Spiekbrief tentamen 1/2

Reliability Exponential
− λt
Weibull
t
Gamma
β P ( A ∪B ) =P ( A )+ P ( B )
R ( t )=P(T ≥ t) R ( t )=e β t β −1 − ( )f ( t )= t γ−1 ∗α−γ∗e −tα P ( A ∪B ) =P ( A )+ P ( B )−P (
T= time to failure
F ( t )=1−R (t)
F ( t )=1−e−λt
λ ( t )= ( ) , R ( t )=e
θ θ Γ ( y)
θ

P ( A ∪B ) =P ( A )+ P ( B )−P ( A
f ( t )= λe− λt
β

() β −1 − t t P ( A ∩B )=P ( A )∗P ( B )
F ( 0 )=0 β t
f ( t )= ( ) e I( ,γ)
θ
1 α P ( A|B )=P( A ∩ B)/ P ( B )
lim F (t )=1 MTTF= , θ θ F ( t )=
λ Γ (γ )
→∞
d −d 2 1 MTTF=θΓ ( 1+ )
1
t
∫ f ( x ) g ' (x )dx=¿ ¿
f ( t )= F ( t )= R ( t )σ = 2 β f ( x ) g ( x ) −∫ g ( x ) f ( x ) dx
a
dt dt λ t
1I ( α , γ )=∫ y
γ−1 −y
t 2 ∗e dy 2 Binominaal
R(t ¿¿ u)=e− λt =u ¿ u
σ =θ ( Γ ( 1+ ) −( Γ ( 1+ ) ) )
2 2
F ( t )=∫ f ( t ' ) dt ' p ( x )= n p x (1− p)n−x E ( X )=
0 t u=
−1
ln ⁡( R)
β
Τ ( x )=( x−1)Γ ( x−1)
β
met y=
t '
0
x ()
∞ λ α n= n!
R ( t )=∫ f ( t ' ) dt '
t
R ( t|T 0 ) =R ( t )
(memoryless)
t =θ (−ln ( u ) )
u
1
β x x ! ( x−n ) !
Data collective t , t ,…, t represents the time of failure of the i unit. The sample
1 2
represents n ind. Values, thus the joint prob: f(t )*f(t )*…*f(t )
n ()
1 2 n
th




P ( a ≤T ≤b )=F ( b )−F ( Failure
a ) modes 1 1
Single censored data: all units have the same test time, test is concluded before all
n t mode=θ (1− ) β for β>1
units have failed (on the left: failures are known to occur before a certain time, on the

R ( t ) =∏ R i ( t ) β right: failures are known to be only after a certain time).
¿ R ( a ) −R ( b ) i=1 ¿ 0 for β ≤ 1
Type I: testing is terminated after a fixed time (t*)
Type II: testing is terminated after a fixed #failures (t ) r
b n
0< β <1 → DFR (Beta = Multiple censored data: testtimes differ among censored units. Censored units are
¿ ∫ f ( t ) dt λ ( t )=∑ λi (t ) shape par.) removed or gone into service at various times from the sample.
Ungrouped complete data: als n aantal failures zijn in een random sample, dan zijn
a i=1 β=1 →CFR (Theta = het aantal surviving units op tijd ti gelijk aan n-i.
R(t ¿¿ u)=u , R(t ¿¿ med)=0.5 ¿ ¿ Two parameter scale/char. life) Grouped complete data: failure times zijn vervangen door intervallen. De individuele
∞ −d β >1→ IFR
R ( t ) ¿ λe−λ(T−t ) observatie is niet beschikbaar, nk zijn het aantal units dat survived op tijdstip tk.
f ( t )= 0

MTTF=E [ T ] =∫ tf ( t ) dt dt Failure modes UGCensoredD product estimator GCensoredD product estimator
0 (0< t 0 ≤T <∞)
k
β t β −1 n+2−i F i=¿ failures ith interval
R ( t i−1 ) HJ =
∞

∫ R(t )dtdata
¿UGComplete Dynamic models
λ ( t )= ∑
i=1 θi θ i
( ) ^
^
R t
n+1 C i=¿ removals( censors) ith
H i=¿ at risk at time t i−1
0 i i
Periodic loads
Als λ ( t ) zijn identiek: ( i )HJ n+ 1−i
R KM ( t i ) =n− =1− Rn =R n
^ =
n n Random loads nβ t β −1
R ( t i−1 ) HJ n+ 2−i
^ H i=H i−1−F i−1−C i−1

