Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
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Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
CHAPTER 2
2.1
1 =.91 (.001)(100) + 1
a) R(100 ) = b) λ ( t ) =
R(1000 ) =
and
1 =.5 (.001)(1000 ) + 1
1 − dR( t ) 1 − d ((.001t + 1)−1 ) ⋅ = ⋅ =.001(.001t + 1)−2 ⋅ (.001t + 1) (.001t + 1)−1 dt R( t ) dt
.001(.001t + 1) .001 = 2 .001t + 1 (.001t + 1) λ ( t ) is decreasing because λ ( t ) goes to zero as t goes to infinity. =
2.2 a) R( t ) = e
z
− 0t λ ( t ')dt '
=e
z
−.4 0t t 'dt '
0
F (1 / 12) = 1 − R(1 / 12) = 1 − e b) R( t ) = e −.2 t =.95
→
2
2.3 a) R( t ) = b) λ ( t ) =
z
z
100
t
− .2(1/12 )2
−.2t 2 = ln(.95)
− ln(.95) =.506 yrs .2
t=
0 ≤ t ≤ 100
t
z
100
0
z
100
0
0 ≤ t ≤ 100
2 100 2 (1−.01t )dt = t 100 0 −.005t 0 = 100−.005(100 ) = 50 days
z
t f ( t )dt − ( MTTF )2 =.01 0 t 2 dt − 502 =.033 t 3
100 2 0
100
σ = σ 2 = 28.9 days e) R( tmedian ) = 1−.01tmedian =.5 2.4
=
→
100
.01 f (t ) − dR( t ) 1 ⋅ = = dt R( t ) R( t ) 1−.01t
z
a) R( t ) =
=.00139
f ( t ′ )dt ′ = .01dt ′ =.01t ′ 100 =.01(100 − t ) = 1−.01t t
c) MTTF = R( t )dt = d) σ 2 =
2 2 = e −.2t ' |t = e −.2t
z
1000 t
.01tmedian =.5
z t ′ dt ′ = LMN101 t' OPQ 1000 t
h
2
1000
3
0
c) R( t ) = 1 −
t3 =.99 109
3 109
tmedian = 50 days
9
z
→
1000
0
t 3dt =
t
0 ≤ t ≤ 1000 hrs
F (100) = 1 − R(100) = 1 − (1 − 1003 / 109 ) =
z
→
− 502 = 8333 . (days )2
3
1 t 10003 − t 3 = 1 − 9 9 10 10
b) MTTF = t ⋅ f ( t ) dt =
0
100
3 f ( t ′ )dt ′ = 9 10
c
→
100
106 1 = 3 9 10 10
3 10 ⋅ 4 9
t3 =.01 109
t4
1000 0
→
2.5 2-1
=
3 10 ⋅ 4 9
10004 − 0 = 750 hrs
t = 3 107 = 215.443 hrs
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
a) R(50) = e-
.001(50)
=.8
1/ 2 1 1 1 − d ( e − (.001t ) ) .0005 b) λ ( t ) = ⋅ − (.001t )1/ 2 == (.001t )−1/ 2 (.001)e − (.001t ) ⋅ − (.001t )1/ 2 = 2 dt .001t e e λ ( t ) is decreasing because λ ( t ) goes to zero as t goes to infinity. 1/ 2
R(T0 + t ) R(T0 )
c) R( t / T0 ) =
→
R(50 + 10) R( 60) e − = = R(10) R(10) e−
R(50 / 10) =
.001( 60 ) .001(10)
=.865
R(t + 10) e − .001( t +10) = − .001(10) =.95 d) R (t / 10) = R(10) e
[ ln .859596] t=
2
e
− .001( t +10)
2.6
z
a) R( t ) =
= .95e
z
10
t
− .001(10)
= .859596
→
10
10
t
t
f ( t ′ )dt ′ = (.2−.02t ′ )dt ′ =.2 t ′
− 10 = 12.9 hrs
.001
−.01 t ′ 2
10 t
= ( 2−.2t ) − (1−.01t ) = 1−.2t +.01t 0 ≤ t ≤ 10 yrs 2
2
f (t ) .2 −.02t .2(1−.1t ) .2 = = = 2 2 R(t ) 1−.2t +.01t (1−.1t ) 1−.1t λ ( 0) =.2 and λ ( t → 10) = ∞ so the hazard rate is, in fact, increasing.
