2
t
P (X 1 < X 2) =x2=t´
x2=0
f 2(x2)dx2
x1=x2´ x1=0
f 1(x1)dx1
=x2=t´
x2=0
x1=x2´ x1=0
f 2(x2)f 1(x1)dx1dx2.
q
C 1
C 2
t
PANDC 1,C 2(t) =
x2=tˆ
x2=t02
x1=x2ˆ
x1=t01
f q,2(x2)f q,1(x1)dx1dx2.
n
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5.2 Portas das Árvores de Falha Dinâmicas 91
PANDC 1,C 2,··· ,C n(t) =
xn=t´ xn=t0n
xn−1=xn´ xn−1=t0n−1
· · ·
· · ·x2=x3´
x2=t02
x1=x2´ x1=t01
f q,n(xn)f q,n−1(xn−1) · · · f q,2(x2)f q,1(x1) dx1dx2 · · ·
· · · dxn−1dxn,
t01 ≤ t02 ≤ · · · ≤ t0n,
PANDC 1,C 2,··· ,C n(t)
PAND
C 1,C 2,··· ,C n(t) =
f q,n(t)xn−1=t´
xn−1=t0n−1
xn−2=xn−1´ xn−2=t0n−2
· · ·
· · ·x2=x3´
x2=t02
x1=x2´ x1=t01
f q,n−1(xn−1)f q,n−2(xn−2) · · · f q,2(x2)f q,1(x1) dx1dx2 · · ·
· · · dxn−2dxn−1,
t01 ≤ t02 ≤ · · · ≤ t0n
n
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92 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
q
q
limt→∞ PANDC 1,C 2,··· ,C n(t) < 1.
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5.2 Portas das Árvores de Falha Dinâmicas 93
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
PAndC 10,C 9(t)
F q,C 10(t)
F q,C 9(t)
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94 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 5 10 15 20 25
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0 4 h 1
0 5 10 15 20 250
0,5
1,
1,5
2,
2,5
t 103h
h q
1 0
5 h 1
0 5 10 15 20 25 30 35
0
0,5
1
1,5
2
2,5
3
3,5
4
t 103h
h q
1 0 4 h 1
0 5 10 15 20 25 30 350
0,2
0,4
0,6
0,8
1,
1,2
1,4
t 103h
h q
1
0 5 h 1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0 3 h 1
0 2 4 6 8 1 0 12 140
0,5
1,
1,5
2,
2,53,
3,5
t 103h
h q
1 0 5 h 1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0 3 h 1
0 2,5 5 7,5 10 12,5 15 17,50
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0 4
h 1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0 3 h 1
0 2,5 5 7,5 10 12,5 15 17,50
0,20,40,60,8
1,1,21,4
t 103h
h q
1 0 5 h 1
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0 3 h 1
0 2 4 6 8 10 120
0,2
0,4
0,6
0,8
t 103h
h q
1 0 4 h 1
C 1
C 2 C 1
C 14 C 1
C 4 C 1
C 5 C 2
C 14 C 11
C 10
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5.2 Portas das Árvores de Falha Dinâmicas 95
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,5
1
1,5
2
2,5
3
3,5
t 103h
h q
1 0 3 h 1
0 2 4 6 8 10 12 140
0,5
1
1,5
2
2,5
t 103h
h q
1 0 7 h 1
0 5 10 15 20 25
0
0,25
0,5
0,75
1
1,25
1,5
1,75
t 103h
h q
1 0 3 h 1
0 2,5 5 7,5 10 12,5 15 17,50
0,2
0,4
0,6
0,8
1,1,2
1,4
t 103h
h q
1 0 5 h 1
0 2 4 6 8 10 12
0
0,5
1
1,5
2
t 103h
h q
1 0 3 h 1
0 2 4 6 8 10 120
1,
2,
3,
4,5
6,
7,
t 103h
h q
1 0 5
h 1
C 12
C 5
C 14 C 10
C 15 C 5
C 10 C 9
C 1
C 2
C 1 C 14
C 1
C 4
C 1
C 5
C 2
C 14
C 11
C 10
C 12
C 5
C 14
C 10
C 15
C 5
C 10
C 9
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96 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
SE QC 1,C 2,··· ,C n(t) =
K SE Q
xn=t
´ xn=t0n
xn−1=xn
´ xn−1=t0n−1
· · ·
· · ·x2=x3´
x2=t02
x1=x2´ x1=t01
f q,n(xn)f q,n−1(xn−1) · · · f q,2(x2)f q,1(x1) dx1dx2 · · ·
· · · dxn−1dxn,
t01 ≤ t02 ≤ · · · ≤ t0n ,
K SE Q
= 1
xn=∞´
xn=t0n
···x2=x
n´
x2=t02
x1=x
2´
x1=t01
f q,n(xn)···f q,2(x2)f q,1(x1)dx1dx2···dxn
.
t01 ≤ t02 ≤ · · · ≤ t0n
K SE Q
SE QC 1,C 2,··· ,C n(t)
SE Q
C 1,C 2,··· ,C n(t) =
K SE Q f q,n(t)xn−1=xn´
xn−1=t0n−1
xn−2=xn´ xn−2=t0n−2
· · ·
· · ·x2=x3´
x2=t02
x1=x2´ x1=t01
f q,n−1(xn−1)f q,n−2(xn−2) · · · f q,2(x2)f q,1(x1) dx1dx2 · · ·
· · · dxn−2dxn−1,
t01 ≤ t02 ≤ · · · ≤ t0n ,
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5.2 Portas das Árvores de Falha Dinâmicas 97
. . .
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98 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
SE QC 10,C 9(t)
F q,C 10(t)
F q,C 9(t)
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5.2 Portas das Árvores de Falha Dinâmicas 99
0 5 10 15 20 25 30
0,1
0,2
0,3
0,4
0,5
0,6
0,7
t 103h
h q
1 0 3 h 1
0 5 10 15 20 25 300
0,1
0,2
0,3
0,4
0,5
0,6
t 103h
h q
1 0 4 h 1
0 2 4 6 8 10
0
0,5
1
1,5
2
2,5
t 103h
h q
1 0
4 h
1
0 5 10 15 20 25 30
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0 3 h 1
0 5 10 15 20 25 300
0,2
0,4
0,6
0,8
t 103h
h q
1 0 4 h 1
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12
0
0,5
1
1,5
2
2,5
3
t 103h
h q
1 0
4 h
1
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0
3 h
1
C 1
C 2 C 1
C 14 C 1
C 4 C 1
C 5 C 2
C 14 C 11
C 10
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100 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 2,5 5 7,5 10 12,5 15 17,5
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0
3 h
1
0 2 4 6 8 10 12
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 103h
h q
1 0
3 h
1
C 12
C 5 C 14
C 10 C 15
C 5 C 10
C 9
C 1 C 2
C 1
C 14
C 1
C 4
C 1
C 5
C 2
C 14
C 11
C 10
C 12
C 5
C 14
C 10
C 15
C 5
C 10
C 9
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5.2 Portas das Árvores de Falha Dinâmicas 101
λ
αλ
α
(0 ≤ α ≤ 1)
α = 1
α = 0
q
ηd
ηd = η
α, 0 < α ≤ 1
ηd
η
S i
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102 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
f S idorα(t) = β i (2 − q i)
ηi
α − t
0iβi
[t − t0i ]βi−1+ expqi −
[t − t0i ]+ηiα
− t0i βi
,
S i
i
α
β i
ηi
t0i
q i
q
i
[x]+
x
S 1dor P
P S 1atv
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5.2 Portas das Árvores de Falha Dinâmicas 103
ta
W SP P 1,S 1(t) = 1 −
1 − W SP S →P P 1,S 1
(t)
1 − W SP P →S P 1,S 1
(t)
,
W SP S →P
P 1,S 1(t)
t
W SP P →S
P 1,S 1(t)
W SP S →P P 1,S 1
(t)
W SP S →P P 1,S 1
(t) =x2=t´
x2=t02
x1=x2´ x1=t01
f q,P (x2)f S 1dorα(x1)dx1dx2, t01 ≤ t02
=x2=t´
x2=t02
f q,P (x2)F S 1dorα(x2)dx2,
f q,P (ta)
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104 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
x2 − x1
f S 1atv(•)
x2
W SP P →S P 1,S 1
(t) = K W SP
x1=t´ x1=t01
f q,P (x1)
x2=t´
x2=x1
f S 1atv(x2 − x1)dx2
dx1
= K W SP
x1=t´ x1=t01
f q,P (x1)F S 1atv(t − x1)dx1,
K W SP = 1
x1=∞´ x1=t01
f q,P (x1)F S 1atv(t − x1)dx1
.
