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CHAPTER 9
LM
N
t '2
t '3
dt ' =
9.1 H ( t ) =
333
3(333)
z
t
1
z
10
MTTR = t ⋅ h( t )dt =
1
t
H ( tmed ) =.5 =
9.2
a)
R( t ) = 1 − Φ
z
10
1
OP = t − 1
Q 999
t
3
1
→
LM
N
t3
t4
dt =
333
4(333)
−1
999
3
med
MTTF = t med e s
tmed =
MTTF
tmode =
t med
9.3
H (t ) = Φ
2
1
g
2
/2
→
= 12.249e.7
tmed =
2
/2
5
.7 ( −1.28 )
e
= 15.650 yrs
/2
2
.
hrs
= 1/ 2 = 1213
e
→
MTTF
s2 / 2
2
/2
=
= 7.940 hrs
→
Φ
FG 1 ln 5 IJ =.10
H .7 t K
med
= 12.249 yrs
m
.
med
ie
h
1213
.
= 1 =.446 hrs
e
MTTR = tmed e s
1/ 3
o
F
1
4 I
4 I
F
= ΦG ln
H ( 4) = ΦG ln
J =.883
J
. K
H s t K cH 1213
FG 1 ln t IJ = ΦFG 1 ln 10 IJ =.90
.
t K
H s t K H 121
c
e
104 − 1
= 7.507 hrs
1332
med
→ tmed =
e
med
9.4 tmed =
=
tmed = 999(.5) + 1
b)
es
10
FG 1 ln t IJ = 1 − ΦFG 1 ln 5 IJ =.90
Hs t K
H .7 t K
1
5
ln
= −128
.
.7 tmed
2
OP
Q
b
→
med
es
63 − 1
H ( 6) =
=.215
999
med
10
1.21(1.28 )
e
= 213
. yrs
= 213
. e1.21 / 2 = 4.43 hrs
2
e
10,000
e2
2
/2
T
= 1353353
.
hrs
c
With PM: T = 100 hrs , n = 5 , and t = 550 hours , and R ( t ) = R(T )n R( t − nT )
A
LM
N
m
FG
H
1
100
R (550) = R(100)5 R(50) = 1 − Φ ln
m
.
2 1353353
IJ OP LM1 − ΦFG 1 ln 50 IJ OP
K Q N H 2 1353353
. KQ
5
= 1−.0968 5 1−.0495 =.5735
Without PM: R(550) = 1 − Φ
FG 1 ln 550 IJ = 1−.3264 =.6736
H 2 1353353
. K
Preventative maintenance is not advantageous in this case.
9-1
9.5
d i
a) R( t ) = e − λt e
−
FG t IJ
HθK
β
c
R(5) = e −.008( 5 )
→
FG 5 IJ
e H 10 K = .9606 .7504 =.7208
h
1. 8
−
b
g
b) Considering reliability at the end of maintenance interval such that 5=nT:
d
Rm (5) = e
−
e
−
− .008( 5 )
FG IJ
H K
5 T 1.8
T 10
ie
−n
FG T IJ
H 10 K
1.8
b
g
= .9606 e
−
FG IJ
H K
5 T
T 10
1.8
=.95
.95
=.9890
.9606
=
5 ⋅ T 1.8
= ln.9890
101.8 ⋅ T
T .8 = −
d
101.8
ln.9890
5
c) R( t ) = e
b
− .008( 5 )
g
ie
−n
FG T IJ
H 10 K
→
T = .8 −
101.8
ln.9890 =.0853 yr → n = 5 / T = 58.6 ≈ 59
5
1.8
b1−.005g
n
m
n
The 1−.005 term greatly reduces R(t) for large n so a .95 reliability is not possible.
