Reliability & Maintainability Engineering Ebeling Chapter 2 Book Solutions - Failure Distribut PDF

 

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  CHAPTER 2

2.1

1

a)  R(100) =

= λ ( t )  is

R(1000) =

(.001)(100) + 1

    − dR( t ) b) λ ( t ) =

  =.91   and

  − d ((.001t + 1)−1 )

1

⋅ R( t )  =

dt

.001

(.001t + 1)

.001t + 1

  =

2

 

−2

⋅ (.001t + 1)  =.001(.00001t + 1) ⋅ (.001t + 1)   −1

 

λ ( t ) goes

decreasing because

  =.5

(.001)(1000) + 1

1

dt

.001( .001t + 1)

1

to zero as t goes to infinity.

  2.2 2.2

  − z 0t   λ ( t ')dt '

a)  R( t ) = e

  −.4   z 0t t 'dt '

=e

0

F (1 / 12) = 1 − R(1 / 12) = 1 − e

b)  R( t ) = e −

.2 t 2

2 2 = e  − .2t ' |t  = e  − .2t   

=.95   →

−.2(1/ 12 )2

=.00139  

−.2t 2 = ln(.95)

t =   − ln(.95) =.506 yrs .2

→

2.3 2.3 a)  R( t ) = b)

z 

z 

100

100

f ( t ′ )dt ′ = .0 1dt ′ =   .0 1t ′   =.01(100 − t ) = 1−.01t   0 ≤ t  ≤ 100

t

t 

    − dR( t )

λ ( t ) =

dt

c)  MTTF = d) σ 

2

100 t 

z

⋅

1 R( t ) 

100

0

R( t ) dt =

=

z 

 f ( t )  R( t )

100

0

=

   m   

   o  

   c  

.01 1−.01t 

 . 

   0 ≤ t ≤ 100      e  

(1−.01t )dt = t 1000 −.005t 2 1000 = 10   i0  −.005(100)2 = 50 days 100

= z 0 t 2 f ( t )dt −  ( MT MTTF )2 =.0 1z 0 t 2 dt − 50   h2   =.033 t 3   − 502 = 833.3 (days)2   0 100

100

= σ 2 = 28.9 days e)  R( tmedian ) = 1−.01tmedian =.5   σ

2.4 2.4 a)  R( t ) =

=

z 

  f ( t ′ )dt ′ =

1000

t

1

3

   e  

.01tmedian =.5

   T  

3

 L 1   t ' O  z  t ′ dt ′ = M N10 PQ

 

2

3

t 

− t 3 h = 1 −

→

100

1000

   e10    9

c100   c0  

   c  

→

 

9

t 

3

t 

 

0 ≤ t  ≤ 1000 hrs  

t median = 50 days  

109

109

   A  

3

106

9

1

F (100) = 1 − R(100) = 1 − (1 − 100 / 10 ) = 109 = 103   1000 1000 3 1000 3 3 3 b)  MTTF = t ⋅ f ( t ) dt = 9 t dt = 9 t 4  = 9 10  004 − 0 = 750 hrs 0 0 10 0 10 ⋅ 4 10 ⋅ 4

z

z 

  t3 c)  R( t ) = 1 − 9 =.99  10

t 3

→

9

10

=.01

→

t  = 3 107 = 215.443 hrs  

2.5 2-1

 

  a)  R(50) = e

- .0 01 01 (5 (5 0) 0)

    − d (e

− (.001t )1/2

b) λ ( t ) =

dt 

λ ( t )  is

=.8   )

1

⋅

 

e

1

1/2

− (.001t )

decreasing because

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

λ ( t ) goes

R(50 + 10)

 

 R(50 / 10) =

→

e

1/2

− (.001t )

=  

.0005 .001t 

 

to zero as t goes to infinity.

