Solution Manual for An Introduction To Reliability and Maintainability Engineering

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 11 11.1 a)

At2 −0 =

A= b)

λ r .1 .02 + 1 − e − ( λ +r ) t2 → A30 = + 1 − e − (.02 +.1)30 =.8784 2 .1+.02 .1+.02 2 30 r + λ r + λ t2

b

gb g

→

A=

r r+λ

b

gb g

.1 =.8333 .1+.02

L λ + λ λ OP + λ P A = P + P = M1 + N r r Q r L .02 + .02 OP + .02 LM1 + .02 + .02 OP =.8065+.2 .8065 =.9678 A = M1 + N .1 .1 Q .1 N .1 .1 Q −1

1

1

1 2 2

2

2

1

1

−1

2

2

−1

2

As = 1 − (1−.8333)2 =.9722 (note: assumes two repair crews are c) As = 1 − (1 − Ai )2 → available. For a single repair crew see problem 11.9) 1 − Ai xMTBFi =.0103 MTBFi Ai . ) = 892 → MTTR ≤ 9.2 hrs prop: MTBF = 10,000Γ (1 + 1 / 17

11.2 Ai = 5 .95 =.98979 and MTTRi ≤

. → MTTR ≤ 3.4 hrs avion: MTBF = 3333 . ) = 1771 → MTTR ≤ 18.2 hrs struc: MTBF = 2000Γ (1 + 1 / 21 . ) = 773 → MTTR ≤ 7.9 hrs elec: MTBF = 870Γ (1 + 1 / 21 env: MTBF = 10,000 → MTTR ≤ 10.3 hrs 11.3 From Eq 11.19:

L λ λ λ OP + λ LM1 + λ + λ λ OP = LM1 + 1 / 10 + 1 / 10(1 / 5) OP A = P + P = M1 + + .5 N r r Q r N r r Q N .5 Q 1 / 10 L 1 / 10 1 / 10(1 / 5) O 1+ + + PQ =.7813+.2 .7813 =.9376 .5 MN .5 .5 −1

1

1

1

2

2

−1

1

1

1

2

−1

2

2

2

−1

2

11.4 From Eq. 11.19:

LM λ + λ λ OP + λ LM1 + λ + λ λ OP = LM1 + 1 + 1 OP LM1 + 1 OP = 3 N r r Q r N r r Q N 2 2Q N 2Q 4 −1

A = P1 + P2 = 1 +

1

1

2

2

−1

1

1

1

2

2

11.5 .4 r 1 / 2.5 = = =.80 r + λ 1 / 2.5+.1 .4 + 1 .4 .1 b) A2 = + 1 − e − (.4 +.1)2 =.80+.1264 =.9264 .4+.1 .4+.1 2 2

a) A =

b gbg

11-1

−1

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

Waveland Press, Inc., Copyright © 2009

r .4 .1 −.5(2 ) λ − ( λ + r )2 e e + = + =.8736 r+λ r+λ .4+.1 .4+.1 d) Two components in series: a) As = A2 =.802 =.64 , b) As = A2 2 =.92642 =.8582 , c) As = A( 2)2 =.87362 =.7632 e) Two components in parallel: a) As = 1 − (1 − A)2 = 1 − (1−.8 )2 =.96 , b) As = 1 − (1 − A2 )2 = 1 − (1−.9264)2 =.9946 c) A( 2) =

c) As = 1 − (1 − A( 2 ))2 = 1 − (1−.8736 )2 =.9840

LM .1 + .1 OP + .1 LM1 + .1 + .1 OP =.7619+.25(.7619) =.9524 N .4 .4 Q .4 N .4 .4 Q 2

f) A = P1 + P2 = 1 +

−1

2

2

−1

2

11.6 1

λ1=2

r=3/2

State 1 2

Condition primary operating (λ1=2) 3 primary failed and secondary operating λ2=4 (λ2=2) 3 both failed (r=3/2) 1 3 4 2 P1 = P3 → P3 = P1 and 2 P1 = 4 P2 → P2 = P1 2 2 3 2

FG H

P1 + P2 + P3 = P1 1 + A = P1 + P2 =

IJ K

LM N

1 4 6 + 3+ 8 = 1 → P1 = + 6 2 3

FG IJ H K

OP = 6 Q 17 −1

6 1 6 9 + = =.529 17 2 17 17

11.7

1

1

State 1 2 3

1

4 2

2

Condition operating degraded failed

3

Steady-state equations where λ 12 = 1, λ 13 = 1, λ 23 = 2,r = 4 : (1) rP3 = λ 12 P1 + λ 13 P1 = ( λ 12 + λ 13 ) P1 (2) λ 12 P1 = λ 23 P2 (3) λ 23 P2 + λ 13 P1 = rP3 (4) P1 + P2 + P3 = 1 1 λ Solving (2) for P2 results in: P2 = 12 P1 = P1 =.5 P1 2 λ 23 λ P + λ 13 P1 2(.5P1 ) + P1 = =.5P1 P3 from (3) is: P3 = 23 2 4 r 1 Solving (4) for P1 results in: P1 +.5 P1 +.5 P1 = 1 → P1 = =.5 2 Therefore, P2 = P3 =.25 and Availability=P1+P2=.75 .

11-2

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

Waveland Press, Inc., Copyright © 2009

11.8 A(T ) =

1 − e − λT 1 − e −.00314 T = λ T + t1 + t2 (1 − e − λt ) .00314 T +.25 + (1 − e −.00314 T )

T in days A(T) 1 .797 2 .884 3 .916 4 .932 6 .948

8 10 11 12 14

.955 .9576 .9582 .9585 .9585

15 16 18 20

.9579 .9574 .9560 .9544

11.9 λ 2

r r

3

4

λ

r

λ

State 1 2 3 4

λ

1

Steady-state equations: (1) rP2 + rP3 = 2λP1 (4) λP2 + λP3 = rP4 From (3): P3 =

FG λ IJ P H λ + rK

Unit 1 operating failed operating failed

Unit 2 operating operating failed failed

(2) λ 1 P1 + rP4 = rP2 + λP2 (5) P1 + P2 + P3 + P4 = 1

(3) λ 1 P1 = rP3 + λP3

1

FG IJ H K λP + λP λ λF λ I Solving (4) for P results in: P = = P + G JP r r r H λ + rK λ L 2λ F λ IJ P OP + λ FG λ IJ P = 2λ P = M P −G H λ + rK Q r H λ + rK r rN r

Solving (1) for P2 results in: P2 =

λ 2λP1 − rP3 2λ P1 − P = r r λ +r 1 2

4

3

4

2

1

2

1

1

F 2λ − λ + λ + 2λ IJ P = 1 → From (5): G1 + H r λ +r λ+r r K

1

2

2

1

2

L 2λ + 2λ OP P = M1 + N r r Q 2

1

1

LM N

−1

2

2λ2 2λ 2λ2 Therefore, Availability = P1 + P2 + P3 = 1 − P4 = 1 − 2 1 + + 2 r r r

OP Q

−1

11.10 λ 1 = λ 2 = λ 1

r

λP1 = rP2 → P2 =

λ 2 2r λ

4

2rP4 = λP2

→

λ r

P4 =

P1

λ 2r

P2 =

λ2 2r 2

P1

L λ λ OP P+ P+ P = 1 → P = M1 + + r 2r N r 2r Q λ

1

λ2

1

11-3

2

2

1

1

2

−1

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

Waveland Press, Inc., Copyright © 2009

LM1 + λ + λ OP Availability = 1 − P = 1 − 2r N r 2r Q λ2

2

2

2

4

11.11 State 1 2 4

Primary on-line failed failed

−1

Backup stand-by on-line failed

Steady-state equations are: (1) r1P2 = λ 1 P1 (4) λ 2 P2 = r2 P4

(2) λ 1 P1 + r2 P4 = r1 P2 + λ 2 P2 (5) P1 + P2 + P4 = 1

Solving (1) for P2 results in: P2 = Solving (4) for P4 results in: P4 =

λ1 r1

λ2 r2

P1

P2 =

λ 1λ 2 r1r2

P1

L λ + λ λ OP P = 1 → P = M1 + From (5): P + P + r rr N r rr Q λλ L λ λλ O Availability = 1 − P = 1 − M1 + r + r r PQ rr N λ1

1

λ 1λ 2

1

1

2

1

1 2

1

1 2

−1

1

1 2

−1

1 2

1

1 2

1 2

1

1 2

4

11.12 From 11.5 the steady state availability is .8 which means that: .799 ≤ A( t ) ≤.801 r .4 .1 −.5( t ) λ − ( λ +r )2 e e A( t ) = + = + =.8+.2e −.5t r+λ r+λ .4+.1 .4+.1 A(t) t 10 .8013 11 .8008 11 days

11.13

r .2 .0031 − (.0031+.2)10 λ − ( λ +r )t e e + → Aps (10) = + =.9867 r+λ r+λ .2+.0031 .2+.0031 λ − ( λ +r )t r .33 .01 − (.01+.33)10 e e + → Atr (10 ) = + =.9716 Atr ( t ) = r+λ r+λ .33+.01 .33+.01 Aps ( t ) =

A( t ) = Aps ( t ) Atr ( t ) =.9867⋅.9716 =.9587 Aps =

.2 .33 r = =.9847 and Atr = =.9706 .33+.01 r + λ .2+.0031

A = Aps Atr =.9847⋅.9706 =.9557 and %difference = High Level Redundancy: As (10) = 1 − (1 − A(10))2 = 1 − (1−.9587)2 =.9983 As = 1 − (1 − A)2 = 1 − (1−.9557)2 =.9980

11-4

.9587−.9557 =.0031 .9557

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

Waveland Press, Inc., Copyright © 2009

.2 .0031 + 1 − e − (.0031+.2)10 =.9913 .2+.0031 (.2+.0031)210 .33 .01 Atr10 = + 1 − e − (.01+.33)10 =.9790 .33+.01 (.33+.01)210 A10 = Aps10 Atr10 =.9913⋅.9790 =.9705 Aps10 =

As10 = 1 − (1 − A10 )2 = 1 − (1−.9705)2 =.9991

Low Level Redundancy: As (10) = (1 − (1 − Aps (10))2 ) × (1 − (1 − Atr (10))2 ) = (1 − (1−.9867)2 ) × (1 − (1−.9716)2 ) =.9990 As = (1 − (1 − Aps )2 ) × (1 − (1 − Atr )2 ) = (1 − (1−.9847 )2 ) × (1 − (1−.9706 )2 =.9989

As10 = (1 − (1 − Aps10 )2 ) × (1 − (1 − Atr10 )2 ) = (1 − (1−.9913)2 ) × (1 − (1−.9790)2 ) =.9995 11.14 MTBF = 400Γ(1 + 1 / 2.4 ) = 354.54 , MTTR = tmed e s / 2 = 4.8e1.2 / 2 = 9.86 MTBF 354.54 = =.9729 a) Ai = MTBF + MTTR 354.54 + 9.86 b) 1 1 1 λMTTR + MPMT 9.86 + 6 TP 200 = 7.39 hrs M= = 354.54 1 1 1 λ+ + 354.54 200 TP MTBM 127.87 1 1 MTBM = = =.945 = = 127.87 , Aa = 1 1 1 1 MTBM + M 127.87 + 7.39 + + MTBF TP 354.54 200 2

11.15

z

2

F TI t F TI T T T− R(T )T + tf ( t )dt GH1 − b JK T + b dt GH1 − b JK T + 2b 2b = = = A(T ) = TO TO T + t + t 1 − R( t ) LT O L L T +t +t M P T +t +t M P T +t +t M P NbQ NbQ NbQ

z

T

2

0

0

1

2

T

2

1

2

1

2

1

2

T2 10,000 A(T ) = T: 300 400 500 600 T A(T): .8863 .8935 .8942 .8915 T + 24 + 72 5,000 An interval of 500 appears optimal. Since availability seems to be insensitive to changes in T in the neighborhood of 500 hours, no further refinement to T is made.