9.6
a) MTTF =
z
1000
0
z
R( t )dt = 0
1000
z
=
(1−.001t )dt = t − (.001t 2 ) / 2
(1−.001t )dt 95
=
= 950 hrs
b) MTTFM
1 − R(100 )
.1
c) R( 225) = 1−.001( 225) =.775
100
0
.c
1000
0
o
= 500 hrs
ib e
h
g
RM ( 225) = R(100)2 (.99)2 R( 225 − 200) =.92 (.99)2 1−.001( 25) =.774
c
d) RM ( 225) = R(50 ) (.99) R( 225 − 200) =.95 (.99) 1−.001( 25) =.763 No improvement.
4
9.7 MTTR = 2, s =.2
4
4
4
T
I
e
tmed
a) tmed = MTTRe −.5 s = 2e −.5(.2 ) = 196
. hrs and tmod e =
2
b) H ( t ) =.95 = Φ
2
t
FG 1 e
ln
J
H .2 196
. K
c
→
=
196
.
= 188
. hrs
e.04
t = 196
. e.2(1.645 ) = 2.72 hrs
A
. hr ) = Φ
c) P( H < 100 min ) = P( H < 167
9.8
es
2
. I
FG 1 ln 167
J = Φ( −.80) =.212
H .2 196
. K
λ ( t ) = at b =.0004521t .8 → a =.0004521 and b =.8
. and θ = 1.8 β / a = 1.8 18
. /.0004521 = 100
β = b + 1 = 18
FG 20 IJ
R( 20 ) = e H 100 K =.9463
1. 8
A:
FG 1 IJ
R( 20) = e H 100 K =.995
1. 8
−20
−
B:
C: Rsys ( 20) = 1 − (1 − R( 20 )) = 1 − (1−.9463)2 =.9971
2
An active redundant system results in the greatest reliability.
9-2
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z
10
10
t
t
9.9 R( t ) = ( 2−.02t ' )dt ' = .2t '−.01t '2
= ( 2 − 1) − (.2t −.01t 2 ) =.01t 2 −.2t + 1
2
. ) = R(.5)2 R(125
. − 2(.5)) = .01(.5)2 −.2(.5) + 1 .01(.25)2 −.2(.25) + 1 =.7758
a) Rm (125
b)
z
MTTF =
z
LM.01t
3
=N
R( t )dt
(.01t 2 −.2t + 1)dt
0
=
1 − R(T ) 1 − .01(.5)2 −.2(.5) + 1
T
0
.5
3
−.1t
2
O
+ tP
Q
.5
0
.0975
= 4.871 yrs
9.10 (a)
MTBFprop = 1000 Γ(1+1/1.7) = 892.4
MTBFstruc = 2000 Γ(1+1/2.1) = 1771.5
MTBFelec = 870 Γ(1+1/1.8) = 773.3
5
MTTRSYS =
∑ q f MTTR
i i
i =1
i
5
∑q f
i =1
=
i i
m
50000
50000
50000
(6.8) + .003(50000)(3.2) +
(5.2) +
(2) + .001(50000)(4.8)
1377
892.4
1771.5
773.3
=
= 3.95
50000
50000 50000
348.9
+ .003(50000) +
+
+ .001(50000)
892.4
1771.5 773.3
(b) For power law intensity # failures= f =
FG 50000IJ
H θ K
.c
β
o
ie
h
.
Therefore fprop = (50000/1000)1.7 = 773, fstruc = (50000/2000)2.1 = 862, and fele = (50000/870)1.8
= 1469.
MTTRsys = [ 773 (6.8) + 150 (3.2) + 862 (5.2) +1469 (2) +50 (4.8)]/[773+150+862+1469+50]
c
e
=
13396.8 / 3304 = 4.05
MTTR is not very sensitive to the maintenance concept but the number of failures is.
9.11
e
T
Replace (renewal process): R (300) = 1 − Φ
c
FG 1 ln t IJ = 1 − ΦFG 1 ln 300 IJ =.9312
H .9 1150K
Hs t K
med
-8
Repair: let a = .4 x 10 and b = 2.8. Expected number of failures in the interval (400,700) is
A
(.4 x 10−8 )(2.8) 2.8 700
⎡⎣t ⎤⎦ = .2928757
(.4 x 10 )(2.8) ∫ t dt =
400
400
2.8
−8
700
1.8
Therefore Pr{N(700) - N(400) = 0} = e-.2928757 = .7461
Conclusion: Replace the part.