  R(T0 + t ) c)  R( t / T 0 ) =  R(T  ) 0

1

1/2

R( 60)

e

− .00 1( 6 0)

= R(10) = e−

 R(10)

.001(10 )

=.865

− .001( t +10)

  R (t + 10) e = − d)  R ( t / 10) =  R(10) e

=.95  

001(10) .001

2

e

− .001( t +10 )

= .95e−

.001(10 )

= .859596

→

t  =

[ ln.859596] .001

− 10 = 12.9 hrs  

2.6 2.6 a)  R( t ) =

z

z 

10

t

10

10

10

 f ( t ′ )dt ′ = (.2−.0 2t ′ ) dt ′ =.2  t ′ t   −.01  t ′ 2 t 

t 

= ( 2−.2t ) − (1−.01t ) = 1−.2t +.01t 0 ≤ t ≤ 1 0 yrs   .2 −.02t    f (t ) .2  (1−.1t ) .2    2 =  2 = = λ ( t ) = 1−.1t   R(t ) 1−.2t +.01t  (1−.1t )   0) = ∞ so the hazard rate is, in fact, increasing. λ ( 0) =.2  and λ ( t → 1 2

b)  MTTF

z

10

= R(t ) = 0

z 

10

0

(1−.2t +.01t 2 )dt = t

c)  R( t median ) = 1−.2tmed + .01t t ( median ) =

2 

2 m ed

=.5 

→

. 2 ± ( −.2)2 − 4(.01)(.5) 2(.01)

=

10 0

−.1 t 2

.01t

2 med

. 2±.1414 2(.01)

10 0

+.00333 t 3

−.2t med +  .5 = 0 . 

   m    10 0

   o  

= 10 − 10 + 3.33 = 3.33 yrs  

   c  

= 17.07,293 .   e        i 

t ( median ) is 2.93 years. (17.07 is outside outside range of values for t.)    h  

d)  f ( t mod e ) =  MAX [ f ( t )]   0≤t ≤10

(0) = .2 and f (10)=0. (   c10)=0. Therefore, the mode mode occurs at t=0 yr.  f (t ) is linearly decreasing: f (0)    e) σ 

2

=

z

10

0

t f (t )dt − MT MTTF = 2

2

z  t (.2−.02t )dt −  3.33 10

0

2

   e  

2

L .2t =M N 3

3

−

.02t 4 O 4

10

P   − 3.33 Q

2

0

 

   T  

= (66.67 − 50) − 0 − 11.09 = 5.58 

→

σ

=

σ 

2

= 5.58 = 2.36 yrs  

   e  

2.7 2.7 100

a)  R(1) =

+

   c   100

 2 =

(1   A   10)

121

=.826  

  −2 ∞ L   O   I  = F   −100 100 2 0 0 1 0   ′ + ( )   t  −3    =   = 0−G  R( t ) =  f ( t ′ )dt ′ = 200  ( t ′ + 10)  dt ′ = M P 2 J  2 2 K  H  − + 10 10 + ( ) ( ) t t  N Q     ∞   −1 ∞ ∞ ∞ L   O 10 ( )   −100   +   t    O L −2 b)  MTTF = R( t )dt = 100 ( t + 10) dt  = 100M P   = MN t + 10 PQ   = 0 − ( −10) = 10 yrs 0 0 1 − 0 N Q0

z

∞

t

z  z  ∞

t 

z

t 

2-2

 

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  c)  R( t ) =

100

  2 =.95 

( t + 10)

→

100

t + 10 =

d) The failure rate is DFR because

λ ( 0)

.95

=.2  and

→

λ ( t →

t  =

100 .95

− 10 =.26 yrs  

∞) = 0 .

200 2

3

  λ ( t ) =

f ( t ) = (  t + 10) = 20 0( t + 10 )3 = 2   100 100( t + 10 ) t + 10  R( t ) 2 ( t + 10)

e) A one year burn-in period will improve improve the reliability reliability because the failure rate is decreasing decreasing.. 100 2   R (1 + 1)  R( 2) ( 2 + 10)2 11 = 2 =.84   = =  R(1 / 1) = 100 12  R(1)  R(1) (1 + 10) 2.8 2.8

2

L 1  O   t  F ( t ) = z   f ( t  ′ )dt ′ = z   dt ′  = M   t ′ P   = ,   b Nb Q b t

t 

0

0

1

t 

0≤ t ≤ b 

0

  t  b t   R( t ) = 1 − F ( t ) = 1 − =   − b b

1

   m   

  f ( t )