W SP
P 1,S 1(t) = f q,P (t)F S 1dorα(t)×1 − K W SP
x1=t´ x1=t01
f q,P (x1)F S 1atv(t − x1)dx1
+
1 −x2=t´
x2=t02
f q,P (x2)F S 1dorα(x2)dx2
×
K W SP
x1=t´
x1=t01
f q,P (x1)f S 1atv(t − x1)dx1 + f q,P (t)F S 1atv(0)
.
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5.2 Portas das Árvores de Falha Dinâmicas 105
0 5 10 15 20 25
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
W SP C 24,C 24(t)
C 24
F q,C 24atv(t)
F q,C 24dor(t)
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106 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
q
W SP C i,C i(t)
hq,C iatv(t)
i
hq,C idor(t)
C 20
i = 20;
C 21
i = 21)
i = 24
0 2,5 5 7,5 10 12,5 15 17,5 20
0
0,1
0,2
0,3
0,4
0,5
t 103
h
h q
1 0
3 h
1
C 20
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5.2 Portas das Árvores de Falha Dinâmicas 107
0 5 10 15 20 25 30
0
0,05
0,1
0,15
0,2
0,25
t 103h
h q
1 0
3 h
1
C 21
0 2,5 5 7,5 10 12,5 15 17,5 20
0
0,1
0,2
0,3
0,4
0,5
t 103h
h q
1 0
3 h
1
C 22
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108 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 5 10 15 20 25 30 35 40
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
C 23
0 5 10 15 20 25 30
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
C 24
15 × 103
20 × 103
tlim = 18.000
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5.2 Portas das Árvores de Falha Dinâmicas 109
W SP C 23,C 24(t)
C 23
C 24
0 5 10 15 20 25
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
W SP C 23,C 24(t)
F q,C 23atv(t)
F q,C 24atv(t)
F q,C 24dor(t)
W SP C 23,C 24
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110 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
0 5 10 15 20 25 30 35
0
0,2
0,4
0,6
0,8
1
1,2
t 103h
h q
1 0
3 h
1
W SP C 23,C 24
C 23
C 24
W SP 2S P 1,S 1,S 2(t) = 1 −
1 − W SP S 2 →[P 1,S 1][P 1,S 1],S 2
(t)
1 − W SP [P 1,S 1]→S 2[P 1,S 1],S 2
(t)
,
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5.2 Portas das Árvores de Falha Dinâmicas 111
W SP
S 2 →[P 1,S 1]
[P 1,S 1],S 2 (t) =
x2=t
´ x2=t02 W SP
P 1,S 1(x2)F S 2dorα(x2)dx2
W SP [P 1,S 1]→S 2[P 1,S 1],S 2
(t) = K W SP 2S
x1=t´ x1=t01
W SP
P 1,S 1(x1)F S 2atv(t − x1)dx1
K W SP 2S =x1=∞´
x1=t01
W SP
P 1,S 1(x1)F S 2atv(t − x1)dx1.
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112 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
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5.2 Portas das Árvores de Falha Dinâmicas 113
0 5 10 15 20 25
0
0,2
0,4
0,6
0,8
1
t 103h
F q
t
W SP 2S C 24,C 24,C 24(t)
F q,C 24atv(t)
F q,C 24dor(t)
0 5 10 15 20 25 30 35
0
0,2
0,4
0,6
0,8
1
t 103h
h q
1 0
3 h
1
W SP 2S C 24,C 24,C 24
C 24
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114 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
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5.3 Conclusões 115
q
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116 Capítulo: 5 q -Weibull aplicada a Árvore de Falha Dinâmica
q
q
q
q
q
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117
q
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118 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
q
t
q
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6.2 Confiabilidade e manutenção com o modelo q -Weibull 119
T
Rq(t)
t ≤ T
Rmq(t) = Rq(t).
t > T
T
T < t ≤ 2T
Rmq(t) = Rq(T )Rq(t − T ), T < t ≤ 2T,
R(t − T )
T
t − T
T
t
Rmq(t) = Rq(T )nRq(t − nT )
nT < t ≤ (n + 1)T
n = 0, 1, 2 . . . ,
Rq(T )n
n
Rq(t−nT )
t
n(t) =
t
T
,
(x)
x
Rmq,T (t) = Rq(T )n(t)Rq [t − n(t)T ] ,
q
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120 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
Rmq,T (t) =
exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
×
exp 1
2−q
−(2 − q ) t−n(t)T −t0
η−t0β
.
β
η, t0
q
t
Rmq,T (t)
T
F mq,T (t) = 1 −
exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
×
exp 1
2−q
−(2 − q )
t−n(t)T −t0
η−t0
β
.
F mq,T (t) = 1 − exp 1
2−
q−(2 − q )T −t0
η−t0 βn(t)
×1 −
´ tn(t)T
β(2−q)(η−t0)
x−n(t)T −t0
η−t0
β−1
×
× expq
−
x−n(t)T −t0η−t0
β dx
,
F mq,T (t) = 1 −
exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
+
´ tn(t)T +t0
exp 1
2−q
−(2 − q )×
×
T −t0η−t0
β
n(t)
× β(2−q)(η−t0)
x−n(t)T −t0
η−t0
β−1
× expq
−
x−n(t)T −t0η−t0
β
dx.
1 −
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6.2 Confiabilidade e manutenção com o modelo q -Weibull 121
exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
n(t)T
t
F mq,T (t)
f mq,T (t) =
exp 1
2−q
−(2 − q )×
×
T −t0η−t0
β
n(t)
β(2−q)(η−t0)
t−n(t)T −t0
η−t0
β−1
×
expq
−
t−n(t)T −t0η−t0
β
.
hmq,T (t) = f mq,T (t)
Rmq,T (t).
hmq,T
T
k
k ∈ R
0 < k < 1
k → 1
Rmkq,T (t) =
k exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
× exp 1
2−q
−(2 − q )
t−n(t)T −t0
η−t0
β
.
F mkq,T (t) = 1 −
k exp 1
2−q
−(2 − q )
T −t0η−t0
βn(t)
× exp 1
2−q
−(2 − q )
t−n(t)T −t0
η−t0
β
,
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122 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
f mkq,T (t) =
k exp 1
2−q
−(2 − q )×
×
T −t0η−t0
β
n(t)
× β(2−q)(η−t0)
t−n(t)T −t0
η−t0
β−1
×
expq
−
t−n(t)T −t0η−t0
β
.
t
T ´ t0
tf (t)dt + T ∞ T
f (t)dt
q
E [T ] =
T
ˆ t0
tf q(t)dt + T
∞
ˆ T
f q(t)dt.
T ˆ
t0
tf q(t)dt = −T Rq(T ) +
T ˆ
t0
Rq(t)dt,
E [T ] = −T Rq(T ) +T ´
t0
Rq(t)dt + T ∞ T
f q(t)dt
= −T Rq(T ) +T ´
t0
Rq(t)dt + T Rq(T )
=T ´
t0
Rq(t)dt.
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6.3 Exemplos 123
E [T ] =
(T − t0) 2F 1q−21−q
, 1β
; 1 + 1β
; (1 − q )
T −t0η−t0
β
, q = 1
(η−t0)β
−Γ
1β
,
T −t0η−t0
β
+ Γ
1β
, 0
, q = 1,
2F 1 (a, b; c; z )
1 + abc 1!
z + a(a+1)b(b+1)c(c+1)2!
z 2 +a(a+1)(a+2)b(b+1)(b+2)
c(c+1)(c+2)3! z 3 + · · ·
E [C ] = C cF q(T ) + C pRq(T ),
C c
C p
Z [T ] =
E [C ]
E [T ] =
C cF q(T ) + C pRq(T )
E [T ]
T
Z [T ] = C cF q(T )+C pRq(T )
E [T ] .