9.12
FG1 − t IJ dt LMt − t OP
R( t )dt
H b K = N 2b Q
MTTF =
=
T
1 − R(T )
F TI
1 − G1 − J
H bK
b
z
T
0
z
T
0
2
T
0
9-3
=
LM
N
OP
Q
b
T2
T
= b−
T−
2b
2
T
9.13
zc
z
LM− 2 e + 1 e OP
hdt = N λ 2λ Q
h 1 − 2e + e
− λt
2e− λt − e −2 λt
R( t )dt
=
MTTF =
1 − R(T ) 1 − 2e − λT − e−2 λT
T
0
=
9.14
−
c
c
hc
− λT
h
c
1
1
1 − e− λT −3 + e − λT
−
−3 + e − λ T
2λ
= 2λ
− λT 2
1 − e − λT
1− e
c
c
h
h
T
−2 λt
h
−2 λT
0
=
−
c
h
1
4e− λT − e−2 λT − 3
2λ
2
1 − e− λT
c
h
3
λ
2
as T → 0, MTTFM → ∞
as T → ∞, MTTFM →
dP1 ( t )
dP2 ( t )
= − λP1 ( t ) + rP2 ( t )
= λP1 ( t ) + ( λ + r ) P2 ( t )
and
dt
dt
sz1 − 1 = − λz1 + rz2
and
sz2 = λz1 − ( λ + r ) z2
1
−r
0 λ +r+s
λ +r+s
λ +r+s
=
= 2
z1 =
−r
s+λ
( s + λ )( λ + r + s ) − λr s + ( 2λ + r )sλ2
−λ λ + r + s
m
.c
o
( 2λ + r )2 − 4λ2
λ + r + x1
λ + r + x2
− ( 2λ + r )
x1 , x2 =
±
; A=
; B=
2
2
x1 − x2
x2 − x1
λ +r+s
λ + r + x1 x1t λ + r + x2 x2t
A
B
z1 =
=
+
→ P1 ( t ) =
e −
e
x1 − x2
x1 − x2
( s − x1 )( s − x2 ) s − x1 s − x2
ie
h
s+λ 1
−λ 0
λ
λ
A
B
z2 = 2
=
=
+
→ P2 ( t ) =
e x1t − e x2t
2
s + ( sλ + r ) + λ
x1 − x2
( s − x1 )( s − x2 ) s − x1 s − x2
1
R( t ) = P1 ( t ) + P2 ( t ) =
( 2λ + r + x1 )e x1t − ( 2λ + r + x2 )e x2t
x1 − x2
z e
c
A
∞
MTTF = R (t )dt =
0
9.15 H ( t ) =
t−a
b−a
MTTR =
z
b
a
c
e
TL
−1 2λ + r + x1 2λ + r + x2
−
x1 − x2
x1
x2
LM
N
MN
t
t2
dt =
b−a
2( b − a )
OP
Q
b
=
a
OP
Q
b2 − a 2 ( b + a )( b − a ) ( b + a )
=
=
2( b − a )
2( b − a )
2
9-4
c
h
9.16 R( t ) =
b−t
b
0 ≤ t ≤ b; TOV = random variable, time between overhauls
F b − t IJ dt = LM −(b − t ) OP
=z G
H b K N 2b Q
2
T0
=
T0
0
E TOV
0
−( b − T0 )2 b2
T2
+
= T0 − 0
2b
2b
2b
As T0 → b, E T0V → b − b / 2 = b / 2
9.17 f ( t ) =
1
100
0 ≤ t ≤ 100; F ( t ) =
Rm ( t ) = (.95) N
100 − ( t − 5 N )
100
t
100 − t
; R( t ) =
;
100
100
T = 5 days
5N ≤ t ≤ 5( N + 1) → Rm (17) =.8402275 ( N = 3)
LM
N
OP
Q
5
−(100 − t )2
952
1002
100 − t
−
+
dt
2(100) 0
2(100) 2(100)
100
MTTFM =
=
= 97.5 hrs
=
.05
1 − R(5)
1−.95
Without maintenance: R (17) =.83 and MTTF = 50 hrs
z
5
0
9.18 Repair initiated only after 3 generators have failed:
a) MTTR = 10 hrs
b) ( n − k + 1) MTTR = ( 6 − 4 + 1) MTTR = (3)10 = 30 hrs
MTTR 10
=
= 3333
.
