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Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.

CHAPTER 2

2.1

1 =.91 (.001)(100) + 1

a) R(100) = b) λ ( t ) =

R(1000) =

and

1 = .5 (.001)(1000 ) + 1

1 − d ((.001t + 1)−1 ) − dR( t ) 1 .001(.001t + 1)−2 ⋅ (.001t + 1) ⋅ = ⋅ −1 = 001 1 dt R( t ) (. 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

− 0tλ (t ')dt '

=e

z

− .4 0t t 'dt '

= e− .2 t ' | t = e−.2 t 2

2

0

2

F (1 / 12 ) = 1 − R(1 /12 ) = 1 − e −.2(1/12 ) = .00139 2

b) R( t) = e −.2t =.95 2.3

−.2t 2 = ln(.95)

→

a) R( t ) =

z

b) λ (t ) =

−dR (t ) 1 f (t ) .01 ⋅ = = dt R( t ) R (t ) 1−. 01t

t

c) MTTF = d) σ 2 =

z

z

100

100

z

100

R( t )dt =

0

z

100

0

(1− .01t )dt = t

0 ≤ t ≤ 100 100 2 100 2 0 − .005t 0 = 100− .005(100 )

z

σ = σ2 = 28.9 days e) R( t median ) = 1−.01t median =.5 2.4

=

0 ≤ t ≤ 100

t

t f ( t )dt − ( MTTF )2 =.01 100 t2 dt − 502 =.033 t3 0

z

− ln(.95) =.506 yrs .2

t=

=.01(100 − t ) = 1−.01t f ( t ′ )dt ′ = .01dt ′ =.01t ′ 100 t

100 2 0

a) R( t ) =

→

1000

f ( t ′ )dt ′ =

t

c

3 109

→

.01t median =.5

z t ′ dt ′ = LMN101 t ' OPQ 1000 t

2

0

3 109

3

c) R( t ) = 1 −

t =.99 10 9

t median = 50 days

100

9

t

0 ≤ t ≤ 1000 hrs

F (100) = 1− R(100 ) = 1− (1− 1003 / 109 ) = 1000

− 50 2 = 8333 . (days ) 2

3

h

z

0

3

t 1 3 − t3 = 1 − 9 9 1000 10 10

b) MTTF = t ⋅ f (t ) dt =

→

100

= 50 days

z

1000 0

t 3 dt =

1 10 6 = 3 9 10 10

3 t4 109 ⋅ 4

1000 0

=

3 1000 4 − 0 = 750 hrs 109 ⋅ 4

3

→

t = .01 10 9

→

2.5 8

t = 3 10 7 = 215.443 hrs

Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.

a) R(50) = e -

.001(50)

−d ( e

b) λ ( t ) =

=.8

−(.001 t )1/ 2

)

dt

.0005 1/ 2 1 1 == (.001t )−1/ 2 (.001)e− (. 001t ) ⋅ −(. 001t )1/ 2 = .001t 2 e

1

⋅ e

− (. 001t )1/ 2

λ( t ) is decreasing because λ ( t ) goes to zero as t goes to infinity.

R( T0 + t) R(T0 )

c) R( t / T0) =

→

R( 50 / 10) =

R( 50 + 10) R(60 ) e − = = R(10 ) R(10) e −

. 001( 60 ) . 001(10)

=.865

t R (t +10 ) e− . 001( + 10) d) R (t /10 ) = = − .001(10) =.95 R (10) e

e

− .001(t + 10)

= .95e −

.001(10)

= .859596

→

t=

[ ln.859596]2 − .001

10 = 12.9 hrs

2.10 a) Wear-out is indicated by an increasing failure rate. 1 1 −d (1 − t / t 0 )2 −dR(t ) 1 λ(t ) = ⋅ = [−2 (1 − t / t 0 )(−1 / t 0 )]⋅ ⋅ = 2 dt R( t ) (1 − t / t0 ) (1 − t / t 0 )2 dt 2 2 2 = = Since λ (0 ) = and λ (t → t 0 ) = ∞ , λ (t ) is IFR. 1 t 0( − t / t 0 ) t 0 − t t0 b)

F t I L F t I F − t JIOP MTTF = R( t )dt = G1 − J dt = M G1 − J G H t K MN H t K H 3 K PQ OP = t − t LF t I M = − − − ( ) 1 1 0 G J PQ 3 3 MH t K N

z

z

t0

0

2

t0

3

t0

0

0

0

0

0

3

0

3

0

0

0

c) R( t ) = (1 − t / 2000 )2 =.90

2.14

z

R( t ) =.1

t = 2000 (1 − .9 ) = 102 .63 hrs

LM (1+.05t′ ) O N −2(.05) PQ −2

∞

∞

(1 +.05 t′ )−3 dt′ =.1

t

R(10 / 1) =

d

i

= 0 − − (1+ .05t ) −2 = (1+ .05t ) −2 t

R(10 + 1) (1+.05(11)) −2 = =.459 R(1 ) ( 1+. 05(1)) −2

z z

∞

MTTFbefore = R (t )dt = 0

MTTFafter

→

z

∞

0

(1+ .05t )

1 = R( t / T0 ) dt = T0 R(T0 ) ∞

2

= (105 . )

z

∞

1

L (1 +.05t ) O dt = M N −.05 PQ −1

−2

z

∞

T0

R (t ) dt =

FG H

∞

=0 − − 0

1 (1+.05(1)) −2

L (1+ .05t ) OP . )M dt = (105 N − (.05) Q

−1 ∞

−2

(1+ .05t )

2

1

9

IJ K

1 = 20 . 05

z

∞

1

(1+.05t ) −2 dt

L MN

=0− −

OP Q

(105 . )2 = 21 .05(1+.05(1))

Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed.

2.18 ( a) f ( t ) =

ka k k +1

(t + a )

( c) MTTF = ∫ ( d ) tmed

(t + a )

k → DFR t+ a − k +1

ak

∞

0

; (b ) λ (t ) =

k

dt =

a k (t + a ) −k +1

∞

= 0

−a k a− k+1 a = −k +1 k −1

⎛ 1 ⎞ = a ⎜⎜ k − 1⎟⎟ ⎝ .5 ⎠ a

⎛ t2 t3 2t t 2 ⎞ 2.21 (a) MTTF = ∫ ⎜ 1 − + 2 ⎟ dt = t − + 2 = a / 3 a a ⎠ a 3a 0 0⎝ ⎛ 2t t 2 ⎞ 2 2 ⎜ 1 − a + a 2 ⎟ = .5; t − 2at + .5a = 0 ⎝ ⎠ a

(b)

t=

2 a ± 4 a2 − 2 a2 a 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

.000064t 3 = .5 (c)

tmed = 3 .5 / .000064 = 19.8425 yr.

(d) λ(t) = .000192t2 / (1 - .000064t3) ; IFR 25

(e) MTTF =

∫ (1− .000064t

3

4 25

0

(f) R(15) / R(10) = .784 / .936 = .8376 25

MTTF (10) = (g)

=

4

)dt = [t − .000016t ]0 = 25− .000016(25) = 18.75 yr .

25 1 1 (1 − .000064t 3) dt = t − .000016t 4 ) ( ∫ 10 .936 R(10) 10

1 ⎡ 15 − .000016( 254 − 104 )⎤ = 9.52 yr ⎦ .936 ⎣

10...