q
C 4 C 3
C 3
C 4
pd = 0, 9
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124 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
q
q
β η t0 q
C 1
C 2
C 3
C 4
T k
T k
C C C P
C 1
C 1
C 2
C 2
PDepC 3,C 4
pd = 0, 9
PDepC 4,C 3
pd = 0, 9
C 1
T = 173
Rq,T (3000) = 0, 9
T
C 1
k = 1
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6.3 Exemplos 125
0 10 20 30 40 50 60
0,5
0,6
0,7
0,8
0,9
1
t 102
h
R m q
, T 1 7 3
t ,
R m q
, T 4 0
0 t ,
R q
t
T = 173h
T = 400h
C 1
C 1
0 10 20 30 40 50 60
0
0,1
0,2
0,3
0,4
0,5
t 102h
f m q ,
T
1 7 3
t ,
f q t 1 0
3
C 1
T = 173h
Rq,T (3000) = 0, 9
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126 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
hmax = 3, 0 × 10−4h−1
T = 399h
0 10 20 30 40 50 60
0
0,2
0,4
0,6
0,8
t 102
h
h m q , T
3 9 9
t ,
h q
t 1 0
3
C 1
T = 399h
C 1
C 1
C c = 10.000
C p = 500
T = 353h
Z q = 2, 40
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6.3 Exemplos 127
0 20 40 60 80 100
2
4
6
8
10
12
t 102h
Z q
t
2 3 4 5 6 7 82,4
2,6
2,83
3,2
3,4
3,6
t 102h
Z q
t
k = 0, 85
T = 548h
T = 700h
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128 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
0 10 20 30 40 50 60
0,2
0,3
0,4
0,5
0,6
0,7
0,8
0,9
1
t 102
h
R m k q
, T 5 4 8
t ,
R m k q
, T 7
5 0
t ,
R q
t
T = 548h
Rmkq,T (1000) = 0, 8
k = 0, 85
T = 700h
C 1
t
C 1
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6.3 Exemplos 129
0 10 20 30 40 50 60
0
0,1
0,2
0,3
0,4
0,5
t 102h
f m k
q ,
T
5 4 8
t ,
f q t
1 0
3
C 1
T = 500h
k = 0, 85
C 1
C 1
T = 692h
hmax = 6, 0 × 10
−4
h
−1
4000h
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130 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
0 10 20 30 40 50 60
0
0,2
0,4
0,6
0,8
t 102
h
h m k
q ,
T 6 9 2
t ,
h q
t
1 0
3
C 1
T = 692h
C 1
C 1
1500h
T = 264h
Rmq,T (1500) = 0, 95
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6.3 Exemplos 131
0 5 10 15 20 25 30
0,5
0,6
0,7
0,8
0,9
1
t 102
h
R m q
, T 2 6 4
t ,
R m q
, T 5 0
0 t ,
R q
t
C 2
T = 264h
T = 500h
C 2
C 2
C 2
0 5 10 15 20 25 30
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
t 102h
f m q ,
T
2 6 4
t ,
f q t 1 0
3
C 2
T = 264h
C 2
C 2
hmax = 5 × 10−4h−1
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132 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
800h
0 5 10 15 20 25 30
0
0,5
1
1,5
2
2,5
3
3,5
t 102
h
h m q ,
T 8 0 0
t ,
h q
t 1 0
3
C 2
T = 800h
C 2
C 2
C c = 10000
C p = 500
T = 443h
Z q = 1, 88
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6.3 Exemplos 133
0 5 10 15 20 25 30
2
4
6
8
10
12
14
16
t 102h
Z q
t
3 3,5 4 4,5 5 5,5 6 6,5
1,9
1,952
2,05
2,1
t 102h
Z q
t
544h
350h
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134 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
0 5 10 15 20 25 30
0,5
0,6
0,7
0,8
0,9
1
t 102
h
R m k q
, T 5 4 4
t ,
R m k q
, T 3
5 0
t ,
R q
t
C 2
T = 544h
Rmkq,T (1500) = 0, 7
k = 0, 9
T = 350h
k
C 2
C 2
C 2
0 5 10 15 20 25 30
0
0,1
0,2
0,3
0,4
0,5
0,6
0,7
t 102h
f m k
q ,
T
5 4 4
t ,
f q t 1 0
3
C 2
T = 544h
k = 0, 9
C 2
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6.3 Exemplos 135
T = 516h
hmax = 2, 5 × 10−4h−1
0 5 10 15 20 25 30
0
0,25
0,5
0,75
1
1,25
1,5
1,75
2
t 102h
h m k
q , T 5 1 6
t ,
h q
t 1 0
3
C 2
T = 516h
k = 0, 9
C 2
C 3 C 4
PDepC 3,C 4 C 4
C 3.
pd = 0, 9
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136 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
0 1 2 3 4 5 6
0,5
0,6
0,7
0,8
0,9
1
t 103h
R m q
, T 5 6 8
t ,
R m q
, T 1 0 0 0
t ,
R q
t
0 1 2 3 4 5 6 7 8
0,5
0,6
0,7
0,8
0,9
1
t 103h
R m q
, T 8 8 4
t ,
R m q
, T 1 2 0 0
t ,
R q
t
T 1
T 2
C 3 C 4
T 1 = 568h
Rmq,T (3000) = 0, 95
T 2 = 1000h
C 4
C 3
T 1 = 884h
Rmq,T (2000) = 0, 9
T 2 = 1200h
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6.3 Exemplos 137
0 1 2 3 4 5 6
0
0,1
0,2
0,3
0,4
0,5
0,6
t 103h
f m q ,
T
5 6 8
t ,
f q t 1 0
3
0 1 2 3 4 5 6 7 8
0
0,1
0,2
0,3
0,4
0,5
t 103h
f m q ,
T
8 8 4
t ,
f q t 1 0
3
T
C 3
C 4
T = 568h
C 4
C 3 T = 884h
0 2 4 6 8 10 12 14
0
0,2
0,4
0,6
0,8
1
1,2
1,4
t 10
3
h
h m q ,
T 1 0 4 2
t ,
h q
t 1 0
3
0 2 4 6 8 10
0
0,5
1
1,5
2
t 10
3
h
h m q ,
T 9 7 5
t ,
h q
t 1 0
3
T
C 3
C 4 T = 1042h
hmax = 4 × 10−4h−1
C 4
C 3
T = 975h
hmax = 3 × 10−4h−1
8700h
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138 Capítulo: 6 Manutenções corretiva e preventiva baseadas no modelo q -Weibull
12500h
q
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6.4 Conclusões 139
q
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141
q
F q(t) = 1 −
expq
−
t − t0η − t0
β2−q
,
expq =
[1 + (1 − q ) x]1
1−q
+ β
q
η
q → 1
q
q
β < 1
β = 1
β > 1
q
q
q
q
β
q
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142 Capítulo: 7 Considerações finais
q
q
β < 1
β = 1
β > 1
q
q
q
q
q
q
β = 0, 65
β = 1, 04
β = 1, 74
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7.1 Conclusões 143
q
q
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144 Capítulo: 7 Considerações finais
q
q
q
q
0 < k < 1