λ (t ) =
b) MTTF =
z
10
0
R(t ) =
z
10
(1−.2t +.01t 2 )dt = t
0
2 c) R( tmedian ) = 1−.2tmed +.01tmed =.5
→
10 0
10
−.1 t 2
0
+.00333 t 3
10 0
= 10 − 10 + 3.33 = 3.33 yrs
2 .01tmed −.2tmed +.5 = 0
.2 ± ( −.2 )2 − 4(.01)(.5) .2±.1414 = = 17.07,2.93 2(.01) 2(.01) t( median ) is 2.93 years. (17.07 is outside range of values for t.) t( median ) =
d) f ( t mod e ) = MAX [ f ( t )] 0≤ t ≤10
f(t) is linearly decreasing: f(0) = .2 and f(10)=0. Therefore, the mode occurs at t=0 yr.
e) σ = 2
z
10
0
t f (t )dt − MTTF = 2
2
z
10
0
t (.2 −.02t )dt − 3.33 2
= (66.67 − 50) − 0 − 1109 . = 558 .
2
L.2t =M N3
3
.02t 4 − 4
PQO
10
− 3.332 0
σ = σ = 558 . = 2.36 yrs
→
2
2.7 a) R(1) =
100 100 = =.826 2 121 (1 + 10)
L 200(t ′ + 10) R( t ) = f ( t ′ )dt ′ = 200 ( t ′ + 10) dt ′ = M N −2 L (t + 10) b) MTTF = R( t )dt = 100 ( t + 10) dt = 100M N −1
z
z
∞
t
∞
−3
t
z
∞
0
z
∞
−2
0
2-2
OP = 0 − FG −100 IJ = 100 H (t + 10) K (t + 10) Q OP = L −100 O = 0 − ( −10) = 10 yrs Q MN t + 10 PQ
−2 ∞
2
t
−1 ∞
∞
0
0
2
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
100 =.95 ( t + 10)2
100 100 → t= − 10 =.26 yrs .95 .95 d) The failure rate is DFR because λ ( 0) =.2 and λ ( t → ∞ ) = 0 .
c) R( t ) =
→
t + 10 =
200 2 f ( t ) ( t + 10 )3 200( t + 10 )2 = = = λ (t ) = 3 100 t + 10 R( t ) 100( t + 10 ) ( t + 10)2 e) A one year burn-in period will improve the reliability because the failure rate is decreasing. 100 R(1 + 1) R( 2) ( 2 + 10)2 112 = 2 =.84 R(1 / 1) = = = 100 R(1) R(1) 12 (1 + 10)2 2.8 F ( t ) =
z
t 0
LM OP N Q
z
1 1 f ( t ′ )dt ′ = dt ′ = t ′ b b t 0
R( t ) = 1 − F ( t ) = 1 −
t b−t = b b
t
0
t = , b
1 f (t ) b 1 = b = = , λ(t ) = − b t R( t ) b( b − t ) b − t b
z
b
MTTF = R( t )dt = 0
R( tmedian ) =
z
b
0
LM N
b−t t2 1 dt = bt − b b 2
b − tmed 1 = b b 2
0≤t ≤b
IFR
OP Q
b
= 0
1 b
LMF b MNGH
2
−
I JK
OP PQ
b2 2b 2 − b 2 b −0 = = 2 2b 2
There is no mode because all failure times are equally likely.