T−

LM N

11.16 A′( t ) = =

z

t

0

OP Q

1 − H ( t ′ ) f ( t ′ )dt ′ ==

LM OP − e Nλ + r Q λ + r λ

λ

z

t

0

z

t

e − rt ′ λe − λt ′ dt ′ = λ e − ( r + λ )t ′ dt ′ 0

− ( r +λ )t

11-5

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

Waveland Press, Inc., Copyright © 2009

A( t ) = 1 − A′( t ) = 1 −

LM λ OP − λ e Nλ + r Q λ + r

− ( r + λ )t

11.17 Steady-state equations are: (1) rP2 + rP3 + rP4 = 2λP1 (4) λP2 + λP3 = rP4

λ

λ +r

(3) λP1 = rP3 + λP3

P1

2λP2 2λ2 = 2 P1 r r r Substituting the above into (5):

From (4): P4 =

λP2 + λP3

r λ + r − λ + λe − ( r + λ ) t λe − ( r + λ ) t = + λ+r λ +r λ +r

(2) λP1 = rP2 + λP2 (5) P1 + P2 + P3 + P4 = 1

From (2): P2 =

From (2)and (3): P2 = P3 ;

=

=

L 2λ + 2λ OP → P = M1 + N λ +r r Q 2λ L 2λ 2λ O Availability = P + P + P = 1 − P = 1 − 1+ + M r N λ + r r PQ λ

λ

2λ2 P1 + P1 + P1 + 2 P1 = 1 r λ +r λ +r 1

2

3

2

1

4

−1

2

2

2

2

2

−1

11.18

L λ + λ λ OP = LM1 + .004 + .004(.002) OP = 0.676 a) P = M1 + N r r Q N .01 .01 Q −1

1

1

1

P2 =

−1

2

2

λ1 r

P1 =

2

.004 .676 = 0.270 → A1 = A2 = A3 = AISOC = P1 + P2 =.946 .01

b) ISOC is in series with all 3 sensors: Asys3 = A1 A2 A3 AISOC = (.946 )4 = 0.801 c) ISOC is in series with 2 out of 3 redundant sensors: 3 3 x 3− x = (.946 )(.14498+.8466 ) = 0.938 .946 1−.946 Asys 2 = AISOC x x =2

b

gLM∑ FGH IJK b g b N

g OP Q

11.19 a) STATES FOR SENSOR 1, ISOC, AND SHARED BACKUP Sensor 1 ISOC Backup State 1 on-line on-line stand-by 2 failed on-line on-line 3 on-line failed on-line 4 failed failed on-line 5 on-line failed failed 6 failed on-line failed 7 failed failed failed

11-6

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

Waveland Press, Inc., Copyright © 2009

λ1

λ2

1 r

r

2

λ2

λ1

r

r

λ3 r

4

3

r

λ3

5

r

λ1

r

λ3

6

r

λ2

7

Availability of sensor 1=P1+P2+P3 Availability of ISOC= P1+P2+P3+P4+P5+P6 Solve using transition matrix: −( λ 1 + λ 2 ) r

OP 0 0 r 0P −( λ + λ ) − r r 0 0 0P −( λ + λ ) − r r r P λ λ 0 0 − λ − 2r rP 0 0 0 λ −λ − r rP P λ 0 0 0 −λ − r r P 1 1 1 1 1 1PQ LM.46468OP LM0OP MM0PP MM.18587 PP .18587 0 M P M P P = T b = M.07435P b = M 0P MM0PP MM.03717PP MM0PP MM.03717PP N.01487Q N1Q

LM MM λ λ M 0 T=M MM 0 MM 0 N 1 LM P OP MM P PP P M P P = MP P MM P PP MM P PP NP Q

1

2

2

0

r

0

3

3

1

2

1

3

3

1

3

1

2

3

2

−1

4 5 6 7

a) Availability of ISOC= P1+P2+P3+P4+P5+P6=1-P7=.9851 b) Availability of sensor 1=A1=P1+P2+P3=.8364 Availability of sensors 2 and 3 same as in problem 11.18: A2 = A3 = 0.946 Availability of system=A1A2A3AISOC=(.8364)(.946)(.946)(.9851)=.7374 11.20

λP1 = rP2 → P2 =

λ r

0

P1

11-7

0

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

Waveland Press, Inc., Copyright © 2009

λP1 + rP3 = ( r + λ ) P2 → P3 = rP4 = λP3 → P4 =

λ

P3 =

r

λ3 r3

FG H

IJ K

(r + λ ) ( r + λ )λ λ λ λ2 P1 = 2 P1 P2 − P1 = − 2 r r r r r

P1

F λ λ λI F λ λ λI 1= P + P + P + P = G1 + + + J P → P = G1 + + + J H r r rK H r r rK 2

1

2

3

4

3

2

1

2

1

3

−1

2

3

3

λ = 2 and r = 5

L 2 4 + 8 OP =.6158, P = 2 P =.2463, P = 4 P =.0985 P = M1 + + 5 5 N 5 25 125 Q −1

1

2

1

3

1

A = P1 + P2 + P3 =.9606 11.21 State 1 2 3 4

Unit 1 operating failed operating failed

Unit 2 operating operating failed failed

Steady-state equations: (1) rP2 + rP3 = 2λP1 (3) λ 1 P1 + rP4 = rP3 + λP3 (5) P1 + P2 + P3 + P4 = 1

(2) λ 1 P1 + rP4 = rP2 + λP2 (4) λP2 + λP3 = 2rP4

From (2) and (3): P2 = P3

λ 2λ P1 = P1 2r r λP + λP3 2λP2 λ λ λ2 P1 = 2 P1 Solving (4) for P4 results in: P4 = 2 = = r r 2r 2r r Solving (1) for P2: rP2 + rP2 = 2λP1 → P2 =

FG IJ H K F λ λI r λ λ λ From (5): P + P + P + P = 1 → P = G1 + 2 + J = r r r H r r K r + 2λr + λ 2

2

1

1

1

2

1

1

2

−1

2

2

2

λ2 λ2 and Availability=1 − P = − 1 4 r 2 + 2λr + λ2 r 2 + 2λr + λ2 2 2 n r (r + λ ) − r λ2 Equation: As ( t ) = 1 − ∏ (1 − Ai ( t )) = 1 − 1 − = 1− = 1− r+λ r+λ ( r + λ )2 i =1

Therefore, P4 =

FG H

IJ K

11.22 (a) rate diagram

11-8

FG H

IJ K

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

Waveland Press, Inc., Copyright © 2009

(b)

4P1 = 4P2 + 4P3 12P3 = 2P1 4P4 = 8P2 + 8P3 P1 + P2 + P3 + P4 = 1

11.23 (a)

(c) P1 =6/24; P2 = 5/24; P3 -1/24; P4 = 12/24; A = P1 + P2 + P3 = 12/24 = .5

1

.1

2

.5 .1

3

(b) .1P1 = .5P3 ; .1P1 = .1P2 ; P1 + P2 + P3 =1 (c) P3 = (1/5) P1 ; P2 = P1 ; P1 = [1 + 1 + 1/5]-1 = 5/11;

A = P1 + P2 = 10/11 = .909

11.24

( 30 )( 24 ) = 8.12; MTTR = 10e.84 /2 = 14.23 100Γ (1 + 1/ 2.4 ) ___ ( 30 )( 24 ) = 64.75; M (8.12 )(14.23) + 3(2) MTBM = =

m(30) =

2

8.12 + 3 8.12 + 3 MTBM 64.75 Aa = = = .8555 ___ + 94.75 10.93 MTBM + M

11.25

m(30) = .51095 ⋅10−7 ( 30 ⋅ 24 )

2.74

= 3.4472

3.4472 ( 6.7 ) + 5(4) 720 = 85.23 = 5.1; MTBM = 8.4472 8.4472 85.23 = .9435 Aa = 85.23 + 5.1

___

M=

11.26 Aa = (.8555) (.9435) = .8072 11.27 M(100) = .00025 (100)2.4 = 15.77 failures; MTTR = 4 Exp[.852 /2] = 5.74 days A = 100 / [100 + 15.77 x 5.74] = .525 11.28 rate diagram: ( 1)→3λ→(2)→2λ→(3)→λ→(4)→r→(1) Equations: 3λP1 =rP4; 2λP2 = 3λP1; λP3 = 2λP2 ; P1 + P2 + P3 + P4 = 1 Solving: P1 = .1639, P2 = .2459, P3 =.4918, Availability = .9016

11-9

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

Waveland Press, Inc., Copyright © 2009

11.29 Steady-state equations: 3P1 = 8P2 + 4P3, 12P2 = 2P1, 8P4 = 4P2 + 2P3 Solving: P1 = 48/85, P4 = 9/85, A = 1 – 9/85 = 76/85 = .894 11.30 2λP1 = rP2, λ+ P2 = rP3, P1 + P2 + P3 = 1

1

2λ r 2

r

λ+

3

P1 = 1 / [1 + 2λ/r + 2λλ2 / r2 ] = .6098, P2 = (2λ/r) P1 = .2439; A = .8537 11.31 (a) m(5000) = .000023(5000)1.7 = 44.67, MTTR = 7exp[.64/2] = 9.64, MTBF = 5,000/44.67 = 111.94; Ainh = 111.94 (111.94 + 9.64) = .921 (b) m(500) = .000023(500)1.7 = .8912 expected nbr failures + 1 PM

M=

.8912(9.64) + 1(12) = 10.888 hr .8912 + 1

MTBM = 500 /1.8912 = 264.38; Aach =

264.38 = .96 264.38 + 10.888

≈4 11.32 MTTR = 3e m(1000,1250) = .0001[12501.825 – 10001.825] = 15.0; .7582 / 2

A = 250 / [250 + 15(4) + 10(1)] = 250/ 320 = .781

11-10

250/T = 250 / 25 = 10 sched. maint.