hrs
c)
n − k +1 3
3
1
1 1
d) MTTR
= MTTR 1 + +
= 10(1833
. ) = 18.33 hrs
2 3
i =1 i
FG
H
∑
IJ
K
.c
m
o
ie
h
9.19 λ 1 =.001, λ 2 =.003, r =.01 , K1 = λ 1 + λ 2 + r =.014, K 2 = λ 1 ⋅ λ 2 = 3 × 10−6
c
− K1 ± K12 − 4 K 2
= −.000218,−.01378
2
( K1 + x1 )e x1t − ( K1 + x2 )e x2t
R( t ) =
=.90
x1 − x2
Solving for t by trial and error results in td = 558 hrs
x1 , x2 =
e
T
e
c
9.20 K1 = λ 1 + λ 2 + r = 1 + 2 + 2 = 5; K 2 = λ 1 ⋅ λ 2 = 2; r = 2
A
− K1 ± K12 − 4 K 2 −5 ± 25 − 8
=
= −.4384, − 4.561
2
2
(5−.4384)e −.4384 t − (5 − 4.561)e −4.561t
R( t ) =
= 11065
.
e −.4384 t −.1065e−4.561t
−.4384 + 4.561
R(1) =.7125
x1 , x2 =
9-5
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9.21 r =.10; λ =.023; eqn 9.18
−(3(.023) + .10) ± .0232 + 6(.023).10 + .102
= −.006511, −.162489
2
( −.006511)e−.162489 t − ( −.162489 )e−.006511t
= −.0417e−.162489 t + 10417
R( t ) =
.
e −.006511t
−.006511+.162489
R( 72) =.65185
x1 , x2 =
MTTFsys = (1.5 + .5 [43.48/10]) (43.48) = 159.7 hr
9.22 r = 0; λ = λ
−(3λ + 0 ) ± λ2 + 6λ ( 0) + 02 −3λ ± λ2
x1 , x2 =
=
= − λ , −2 λ
2
2
( x )e x2t − ( x2 )e x1t
− λ −2 λt −2λ − λt
R( t ) = 1
=−
e +
e = − e−2 λt + 2e − λt [eqn. 3.24]
x1 − x2
−λ
−λ
9.23 NHPP:
(a) m(0,1000) =
z
1000
0
e
−7 .5 + .003t
e −7.5+ .003t
(b) m(0,100) =
.003
OP
Q
e −7.5+ .003t
dt =
.003
OP
Q
1000
= 352
.
.c
0
100
m
o
= .0645 ; R(100) = e-.0645 = .9375
ie
h
0
(c) Since ρ(t) is increasing, the automobile is deteriorating.
c
9.24 The CDF for the cycle time is G(t) = 1 −
e
T
e − .1t −.1e − t
.