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7.2 Sugestões para trabalhos futuros 145
q
q
q
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146 Capítulo: 7 Considerações finais
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147
q
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148 Referências
q
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Referências 149
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150 Referências
pi
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Referências 151
q
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152 Referências
q
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Referências 153
q
q
q
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154 Referências
t
r
f q
q
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Referências 155
q
q
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157
f (t) = β (t − t0)β−1
(η − t0)β exp
−
t − t0η − t0
β
t ≥ t0
q
f q(t) = β (2 − q )
η − t0 t − t0
η − t0
β−1
expq − t − t0
η − t0
β
, t ≥ t0
q
expq(x) =
exp(x),
q = 1
(1 + (1 − q ) x)1
1−q , q = 1 ∧ (1 + (1 − q ) x) > 0
0,
q = 1 ∧ (1 + (1 − q ) x) ≤ 0
Rq(t) =
∞ t
f q(x)dx
Rq(t) =
∞
ˆ t
β (2 − q )
η − t0t − t0
η − t0
β−1
expq −t − t0
η − t0
β
dt
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158 Apêndice A
∞
t
expq(ζ )dζ = 1
(2 − q )
expq(ζ )
2−q
ζ = −
t−t0η−t0
β
dζ = −β
1
η−t0
t−t0η−t0
β−1
dt
Rq(t) = − (2 − q )∞ t
expq [ζ ] dζ ,
q
Rq(t) = − (2 − q ) 1(2−q)
expq(ζ )2−q
∞t
= −expq(ζ )
2−q∞
t
= −
expq
−
t−t0η−t0
β2−q
∞
t
Rq(t) =
expq
−
t − t0η − t0
β2−q
, t ≥ t0
F q(t) = 1 −
expq
−
t − t0η − t0
β2−q
, t ≥ t0
hq(t) = f (t)
R(t) =
β(2−q)η−t0
t−t0η−t0
β−1
expq
−
t−t0η−t0
β
expq
−
t−t0η−t0
β2−q
hq(t) = β (2 − q )
η − t0
t − t0η − t0
β−1
expq
−
t − t0η − t0
βq−1
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A.1 Funções de confiabilidade e derivadas 159
q
d expq x
dx =
expq x
q
k = β(2−q)η−t0
ζ =
t−t0η−t0
dζ = 1η−t0
dt
hq(t) = kζ β−1
expq
−ζ β
q−1
hq(t)k
= (β − 1) ζ β−2 1η−t0
expq
−ζ β
q−1+
+ζ β−1 (q − 1)
expq
−ζ β
q−2 expq
−ζ β
q(−β )
ζ β−1
1η−t0
hq(t)
k = (β − 1) ζ β−2 1
η − t0 expq
−ζ β
q−1
+ζ 2β−2 (q − 1)
expq
−ζ β
2q−2
(−β ) 1
η − t0
hq(t) =
k
η − t0
expq
−ζ β
q−1ζ β−2
(β − 1) + ζ β (q − 1)
expq
−ζ β
q−1(−β )
l =
(β − 1) + ζ β (q − 1)
expq
−ζ β
q−1(−β )
l = (β − 1) + ζ β (q − 1)
1 + (1 − q )
−ζ β 1
1−q
+
q−1
(−β )
l = (β − 1) + ζ β (q − 1)
1 + (1 − q )
−ζ β−1
+
(−β )
l = (β − 1) + ζ β (q − 1) (−β )
[1 + (1 − q ) (−ζ β)]+
l =(β − 1)
1 + (1 − q )
−ζ β
+
+ ζ β (q − 1) (−β )
[1 + (1 − q ) (−ζ β)]+
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160 Apêndice A
l = (β − 1) + (β − 1)(1 − q )
−ζ β
+ ζ β (q − 1) (−β )
[1 + (1 − q ) (−ζ β)]+
l = (β − 1) + (β − 1) (q − 1)
ζ β
+ ζ β (q − 1) (−β )
[1 + (1 − q ) (−ζ β)]+
l = (β − 1) + (q − 1)
ζ β
[(β − 1) + (−β )]
[1 + (1 − q ) (−ζ β)]+
l = (β − 1) + (1 − q ) ζ β
[1 + (1 − q ) (−ζ β
)]+
l =(β − 1)
1 +
1−qβ−1
ζ β
[1 + (1 − q ) (−ζ β)]+
hq(t) =
k (β − 1)
η − t0expq −ζ β
q−1ζ β−2
1 +
1−qβ−1
ζ β
[1 + (1 − q ) (−ζ β
)]+
hq(t) =
k (β − 1)
η − t0
ζ β−2
[1 + (1 − q ) (−ζ β)]+
1 +
1−qβ−1
ζ β
[1 + (1 − q ) (−ζ β)]+
hq(t) =
k (β − 1) ζ β−2
η − t0
1 + 1−qβ−1 ζ β
[1 + (1 − q ) (−ζ β)]2+
hq(t) =
β (2 − q )
η − t0
(β − 1) ζ β−2
η − t0
1 +
1−qβ−1
ζ β
[1 + (1 − q ) (−ζ β)]2+
hq(t) = (2 − q ) β (β − 1)
t−t0
η−t0β−2
(η − t0)2
1 + 1−q
β−1 t−t0
η−t0β
1 + (1 − q )
−
t−t0η−t0
β2
+
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A.1 Funções de confiabilidade e derivadas 161
limq→1
hq(t) = β(β−1)
(η−t0)2
t−t0η−t0
β−2
h
q(t) = 0
1 +
1 − q
β − 1
t − t0η − t0
β
= 0
1 − q
1 − β
t − t0η − t0
β
= 1
t − t0η − t0
β
=
11−q1−β
t − t0η − t0
=
1 − β
1 − q
1/β
t∗ = t0 + (η − t0)
1 − β
1 − q
1/β
hq(t∗) = β (2 − q )
η − t0
t∗ − t0η − t0
β−1
expq
−
t∗ − t0η − t0
βq−1
hq(t∗) = β(2−q)η−t0
t0+(η−t0)( 1−β
1−q )1/β
−t0
η−t0
β−1
×
expq
−
t0+(η−t0)(1−β1−q )
1/β
−t0η−t0
βq−1
hq(t∗) = β(2−q)η−t0
1−β1−q
1/ββ−1
×expq
−
1−β1−q
1/ββq−1
hq(t∗) = β(2−q)η−t0
1−β1−q
β−1β
expq
−1−β1−q
q−1
q
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162 Apêndice A
hq(t∗) = β(2−q)η−t0
1−β1−q
β−1
β
1 − (1 − q )
1−β1−q
1
1−q
+
q−1
hq(t∗) = β(2−q)η−t0
1−β1−q
β−1β
[β ]
1
1−q
+
q−1
hq(t∗) = β(2−q)η−t0
1−β1−q
β−1β
β −1
hq(t∗) = 2−qη−t0 1−β
1−qβ−1
β
m = (2−q)β(β−1)
(η−t0)2
u =
t−t0η−t0
β−2
v =1 +
1−qβ−1
t−t0η−t0
β
1 + (1 − q )
−
t−t0η−t0
β2+
hq (t) = m(uv)
hq (t) = m(uv + uv)
u =
β − 2
η − t0
t − t0η − t0
β−3
r = 1+ 1−q
β−1 t−t0
η−t0β
s = 1 + (1 − q )− t−t0
η−t0β
2
+
v = r
s
v =
rs
= rs−rs
s2
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A.1 Funções de confiabilidade e derivadas 163
r =
1−qβ−1
βη−t0
t−t0η−t0
β−1
s = 2
1 + (1 − q )
−
t − t0η − t0
β
(−β ) (1 − q )
η − t0
t − t0η − t0
β−1
vs2 =
1−qβ−1
βη−t0
t−t0η−t0
β−1
1 + (1 − q )
−
t−t0η−t0
β2
−
1 +
1−q
β−1 t−t0
η−t0β
2
1 + (1 − q )
− t−t0
η−t0β (−β)(1−q)
η−t0 t−t0
η−t0β−1
vs2 =
β(1−q)η−t0
t−t0η−t0
β−1
1 + (1 − q )
−
t−t0η−t0
β
×
1β−1
1 + (1 − q )
−
t−t0η−t0
β
+1 +
1−qβ−1
t−t0η−t0
β
2
p =
1β−1
1 + (1 − q )
−
t−t0η−t0
β
+
1 +
1−qβ−1
t−t0η−t0
β
2
p =
1β−1
+
1−qβ−1
−
t−t0η−t0
β
+
2 + 2
1−qβ−1