z
σ = t f ( t )dt − MTTF = 2
b 2 0
2
z
b 0
FG IJ = LM t OP − b = b − b = b → σ = H K N 3b Q 4 3 4 12
b t2 dt − b 2
2
3
b
2
2
2
2
0
2.9
t t 20 − t 1 = ; R( t ) = ; λ(t ) = b 20 20 20 − t 20 20 20 MTTF = = 10 yrs; tmedian = = 10 yrs; σ = = 5.77 yrs 2 2 12
F (t ) =
2.10 a) Wear-out is indicated by an increasing failure rate. − dR( t ) 1 − d (1 − t / t0 )2 1 1 ⋅ = ⋅ = [−2(1 − t / t0 )( −1 / t0 )] ⋅ λ (t ) = 2 dt R( t ) dt (1 − t / t0 ) (1 − t / t0 )2 2 2 2 Since λ ( 0) = and λ ( t → t0 ) = ∞, λ ( t ) is IFR. = = t0 (1 − t / t0 ) t0 − t t0
2-3
b 12
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
b) MTTF =
=
z
t0
0
− t0 3
F t I LF t I F −t IJ OP R( t )dt = G1 − J dt = MG1 − J G H t K MNH t K H 3 K PQ O LMF t I GMH1 − t JK − (1 − 0) PP = t3 Q N
z
2
t0
3
t0
0
0
0
0
0
3
3
0
0
0
c) R( t ) = (1 − t / 2000 )2 =.90
→
t = 2000(1 − .9 ) = 102.63 hrs
2.11
R(T0 + t ) ( t + T0 + 1)−3/ 2 = R(T0 ) (T0 + 1)−3/ 2
R( t / T0 ) =
where t = 2, T0 =.5
R( 2.5) (35 . )−3/ 2 = =.28 R(.5) (15 . )−3/ 2 .28-.19 An improvement of = 47% is obtained with a 6 month burn - in. .19 R( 2/.5) =
2.12
R( t ) = e z
=e z
t
− λ ( t ′ ) dt ′
t
− at ′dt ′
0
0
=e
− at 2 / 2
dR( t ) d (e =− dt dt
f (t ) = −
)
− a t ′2 / 2
− e − at
t 0
2
/2
= e − at
2
/2
a > 0 and t ≥ 0
( − at ) = ate − at
2
/2
R( t median ) = e − ( atmed )/ 2 =.5 2
2 at med = ln(.5) 2 −2 ln(.5) 118 . t median = = a a The mode is found by setting the first derivative of f(t) equal to 0 and solving for t. 2 2 2 2 df ( t ) d ( e − ( atmed )/ 2 ) = = a[te − at / 2 ( − at ) + e − at / 2 ] = e − at / 2 [− at 2 + 1] = 0 dt dt 2 [− at + 1] = 0
−
t mod e = 1 / a
2.13 AFR = = 2.14
ln R( t1 ) − ln R( t2 ) ln(1 − t13 / 109 ) − ln(1 − t23 / 109 ) = where t1 = 0 and t 2 = 500 t2 − t1 t2 − t1 ln(1) − ln(1 − 5003 / 109 ) 0 − ln(.875) = =.000267 failures / hr 500 − 0 500
z
R( t ) =.1
∞
t
(1+.05t ′ )
−3
L (1+.05t ′ ) OP dt ′ =.1M N −2(.05) Q
−2 ∞
d
i
= 0 − −(1+.05t )−2 = (1+.05t )−2
t
2-4
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
R(10 + 1) (1+.05(11))−2 = =.459 R(1) (1+.05(1))−2
R(10 / 1) =
z z
∞
MTTFbefore = R( t )dt = 0
z
∞
0
(1+.05t )−2 dt =
∞
1 = R(t / T0 )dt = T0 R (T0 )
MTTFafter
= (105 . )
2
z
∞
1
(1+.05t )
−2
z
∞
T0
LM (1+.05t ) OP N −.05 Q
−1 ∞
R (t )dt =
FG H
= 0− −
0
1 (1+.05(1)) −2
L (1+.05t ) OP dt = (105 . ) M N −(.05) Q −1
IJ K
1 = 20 .05
z
∞
1
(1+.05t ) −2 dt
∞
LM N
= 0− −
2
1
OP Q
(105 . )2 = 21 .05(1+.05(1))
2.15 λ ( t ) is decreasing. R(t) is decreasing as R(0) = 1 and R( ∞ ) = 0. f(t) = λ (t)R(t). Since λ ( t ) and R( t ) are decreasing, f( t ) is decreasing and the mode must occur at t = 0. 2.16
R( t ) = e z
t
− λ ( t ′ ) dt ′ 0
=e
−a
z
t
0
et ′ dt ′
=e
− a et ′
t
=e
0
− a et −1
=e
a 1− et
f ( t ) = λ ( t ) R( t ) = ( ae t )( e a − ae ) = ( ae t )( e a )( e − ae ) = ae( t + a −ae ) t