F KM ( t i )=1−R
^ ^ ( t i ) = i −k ∞ n
Constant strength k Constant stress s
−s
∞
λ ( t ) =
n
Random stress & strength
θ n eθ−αt −(1−R ) αt ^
μ
() 1 R ( t ) =
n+ 1−i ^
R ( t ) H
'
i =H i −
Ci
, adjusted ¿ risk
n R ( t ) =∑ R Pn (t )=∑ R ( αtR=
Exponential
Dit klopt niet want n/n=1
R=1−exp ( )
μ x n=0
R=exp
μn=0 ( ) ) −n t =e
x R ( t ) =e μ x + μ y
En n (!θ )
y
=
1+ μ x /μ
β Statici HJ models
n+ 2−i i−1 HJ
X is de stress, Y is de strength (capaciteit). De kans dat de stress niet
If censoring
y rather than failure
^
(assuming that censors occur
uniformly)
x
2
Improved by Johnson
B ThreeB parameter ( Takes place at t : R P ¿ ( X ≤ x )i
=F x F i ∫ f x x dx '
( x ) = ( ')
k s groter is dan x:
i Random

[( )]fixed stress and

[( )]
x y '

F Weibull
^ HJ ( t i) =
R=1−exp strenght − R=exp − t 0 is minimumlife ¿
Met getallen oplossen δ =1 if failure occurs at t i ' : the 0 conditional prob of failure
n+1 θ −αtθ β
H
R t =R+(1−R) e
( ) x y
−(
t −t
) μ −μ 0 De kans dat de capaciteit niet groter is dan
y
y:
i
i n+1−i R (yt )=e θ
= k −μx s−μ in '
R HJ ( t i ) =1−
^ 0 ifP censoring occursat t
Normal

− ^
R (
R=Φ n+1
n+1
t )−
( )
σ
^ ( t i )x
R
R=1−Φ
σ λy( t )=
( ) ( )
β
R=Φ( β−12 2 )
t−t
y

√State-dependent
0 σ +σ
x
x

y
^
R t
Static
( Y ≤ y )=F y ( y )=∫ f yi ' (The
=¿
( i )HJreliability: kans dat de stress nietpgroter 0
y ith ) dyinterval
=1−
' given survival to t
Fi
is dan decond strength. . prob of su
t-1




^f HJ ( t i )= i +1 θ θ ln ⁡(m / m ) ^ R
systems
(
Bij0 ) =1
random
with repair
stress en constant strength:
i
H '
t i+1−t i1 ln k 1 s
th
Identical standby units time to k failure



¿
Lognormal
1
R=Φ
s ( ( ))
x m x
R=1−Φ
s y
ln
m ( ( )) (
y
R=Φ d P21y(t )2=−2
s
√ dtx y + s
x
Kolom
KolomR=
λ P
1:
2: t
k
t
f)
i’s((aantal
) +
( x
∫ x R ( t )Grouped
1 r
failure+censor)
Parallel
P
failures)
dx=F
2
i
(t )
Configuration
times
=1− x (k )
∏
n
R
Complete
+
R(1−R
^ (
k= 1− t ) =e
i data ( t )) '(i=0
−F λt

(
i
k−1

∑
i
∗R
( λt )i
^i!i−1, MTTF=
independent ) )
(t i+1−t i )(n+1) d P2 (t ) 0 S
Hoe standby/ switching failure sommen i Hconfiguration
^ uitwerken:
i switching failures
=2 λ P1 ( t )−(r + λ) P^2 (t )∞ n∞iKolom i=1 Standby Combined system f ( t(ti ) , t ) ni−ni +1
with
^ dt R ( t i )= d, ^λk-out-of-n t1:i )=interval
^λ HJ ( t i) = f ( t ) = 1 ( redundancy = i-1 i

MTTF MTTF= System= ∫ R ∫n ( R
t
Kolom ) P(
dt t )
2:
1 dt
(
for t )
#failures^
=−¿
Rtwo (nt F) ( t i+1−tin−
component )∗n (exp
Ri k (1−R) k i
d P3 (t )
R
^ (t ) ( t i +1−t i ) ( n+ 1−i )
n
t dt
=λ P2 ( t )
∀ t , tϵ∞[ 0 , ∞ ) : R (^t )=P
0
0 dt P( k)=
s
Kolom 3: #censorskC i
d∞ 1 ( t ) + P2 (t )∞+ P3 ( t )
P−λ4:2− ttR 1−
i
()
^ MTTF HJ =∑ i Normal plots
λ+ r + x 1 x P −λ t R (dtt i+1
Kolom N= ( #at
) total ^)= risk H i pe)n−(λ
(opt(icomponents
#components
)tijdstip λ 1−n P 1 ( tλ ) )t−λdt 2
t i ( tλ+r x2t dat ¿ ¿1+
¿) :∫ De+ ekans
^
f dt
( ei ) Kolom
x t +¿ ∫ e 1
K= working dt−¿ 2
∫ i needed i+1
I¿
2