t−t0η−t0
β
p =
1β−1
−
1−qβ−1
t−t0η−t0
β
+ 2 + 2
1−qβ−1
t−t0η−t0
β
p =
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
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164 Apêndice A
vs2 = β(1−q)η−t0
t−t0η−t0
β−1
1 + (1 − q )
−
t−t0η−t0
β
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
v =
β(1−q)η−t0
t−t0η−t0
β−1
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
1+(1−q)
−t−t0η−t0
β3+
hq (t) = m
β−2η−t0
t−t0η−t0
β−3 1+( 1−qβ−1)
t−t0η−t0
β1+(1−q)
−t−t0η−t0
β2+
+
t−t0η−t0
β−2
β(1−q)η−t0
t−t0η−t0
β−1
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
1+(1−q)
−t−t0η−t0
β3+
hq (t) = m 1
(η−t0)
1+(1−q)
−t−t0η−t0
β2+
t−t0η−t0
β−3
×
(β − 2) + (β − 2)
1−qβ−1
t−t0η−t0
β
+
β (1 − q )
t−t0η−t0
β
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
1+(1−q)
−t−t0η−t0
β
g =
(β − 2) + (β − 2)
1−qβ−1
t−t0η−t0
β+
β (1 − q )
t−t0η−t0
β
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
1+(1−q)
−t−t0η−t0
β
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A.1 Funções de confiabilidade e derivadas 165
g =
(β − 2) + (β − 2)
1−qβ−1
t−t0η−t0
β
1 + (1 − q )
−
t−t0η−t0
β
+
β (1 − q )
t−t0η−t0
β
1β−1
+
1−qβ−1
t−t0η−t0
β + 2
1+(1−q)
−t−t0η−t0
β
g =
(β − 2) + (β − 2)
1−qβ−1
t−t0η−t0
β
+
(β − 2) + (β − 2) 1−qβ−1 t−t0
η−t0β
(1 − q ) − t−t0η−t0
β
+
β (1 − q )
t−t0η−t0
β
1β−1
+
1−qβ−1
t−t0η−t0
β
+ 2
1+(1−q)
−t−t0η−t0
β
g =
(β − 2) + (β − 2)
1−qβ−1
t−t0η−t0
β
− (β − 2)(1 − q )
t−t0η−t0
β
+
− (β − 2)(1 − q )t−t0η−t0
2β
1−qβ−1 + β
1−qβ−1
t−t0η−t0
β
+
β (1 − q )
1−qβ−1
t−t0η−t0
2β + 2β (1 − q )
t−t0η−t0
β
1 + (1 − q )
−
t−t0η−t0
β
g =
(β − 2) + (2β − 2)
1−qβ−1
t−t0η−t0β
+ (β + 2) (1 − q )
t−t0η−t0β
+
2 (1 − q )
1−qβ−1
t−t0η−t0
2β
1 + (1 − q )
−
t−t0η−t0
β
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166 Apêndice A
g =
(β − 2) + 2 (1 − q )t−t0η−t0
β
+ (β + 2) (1 − q )t−t0η−t0
β
+ 2(1 − q )1−qβ−1
t−t0η−t0
2β
1 + (1 − q )
−
t−t0η−t0
β
g =
(β − 2) + (β + 4) (1 − q )
t−t0η−t0
β
+ 2(1 − q )
1−qβ−1
t−t0η−t0
2β
1 + (1 − q )
−
t−t0η−t0
β
g
m
hq (t) = (2−q)β(β−1)
(η−t0)3
t−t0η−t0
β−3
×
(β − 2) + (β + 4) (1 − q )
t−t0η−t0
β
+ 2(1 − q )
1−qβ−1
t−t0η−t0
2β
1+(1−q)
−t−t0η−t0
β3+
q
(1 + (1 − q ) x) = 0 ⇒ 1 − (1 − q )
t
− t0η − t0
β
= 0
(1 − q )
t
− t0η − t0
β
= 1 ⇒
t
− t0η − t0
β
= 1
1 − q
t
− t0η − t0
=
1
1 − q
1β
⇒ t
− t0 = (η − t0)
1
1 − q
1β
t
= t0 + (η − t0) (1 − q )−1β
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A.3 Momentos da distribuição q -Weibull para t0 = 0 167
q t0 = 0
n q > 1
µn =
∞
0
tnf q(t)dt =
∞
t0
tn(2 − q ) β
η − t0
t − t0η − t0
β−1
expq
−
t − t0η − t0
β
dt
τ = t−t0η−t0
t = τ (η − t0) + t0
dt = (η − t0) dτ
µn =
∞ t0
[(η − t0) τ + t0]n (2 − q ) βη−t0
(τ )β−1 expq
− (τ )β
(η − t0) dτ
=∞ t0
n j=0
n j
[(η − t0) τ ] j tn− j
0
(2 − q ) β
η−t0(τ )β−1 expq
− (τ )β
(η − t0) dτ
=n
j=0
n j
tn− j0 (η − t0) j (2 − q )
∞ t0
τ jβ (τ )β−1 expq
− (τ )β
βdτ
=
n j=0
n j
tn− j0 (η − t0)
j
(2 − q )
∞´ t0
β (τ )β+ j−1
expq
− (τ )β
βdτ
τ β = x
dx = βτ β−1dτ
τ = x
1
β
dτ = 1
β
x1β
β−1 dx
dτ = 1
βxβ−1β
dx
µn
µ
n =
n j=0
n
j
t
n− j
0 (η − t0)
j
(2 − q )
∞´ 0 β
x
1ββ+ j−1
expq
−
x
1ββ 1
βxβ−1β dx
=
n j=0
n j
tn− j0 (η − t0) j (2 − q )
∞ 0
xβ+j−1
β expq [−x] x1−ββ dx
=n
j=0
n j
tn− j0 (η − t0) j (2 − q )
∞ 0
xjβ expq [−x] dx
q
expq(−x) = 1
Γ
1q−1
∞
0
u 1q−1
−1 e−u e−(q−1)xu du (q > 1, x > 0).
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168 Apêndice A
n
µn = n
j=0
n j
tn− j0 (η − t0) j (2 − q ) ∞
0
xjβ dx expq [−x]
=
n j=0
n j
tn− j0 (η − t0) j (2 − q )
∞ 0
xjβ dx 1
Γ( 1q−1)
∞ 0
u 1q−1
−1 e−u e−(q−1)xu du
µn =
n j=0
n j
t
n−j
0 (η−t0)
j
(2−q)Γ( 1
q−1)
∞´ 0
du u 1
q−1−1 e−u
∞´ 0
dxxj
β e−(q−1)xu
Γ
Γ(z ) =
∞
0
e−vvz−1dv
v = (q − 1)xu x = v(q−1)u
dx = dv(q−1)u
z − 1 = j/β
z = 1 + j/β
∞ 0
dxxjβ e−(q−1)xu =
∞ 0
dv(q−1)u
v(q−1)u
jβ
e−v
=∞ 0
dv
(q−1)1+
jβ u
1+ jβ
vjβ e−v
= 1
(q−1)1+
jβ u
1+ jβ
∞
´ 0v
jβ e−vdv
= 1
(q−1)1+
jβ u
1+ jβ
Γ
1 + jβ
µn =
n j=0
n j
tn−j0 (η−t0)
j(2−q)
Γ( 1q−1)
∞ 0
du u 1q−1
−1 e−u Γ(1+ jβ )
(q−1)1+
jβ u
1+ jβ
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A.3 Momentos da distribuição q -Weibull para t0 = 0 169
µn =
n j=0
n j
tn−j0 (η−t0)
j(2−q)Γ(1+ jβ )
(q−1)1+
jβ Γ( 1
q−1)
∞ 0
du u 1q−1
−1 e−u 1
u1+
jβ
=n
j=0
n j
t
n−j
0 (η−t0)
j
(2−q)Γ(1+ j
β )(q−1)
1+ jβ Γ( 1
q−1)
∞´ 0
du u 1
q−1−1−1− j
β e−u
=n
j=0
n j
tn−j0 (η−t0)
j(2−q)Γ(1+ jβ )
(q−1)1+
jβ Γ( 1
q−1)
∞ 0
du u 1q−1
−2− jβ e−u
Γ
µn =
n
j=0
n j
tn−j0 (η−t0)
j(2−q)Γ(1+ jβ )
(q−1)
1+ jβ
Γ( 1
q−1)
Γ 1q−1
− 1 − jβ
1q−1
> 0 ⇒ q − 1 > 0 ⇒ q > 1
1q−1
− 1 − jβ
> 0 ⇒ 1q−1
> 1 + jβ
⇒ q − 1 < ββ+ j
⇒ q < 1 + ββ+ j
1 < q < 1 + β
β+ j
j = 1, 2, · · · , n
1 < q < 1 + ββ+n
1q−1
− 1 = 2−qq−1
Γ(m + 1) = mΓ(m)
m + 1 = 1
q−1
Γ
1q−1
=
1q−1
− 1
Γ
1q−1
− 1
=2−qq−1
Γ2−qq−1
µn =
n j=0
n j
tn−j0 (η−t0)
j(2−q)Γ(1+ jβ )
(q−1)1+
jβ ( 2−q
q−1)Γ( 2−qq−1)
Γ
1q−1
− 1 − jβ
µn =
n j=0
n j
tn− j0 (η − t0) j Γ(1+ j
β )
(q−1)jβ Γ( 2−q
q−1)Γ
1q−1
− 1 − jβ
=n
j=0
(nj)tn−j
0 (η−t0)
j
Γ( 2−qq−1)
1
(q−1)
jβ
Γ1 + jβΓ 1
q−1 − 1 − j
β=
n j=0
(nj)tn−j
0 (η−t0)
j
Γ( 2−qq−1)
1
(q−1)jβ
Γ
1 + jβ
Γ2−qq−1
− jβ
µn =
n
j=0n jtn− j
0 (η − t0) j Γ1 + jβ