2.17 Using (a) R (t ) = (b) R (t ) =
n ∫ ( x + c) dx =
∫
∞
t
t
( x + c) n +1 , n ≠1 ; n +1
a ( t '+ a ) dt ' = 2 −1 ( t '+ a ) a
−1 ∞
= t
t
dx
∫ x + c = ln ( x + c )
a 10 ; R(8) = .55 ; R (t ) = t+a t + 10
10 10 = .95; t = − 10 = .526 yr = 6.3 mo. t + 10 .95
(c) R(5|5) = .75 R(10) / R(5) = (10/20) / (10/15) = 15/20 = .75 (d) F(90 days) = 1 - 10 / (10 + .2465) = .024
a
(t + a ) (e) λ (t ) =
2
a (t + a )
=
1 t+a
(f) R(t) = .5 = a / (t + a) or tmed = (a / .5) – a = 2a –a = a ∞
(g) MTTF =
a
∫ t + a dt = a ln(t + a)
∞ 0
→ ∞ ; does not exist
0
2-5
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
2.18 (a ) f (t ) =
ka k
(t + a )
(c) MTTF = ∫
k +1
(t + a )
k → DFR t+a − k +1 ∞
ak
∞
0
; (b) λ (t ) =
k
ak (t + a ) dt = −k + 1
= 0
− a k a − k +1 a = −k + 1 k −1
⎛ 1 ⎞ (d )tmed = a ⎜⎜ k − 1⎟⎟ ⎝ .5 ⎠
2.19
dR(t ) 3t 2 (a ) f (t ) = = 3 , 0≤t ≤k dt k 2 3t ⎛ t 3 ⎞ 3t 2 (b) λ (t ) = 3 / ⎜1 − 3 ⎟ = 3 3 k ⎝ k ⎠ k −t k
(c) MTTF = ∫
k
0
(d ) 1 −
3t 3 3t 4 3 dt = 3 = k 3 k 4k 0 4
t3 = .5 or tmed = k 3 .5 = .7937 k k3
2.20 rewrite f(t) = .003(10-t)2 (a) R(t) = .001(10 – t)3 ; R(1) = .729 (b) .001(10-t)3 = .90; solving for t = 10 – (.9/.001)1/3 = .345 yr. (c) λ(t) = f(t) /R(t) = 3/(10-t); IFR 10
(d) MTTF = R (t ) dt = .001(10 − t ) dt = 2.5 yr. ; R (2.5) = .422
∫
∫
3
0
(e) F(1/12) = .02479; R(11/12 | 1/12) = .729/.9752 = .7475
⎛ 2t t 2 ⎞ t2 t3 2.21 (a) MTTF = ∫ ⎜1 − + 2 ⎟ dt = t − + 2 a a ⎠ a 3a 0⎝ ⎛ 2t t 2 ⎞ 2 2 ⎜1 − + 2 ⎟ = .5; t − 2at + .5a = 0 a a ⎠ ⎝ (b) a
t=
a
= a/3 0
a 2 2a ± 4 a 2 − 2a 2 =a± = a (1 ± .7071) = .293a 2 2
Why not the positive root? (c) R(a/3) = 4/9 = .444 (d) λ (t ) =
2(a − t ) 2 / a − 2t / a 2 2 = = ; IFR 2 2 2 1 − 2t / a − 2t / a (a − t ) (a − t )
(e) R(1) = .81; R(1|1) = R(2) /R(1) = .64 / .81 = .79 2.22
(a) R(t) = 1 - .000064t3 ; R(15) = .784; (b) F(10) – F(2) = .000064(1000 - 8) = .063488 2-6
Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009
.000064t 3 = .5 (c)
tmed = 3 .5 / .000064 = 19.8425 yr.
(d) λ(t) = .000192t2 / (1 - .000064t3) ; IFR 25
(e) MTTF =
∫ (1 − .000064t
3
)dt = [t − .000016t 4 ]025 = 25 − .000016(25) 4 = 18.75 yr.
0
(f) R(15) / R(10) = .784 / .936 = .8376 25
(g)
25 1 1 3 4 MTTF (10) = − t dt = t − t (1 .000064 ) .000016 ( ) 10 R(10) 10∫ .936
=
1 ⎡15 − .000016 ( 254 − 104 ) ⎤ = 9.52 yr ⎦ .936 ⎣ 20
2.23 MTTF (10) = ∞
2.24 MTTF =
1 1 (1 − .000125t 3 )dt = (5.3125) = 6.07 yr. ∫ R(10) 10 .875 ∞
225
∫ ( t + 15) 0
2
225 dt = −1( t + 15 )
= 0
225 = 15 yr. (15 )
∞ ⎞ 1 225 1 ⎛ 225 = MTTF (10) = dt ⎜ ⎟ 2 .36 ⎜⎝ −1( t + 15) ) ⎟⎠ R(10) 10∫ ( t + 15 )
∞
= 10
1 [ 0 + 9] = 25 yr .36
λ(t) = 2 / (t+15) ; DFR
2.25
⎡ t ⎤ R (t ) = exp ⎢ − ∫ (1 + .5t '+ 10 / t ') dt '⎥ = exp[−(t + .25t 2 + 10 ln t − 1.25)] ⎣ 1 ⎦ dR(t ) f (t ) = − = (1 + .5t + 10 / t ) exp[−(t + .25t 2 + 10 ln t − 1.25)] dt
2-7