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 11 11.1 a)

At2 −0 =

A= b)

λ r .1 .02 + 1 − e − ( λ +r ) t2 → A30 = + 1 − e − (.02 +.1)30 =.8784 2 .1+.02 .1+.02 2 30 r + λ r + λ t2

b

gb g

→

A=

r r+λ

b

gb g

.1 =.8333 .1+.02

L λ + λ λ OP + λ P A = P + P = M1 + N r r Q r L .02 + .02 OP + .02 LM1 + .02 + .02 OP =.8065+.2 .8065 =.9678 A = M1 + N .1 .1 Q .1 N .1 .1 Q −1

1

1

1 2 2

2

2

1

1

−1

2

2

−1

2

As = 1 − (1−.8333)2 =.9722 (note: assumes two repair crews are c) As = 1 − (1 − Ai )2 → available. For a single repair crew see problem 11.9) 1 − Ai xMTBFi =.0103 MTBFi Ai . ) = 892 → MTTR ≤ 9.2 hrs prop: MTBF = 10,000Γ (1 + 1 / 17

11.2 Ai = 5 .95 =.98979 and MTTRi ≤

. → MTTR ≤ 3.4 hrs avion: MTBF = 3333 . ) = 1771 → MTTR ≤ 18.2 hrs struc: MTBF = 2000Γ (1 + 1 / 21 . ) = 773 → MTTR ≤ 7.9 hrs elec: MTBF = 870Γ (1 + 1 / 21 env: MTBF = 10,000 → MTTR ≤ 10.3 hrs 11.3 From Eq 11.19:

L λ λ λ OP + λ LM1 + λ + λ λ OP = LM1 + 1 / 10 + 1 / 10(1 / 5) OP A = P + P = M1 + + .5 N r r Q r N r r Q N .5 Q 1 / 10 L 1 / 10 1 / 10(1 / 5) O 1+ + + PQ =.7813+.2 .7813 =.9376 .5 MN .5 .5 −1

1

1

1

2

2

−1

1

1

1

2

−1

2

2

2

−1

2

11.4 From Eq. 11.19:

LM λ + λ λ OP + λ LM1 + λ + λ λ OP = LM1 + 1 + 1 OP LM1 + 1 OP = 3 N r r Q r N r r Q N 2 2Q N 2Q 4 −1

A = P1 + P2 = 1 +

1

1

2

2

−1

1

1

1

2

2

11.5 .4 r 1 / 2.5 = = =.80 r + λ 1 / 2.5+.1 .4 + 1 .4 .1 b) A2 = + 1 − e − (.4 +.1)2 =.80+.1264 =.9264 .4+.1 .4+.1 2 2

a) A =

b gbg

11-1

−1

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

Waveland Press, Inc., Copyright © 2009

r .4 .1 −.5(2 ) λ − ( λ + r )2 e e + = + =.8736 r+λ r+λ .4+.1 .4+.1 d) Two components in series: a) As = A2 =.802 =.64 , b) As = A2 2 =.92642 =.8582 , c) As = A( 2)2 =.87362 =.7632 e) Two components in parallel: a) As = 1 − (1 − A)2 = 1 − (1−.8 )2 =.96 , b) As = 1 − (1 − A2 )2 = 1 − (1−.9264)2 =.9946 c) A( 2) =

c) As = 1 − (1 − A( 2 ))2 = 1 − (1−.8736 )2 =.9840

LM .1 + .1 OP + .1 LM1 + .1 + .1 OP =.7619+.25(.7619) =.9524 N .4 .4 Q .4 N .4 .4 Q 2

f) A = P1 + P2 = 1 +

−1

2

2

−1

2

11.6 1

λ1=2

r=3/2

State 1 2

Condition primary operating (λ1=2) 3 primary failed and secondary operating λ2=4 (λ2=2) 3 both failed (r=3/2) 1 3 4 2 P1 = P3 → P3 = P1 and 2 P1 = 4 P2 → P2 = P1 2 2 3 2

FG H

P1 + P2 + P3 = P1 1 + A = P1 + P2 =

IJ K

LM N

1 4 6 + 3+ 8 = 1 → P1 = + 6 2 3

FG IJ H K

OP = 6 Q 17 −1

6 1 6 9 + = =.529 17 2 17 17

11.7

1

1

State 1 2 3

1

4 2

2

Condition operating degraded failed

3

Steady-state equations where λ 12 = 1, λ 13 = 1, λ 23 = 2,r = 4 : (1) rP3 = λ 12 P1 + λ 13 P1 = ( λ 12 + λ 13 ) P1 (2) λ 12 P1 = λ 23 P2 (3) λ 23 P2 + λ 13 P1 = rP3 (4) P1 + P2 + P3 = 1 1 λ Solving (2) for P2 results in: P2 = 12 P1 = P1 =.5 P1 2 λ 23 λ P + λ 13 P1 2(.5P1 ) + P1 = =.5P1 P3 from (3) is: P3 = 23 2 4 r 1 Solving (4) for P1 results in: P1 +.5 P1 +.5 P1 = 1 → P1 = =.5 2 Therefore, P2 = P3 =.25 and Availability=P1+P2=.75 .

11-2

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

Waveland Press, Inc., Copyright © 2009

11.8 A(T ) =

1 − e − λT 1 − e −.00314 T = λ T + t1 + t2 (1 − e − λt ) .00314 T +.25 + (1 − e −.00314 T )

T in days A(T) 1 .797 2 .884 3 .916 4 .932 6 .948

8 10 11 12 14

.955 .9576 .9582 .9585 .9585

15 16 18 20

.9579 .9574 .9560 .9544

11.9 λ 2

r r

3

4

λ

r

λ

State 1 2 3 4

λ

1

Steady-state equations: (1) rP2 + rP3 = 2λP1 (4) λP2 + λP3 = rP4 From (3): P3 =

FG λ IJ P H λ + rK

Unit 1 operating failed operating failed

Unit 2 operating operating failed failed

(2) λ 1 P1 + rP4 = rP2 + λP2 (5) P1 + P2 + P3 + P4 = 1

(3) λ 1 P1 = rP3 + λP3

1

FG IJ H K λP + λP λ λF λ I Solving (4) for P results in: P = = P + G JP r r r H λ + rK λ L 2λ F λ IJ P OP + λ FG λ IJ P = 2λ P = M P −G H λ + rK Q r H λ + rK r rN r

Solving (1) for P2 results in: P2 =

λ 2λP1 − rP3 2λ P1 − P = r r λ +r 1 2

4

3

4

2

1

2

1

1

F 2λ − λ + λ + 2λ IJ P = 1 → From (5): G1 + H r λ +r λ+r r K

1

2

2

1

2

L 2λ + 2λ OP P = M1 + N r r Q 2

1

1

LM N

−1

2

2λ2 2λ 2λ2 Therefore, Availability = P1 + P2 + P3 = 1 − P4 = 1 − 2 1 + + 2 r r r

OP Q

−1

11.10 λ 1 = λ 2 = λ 1

r

λP1 = rP2 → P2 =

λ 2 2r λ

4

2rP4 = λP2

→

λ r

P4 =

P1

λ 2r

P2 =

λ2 2r 2

P1

L λ λ OP P+ P+ P = 1 → P = M1 + + r 2r N r 2r Q λ

1

λ2

1

11-3

2

2

1

1

2

−1

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

Waveland Press, Inc., Copyright © 2009

LM1 + λ + λ OP Availability = 1 − P = 1 − 2r N r 2r Q λ2

2

2

2

4

11.11 State 1 2 4

Primary on-line failed failed

−1

Backup stand-by on-line failed

Steady-state equations are: (1) r1P2 = λ 1 P1 (4) λ 2 P2 = r2 P4

(2) λ 1 P1 + r2 P4 = r1 P2 + λ 2 P2 (5) P1 + P2 + P4 = 1

Solving (1) for P2 results in: P2 = Solving (4) for P4 results in: P4 =

λ1 r1

λ2 r2

P1

P2 =

λ 1λ 2 r1r2

P1

L λ + λ λ OP P = 1 → P = M1 + From (5): P + P + r rr N r rr Q λλ L λ λλ O Availability = 1 − P = 1 − M1 + r + r r PQ rr N λ1

1

λ 1λ 2

1

1

2

1

1 2

1

1 2

−1

1

1 2

−1

1 2

1

1 2

1 2

1

1 2

4

11.12 From 11.5 the steady state availability is .8 which means that: .799 ≤ A( t ) ≤.801 r .4 .1 −.5( t ) λ − ( λ +r )2 e e A( t ) = + = + =.8+.2e −.5t r+λ r+λ .4+.1 .4+.1 A(t) t 10 .8013 11 .8008 11 days

11.13

r .2 .0031 − (.0031+.2)10 λ − ( λ +r )t e e + → Aps (10) = + =.9867 r+λ r+λ .2+.0031 .2+.0031 λ − ( λ +r )t r .33 .01 − (.01+.33)10 e e + → Atr (10 ) = + =.9716 Atr ( t ) = r+λ r+λ .33+.01 .33+.01 Aps ( t ) =

A( t ) = Aps ( t ) Atr ( t ) =.9867⋅.9716 =.9587 Aps =

.2 .33 r = =.9847 and Atr = =.9706 .33+.01 r + λ .2+.0031

A = Aps Atr =.9847⋅.9706 =.9557 and %difference = High Level Redundancy: As (10) = 1 − (1 − A(10))2 = 1 − (1−.9587)2 =.9983 As = 1 − (1 − A)2 = 1 − (1−.9557)2 =.9980

11-4

.9587−.9557 =.0031 .9557

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

Waveland Press, Inc., Copyright © 2009

.2 .0031 + 1 − e − (.0031+.2)10 =.9913 .2+.0031 (.2+.0031)210 .33 .01 Atr10 = + 1 − e − (.01+.33)10 =.9790 .33+.01 (.33+.01)210 A10 = Aps10 Atr10 =.9913⋅.9790 =.9705 Aps10 =

As10 = 1 − (1 − A10 )2 = 1 − (1−.9705)2 =.9991

Low Level Redundancy: As (10) = (1 − (1 − Aps (10))2 ) × (1 − (1 − Atr (10))2 ) = (1 − (1−.9867)2 ) × (1 − (1−.9716)2 ) =.9990 As = (1 − (1 − Aps )2 ) × (1 − (1 − Atr )2 ) = (1 − (1−.9847 )2 ) × (1 − (1−.9706 )2 =.9989

As10 = (1 − (1 − Aps10 )2 ) × (1 − (1 − Atr10 )2 ) = (1 − (1−.9913)2 ) × (1 − (1−.9790)2 ) =.9995 11.14 MTBF = 400Γ(1 + 1 / 2.4 ) = 354.54 , MTTR = tmed e s / 2 = 4.8e1.2 / 2 = 9.86 MTBF 354.54 = =.9729 a) Ai = MTBF + MTTR 354.54 + 9.86 b) 1 1 1 λMTTR + MPMT 9.86 + 6 TP 200 = 7.39 hrs M= = 354.54 1 1 1 λ+ + 354.54 200 TP MTBM 127.87 1 1 MTBM = = =.945 = = 127.87 , Aa = 1 1 1 1 MTBM + M 127.87 + 7.39 + + MTBF TP 354.54 200 2

11.15

z

2

F TI t F TI T T T− R(T )T + tf ( t )dt GH1 − b JK T + b dt GH1 − b JK T + 2b 2b = = = A(T ) = TO TO T + t + t 1 − R( t ) LT O L L T +t +t M P T +t +t M P T +t +t M P NbQ NbQ NbQ

z

T

2

0

0

1

2

T

2

1

2

1

2

1

2

T2 10,000 A(T ) = T: 300 400 500 600 T A(T): .8863 .8935 .8942 .8915 T + 24 + 72 5,000 An interval of 500 appears optimal. Since availability seems to be insensitive to changes in T in the neighborhood of 500 hours, no further refinement to T is made.