i=1 n t−u P1 ( t )= e = het system zich = t in ' toestand

( )
1 2

F ( t )=Φ =Φ ( z )systems − 0 d0t 5:
−t adjusted n #risks
0 (t H −t )∗n
2
σ
Standby x 1−x 2 bevindt x 1−x 2 i+1P ( t )∑ i=λ P ( ti+1
2 1 ), )if−λ
i i
1 P3(t )
n
( t i− ^ MTTF ) dti=2 R 3=
ofP(k p
2 d P1 ( t ) Alle i’s definiëren: i=1 dan..,Kolom k−1 dan… s (prob
6: n −n survive )
exponential:
i t i +t i+1
s HJ =∑ The inverse function can be written as
=−¿ ∀ t , tϵ [ 0 ,^∞
MTTF= ) : P ( t )∑+ P t´ ( t ) +i
k=x P i
( t
+1
) + ,Pt´ = ( t ) =1
n−1 tP dt = tWeibull 2 λ 2 λ 1 2 i n 3 ^
R i4
ex t − e x tLoad sharing i=0system n n i −λkt 2
Kolom 7: Reliability
i=1
i −μ2 ( t )= i μplots 1 2
−1
z i=Φ [ F (t ) ] =Markov
dσP22-good
1: 1-good,
analysis x 1−
) σln σ
( t2-good
−x 2 1
[ =
x 1−x
]()t β
2
S2=∑
d k−1
P1t(´2ti )=−(λ
n R
( i k=x
Serial s configuratione
−n = ∑ i+ 1 ) k
1+ λ −of2)^ P1 ( t ) 2 1
MTTF
()
(1−e−λt )
Which is linear in t. Plot ( t i , F
2: 1-fail,^ ( t i )= λ 1−F
P ( t
) 1 x12 x 2t 2 x1 x i=0 ) (
−λt ) P θ( t ) dt
R = reliability
n
1
n
component

P3dt
3: 1-good, 2-fail
( t2-fail 1e − ed t R t ∏ ) −λR+¿ i ( t ) ( independ
) =1+ 1 2
Least-squares:

exponentially distributed:
4: 1-fail,
Time of the kth when T is Serial: R ( t ) =P 1 ( t ) 2
d P3 (st ) −¿1P1−F
Parallel: R p =
i


(
ln ln
t )λ=¿
x −x
[ ]
=β x 1−x
( t ) − λ P( t( t)) P ( t ) =e ¿
1 1
ln t−β
2
3
dt
d 2
P
lnθ
1
2 ( t )=λ 1 P1 ( ti=1
−¿¿
S ( ) =

Ri ( t ) =e− λ t (if+¿exponential)
2
P (t ) ¿

i
2




k
dt 1
Plot 2 2 P ( t ) =λ 2 P1 ( t ) −λ
P (t ) ¿ 3


Y K =∑ T i , x , x = [
P1 ( t ) + P2 2( t ) + P3 ( t )
1 2 − ( 3 λ−r ) ± √ 1
λ + 6 λr+ dt r 1]
MTTF=
1 1
, MTTF
1

f y ( t )=
i =1
λ k k−1 −λtP1 ( t ) + P2 ( t ) + P3 1( t ) + P
t e P2 ( t ) =
dR ( t )=1−P λ
lnt i ,λln ln

+ λ
−¿−λ
( ¿ ¿
1−
x
¿
4 ( tF
1
^)=1
2 x t
[
( t i ) werk doen en fandersom
λ
x
+ ¿=aangepaste
2
])
Normal
Als 1 kapot gaat moet 2 meerλ

( t ) =
x t faalintensiteit ¿
distribution



√ 2
1
π σ
e
(IFR)
−1
2
¿¿
s= n

∑
Τ (k ) P1 ( t )=−( 3λ1 + λ 2x) P
Or 1 Weibull 2( t ) =
paper ( e −
t1i (,tF) ( t i ) ) x −x −( λ + λ ) t
^ 1
2
e 1 i=1
Lognormal plots
dt −x ∞ −1
1 1 k
1 2
P1 ( t )=e
1 2 1
12
¿¿
1 λR1 ( t )= ∫ e2 ¿
z=Φ−1 [ F ( t ) ] = ln ( t )− ln ( t med )
s
plot ( ln t i , z i ) lognormal paper or paper
s
∫ t
E [ Y K ]=
n
dt=
t n+1λ
for n≠
d
dt
1
RP ( t2)=P
( t )=λ
^
1 ( t1)P
β=
+lnP(ln
1 t2)(−λ t 1−F [
) +2PP3 2((t(tt))=e )
−λ t
+ ]
λ1 + λ−¿−
P2 ( t )= z=( 2
1
λ
¿Tλ¿−μ
√2 π σ t
1
2
2


n+1 d ln t−lnPθ( t ) λ 1+ λ2−λ +¿ )
P ( t ) =λ
Exponential P ( t
plots ) −λ σ 2 ¿¿¿
( ti , F^ ( ti ) ) dt dt 3
12 1 λ11 3


∫ =ln t MTTF= −ln λ+
− ( λ +[ 1−F )t 1 ]=ln 1 =λt F ( t )=P ( T ≤ t )=Φ( t −μ )
( t ) −¿−λ 2 2