1−.1
e −.1 −.1e −1
Therefore Pr{T<1} = G(1) = 1 −
=.0355
.9
e
9.25 For this distribution, MTTF = 3.026 and σ = 3.435.
c
A
Pr{T35 > 130} ≈ 1 − Φ
FG 130 − 35(3.026) IJ = 1 − Φ(119
. ) =.117
H 3.435 35 K
9.26 (a) R(750) = exp[-(750/500)2.1] = .096; Rm(750) = R(75)10 = .98110 = .825
(b) MTTR = 5 exp[.72 / 2] = 6.388 hr; H(8) = Φ[(1/.7) ln(8/5)] = Φ(.67) = 74.9 %
(c) ) (i) m(750) = 8.4/2.1 x 10-6 (750)2.1 = 4.362; (ii) m(675,750) = .866 and 1- exp[-.866] = .579
(iii) ρ(750) = 8.4 x 10-6 x 7501.1 = .0122 ; MTBFi = 1/.0122 = 81.875
9-6
9.27 (a) m(100) = 100/20.076 + .00025(100)2.4 = 20.75
(b) mB (120) = .00025(120) 2.4 = 24.433; MTTRB = 4e.85
2
/2
= 5.74
⎛ 120 ⎞
⎜
⎟ (2) + 24.433(5.74)
20.76 ⎠
⎝
MTTR =
= 5.005days
30.41
1.7
(c) mB (10) = .00025(10)
2.4
= .0628; Rs (10) = e
⎛ 10 ⎞
−⎜
⎟
⎝ 22.5 ⎠
⋅ e −.0628 = .73
⎡ −⎛⎜ 5 ⎞⎟
= .01189; RM (10) = ⎢e ⎝ 22.5 ⎠
⎢⎣
1.7
(d ) mB (5) = .00025(5) 2.4
2
⎤
2
⎥ ⋅ ⎡⎣e −.01189 ⎤⎦ = .8362
⎥⎦
2⎞
⎛ 2−2⎞ ⎛ 1
(e) P {T ≤ 2} = Φ ⎜
ln ⎟ = .10375
⎟Φ⎜
⎝ .5 ⎠ ⎝ .85 4 ⎠
9.28 a. find the .98 design life: td = 500,000 (-ln .98)1/1.841 = 60,048 miles
b. R(100,000) = .94965; Rm(100,000) = R(60048) R(100,000-60048) = .9707
R(150,000) = .89674; Rm(150,000) = R(60048)2 R(150,000 - 2 x 60048) = .9551
.c
m
o
6.25 ⎞
⎛ 1
ln
tmed = 5 / exp (.6882 / 2 ) = 4; H (6.25) = Φ ⎜
⎟ = .7479
4 ⎠
⎝ .668
1
t⎞
t
⎛ 1
ln ⎟ = .95; z = 1.645 =
ln ; t = 12 hrs
(b) H (t ) = Φ ⎜
.668 4
⎝ .668 4 ⎠
9.29 (a)
ie
h
12 work hrs or noon the next day
c
9.30 (a) m(500,750) = .0001[ 7501.7635 – 5001.7635 ] = 6; (b) 1 – p(0) – p(1) – p(2) = 1 – e-6 – 6e-6 –
36e-6/2 = .938
e
T
9.31 R(t) = exp[-(t/5000)1.45] ; R(30) = .9994 ; Rm(1825) = .999460 R(25) = .9642
e
9.32
c
d 2 ρ (t )
= ab(b − 1)(b − 2)t b −3
power law:
2
dt
2
d ρ (t )
= b 2 e a +bt
log-linear:
dt 2
d 2 ρ (t )
= −ab 2 e − bt
bounded:
2
dt
A
For the power law process, the intensity function is convex for b > 2 and concave for 1 < b < 2.
For 0 < b <1, the function is decreasing (concave). For the log-linear process, the intensity
function is convex and the bounded intensity function is concave. The rate of increase in failures
of a system would be a consideration in the selection of an appropriate minimal repair model.
9-7
9.33 Graphing the expected number of failures as a function of time for both machines indicates
that the cross-over point occurs around 11 years. E[N(11)] = 20.3 for both machines. If the
machine is to be used less than 11 years, then machine B is preferred; otherwise machine A is
preferred.
⎧λ t
t
∫
⎩λt + a (t − c) for t > c
Then E[N(5,000)] = .001 (5000) + .0001 (5000 – 1000)1.4 = 5 + 11.0 = 16.0
b
0
.c
ie
h
e
c
e
T
m
for t ≤ c
9.34 Since E[ N (t )] = ρ (t ') dt ' = ⎨
c
A
9-8
o
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