T−

LM N

11.16 A′( t ) = =

z

t

0

OP Q

1 − H ( t ′ ) f ( t ′ )dt ′ ==

LM OP − e Nλ + r Q λ + r λ

λ

z

t

0

z

t

e − rt ′ λe − λt ′ dt ′ = λ e − ( r + λ )t ′ dt ′ 0

− ( r +λ )t

11-5

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

Waveland Press, Inc., Copyright © 2009

A( t ) = 1 − A′( t ) = 1 −

LM λ OP − λ e Nλ + r Q λ + r

− ( r + λ )t

11.17 Steady-state equations are: (1) rP2 + rP3 + rP4 = 2λP1 (4) λP2 + λP3 = rP4

λ

λ +r

(3) λP1 = rP3 + λP3

P1

2λP2 2λ2 = 2 P1 r r r Substituting the above into (5):

From (4): P4 =

λP2 + λP3

r λ + r − λ + λe − ( r + λ ) t λe − ( r + λ ) t = + λ+r λ +r λ +r

(2) λP1 = rP2 + λP2 (5) P1 + P2 + P3 + P4 = 1

From (2): P2 =

From (2)and (3): P2 = P3 ;

=

=

L 2λ + 2λ OP → P = M1 + N λ +r r Q 2λ L 2λ 2λ O Availability = P + P + P = 1 − P = 1 − 1+ + M r N λ + r r PQ λ

λ

2λ2 P1 + P1 + P1 + 2 P1 = 1 r λ +r λ +r 1

2

3

2

1

4

−1

2

2

2

2

2

−1

11.18

L λ + λ λ OP = LM1 + .004 + .004(.002) OP = 0.676 a) P = M1 + N r r Q N .01 .01 Q −1

1

1

1

P2 =

−1

2

2

λ1 r

P1 =

2

.004 .676 = 0.270 → A1 = A2 = A3 = AISOC = P1 + P2 =.946 .01

b) ISOC is in series with all 3 sensors: Asys3 = A1 A2 A3 AISOC = (.946 )4 = 0.801 c) ISOC is in series with 2 out of 3 redundant sensors: 3 3 x 3− x = (.946 )(.14498+.8466 ) = 0.938 .946 1−.946 Asys 2 = AISOC x x =2

b

gLM∑ FGH IJK b g b N

g OP Q

11.19 a) STATES FOR SENSOR 1, ISOC, AND SHARED BACKUP Sensor 1 ISOC Backup State 1 on-line on-line stand-by 2 failed on-line on-line 3 on-line failed on-line 4 failed failed on-line 5 on-line failed failed 6 failed on-line failed 7 failed failed failed

11-6

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

Waveland Press, Inc., Copyright © 2009

λ1

λ2

1 r

r

2

λ2

λ1

r

r

λ3 r

4

3

r

λ3

5

r

λ1

r

λ3

6

r

λ2

7

Availability of sensor 1=P1+P2+P3 Availability of ISOC= P1+P2+P3+P4+P5+P6 Solve using transition matrix: −( λ 1 + λ 2 ) r

OP 0 0 r 0P −( λ + λ ) − r r 0 0 0P −( λ + λ ) − r r r P λ λ 0 0 − λ − 2r rP 0 0 0 λ −λ − r rP P λ 0 0 0 −λ − r r P 1 1 1 1 1 1PQ LM.46468OP LM0OP MM0PP MM.18587 PP .18587 0 M P M P P = T b = M.07435P b = M 0P MM0PP MM.03717PP MM0PP MM.03717PP N.01487Q N1Q

LM MM λ λ M 0 T=M MM 0 MM 0 N 1 LM P OP MM P PP P M P P = MP P MM P PP MM P PP NP Q

1

2

2

0

r

0

3

3

1

2

1

3

3

1

3

1

2

3

2

−1

4 5 6 7

a) Availability of ISOC= P1+P2+P3+P4+P5+P6=1-P7=.9851 b) Availability of sensor 1=A1=P1+P2+P3=.8364 Availability of sensors 2 and 3 same as in problem 11.18: A2 = A3 = 0.946 Availability of system=A1A2A3AISOC=(.8364)(.946)(.946)(.9851)=.7374 11.20

λP1 = rP2 → P2 =

λ r

0

P1

11-7

0

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

Waveland Press, Inc., Copyright © 2009

λP1 + rP3 = ( r + λ ) P2 → P3 = rP4 = λP3 → P4 =

λ

P3 =

r

λ3 r3

FG H

IJ K

(r + λ ) ( r + λ )λ λ λ λ2 P1 = 2 P1 P2 − P1 = − 2 r r r r r

P1

F λ λ λI F λ λ λI 1= P + P + P + P = G1 + + + J P → P = G1 + + + J H r r rK H r r rK 2

1

2

3

4

3

2

1

2

1

3

−1

2

3

3

λ = 2 and r = 5

L 2 4 + 8 OP =.6158, P = 2 P =.2463, P = 4 P =.0985 P = M1 + + 5 5 N 5 25 125 Q −1

1

2

1

3

1

A = P1 + P2 + P3 =.9606 11.21 State 1 2 3 4

Unit 1 operating failed operating failed

Unit 2 operating operating failed failed

Steady-state equations: (1) rP2 + rP3 = 2λP1 (3) λ 1 P1 + rP4 = rP3 + λP3 (5) P1 + P2 + P3 + P4 = 1

(2) λ 1 P1 + rP4 = rP2 + λP2 (4) λP2 + λP3 = 2rP4

From (2) and (3): P2 = P3

λ 2λ P1 = P1 2r r λP + λP3 2λP2 λ λ λ2 P1 = 2 P1 Solving (4) for P4 results in: P4 = 2 = = r r 2r 2r r Solving (1) for P2: rP2 + rP2 = 2λP1 → P2 =

FG IJ H K F λ λI r λ λ λ From (5): P + P + P + P = 1 → P = G1 + 2 + J = r r r H r r K r + 2λr + λ 2

2

1

1

1

2

1

1

2

−1

2

2

2

λ2 λ2 and Availability=1 − P = − 1 4 r 2 + 2λr + λ2 r 2 + 2λr + λ2 2 2 n r (r + λ ) − r λ2 Equation: As ( t ) = 1 − ∏ (1 − Ai ( t )) = 1 − 1 − = 1− = 1− r+λ r+λ ( r + λ )2 i =1

Therefore, P4 =

FG H

IJ K

11.22 (a) rate diagram

11-8

FG H

IJ K

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

Waveland Press, Inc., Copyright © 2009

(b)

4P1 = 4P2 + 4P3 12P3 = 2P1 4P4 = 8P2 + 8P3 P1 + P2 + P3 + P4 = 1

11.23 (a)

(c) P1 =6/24; P2 = 5/24; P3 -1/24; P4 = 12/24; A = P1 + P2 + P3 = 12/24 = .5

1

.1

2

.5 .1

3

(b) .1P1 = .5P3 ; .1P1 = .1P2 ; P1 + P2 + P3 =1 (c) P3 = (1/5) P1 ; P2 = P1 ; P1 = [1 + 1 + 1/5]-1 = 5/11;

A = P1 + P2 = 10/11 = .909

11.24

( 30 )( 24 ) = 8.12; MTTR = 10e.84 /2 = 14.23 100Γ (1 + 1/ 2.4 ) ___ ( 30 )( 24 ) = 64.75; M (8.12 )(14.23) + 3(2) MTBM = =

m(30) =

2

8.12 + 3 8.12 + 3 MTBM 64.75 Aa = = = .8555 ___ + 94.75 10.93 MTBM + M

11.25

m(30) = .51095 ⋅10−7 ( 30 ⋅ 24 )

2.74

= 3.4472

3.4472 ( 6.7 ) + 5(4) 720 = 85.23 = 5.1; MTBM = 8.4472 8.4472 85.23 = .9435 Aa = 85.23 + 5.1

___

M=

11.26 Aa = (.8555) (.9435) = .8072 11.27 M(100) = .00025 (100)2.4 = 15.77 failures; MTTR = 4 Exp[.852 /2] = 5.74 days A = 100 / [100 + 15.77 x 5.74] = .525 11.28 rate diagram: ( 1)→3λ→(2)→2λ→(3)→λ→(4)→r→(1) Equations: 3λP1 =rP4; 2λP2 = 3λP1; λP3 = 2λP2 ; P1 + P2 + P3 + P4 = 1 Solving: P1 = .1639, P2 = .2459, P3 =.4918, Availability = .9016

11-9

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

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11.29 Steady-state equations: 3P1 = 8P2 + 4P3, 12P2 = 2P1, 8P4 = 4P2 + 2P3 Solving: P1 = 48/85, P4 = 9/85, A = 1 – 9/85 = 76/85 = .894 11.30 2λP1 = rP2, λ+ P2 = rP3, P1 + P2 + P3 = 1

1

2λ r 2

r

λ+

3

P1 = 1 / [1 + 2λ/r + 2λλ2 / r2 ] = .6098, P2 = (2λ/r) P1 = .2439; A = .8537 11.31 (a) m(5000) = .000023(5000)1.7 = 44.67, MTTR = 7exp[.64/2] = 9.64, MTBF = 5,000/44.67 = 111.94; Ainh = 111.94 (111.94 + 9.64) = .921 (b) m(500) = .000023(500)1.7 = .8912 expected nbr failures + 1 PM

M=

.8912(9.64) + 1(12) = 10.888 hr .8912 + 1

MTBM = 500 /1.8912 = 264.38; Aach =

264.38 = .96 264.38 + 10.888

≈4 11.32 MTTR = 3e m(1000,1250) = .0001[12501.825 – 10001.825] = 15.0; .7582 / 2

A = 250 / [250 + 15(4) + 10(1)] = 250/ 320 = .781

11-10

250/T = 250 / 25 = 10 sched. maint.