Γ( 2−qq−1
− jβ )
(q−1)jβ Γ( 2−q
q−1) , 1 < q < 1 + β
β+n
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170 Apêndice A
q→1µn =
n j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
n
q max = 1 + β
j+β β→0q max = 1
β→∞q max = 2 j→∞q max = 1
n
q < 1
q
q < 1
expq(−x) = 12π
Γ2−q1−q
∞
´ −∞
e1+iu
(1+iu)2−q1−q
e−(1−q)(1+iu)xdu,
(q < 1, 1 − (1 − q )x > 0)
n
µn =
n j=0
n
j
tn− j0 (η − t0) j (2 − q )
∞
0
dx xjβ
1
2πΓ
2 − q
1 − q
∞
−∞
e1+iu
(1 + iu)2−q1−q
e−(1−q)(1+iu)xdu
µn =
n j=0
n
j
tn− j0 (η − t0) j (2 − q )
Γ2−q1−q
2π
∞
−∞
du e1+iu
(1 + iu)2−q1−q
∞
0
dx xjβ e−(1−q)(1+iu)x
Γ(z ) =
∞
´ 0 dv v
z−1
e
−v
v = (1 − q )(1 + iu)x
dv = (1 − q )(1 + iu)dx
∞ 0
dx xjβ e−(1−q)(1+iu)x =
∞ 0
dv
1(1−q)(1+iu)
v
(1−q)(1+iu)
jβ
e−v
=
1(1−q)(1+iu)
1+ jβ
∞ 0
dv (v)jβ e−v
= 1
(1−q)1+
j
β (1+iu)1+
j
β
Γ1 + jβ ,
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A.3 Momentos da distribuição q -Weibull para t0 = 0 171
µn =
n
j=0
n j
tn− j0 (η − t0) j (2 − q )
Γ( 2−q1−q )2π
×∞
´ −∞
du e1+iu
(1+iu)2−q1−q
1
(1−q)1+
jβ (1+iu)
1+ jβ
Γ1 + jβ
µn =
n j=0
n
j
tn− j0 (η − t0) j (2 − q )
Γ2−q1−q
2π
∞
−∞
du e1+iu
(1 + iu)2−q1−q
1
(1 + iu)1+ jβ
Γ
1 + jβ
(1 − q )1+
jβ
µn =
n j=0
n
jtn− j0 (η − t0) j Γ2−q
1−q (2−q)(1−q)
Γ(1+ jβ)
(1−q)
j
β
12π
×
∞ −∞
du e1+iu
(1+iu)2−q1−q
1
(1+iu)1+
jβ
µn =
n j=0
n
j
tn− j0 (η − t0) j Γ
2 − q
1 − q
(2 − q )
(1 − q )
Γ
1 + jβ
(1 − q )
jβ
1
2π
∞
−∞
du e1+iu
(1 + iu)2−q1−q+1+ j
β
,
2πe−abbz−1
Γ(z ) =
∞
−∞
du ebui
(a + iu)z
a = 1
z = 2−q
1−q + 1 + j
β b = 1
∞ −∞
du e1+iu
(1+iu)2−q1−q+1+
jβ
= e∞
−∞
du eui
(1+iu)2−q1−q+1+
jβ
= e 2πe−11z−1
Γ( 2−q1−q+1+ j
β )
= 2π
Γ( 2−q1−q+1+ j
β )
n
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172 Apêndice A
µn =
n
j=0
n
jtn− j
0 (η − t0) j Γ2 − q
1 − q
(2 − q )
(1 − q )
Γ
1 + j
β
(1 − q )
j
β
1
2π
2π
Γ2−q1−q + 1 +
jβ
µn =
n j=0
n
j
tn− j0 (η − t0) j (2 − q )
(1 − q )Γ
2 − q
1 − q
Γ
1 + jβ
(1 − q )
jβ
1
Γ2−q1−q
+ 1 + jβ
Γ(z + 1) = z Γ(z )
µn =
n j=0
n
j
tn− j0 (η − t0) j Γ
2 − q
1 − q + 1
Γ
1 + jβ
(1 − q )
jβ
1
Γ2−q1−q
+ 1 + jβ
µn =
n j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
Γ
2−q
1−q + 1
(1 − q )jβ Γ
2−q1−q
+ 1 + jβ
µn =
n j=0
n
j
tn− j0 (η − t0) j Γ
1 +
j
β
Γ3−2q1−q
(1 − q )
jβ Γ
3−2q1−q
+ jβ
n
µn =
n j=0
n j
tn− j0 (η − t0) j Γ
1 + j
β
Γ( 3−2q
1−q )
(1−q)jβ Γ( 3−2q
1−q + jβ )
,
q < 1
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A.3 Momentos da distribuição q -Weibull para t0 = 0 173
µn =
∞
0
(t − µ1)
nf q(t)dt =
∞
t0
(t − µ1)
n(2 − q )
β
η − t0
t − t0η − t0
β−1
expq
−
t − t0η − t0
β
dt
µn =
∞´ t0
[(η − t0) τ + t0 − µ1]n
(2 − q ) βη−t0 (τ )
β−1
expq
− (τ )β
(η − t0) dτ
=n
j=0
n j
(t0 − µ
1)n− j (η − t0) j (2 − q )∞
t0
β (τ )β+ j−1 expq
− (τ )β
βdτ
µn =
n j=0
n j
(t0 − µ1)n− j (η − t0) j Γ
1 + jβ Γ
(
2−q
q−1
− j
β )(q−1)
jβ Γ( 2−q
q−1)
, 1 < q < 1 + ββ+n
µn =n
j=0
n j
(t0 − µ
1)n− j (η − t0) j Γ
1 + j
β
Γ( 3−2q
1−q )
(1−q)jβ Γ( 3−2q
1−q + jβ )
,
q < 1
limq→1
µn =n
j=0
n j
(t0 − µ
1)n− j (η − t0) j Γ
1 + jβ
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174 Apêndice A
q t0 = 0
n q > 1
µn =
∞
0
tnf q(t)dt =
∞
0
tn(2 − q )β
η
t
η
β−1
expq
−
t
η
β
dt
τ = t/η
dt = ηdτ
µn =∞ 0
ηnτ n(2 − q ) βη
(τ )β−1 expq
− (τ )β
ηdτ
= (2 − q )ηn∞ 0
βτ β+n−1 expq
−τ β
dτ
τ β = x
dx = βτ β−1dτ
τ = x
1β
dτ = 1
β
x1β
β−1 dx
dτ = 1
βxβ−1β
dx
µn
µn = ηn(2 − q )∞ 0
β
x1β
β+n−1
expq
−
x1β
β
1
βxβ−1β
dx
= ηn(2 − q )∞ 0
xβ+n−1
β expq [−x] x1−ββ dx
= ηn(2 − q )∞ 0
xnβ expq [−x] dx
q
expq(−x) = 1
Γ
1q−1
∞
0
u 1q−1
−1 e−u e−(q−1)xu du (q > 1, x > 0).
n
µn = ηn(2 − q )∞
´ 0
xnβ dx expq [−x]
= ηn(2 − q )∞ 0
xnβ dx 1
Γ( 1q−1)
∞ 0
u 1q−1
−1 e−u e−(q−1)xu du
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A.4 Momentos da distribuição q -Weibull para t0 = 0 175
µn = ηn(2−q)
Γ( 1
q−1)
∞
´ 0
du u 1
q−1−1 e−u
∞
´ 0
dxxnβ e−(q−1)xu
Γ
Γ(z ) =
∞
0
e−vvz−1dv
v = (q − 1)xu
x = v
(q−1)u
dx = dv(q−1)u
z − 1 = n/β
z = 1 + n/β
∞ 0
dxxnβ e−(q−1)xu =
∞ 0
dv(q−1)u
v(q−1)u
nβ
e−v
=∞ 0
dv
(q−1)1+
nβ u
1+nβ
vnβ e−v
= 1
(q−1)1+
nβ u
1+nβ
∞ 0
vnβ e−vdv
= 1
(q−1)1+
nβ u
1+nβ
Γ1 + nβ
µn = ηn(2−q)
Γ( 1q−1)
∞ 0
du u 1
q−1−1 e−u Γ(1+n
β )(q−1)
1+nβ u
1+nβ
µn = ηn(2−q)Γ(1+n
β )Γ( 1
q−1)(q−1)1+
nβ
∞ 0
du u 1
q−1−1 e−u 1
u1+
nβ
= ηn(2−q)Γ(1+n
β )Γ( 1
q−1)(q−1)1+
nβ
∞ 0
u 1q−1
−2−nβ e−udu
Γ
µn = ηn(2−q)Γ(1+n
β )Γ( 1
q−1)(q−1)1+
nβ
Γ
1q−1
− 1 − nβ
1q−1
> 0 ⇒ q − 1 > 0 ⇒ q > 1
1q−1
− 1 − nβ
> 0 ⇒1
q−1 > 1 + n
β ⇒ q − 1 < β
β+n ⇒ q < 1 + β
β+n
1 < q < 1 + ββ+n
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176 Apêndice A
1q−1
− 1 = 2−qq−1
Γ(m + 1) = mΓ(m)
m + 1 = 1
q−1
Γ
1q−1
=
1q−1
− 1Γ 1q−1
− 1 =2−qq−1Γ2−q
q−1
µn = ηnΓ(1+n
β )( 2−qq−1)Γ( 2−q
q−1)
2−qq−1
1
(q−1)nβ
Γ
1q−1
− 1 − nβ
= ηn
Γ( 2−qq−1)
1
(q−1)nβ
Γ
1 + nβ
Γ2−qq−1
− nβ
µn = ηn
Γ
1 +
n
β Γ( 2−q
q−1−n
β )(q−1)nβ Γ( 2−q
q−1) , 1 < q < 1 +
β
β+n
q→1µn = ηnΓ
1 + nβ
n
q max = 1 + β
n+β β→0q max = 1
β→∞q max = 2
η→∞q max = 1
n
q < 1
q q < 1
expq(−x) = 12π
Γ2−q1−q
∞ −∞
e1+iu
(1+iu)2−q1−q
e−(1−q)(1+iu)xdu,
(q < 1, 1 − (1 − q )x > 0)
n
µn = ηn(2 − q )∞ 0
dx xnβ 12π
Γ2−q1−q
∞ −∞
e1+iu
(1+iu)2−q1−q
e−(1−q)(1+iu)xdu .