1 b = b  =   = ,  λ ( t ) =  R( t ) b − t  b( b − t ) b − t   b

L   O M P N Q

IFR      o   b

2 1 1   t    b  b − t   = dt  = bt −  MTTF = R( t )dt  = 0 0 2 0 b b b

z

b

z 

  b  − t med  1  R( t median ) = = b  2 b

LF  2    b2 I  O 2b  2   c −  b2 b  − 0P =  .  =   MGH b − 2 J K  − 2b 2 MN PQ    e      i 

There is no mode because all failure times are equally likely.

σ 

2

   h   b

 F   I   L t   O b dt − MTTF   = z   dt − G  J   = M P   − = z  t f ( t ) dt H 2 K    c   N 3b Q 4 b b 2 0

 

2

b

t 2

2

  b

3

2

0

=

0

b2

3

−

b2

4

=

b2

12

b

→ σ  =

12

   e  

2.9 2.9

  20 1   t  t     −   T t  ;    R( t ) = ; F ( t ) = = λ ( t ) = 20 20 − t  b 20    e     20 20 20 = 10 yrs; tmedian = = 10 yrs; σ  = = 5.77 yrs  MTTF =    c   2 2 12    A   2.10 2.10 a) Wear-out is indicated indicated by an increasing increasing failure rate.

   − dR( t )

λ ( t ) =

dt

2

=

1

⋅

=

R( t ) 

 

t 0 (1 − t / t 0 )

 − d (1 − t / t 0 )2

  Since λ ( 0) =

 

t0 − t

 

= [−2(1 − t / t0 )( −1 / t 0 )] ⋅

2   (1 − t / t 0 )

dt 

2

=

1

⋅

2 t0

and λ ( t → t0 ) = ∞,  

λ ( t )

1 (1 − t / t 0 )2

 

is IFR.  

2-3

 

  b)  MTTF =

z

t0

LF     t  I 3 F − t 0 I O F 1 −  t  I 2 dt  G t  J K    = MMGH 1 − t  J K  GH  3 J K PP   0 H  0 N 0 Q0

z 

R( t )dt  =

0

t 0

t 0 

3

=   − t0 MLF  G1 −  t 0 J I   − (1 − 0)3 PO = t 0 3

MNH 

t 0

K 

c)  R( t ) = (1 − t / 2000)

PQ

2

=.90 

3

→

t  = 2000(1 − .9 ) = 102.63 hrs  

2.11   R(  T0 + t )

 R( t / T 0 ) =

 R(T 0 )

=

( t + T 0 + 1)−3 / 2 (T 0 + 1)−3 / 2

  wher wh eree t = 2, T0 =.5  

−3 / 2

  R( 2.5) (3.5) = =.28    R( 2/.5) =  R(.5) (15 . )−3 / 2

An improvement of .28-.19 = 47%  is obtained obtained with with a 6 month burn - in. .19 2.12

z   R( t ) = e −

t

0

 

 f ( t ) = −  R( t

λ ( t ′ ) dt ′

dR( t ) dt 

)=e

= e z   

=−

t 

 

−  at ′dt ′      − a t ′ 2 / 2

=e

0

d (e

2 − ( at med  )/ 2

− at 2 / 2

dt 

=.5

)

t  0

2

   m   

= e − at   / 2   a > 0 and t  ≥ 0   2

− e − at  / 2 ( −at ) = ate − at

2

/ 2 

   o  

     c  

 

 . 

median

2

−

at med 

= ln(.5)

2

   e  

  −2 ln(.5)

t median =

1.18

=

 

   i 

a   a    h   The mode is found by setting the first derivative of f (t) (t) equal to 0 and solving for t. df ( t )

d (e

=

2 − ( at med  )/ 2

)

dt  dt  2 [− at  + 1] = 0   t mod e = 1 / a

2

2

   c  

2

= a[te − at  / 2 ( − at ) + e − at / 2 ] = e − at / 2  [− at 2  + 1] = 0      e      T      e  