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 12

12.1

i 1 2 3 4 5

1 /n 0 .2 0 .4 0 .6 0 .8 1

i/( n + 1 ) 0 .1 6 6 6 6 7 0 .3 3 3 3 3 3 0 .5 0 .6 6 6 6 6 7 0 .8 3 3 3 3 3

m e d ia n 0 .1 2 9 0 .3 1 3 0 .5 0 .6 8 6 0 .8 7

( i- .3 ) /( n + .4 ) 0 .1 2 9 6 2 9 6 0 .3 1 4 8 1 4 8 0 .5 0 .6 8 5 1 8 5 2 0 .8 7 0 3 7 0 4

1 0 .8 i/n 0 .6

i/( n + 1 )

0 .4

m e d ia n ( i- .3 ) /( n + .4 )

0 .2 0 12

243

318

502

771

12.2 EMPIRICAL RELIABILITY CALCULATION FOR UNGROUPED DATA - complete or singly censored TIME RELIABILITY 0 1 31.7 .9375 87.9 .875 89.7 .8125 104.2 .75 111.9 .6875 130.3 .625 131.8 .5625 159 .5 160.4 .4375 171.8 .375 173.5 .3125 178.9 .25 187.1 .1875 244.1 .125 437.1 .0625

CUM PROB (CDF) 0 .0625 .125 .1875 .25 .3125 .375 .4375 .5 .5625 .625 .6875 .75 .8125 .875 .9375

DENSITY (PDF) HAZARD RATE 1.971609E-03 1.971609E-03 .0011121 1.18624E-03 3.472231E-02 3.968264E-02 4.310345E-03 5.30504E-03 8.116879E-03 .0108225 3.396739E-03 4.940711E-03 4.166667E-02 6.666667E-02 2.297794E-03 4.084968E-03 4.464305E-02 .0892861 5.482452E-03 1.253132E-02 3.676477E-02 9.803939E-02 1.157409E-02 3.703708E-02 7.62194E-03 3.048776E-02 1.096491E-03 5.847953E-03 3.238342E-04 2.590674E-03

MTTF= 159.96 VARIANCE= 8498.385 STD DEV= 92.18668

9219 . . s 9219 ±t . .

MTTF → 108.90,21102 : 159.96 ± t .025,14 : 159.96 ± 2145 α / 2 , n −1 15 n 15

12-1

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

Waveland Press, Inc., Copyright © 2009

12.3 GROUPED DATA CALCULATIONS UPPER BOUND 0.0000 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000

NO. SURV 300.0000 285.0000 265.0000 247.0000 220.0000 185.0000 154.0000 109.0000 66.0000 0.0000

RELIABILITY 1.0000 0.9500 0.8833 0.8233 0.7333 0.6167 0.5133 0.3633 0.2200 0.0000

DENSITY 0.0500 0.0667 0.0600 0.0900 0.1167 0.1033 0.1500 0.1433 0.2200

HAZARD RATE 0.0500 0.0702 0.0679 0.1093 0.1591 0.1676 0.2922 0.3945 1.0000

COMPUTED MTTF= 5.603333 COMPUTED STANDARD DEVIATION= 2.454788 The failure rate is IFR which is most likely due to wearout.

12.4 a) I 1 2 3 4 5 6 7 8 9 10 11 12

EMPIRICAL RELIABILITY CALCULATIONS - ungrouped multiply censored data adjusted rank method TIME 5 12 15 + 22 27 35 + 49 71 + 73 81 112 + 117

1 1

RANK INCR 1 2

ADJ RANK RELIABILITY .9435484 .8629032

1.1 1.1

3.1 4.2

.7741936 .6854839

1.257143

5.457143

.5841014

1.508571 1.508571

6.965714 8.474286

.4624424 .3407834

2.262857

10.73714

.158295

MULTIPLY CENSORED RELIABILITIES product limit method I 1 2 3 4 5 6 7 8 9 10 11 12

TIME 5 12 15 + 22 27 35 + 49 71 + 73 81 112 + 117

FACTOR .9230769 .9166667 1 .9 .8888889 1 .8571429 1 .8 .75 1 .5

RELIABILITY .9230769 .8461539 .7615384 .6769231 .5802198 .4641759 .3481319 .1740659

12-2

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

Waveland Press, Inc., Copyright © 2009

b) There are three non-failed units at time 117 when the 8 failure occurs and the test is presumed to have ended. These three units can be considered censored at 117 so their times are 117+, 117+, and 117+.

MULTIPLY CENSORED RELIABILITIES product limit method I TIME 1 5 2 12 3 15 + 4 22 5 27 6 35 + 7 49 8 71 + 9 73 10 81 11 112 + 12 117 13 117 + 14 117 + 15 117 +

FACTOR .9375 .9333333 1 .9230769 .9166667 1 .9 1 .875 .8571429 1 .8 1 1 1

RELIABILITY .9375 .875 .8076923 .7403846 .6663461 .5830529 .4997596 .3998077

EMPIRICAL RELIABILITY CALCULATIONS - ungrouped multiply censored data adjusted rank method I TIME 1 5 2 12 3 15 + 4 22 5 27 6 35 + 7 49 8 71 + 9 73 10 81 11 112 + 12 117 13 117 + 14 117 + 15 117 +

RANK INCR ADJ RANK RELIABILITY 1 1 .9545454 1 2 .8896104 1.076923 1.076923

3.076923 4.153846

.8196803 .7497503

1.184615

5.338462

.6728271

1.332692 1.332692

6.671154 8.003846

.5862887 .4997503

1.599231

9.603077

.3959041

12-3

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Waveland Press, Inc., Copyright © 2009

12.5 LIFE TABLE INTVL

FAILURE

1.000 2.000 3.000 4.000 5.000 6.000 7.000 8.000 9.000 10.000

5.000 6.000 12.000 20.000 18.000 25.000 27.000 33.000 31.000 38.000

WITHDRAW

RISK

ADJ-RISK

P-DIE

P-SURVIVE

15.000 500.000 492.500 26.000 480.000 467.000 14.000 448.000 441.000 23.000 422.000 410.500 27.000 379.000 365.500 32.000 334.000 318.000 46.000 277.000 254.000 38.000 204.000 185.000 34.000 133.000 116.000 30.000 68.000 53.000

0.010 0.013 0.027 0.049 0.049 0.079 0.106 0.178 0.267 0.717

0.990 0.987 0.973 0.951 0.951 0.921 0.894 0.822 0.733 0.283

RELIABILITY

0.990 0.977 0.951 0.904 0.860 0.792 0.708 0.582 0.426 0.121

Reliability over a 5-year period if engines overhauled every 2 years: 2 R(5) = R( 2 ) ⋅ R( 2 ) ⋅ R(1) = .977 .99 =.945

b g

→ R( 200) = 1 − 2 / 30 =.9333 12.6 R( t0 ) = 1 − r / n a) and F2 = Fα / 2, 2 r + 2, 2 n −2 r = F.05, 6,56 = 2.25 F1 = Fα / 2, 2 n − 2 r + 2, 2 r = F.05,58, 4 = 5.69 RL =

1 1 = =.8058 3 r +1 1+ F2 1 + 2.25 28 n−r

Ru =

and

F1 F1 +

r n − r +1

F1

=

5.69 +

2 29

=.9880

b)

F2 = Fα / 2, 2 r + 2, 2 n−2 r = F.10, 6,56 = 187 . →

RL =

1 1 = =.8331 3 r +1 1+ .

F2 1 + 187 28 n−r

12.7 LIFE TABLE INTVL FAILURE WITHDRAW RISK ADJ-RISK P-DIE 1.000 5.000 2.000 100.00 99.000 0.051 2.000 8.000 4.000 93.000 91.000 0.088 3.000 12.000 2.000 81.000 80.000 0.150 4.000 18.000 10.000 67.000 62.000 0.290 5.000 24.000 12.000 39.000 33.000 0.727

P-SURVIVE 0.949 0.912 0.850 0.710 0.273

RELIABILITY 0.949 0.866 0.736 0.522 0.142

Probability a patient will survive at least 5 years: 0.142 12.8 GROUPED DATA CALCULATIONS UPPER BOUND 0.0000 100.0000 200.0000 300.0000 400.0000 500.0000

NO. SURV

RELIABILITY

44.0000 41.0000 36.0000 28.0000 18.0000 6.0000

CUM PROB(CDF)

1.0000 0.9318 0.8182 0.6364 0.4091 0.1364

0.0000 0.0682 0.1818 0.3636 0.5909 0.8636

DENSITY

HAZARD RATE

0.0007 0.0011 0.0018 0.0023 0.0027

0.0007 0.0012 0.0022 0.0036 0.0067

The hazard rate is increasing. The MTTF cannot be calculated as the data is incomplete (all 44 units did not fail).

12-4

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12.9 EMPIRICAL RELIABILITY CALCULATIONS - UNGROUPED MULTIPLY CENSORED DATA

product limit method TIME 8 12 13 + 15 + 18 24 + 25 27 29 + 30 33 37

RELIABILITY .9230769 .8461539 1 .7521368 1 .6446887 .5372406 1 .4029304 .2686203 .1343102

rank adjustment

Kaplan Meier method

RELIABILITY .9435484 .8629032

RELIABILITY .9166667 .8333334

.7643369

.7291667

.6516897 .5390425

.6076389 .4861111

.3982335 .2574245 .1166155

.3240741 .162037 0

The reliability at the end of 2-year warranty as calculated by the 3 methods: .752, .764, & .729.

12.10 EMPIRICAL RELIABILITY CALCULATIONS - UNGROUPED MULTIPLY CENSORED DATA

MULTIPLY CENSORED RELIABILITIES product limit method I TIME 1 10 2 15 + 3 33 4 36 5 42 6 42 + 7 50 + 8 55 9 59 10 61 11 62 12 65 13 68 14 71 15-25 72 + 1

FACTOR .9615384 1 .9583333 .9565217 .9545454 1 1 .9473684 .9444444 .9411765 .9375 .9333333 .9285714 .9230769

RELIABILITY .9615384 .9214743 .8814102 .8413461

.7970648 .7527834 .708502 .6642206 .6199393 .5756579 .5313765

Estimate of reliability based on Equation12-18: .5314 12.11 R( t0 ) = 1 − r / n → R( 2000) = 1 − 15 / 500 =.97; F2 = Fα / 2, 2 r + 2, 2 n − 2 r = F.05,32,970 = 146 .