µn = ηn(2 − q )Γ( 2−q
1−q)2π
∞ −∞
du e1+iu
(1+iu)2−q1−q
∞ 0
dx xnβ e−(1−q)(1+iu)x.
Γ(z ) =∞ 0
dv vz−1e−v
v = (1 − q )(1 + iu)x
dv = (1 − q )(1 + iu)dx
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A.4 Momentos da distribuição q -Weibull para t0 = 0 177
∞ 0
dx xnβ e−(1−q)(1+iu)x =
∞ 0
dv
1
(1−q)(1+iu) v
(1−q)(1+iu)nβ
e−v
=
1(1−q)(1+iu)
1+
n
β∞ 0
dv (v)nβ e−v
= 1
(1−q)1+
nβ (1+iu)
1+nβ
Γ
1 + nβ
,
µn = ηn(2 − q )Γ( 2−q
1−q )2π
∞ −∞
du e1+iu
(1+iu)2−q1−q
1
(1−q)1+
nβ (1+iu)
1+nβ
Γ
1 + n
β= ηnΓ
2−q1−q
2−q1−q
Γ(1+nβ )(1−q)
nβ
12π
∞ −∞
du e1+iu
(1+iu)2−q1−q+1+
nβ
,
2πe−abbz−1
Γ(z ) =
∞
−∞
du ebui
(a + iu)z,
a = 1
z = 2−q
1−q + 1 + n
β b = 1
∞ −∞
du e1+iu
(1+iu)2−q1−q+1+
nβ
= e∞
−∞
du eui
(1+iu)2−q1−q+1+
nβ
= e 2πe−11z−1
Γ( 2−q1−q+1+n
β )
= 2π
Γ( 2−q1−q
+1+nβ )
,
n
µn = ηnΓ2−q1−q
2−q1−q
Γ(1+n
β )(1−q)
nβ
12π
2π
Γ( 2−q1−q+1+n
β )
= ηn2−q1−q
Γ2−q1−q
Γ(1+n
β )(1−q)
nβ
1
Γ( 2−q1−q+1+n
β ).
Γ(z + 1) = z Γ(z )
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178 Apêndice A
µn = ηnΓ2−q1−q
+ 1 Γ(1+n
β )(1−q)
nβ
1
Γ( 2−q1−q+1+n
β )
= ηnΓ1 + nβ Γ( 2−q
1−q+1)
(1−q)
nβ
Γ(2−q
1−q+1+
n
β )= ηnΓ
1 + n
β
Γ( 3−2q
1−q )(1−q)
nβ Γ( 3−2q
1−q +nβ )
.
n
µn = ηnΓ
1 + nβ
Γ( 3−2q
1−q )(1−q)
nβ Γ( 3−2q
1−q +nβ )
, q < 1
ηq
ηq
F q (ηq) = 1 − e−1
1 −
expq
−
ηq − t0η − t0
β2−q
= 1 − e−1
expq
−
ηq − t0η − t0
β2−q
= e−1
expq −ηq − t0η − t0
β
= e−1 1
2−q = e 1
q−2
−
ηq − t0η − t0
β
= lnq e 1q−2
ηq − t0η − t0
=
− lnq e 1q−2
1β
ηq = t0 + (η − t0)
− lnq e 1q−2 1
β
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A.6 Média ou MTBF 179
µ1 = MTB F =
1 j=0
t1− j0 (η − t0) j Γ
1 + j
β
Γ( 2−q
q−1− j
β )
(q−1)jβ Γ( 2−q
q−1)
, 1 < q < 1 + β
β+n
µ1 = MTBF =
1 j=0
tn− j0 (η − t0) j Γ
1 + j
β
Γ(
3−2q1−q )
(1−q)jβ Γ( 3−2q
1−q + j
β )
,
q < 1
Rq(t) = exp 1
2−q
− (2 − q )
t − t0η − t0
β
, t ≥ t0
Rq(
) = 0.5
exp 12−q
− (2 − q )
− t0η − t0
β = 0.5
− (2 − q )
− t0η − t0
β
= ln 1
2−q0.5
−
− t0η − t0
β
=ln 1
2−q0.5
2 − q
− t0
η − t0
β
=ln 1
2−q0.5
q − 2
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180 Apêndice A
− t0η − t0
=
ln 1
2−q0.5
q − 2
1β
= t0 + (η − t0)
ln 1
2−q0.5
q − 2
1β
limq→1
= t0 + (η − t0)(ln2)
1
β
t
df q(t)dt
= 0
df q(t)
dt
=
d
β(2−q)
η−t0
t−t0η−t0
β−1
expq
−
t−t0η−t0
β
dt
= 0
df q(t)
dt =
β (2 − q )
η − t0
t − t0η − t0
β−1
expq
−
t − t0η − t0
β
= 0
df q(t)
dt =
β (2 − q )
(η − t0)β
(t − t0)β−1 expq
−
t − t0η − t0
β
= 0
df q(t)
dt =
(t − t0)β−1 expq
−
t − t0η − t0
β
= 0,
(β − 1) (t − t0)β−2 expq
−
t − t0η − t0
β
+ (t − t0)β−1 expq
−
t − t0η − t0
β
= 0
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A.8 Moda 181
(β − 1) (t − t0)β−2 expq
−
t−t0η−t0
β
+
(t − t0)β−1 expq
q− t−t0
η−t0β (−β)
(η−t0)β (t − t0)β−1 = 0
(β − 1) (t − t0)β−2 expq
−
t − t0η − t0
β
+ (t − t0)2β−2 expqq
−
t − t0η − t0
β
−β
(η − t0)β = 0
(t − t0)β−2 expq
−
t − t0η − t0
ββ − 1 + (−β )
t − t0η − t0
β
expq−1q
−
t − t0η − t0
β = 0
β − 1 + (−β )
t − t0η − t0
β
expq−1q
−
t − t0η − t0
β
= 0
a =
t−t0η−t0
β
β − 1 + (−β ) a expq−1q [−a] = 0
(−β ) a expq−1q [−a] = 1 − β
expq−1q [−a] =
β − 1
aβ
expq [−a] =
β − 1
aβ
1q−1
, β > 1
[1 + (1 − q ) (−a)]1
1−q
+ =
β − 1
aβ
1q−1
, q = 1
[1 − (1 − q ) a]1
1−q =
β − 1
aβ
1q−1
, q = 1 ∧ (1 − (1 − q ) a) > 0
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182 Apêndice A
1 − (1 − q ) a = aβ
β − 1
1 − (1 − q ) a − aβ
β − 1 = 0
β − 1 − (1 − q ) (β − 1) a
β − 1 −
aβ
β − 1 = 0
β − 1 − (1 − q ) (β − 1) a − aβ
β − 1 = 0
β − 1 − (1 − q ) (β − 1) a − aβ = 0
(1 − q ) (β − 1) a + aβ = β − 1
a [(1 − q ) (β − 1) + β ] = β − 1
a [β − 1 − qβ + q + β ] = β − 1
a [β (2 − q ) + q − 1] = β − 1
a = β − 1
β (2 − q ) + q − 1
a
t − t0η − t0
β
= β − 1
β (2 − q ) + q − 1
t − t0η − t0
=
β − 1
β (2 − q ) + q − 1
1β
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A.9 Árvore de Falha Dinâmica 183
t = t0 + (η − t0)
β − 1
β (2 − q ) + q − 1
1β
= t0 + (η − t0)
β − 1
β (2 − q ) + q − 1
1β
, β > 1
= t0 + (η − t0)
β − 1
β + β − βq + q − 1
1
β
, β > 1
= t0 + (η − t0)
β − 1
β + β (1 − q ) − (1 − q )
1β
, β > 1
= t0 + (η − t0)
β − 1
β + (β − 1)(1 − q )
1β
, β > 1
limq→1
= t0 + (η − t0)
β − 1
β
1
β, β > 1
W SP P 1,S 1(t) = 1 −
1 − W SP S →P P 1,S 1
(t)
1 − W SP P →S P 1,S 1
(t)
.
W SP
P 1,S 1(t) = −
1 − W SP S →P P 1,S 1 (t)
1 − W SP P →S P 1,S 1 (t)
+
−
1 − W SP S →P P 1,S 1
(t)
1 − W SP P →S P 1,S 1
(t)
.