2.13

   c  

 AFR =

=

 ln R  ( t1 ) − ln R( t2  )    A   t

 

− t 1

2

=

ln (1 − t13 / 109 ) − ln(1 − t 23 / 109 )

ln (1) − ln(1 − 5003 / 109 )

 

500 − 0

   w  wh here t1 = 0 and t 2 = 500

t2 − t 1

=

0  − ln(.875) 500

 failu lure ress / hr =.000267  fai

2.14

z 

 R( t ) =.1

∞

(1+.0 5t ′ )

t 

−3

L ( 1+.05  t ′ )−2  O ∞   = 0 − d −(1+.05t )−2 i = (1+.05t )−2   dt ′ =.1M P N −2(.05) Q  

t 

2-4

 

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

  ∞R( t ) dt

∞ (1+.05t )

−2

z z  1 1 = z R (t / T )dt  = =   ( ) (1+.05t ) dt   R t dt  R (T  ) z (1+.05(1)) z    t ) O   = 0 − L− (1. 05) O = 21     L (1+.05 = (1.05) z  (1+.05t ) dt  = (1.05) M M .05(1+.05(1)) P P Q N N −(.05) Q

 MTTFbefore   =

0

=

0

∞

 MTTFafter 

  +.05t ) dt  =  (1 −.05

ML N

 

1 J  PO   = 0 − GF H − . 05 K I   = 20 Q0

∞

0

T0

0

2

∞

−2

T 0

−1   ∞

 

∞

−2

1

−2

 

2

2

1

 

1

2.15 λ ( t )  is decreasing. R( R(t) t) is de decr crea easi sing ng as R( R(0) 0) = 1 aand nd R(∞ ) = 0. f(t) = λ (t)R(t).

 

Since Sin ce λ ( t ) and R(t ) are decrea decreasin sing, g, f(t ) is de decre creasi asing ng and the mode mode must must occur occur at t = 0. 2.16 −

t 

λ    ( t ′ ) dt ′

 

−a

t 

t  e   ′ dt ′  

  

a et ′

  t 

t

t 

 

 R( t ) = e

z

z 

=e

0

 

 f ( t ) = λ ( t ) R( t ) = ( ae )( e t

2.17 Using

(a)  R (t ) = (b)  R (t ) =

∫ ∫

a − ae t

( x + c ) dx =

t 

( t '+ a )

10 t + 10

2

t

a

10 .95

=e

− aet 

−1 ∞

= t 

a1 e

) = ae(

t + a − aet  )

∫ x+c a

t+a

   m   

 

dx

, n ≠1;

−1

= .95; t =

1

+1

a ( t '+ a )

dt ' =

a e

) = ( ae )( e )( e

n +1

 

a

=e

0

n   ( x + c)

n

∞

   

=e

0

 

= ln ( x + c )    . 

; R (t )

=    i 

   e   10

t + 10

   h  

−  10 = .526 yr  = 6.3 mo.      c  

(c) R(5|5) = .75 R(10) / R(5) = (10/20) (10/20)   e  / (10/15) = 15/20 = .75    T   (d) F(90 days) = 1 - 10 / (10 + .2465) = .024

a (e)  λ (t ) =

(t + a )

   e  

2

a

   c  =

 

1 t+a

 

(t + a )

   A  

(f) R(t) = .5 = a / (t + a) or tmed = (a / .5) – a = 2a –a = a ∞

(g)  MTTF

=∫ 0

a t+a

∞

dt = a  ln(t + a ) 0 → ∞ ; does not exist

2-5

 

   o      c  

; R(8) = .55

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  2.18 ka k 

( a) f (t ) =

; (b )

λ (t )

( t + a ) k +1 (c ) MTTF =

∫

(t + a )

0

k 

→ DFR

t+a

ak

∞

=

k 

dt = 

a k  ( t + a )

− k +1 ∞

−k + 1

 

= 0

⎛ 1 ⎞ ⎜ .5 − 1⎟⎟ ⎝ ⎠

(d )tmed  = a ⎜ k 

2.19

(a ) f (t ) =

dR(t ) dt 2

(b )