RL =

1 1 = =.954 16 r +1 146 1+ .

F2 1 + 485 n−r

Therefore, the product specification is being met.

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12.12

EMPIRICAL RELIABILITY CALCULATIONS FOR UNGROUPED DATA - complete or singly censored

TIME RELIABILITY 0 1 134 .952381 220 .9047619 251 .8571429 286 .8095238 365 .7619048 465 .7142857 543 .6666667 752 .6190476 832 .5714286 912 .5238096

CUM PROB (CDF)

DENSITY (PDF)

0 4.761904E-02 9.523809E-02 .1428571 .1904762 .2380952 .2857143 .3333333 .3809524 .4285714 .4761904

3.55366E-04 5.537098E-04 1.536098E-03 1.360544E-03 6.027727E-04 4.761904E-04 6.105006E-04 2.278423E-04 5.952381E-04 5.952381E-04

HAZARD RATE 3.55366E-04 5.813953E-04 1.697793E-03 1.587302E-03 7.446017E-04 .000625 8.547009E-04 3.417635E-04 9.615385E-04 1.041667E-03

Probability an alternator fails during warranty period: 1-.762=.238.

Estimated number of failed alternators at end of warranty period: 5000(.238)=1190.

12.13 EMPIRICAL RELIABILITY CALCULATIONS FOR UNGROUPED DATA - singly censored Design A: TIME

RELIABILITY

CUM PROB (CDF)

DENSITY (PDF)

HAZARD RATE

0 44 77 218 251 317 380 438 739 758 1115

1 .9375 .875 .8125 .75 .6875 .625 .5625 .5 .4375 .375

0 .0625 .125 .1875 .25 .3125 .375 .4375 .5 .5625 .625

1.420455E-03 1.893939E-03 4.432624E-04 1.893939E-03 9.469697E-04 9.920635E-04 1.077586E-03 2.076412E-04 3.289474E-03 1.7507E-04

1.420455E-03 2.020202E-03 5.065856E-04 2.331002E-03 1.262626E-03 1.443001E-03 1.724138E-03 3.691399E-04 6.578947E-03 4.001601E-04

Design B: TIME

RELIABILITY

CUM PROB (CDF)

DENSITY (PDF)

HAZARD RATE

0 32 63 211 248 327 404 476 877 903 1416

1 .9375 .875 .8125 .75 .6875 .625 .5625 .5 .4375 .375

0 .0625 .125 .1875 .25 .3125 .375 .4375 .5 .5625 .625

1.953125E-03 2.016129E-03 4.222973E-04 1.689189E-03 7.911393E-04 8.116883E-04 8.680556E-04 1.558604E-04 2.403846E-03 1.218324E-04

1.953125E-03 2.150538E-03 4.826255E-04 2.079002E-03 1.054852E-03 1.180638E-03 1.388889E-03 2.770851E-04 4.807692E-03 2.78474E-04

Design B provides the best long term reliability - there is a .375 probability of surviving 1416 thousand cycles versus 1115 thousand cycles for design A. Design A has slightly better short term reliability there is a .8125 probability of surviving 218 thousand cycles versus 211 thousand cycles for design A. However, the median time to failure for design A is 739 thousand cycles vs 877 thousand cycles for design B. Select design B!

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12.14 EMPIRICAL CALCULATIONS FOR UNGROUPED DATA - complete or singly censored TIME 0 4 4.4 4.5 5 5.1 5.5 5.7 5.8 6 6.5 7 7.5 7.7 8 8.5 8.9 10 10.2 11.5 14

1-H(T) H(t)-CUM PROB (CDF) DENSITY (PDF) 1 0 1.190476E-02 .952381 4.761904E-02 .1190476 .9047619 9.523809E-02 .4761909 .8571429 .1428571 9.523809E-02 .8095238 .1904762 .4761909 .7619048 .2380952 .1190476 .7142857 .2857143 .2380954 .6666667 .3333333 .4761886 .6190476 .3809524 .2380954 .5714286 .4285714 9.523809E-02 .5238096 .4761904 9.523815E-02 .4761905 .5238096 9.523809E-02 .4285714 .5714285 .2380954 .3809524 .6190476 .15873 .3333333 .6666666 9.523809E-02 .2857143 .7142857 .1190478 .2380952 .7619048 4.329003E-02 .1904762 .8095238 .2380954 .1428571 .8571429 3.663003E-02 .0952381 .9047619 1.904762E-02 4.761905E-02 .952381

HAZARD RATE 1.190476E-02 .125 .5263163 .1111111 .5882359 .15625 .3333336 .714283 .3846157 .1666667 .1818182 .2 .5555561 .4166664 .2857143 .4166671 .1818181 1.250001 .2564102 .2

MTTF= 7.2899 VARIANCE= 6.876747 STD DEV= 2.622355

s 2.622 2.622 ±t ; 7.2899 ± t .025,19 → 6.063,8.517 MTTR ; 7.2899 ± 2.093 α / 2 , n −1 n 20 20 Goal of MTTR = 4 hours not being met and only 80.95% of repairs completed in 10 hours.

12.15

GROUPED DATA CALCULATIONS

UPPER BND

NO. SURV

0.0000 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000

40.0000 33.0000 28.0000 21.0000 15.0000 7.0000 4.0000 1.0000 0.0000

1-H(t)

CDF

DENSITY

HAZARD RATE

1.0000 0.8250 0.7000 0.5250 0.3750 0.1750 0.1000 0.0250 0.0000

0 .175 .3 .475 .625 .825 .9 .975 1

0.1750 0.1250 0.1750 0.1500 0.2000 0.0750 0.0750 0.0250

0.1750 0.1515 0.2500 0.2857 0.5333 0.4286 0.7500 1.0000

COMPUTED MTTF= 3.225 COMPUTED STANDARD DEVIATION= 1.93633 The MTTR is 3.225 hours.

The probability of the production line being down for more than 6 hours is .1.

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12.16 Treating the heating failure times as censored times, the product limit estimator is found: MULTIPLY CENSORED RELIABILITIES product limit method I

TIME

FACTOR

1 2 3 4 5 6 7 8 9

190 + 250 432 + 673 + 780 891 922 1020 1500 +

1 .8888889 1 1 .8333333 .8 .75 .6666667 1

RELIABILITY

.8888889

.7407407 .5925926 .4444444 .2962963

Kaplan Meier method I TIME 1 190 + 2 250 3 432 + 4 673 + 5 780 6 891 7 922 8 1020 9 1500 +

RELIABILITY .875

.7 .525 .35 .175

From either estimate, R(500) < .88 and the suppliers claim is not supported.

12.17 TIME CUM PROB – H(t)(mean plotting position) 5.3 9.090906E-02 6.5 .1818182 7.8 .2727273 8.2 .3636364 10.4 .4545454 12.6 .5454545 13.2 .6363636 15.3 .7272727 18.4 .8181818 20.1 .9090909 MTTF= 11.78 VARIANCE= 25.39512 STD DEV= 5.039357

Goal is not supported. Ninety percent of the repairs are completed in approximately 20 hr.

95% CI: 8.174 hr < MTTR < 15.386 hr 12.18 Life Table year 1.000 2.000 3.000 4.000 5.000

failure censor risk adj-risk p-die 5.000 6.000 150.000 147.000 0.034 10.000 4.000 139.000 137.000 0.073 16.000 8.000 125.000 121.000 0.132 20.000 4.000 101.000 99.000 0.202 30.000 10.000 77.000 72.000 0.417

p-survive rel 0.966 0.966 0.927 0.895 0.868 0.777 0.798 0.620 0.583 0.362 = survival prob

12.19 (a) From the 4 percent and 96 percent ranked distributions, a 92 percent confidence interval on the cumulative probability of the 3rd failure time is given by: .052 < F(50) < .3772. (b) 7 < t(29%) < 192.

12-8

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That is, there a 92 percent confidence that approximately 29 percent of the failures will occur between time 7 and time 192.

12.20 The following table is computed n = 6. Consistent results are obtained for the middle observations.

0 1 2 3 4 5 6

(i-.3)/(n+0.4) (i-.05)/n i/(n+1) 0 0 0 0.1094 0.0833 0.143 0.2656 0.2500 0.286 0.4219 0.4167 0.429 0.5781 0.5833 0.571 0.7344 0.7500 0.714 0.8906 0.9167 0.857

12-9

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12.21 n = 70. Reliability with all failure modes: R(t) = RA(t) RB(t) RC(t)

i 0 1 2 3 4 5 6 7 8 9 10

Failure Failure Failure time Mode mode A mode B 1 3 23 38 66 95 100 130 136 138

A A A A C B C C A A

1 3 23 38 -66 -95 -100 -130 136 138

-1 -3 -23 -38 -66 95 -100 -130 -136 -138

Failure mode C

(n+1-i)/ (n+2-i)

-1 0.9859 -3 0.9857 -23 0.9855 -38 0.9853 66 0.9851 -95 0.9848 100 0.9846 130 0.9844 -136 0.9841 -138 0.9839

RA(t) 1.0

RB(t) 1.0

RC(t) 1.0

0.986 0.972 0.958 0.944 0.944 0.944 0.944 0.944 0.929 0.914

1.000 1.000 1.000 1.000 1.000 0.985 0.985 0.985 0.985 0.985

1.000 1.000 1.000 1.000 0.985 0.985 0.970 0.955 0.955 0.955

R(t)

1 0.986 0.972 0.958 0.944 0.930 0.915 0.901 0.887 0.873 0.859

About 10 percent will fail with the first 10,000 hours. Eliminating failure mode A, the probability of failing under warranty is now 0.045.

i 0 1 2 3 4 5 6 7 8 9 10

Failure Failure Failure time Mode mode A mode B 1 3 23 38 66 95 100 130 136 138

A A A A C B C C A A

1 3 23 38 -66 -95 -100 -130 136 138

-1 -3 -23 -38 -66 95 -100 -130 -136 -138

Failure mode C

(n+1-i)/ (n+2-i)

-1 0.9859 -3 0.9857 -23 0.9855 -38 0.9853 66 0.9851 -95 0.9848 100 0.9846 130 0.9844 -136 0.9841 -138 0.9839

12-10

RA(t) 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

RB(t) 1.0

RC(t) 1.0

1.000 1.000 1.000 1.000 1.000 0.985 0.985 0.985 0.985 0.985

1.000 1.000 1.000 1.000 0.985 0.985 0.970 0.955 0.955 0.955

R(t)