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184 Apêndice A
W SP
P 1,S 1(t) = W S P aP 1,S 1(t) + W S P bP 1,S 1(t),
W S P aP 1,S 1(t) = −
1 − W SP S →P P 1,S 1
(t)
1 − W SP P →S P 1,S 1
(t)
W S P bP 1,S 1(t) = − 1 − W SP S →P P 1,S 1
(t) 1 − W SP P →S P 1,S 1
(t) .
W SP S →P
P 1,S 1(t)
= f q,P (t)F S 1dorα(t).
d t
´ a
g(t) p(t − x1) dx1dt
=
tˆ a
g(t) p(t − x1) dx1 + g(t) p(0)
W SP P →S
P 1,S 1(t)
= K W SP
x1=tˆ
x1=t01
f q,P (x1)f S 1atv(t − x1)dx1 + f q,P (t)F S 1atv(0)
.
W S P aP 1,S 1(t) = f q,P (t)F S 1dorα(t)
1 − W SP P →S P 1,S 1
(t)
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A.10 Tempo esperado por ciclo de manutenção 185
W S P bP 1,S 1(t) =
1 − W SP S →P
P 1,S 1(t)
K W SP ×
× x1=t´
x1=t01f q,P (x1)f S 1atv(t − x1)dx1 + f q,P (t)F S 1atv(0)
,
W SP
P 1,S 1(t) = f q,P (t)F S 1dorα(t)
1 − W SP P →S
P 1,S 1(t)
+
+
1 − W SP S →P P 1,S 1
(t)
×
×K W SP x1=t
´ x1=t01
f q,P (x1)f S 1atv(t − x1)dx1 + f q,P (t)F S 1atv(0) ,
W SP
P 1,S 1(t) = f q,P (t)F S 1dorα(t)×
×
1 − K W SP
x1=t
´ x1=t01
f q,P (x1)F S 1atv(t − x1)dx1
+
+
1 −
x2=t´ x2=t02
f q,P (x2)F S 1dorα(x2)dx2
×
×K W SP
x1=t´
x1=t01
f q,P (x1)f S 1atv(t − x1)dx1 + f q,P (t)F S 1atv(0)
.
T ´ t0
tf q (t)dt
T ´ t0
tf q(t)dt
ˆ udv = uv −
ˆ vdu,
u = t
du = dt
dv = f q(t)dt
v =´
f q(t)dt
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186 Apêndice A
v =´
β(2−q)(η−t0)
t−t0η−t0
β−1
expq −
t−t0η−t0
β
dt
= −(2 − q )´
−β
1η−t0
t−t0η−t0
β−1 expq
−
t−t0η−t0
β
dt,
α = −
t−t0η−t0
β
dα = −β
t−t0η−t0
β−1 1
η−t0
dt
v = −(2 − q )
ˆ expq αdα.
q
ˆ expq xdx =
1
(2 − q )
expq x
2−q+ C,
v = −(2 − q ) 1(2−q) expq α
2−q+ C
= −
expq α2−q + C.
q
expq x
a= exp1− 1−q
a(ax),
v = − exp1− 1−q2−q
[α(2 − q )] + C
= −
exp 1
2−q[α(2 − q )]
+ C
= − exp 1
2−q
−(2 − q )
t−t0η−t0
β
+ C.
´ tf q(t)dt = −t
exp 1
2−q
− (2 − q ) ×
t−t0
η−t0β
+ C
+
´ exp 1
2−q
− (2 − q ) ×
t−t0η−t0
β
+ C
dt,
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A.10 Tempo esperado por ciclo de manutenção 187
T ´ t0
tf q(t)dt =−t exp 1
2−q
− (2 − q ) ×t−t0η−t0
β − tC
t=T
t=t0
+
T ´ t0
exp 1
2−q
− (2 − q ) ×
t−t0η−t0
β
+ C
dt
T ´ t0
tf q(t)dt = −T exp 1
2−q
− (2 − q ) ×
T −t0η−t0
β
− T C +
T ´ t0
exp 1
2−q
− (2 − q ) ×t−t0η−t0
β dt + T C,
T ˆ
t0
tf q(t)dt = −T Rq(T ) +
T ˆ
t0
Rq(t)dt
T ´ t0
Rq (t)dt
T ´ t0
Rq(t)dt
q = 1
IRq(T ) =
t=T
ˆ t=t0
1 + (q − 1) t − t0
η − t0
β
2−q1−q
+
dt
b = q − 1
x =
t−t0η−t0
β
−v = 2−q1−q
dx = β
t−t0η−t0
β−11
η−t0dt
dt = 1β
t−t0η−t0
1−β
(η − t0) dx
IRq(T ) =
t=T ˆ
t=t0
[1 + bx]−v+
1
β
t − t0η − t0
1−β
(η − t0) dx
t−t0η−t0
= x1β
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188 Apêndice A
IRq(T ) =t=T ´
t=t0
[1 + bx]−v+
1β
x1β−1 (η − t0) dx
= (η−t0)
β
t=T ´ t=t0 x
1
β−1
[1 + bx]
−v
+ dx
1β
= a
IRq(T ) = (η − t0) a
t=T ˆ
t=t0
xa−1 [1 + bx]−v+ dx
t = T x =
T −t0η−t0
β
t = t0 x = 0
IRq(T ) = (η − t0) a
T −t0η−t0
βˆ
0
xa−1 [1 + bx]−v+ dx
u =
T −t0η−t0
β
IRq(T ) = (η − t0) a
uˆ
0
xa−1 [1 + bx]−v+ dx
uˆ
0
xa−1 (1 + bx)−v dx = ua
a 2F 1 (v, a ; 1 + a; −bu)
IRq(T ) = (η − t0) a ua
a 2F 1 (v, a ; 1 + a; −bu)
= (η − t0) ua2F 1 (v, a ; 1 + a; −bu)
a = 1
β b = q − 1
v = q−2
1−q u =
T −t0η−t0
β
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A.10 Tempo esperado por ciclo de manutenção 189
IRq(T ) = (η − t0) T −t0η−t0
β
1β
2F 1q−21−q
, 1β
; 1 + 1β
; − (q − 1)T −t0η−t0
β
= (T − t0) 2F 1
q−21−q
, 1β
; 1 + 1β
; (1 − q )
T −t0η−t0
β
.
q = 1
IR(T ) =
T ˆ
t0
R(t)dt =
T ˆ
t0
exp
−
t − t0η − t0
β
dt
x = t−t0
η−t0
dx = 1η−t0
dt dt = (η − t0) dx
IR(T ) =
t=T ˆ
t=t0
exp
−xβ
(η − t0) dx
IR(T ) = (η − t0)
x=T −t0η−t0
ˆ x=0
exp
−xβ
dx
ˆ xm exp(−axn) dx =
−Γ (γ,axn)
naγ , γ =
m + 1
n , a = 0, n = 0
m = 0 a = 1
ˆ exp(−xn) dx = −
Γ1n
, xn
n , n = 0
IR(T ) = (η − t0)
β
−Γ 1
β , xβx=
T −t0η−t0
x=0
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190 Apêndice A
IR(T ) = (η − t0)
β
−Γ
1
β ,
T − t0η − t0
β
+ Γ
1
β , 0
.
IRq(t) =
(T − t0) 2F 1
q−21−q
, 1β
; 1 + 1β
; (1 − q )
T −t0η−t0
β
, q = 1
(η−t0)β
−Γ
1β
,
T −t0η−t0
β
+ Γ
1β
, 0
, q = 1.
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191
q
q
q
q
q
u
u
−u
u
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192 Apêndice B
u = 0, 0063
log(
) > u
log(
) < u
q-Weibull
q-exponencial
Weibull
exponencial
102 103 104103
102
101
100
t min
R q
Weibull
q-Weibull
q-exponencial
exponencial
102 103 104103
102
101
100
t min
R q
q
q
|log(
)| >u
log(
) > u
log(
) < −u
|log(
)| > u
q
q
β 1, 00 0, 58 1, 00 0, 82θ
1525 742 278 318
η 1102 756 279 329t0 −423 14 1 11
q 1, 00 1, 00 1, 41 1, 31R2 0, 7490 0, 9773 0, 9919 0, 9944
q
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B.1 O modelo q -Weibull em mercado de ações 193
log(
) > u
log(
) < −u
q
q
β 1, 00 0, 59 1, 00 0, 74θ
1445 813 319 416
η
1123 826 306 426t0 −322 13 −13 10
q 1, 00 1, 00 1, 40 1, 24R2 0, 8214 0, 9896 0, 9914 0, 9964