λ (t )

=

3t 2  3

k 

3

, 0≤t ≤k 2

= 3t3 / ⎛⎜ 1 − t 3 ⎟⎞ = 33t 

3

 

a   −a k a − k +1 = k − 1 −k + 1

⎝

k

∫

(c) MTTF =

k 

0

k ⎠

3t k

k

3

3

4

dt =

3t 

t k 

 

3

= k  

3

4k 

 

0

4

3

(d ) 1 −

t 

3

k 

= .5 or tmed  = k 3 . 5 = .7937k 

   m       o  

2

2.20 rewrite f(t) = .003(10-t)   3 (a) R(t) = .001(10 – t)  ; R(1) = .729 3

   c  

1/3

(b) λ .001(10-t)  = .90; solving for t = 10 – (.9/.001)  = .345 yr. (c) (t) = f(t) /R(t) = 3/(10-t); IFR 10

(d)  MTTF

 . 

   e  

= ∫ R (t ) dt = ∫ .001(10 − t ) dt = 2.5 yr. ; R (2.   i5  ) = .422 3

0

(e) F(1/12) = .02479; R(11/12 | 1/12) = .729/.9752   h =  .7475 a

2 3 ⎛ 2t t 2 ⎞ t   c   t   + 2 ⎟ dt = t − + 2 = a / 3   2.21 (a)  MTTF = ∫ ⎜1 − a a ⎠    e   a 3a 0 ⎝ 0 ⎛ 2t t 2 ⎞ 2 2 ⎜1 − + 2 ⎟ = .5; t − 2at   T+   .5a = 0 a a ⎠   (b) ⎝    e   2 2 a 2 2a ± 4 a − 2 a = = a (1 ± .7071)  = .293a t= a±    c   a

2

2

Why not the positive root?    A  

(c) R(a/3) = 4/9 = .444 (d) λ (t ) =

2 / a − 2t / a 2 1 − 2t / a − 2t 2 / a 2 

=

2 ( a − t )

=  

2

( a − t )   ( a − t ) 2

; I FR    

(e) R(1) = .81; R(1|1) = R(2) /R(1) = .64 / .81 = .79 2.22

3

(a) R(t) = 1 - .000064t ; R(15) = .784; (b) F(10) – F(2) = .000064(1000 .000064(1000 - 8) = .063488 2-6

 

  3

(c)

=

.0000634t  .5   000064 = 19.84  25 yr . tmed  = .5 / ..0

(d) λ(t) = .000192t2 / (1 - .000064t 3) ; 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

1

25

 R (10)

10

 MTTF (10) = (g (g))

=

∫

1

(1 − .000064t 3 )dt =

(

t − .000016t 

4

.936

)

25

10

 

1

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

2.23 2.23  MTTF (10) = ∞

2.24 2.24  MTTF

=∫

1

20

∫

 R(10) 10

(1 − .000125t 3 )dt =

1 .875

(5.3125) = 6.07 yr.

∞

225

0 ( t + 15 )

2

225

dt =

  −1( t + 15 ) 0

=

225

(15 )

= 15 yr .  

   m       o  

∞

 M TTF (10) =

∞

1 225 1 ⎛ 225 1   dt = [ 0 + 9 . ] =   c  25 yr   ⎜⎜ ⎟⎟⎞ =  2 .36  R (10) 10 ( t + 15)   . 36 ⎝ −1( t + 15) ) ⎠ 10

∫

   e  

λ(t) = 2 / (t+15) ; DFR 

   i 

2.25 2.25

   h   ⎡ t  ⎤ 2  R (t ) = exp ⎢− ∫ (1 + .5t '+ 10 / t ') dt '⎥ = exp[−(t + .25t + 10 ln t − 1.25)]    c   ⎣ 1 ⎦  

 f (t ) = −

dR (t ) dt 

= (1 + .5t + 10 / t ) exp[   e− (  t + .25t 2 + 10 ln t − 1.25)]    T      e      c  

   A  

2-7