1 1.000 1.000 1.000 1.000 0.985 0.970 0.955 0.940 0.940 0.940

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 15

15.1 a) MTTF= 36.2, VARIANCE= 1682.443, STD DEV= 41.0176 b) R-SQUARED

EXPONENTIAL .7423481

WEIBULL .8911204

NORMAL .5958745

LEAST-SQUARES PROBABILITY PLOT FOR WEIBULL FAILURE/REPAIR TIME 1.5 9.5 14.2 18.7 19.1 26.8 27.2 30.8 31.3 33.1 35.9 57.9 164.6

F(ti) .0522388 .1268657 .2014925 .2761194 .3507463 .4253731 .5 .5746269 .6492537 .7238806 .7985075 .8731343 .9477612

INTERCEPT - A SLOPE - B ESTIMATED BETA ESTIMATED THETA CORRELATION R-SQUARED T-STAT FOR SLOPE B

LN(LN(1/(1-F))) -2.925223 -1.99756 -1.491606 -1.129704 -.8394879 -.5905285 -.3665129 -.1569012 4.658983E-02 .2522532 .4712546 .7249492 1.082459

-3.6821 1.001316 1.001316 39.53795 .9439917 .8911204 9.488357 with 11 deg of freedom

c) best fit is for the Weibull distribution MLE FOR THE WEIBULL DISTRIBUTION SAMPLE SIZE-NBR AT RISK NBR OF FAILURES BETA THETA

13 13 1.085579 37.48685

EST MEDIAN TIME TO FAILURE 26.74572 EST MODE OF FAILURE TIME 3.610448 MEAN (MTTF or MTTR) 36.332 STD DEVIATION 33.49884

15-1

LOGNORMAL .8507914

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

Waveland Press, Inc., Copyright © 2009

15.2 a) MTTF= 49.88889, VARIANCE= 291.906, STD DEV= 17.08526 b) R-SQUARED

EXPONENTIAL .5967343

WEIBULL .9190964

NORMAL .953867

LOGNORMAL .854711

LEAST-SQUARES PROBABILITY PLOT FOR NORMAL FAILURE/REPAIR TIME 18.3 38.7 43.1 45.4 50.6 51 58.8 63.6 79.5

F(ti) 7.446808E-02 .1808511 .287234 .393617 .5 .606383 .7127659 .819149 .9255319

INTERCEPT - A SLOPE - B ESTIMATED SIGMA ESTIMATED MEAN CORRELATION R-SQUARED T-STAT FOR SLOPE B SAMPLE MEAN

Zi -1.426255 -.8998006 -.5628688 -.2738667 0 .2738667 .5628688 .8998007 1.426255

-2.564947 5.141319E-02 19.45026 49.88889 .9766611 .953867 12.03059 with 7 deg of freedom 49.88889

c) the best fit is Normal distribution MLE FOR NORMAL DISTRIBUTION MEAN - MU STD DEV - SIGMA NBR AT RISK NBR OF FAILURES

49.88889 16.10814 9 9

15.3 a) MTTF= 98.96667, VARIANCE= 12.77232, STD DEV= 3.573838 b) R-SQUARED

EXPONENTIAL 7.291547E-02

WEIBULL .9510927

NORMAL .9670295

LEAST-SQUARES PROBABILITY PLOT FOR NORMAL FAILURE/REPAIR TIME 93.3 94.2 95 97.5 98.7 98.8 98.9

F(ti) 5.645161E-02 .1370968 .2177419 .2983871 .3790323 .4596774 .5403226

15-2

Zi -1.578393 -1.073978 -.7734843 -.531187 -.3121619 -.1031069 .103107

LOGNORMAL .9659423

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

Waveland Press, Inc., Copyright © 2009

100.1 101 101.7 104.1 104.3

.6209677 .7016129 .782258 .8629032 .9435484

INTERCEPT - A SLOPE - B ESTIMATED SIGMA ESTIMATED MEAN CORRELATION R-SQUARED T-STAT FOR SLOPE B SAMPLE MEAN

.3121619 .531187 .7734842 1.073978 1.578393

-24.99501 .2525599 3.959457 98.96667 .9833766 .9670295 17.1267 with 10 deg of freedom 98.96667

c) The best fit is for the Normal distribution.

MLE FOR NORMAL DISTRIBUTION MEAN - MU STD DEV - SIGMA NBR AT RISK NBR OF FAILURES

98.96667 3.421825 12 12

15.4 a) MTTF= 245.5263 VARIANCE= 47186.15 STD DEV= 217.2237 b) R-SQUARED TIME 68 79 90 93 120 123 134 162 167

EXPONENTIAL .8858965

WEIBULL .8793554

NORMAL .6805078

LEAST-SQUARES PROBABILITY PLOT FOR LOGNORMAL F(ti) Zi 3.608247E-02 -1.813462 204 .5515464 8.762886E-02 -1.334939 205 .6030928 .1391753 -1.064844 236 .6546392 .1907216 -.8645921 262 .7061856 .242268 -.6959938 330 .757732 .2938144 -.544123 386 .8092784 .3453608 -.4018556 396 .8608248 .3969072 -.2652659 421 .9123712 .4484536 -.1318989 1018 .9639176

INTERCEPT - A SLOPE - B ESTIMATED S ESTIMATED TMED CORRELATION R-SQUARED T-STAT FOR SLOPE B

-7.096882 1.350514 .7404586 191.5114 .9812422 .9628362 20.98638 with 17 deg of freedom

c) The best fit is for the LogNormal distribution.

MLE FOR LOGNORMAL DISTRIBUTION

15-3

LOGNORMAL .9628362

.1318989 .2652659 .4018556 .544123 .6959938 .8645922 1.064844 1.334939 1.813462

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

Waveland Press, Inc., Copyright © 2009

Tmed S

191.5115 .6659786

NBR AT RISK NBR OF FAILURES MEAN VARIANCE STD DEV MODE

19 19 239.0592 31900.4 178.6068 122.906

15.5 It is unlikely that the data came from an exponential distribution because the R-Squared is .1638808 which is very low. The LogNormal distribution provides the best fit.

b−t 1 , R( t ) = b b n −r r n−r r 1 1 b − tr b − tr a) L = ∏ = i =1 b b b b ln L = − r ln b + n − r ln( b − tr ) − ln b ln L − r n − r n − r − = − r b − tr + b n − r − n − r b − t r d = + db b b − tr b = − rb + rtr + bn − br − nb + ntr + rb − rtr = − rb + ntr = 0 n ⋅ tr b = r n ⋅ tr 20 ⋅ 90 b) b = = = 180 r 10

15.6 f ( t ) =

LM OPFG N QH

IJ K

b g

FG IJ FG H KH

IJ K

b

15.7 a)

g b g b gb

EMPIRICAL RELIABILITY CALCULATIONS

TIME 0 8 12 22 51 73 85

RELIABILITY CUM PROB (CDF) DENSITY (PDF) HAZARD RATE 1 0 4.032262E-03 4.032258E-03 .9677419 3.225809E-02 8.064508E-03 8.333334E-03 .9354839 6.451613E-02 3.225809E-03 3.448276E-03 .9032258 9.677422E-02 1.112346E-03 1.231527E-03 .8709677 .1290323 1.466277E-03 1.683502E-03 .8387097 .1612903 2.688169E-03 3.205128E-03 .8064516 .1935484

The empirical reliability at 100 days is .8065. b)

g

MLE FOR THE EXPONENTIAL DISTRIBUTION NBR AT RISK 30 NBR OF FAILURES 6 FAILURE OR REPAIR RATE 2.263297E-03 MEAN TIME TO FAILURE/REPAIR 441.8333 Median Time 306.2568

15-4

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

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R(100) = e − λt = e − t / MTTF = e−100/ 441.833 =.797 c)

b gb g

T = 8 + 12 + 22 + 51 + 73 + 85 + 24 100 = 2651 days, ℵ.205,14 = 23.7,ℵ.295,12 = 5.23

2T 2 ⋅ 2651 2T 2 ⋅ 2651 = 1013.77 = = 223.71, MTTFU = 2 = 2 ℵ.05,14 ℵ.95,12 23.7 5.23 223.71 < MTTF < 1013.77 MTTFL =

d) LEAST-SQUARES PROBABILITY PLOT FOR EXPONENTIAL SLOPE - FAILURE/REPAIR RATE 2.489536E-03 L-S ESTIMATE OF MEAN 401.6813 CORRELATION .947015 R-SQUARED .8968375 7.350833 with 4 deg of freedom T-STAT FOR SLOPE B

Therefore R(100) = e − t / MTTF = e −100/ 401.681 =.780 e)

t 0 = t1 = 8 r

∑ ti + ( n − r )tk − nt1

251 + ( 24)100 − 30 ⋅ 8 . = 401833 r 6 R(100) = e − ( t −t0 )/ MTTF = e− (100−8 )/ 401.833 =.795 = i=1 MTTF

=

15.8 a)

. +.5 + 10 + 2.4 + 15 . +.6 + 4.3 + 3 + 12 . + 18 . + 2.8 = 3 + 38 MTTR = 2.91 12 n 2 2 ∑ ti − n ⋅ MTTF 172.07 − 12 ⋅ 2.912 2 i =1 = = 6.40 → s = 2.53 s = 11 n −1 s s −t < MTTR +t < MTTR MTTR α / 2 , n −1 α / 2 , n −1 n n 2.53 2.53 < 2.91 + 1796 < MTTR . = 4.22 160 . = 2.91 − 1796 .

12 12

b) The LogNormal distribution is the best fit.

LEAST-SQUARES PROBABILITY DATA FOR LOGNORMAL INTERCEPT - A -.8185162 SLOPE - B 1.071109 ESTIMATED S .9336115 ESTIMATED TMED 2.147225 CORRELATION .9782265 R-SQUARED .9569271 T-STAT FOR SLOPE B 14.90518 with 10 deg of freedom

15-5

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c) F (3) = Φ

FG 1 ln 3 IJ = ΦFG 1 ln 3 IJ = Φ(.3582) =.64 K .

H s t K H .9336 21472 med

The maintainability objective of 90% is not being met as only 64% of the repairs are completed in 3 hours.

15.9 t 0 = t1 = 52 r

∑ ti + ( n − r )tk − nt1

= i=1 MTTF

=

1184 + (10 )100 − 25 ⋅ 52 = 58.933 15

r

58.933 MTTF ~ = 52 − = 49.643 t0 = t1 − 25 n

2 ⋅14 ⋅ 58.933 2( r − 1) MTTF < 2( r − 1) MTTF = 2 ⋅14 ⋅ 58.933 = 97.64 < MTTF 39.95 = = 2 2 ℵ1−α / 2, 2 r −2 413 16.9 . ℵα / 2, 2 r −2

MTTF 58.933 Fα , 2, 2 r −2 ≤ t0 ≤ t1 = 52 2.5 = t0 L = t1 − . = 52 − 4611 n 25 ( − ln R( t )) + t = 58.933( − ln.95) + 49.643 = 52.666 hrs td = MTTF 0

⋅ r = 120 ⋅ 40 = 4800 hrs 15.10 T = MTTF

2T 2 ⋅ 4800 2T 2 ⋅ 4800 = = 99.402, MTTFU = 2 = = 149.351 2 ℵ.1,80 96.578 ℵ.90,80 64.278 51 . = −99.4 ln.95 = − MTTFL ln( R ) ≤ td ≤ − MTTFU ln( R ) = −149.4 ln.95 = 7.7 MTTFL =

FG IJ F I H K , r=n 15.11 f ( t ) = G J e θ Hθ K F I FG IJ FG IJ 2 F t I GH JK F 2 I F 1 I 1I F H K = 2 G J ∏t e H K = G J G J ∏t e L(θ ) = ∏ G J e H K H K H K Hθ K θ θ θ θ −

2 t

n

i

−

ti

t

2

θ

2

n

n

θ

i =1

−

n

i =1

ti

2

θ

i

2n

n

−

n

i =1

ti

2

θ

i

n n t2 ln L = n ln 2 − 2n ln θ + ∑ ln ti − ∑ i2 i =1

n

i =1

θ

t n d ln L −2n 2i∑ = + =13 = −2nθ 2 + 2∑ ti2 = 0 i =1 dθ θ θ

FH ∑ t IK θ= n

i =1

n

2 i

2 i

.5

→

eqn 15.12 with β = 2 and n = r results in same θ equation

15-6

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

Waveland Press, Inc., Copyright © 2009

15.12

LEAST-SQUARES PROBABILITY PLOT FOR EXPONENTIAL

FAILURE/REPAIR TIME 6 8 12 18 36 37 49 55 67 78

F(ti) 2.755906E-02 6.692913E-02 .108011 .1490928 .1901746 .2312564 .2771714 .3230864 .372281 .4214756

SLOPE - FAILURE/REPAIR RATE L-S ESTIMATE OF MEAN CORRELATION R-SQUARED T-STAT FOR SLOPE B

LOG(1/(1-F)) 2.794593E-02 6.927412E-02 .1143014 .1614522 .2109366 .2629978 .3245831 .3902116 .4656627 .5472746 6.945575E-03 143.9766 .9914783 .9830291 22.19329 with 8 deg of freedom

MLE FOR THE EXPONENTIAL DISTRIBUTION NBR AT RISK NBR OF FAILURES NBR MULTIPLY CENSORS FAILURE OR REPAIR RATE MEAN TIME TO FAILURE/REPAIR Median Time

25 10 15 7.251631E-03 137.9 95.58539

15.13 Two-parameter Weibull: LEAST-SQUARES PROBABILITY PLOT FOR WEIBULL TIME F(ti) LN(LN(1/(1-F))) TIME F(ti) 51 2.302632E-02 -3.759492 142 .3519737 60 5.592105E-02 -2.85518 143 .3848684 61 8.881579E-02 -2.375046 182 .4177631 66 .1217105 -2.041922 193 .4506579 67 .1546053 -1.784079 200 .4835526 72 .1875 -1.571953 201 .5164474 86 .2203947 -1.390433 204 .5493421 94 .2532895 -1.230736 217 .5822368 108 .2861842 -1.087286 254 .6151316 127 .319079 -.9563086 374 .6480263

INTERCEPT - A SLOPE - B ESTIMATED BETA ESTIMATED THETA CORRELATION R-SQUARED T-STAT FOR SLOPE B

-8.997745 1.628538 1.628538 250.8973 .9243651 .8544508 10.27966 with 18 deg of freedom

15-7

LN(LN(1/(1-F)) -.8351164 -.7217132 -.6145616 -.5124371 -.4143314 -.3193859 -.2268419 -.1360024 -4.619714E-02 4.324988E-02

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

Waveland Press, Inc., Copyright © 2009

Three-parameter Weibull (for censored data set n=r) :

p =.8829( 20 )−.3437 =.315, t0 =

TIME 4.8 13.8 14.8 19.8 20.8 25.8 39.8 47.8 61.8 80.8

j = np + = 20⋅.315 + = 7, t1 = 51, t20 = 374, t j = t7 = 86 t1tn − t 2j t1 + tn − 2t j

(51)(374 ) − 862 = 46.2 51 + 374 − 2(86 )

=

LEAST-SQUARES PROBABILITY PLOT FOR WEIBULL (ti’=ti-46.2) F(ti) LOG(1/(1-F)) 2.302632E-02 -3.759492 95.8 .3519737 -.8351164 5.592105E-02 -2.85518 96.8 .3848684 -.7217132 8.881579E-02 -2.375046 135.8 .4177631 -.6145616 .1217105 -2.041922 146.8 .4506579 -.5124371 .1546053 -1.784079 153.8 .4835526 -.4143314 .1875 -1.571953 154.8 .5164474 -.3193859 .2203947 -1.390433 157.8 .5493421 -.2268419 .2532895 -1.230736 170.8 .5822368 -.1360024 .2861842 -1.087286 207.8 .6151316 -4.619714E-02 .319079 -.9563086 327.8 .6480263 4.324988E-02

INTERCEPT - A SLOPE - B ESTIMATED BETA ESTIMATED THETA CORRELATION R-SQUARED T-STAT FOR SLOPE B

-4.791591 .8829869 .8829869 227.3681 .981353 .9630537 21.66087 with 18 deg of freedom

The “fit” has been improved by taking into account a guarantee life.

15.14 p( x ) =

e − λ λx x!

e − λ λxi i =1 xi ! n

L( λ ) = ∏

→ n

n

ln L( λ ) = − λn + ∑ xi ln λ − ∑ ln xi ! i =1

i =1

n

∑ xi

n

d ln L( λ ) = − n + i =1 = 0 λ dλ 15.15

(a)

→

∑ xi

λ = i =1

n

MLE FOR THE EXPONENTIAL DISTRIBUTION

NBR AT RISK NBR OF FAILURES NBR MULTIPLY CENSORS FAILURE OR REPAIR RATE MEAN TIME TO FAILURE/REPAIR Median Time

12 8 4 1.589825E-03 629 435.9913

15-8

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

Waveland Press, Inc., Copyright © 2009

(b)

MLE FOR THE WEIBULL DISTRIBUTION SAMPLE SIZE-NBR AT RISK NBR OF FAILURES BETA THETA

12 8 .7699994 644.3325

EST MEDIAN TIME TO FAILURE EST MODE OF FAILURE TIME MEAN (MTTF or MTTR) STD DEVIATION (c)

400.3059 0 751.1656 987.1099

MLE FOR THE LOGNORMAL DISTRIBUTION Tmed = 340.3 and S = 1.948

1 = 12 . , tmed = e −1.2( −5.46 ) = 700.64 .833 1 50 = 1 − Φ −2.20 =.9861 R(50) = 1 − Φ ln 12 700.64 .

15.16 a = −5.46, b =.833, s =

FG H

15.17

IJ IJ KK

FG H

b

g

( ai + bi + ci )! ai bi qi qi (1 − q0 − qi )ci i =1 ai !bi !ci ! k

L=∏

k

log L = ∑ ( log( ai + bi + ci )!− log( ai !) − log( bi !) − log( ci !) + ai log q0 i =1

+ bi log qi + ci log(1 − q0 − q1 ) )

ci ∂ log L bi ( −1) = 0 = + ∂q1 qi 1 − q0 − qi

→

qik =

(1 − q0 )bi bi + ci

L OP M L a c ( −1)OP = 0 = ∑ M a − ∂ log L c = ∑M + ∂q MM q 1 − q − (1b− +q c)b PPP Nq 1− q − q Q N Q La OP = 1 ∑ a − ∑ c (b + c ) c (b + c ) = ∑M − c (1 − q ) N q (b + c )(1 − q ) − (1 − q )b Q q k

0

i =1

i

k

i

0

0

i

i =1

i

i

0

0

i

k

i =1

i

0

i

i

i

0

i

k

i

0

i

i

0

0 i =1

k

i

i

i

i =1

i

i

i

0

k

k

k

i =1

i =1

= ∑ ai (1 − q0 ) − q0 ∑ ( bi + ci ) = 0

→

∑ ai q0 = k i=1 ∑ ai + bi + ci i =1

15.18 From the 30 failure times, β-hat = 2.01 and θ-hat = 204.8. With z.025 = 1.96, 1.52 ≤ β ≤ 2.66 and 169.9 ≤ θ ≤ 246.9. (note: data was generated from a Weibull distribution with β = 2 and θ = 200.)

15.19 (a) no, R(35) = .8461 → 15 percent of the failures occur within 35 hr

15-9

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

Waveland Press, Inc., Copyright © 2009 empirical

Product Limit

Number

Time

R(t)

0 1 2 3 4 5 6 7 8 9 10 11 12

0 28.0 35.0 65.0+ 110.0 183.0+ 220.0 380.0 415.0 500.0+ 500.0+ 500.0+ 500.0+

1.0 .9231 .8462 .7615 .6663 .5712 .476

(b) λ-hat = .001746, R(50) = exp[-.001746 (50)] = .9164; t90 = 60.3 hr. (c) β = .935, θ = 590.569, R(50) = .9054; t90 = 53.2 hr.

15.20 (a) Exponential (b) Weibull

MTTF 731 541

Least-Squares R(50) parameters .934 .947 β = 1.1941 θ = 574.4

MLE MTTF 746.2 565.3

R(50) .935 .957

parameters β = 1.256 θ = 607.5

15.21 T = 669 + (25-14) 72 = 1461. MTTF = 1461/11 = 132.82 (a) MTTFL = 2(1461) / χ2 .025, 24 = 2(1461) / 39.4 = 74.16 MTTFU = 2(1461) / χ2 .975, 22 = 2(1461) / 11 = 265.64 (b) e

−10/74.16

= .874 ≤ R(10) = .963 = e −10/265.64

(c) 7.81 = −74.16 ln .90 ≤ t0.90 ≤ −265.64 ln .90 = 27.99 15.22 Least-squares fit: Plotting position is approx median. Intercept – a = -4.80912 Slope – b = 0.4087111 Scale Parameter – Alpha = 2.44672 Location Parameter – Mu =11.76654 and Rsquared = 0.97839. MLE: Alpha = 2.2605 and Mu = 11.734 yr. MOM: Alpha = 2.2446 and Mu = 11.7011 yr. (sample was taken from a min extreme value distribution having alpha = 2 and mu = 12) 15.23 MLE: γ = 1.6551, α = 993.021, MOM: γ = 1.234, α 1331.9.5. (Sample came from a gamma distribution with γ = 2, α = 1000).

15.24 Using the Excel template for Weibull interval data, β = 2.6 and θ = 9.0 yr. Therefore an

estimate for the 90 percent design life is t0.90 = 9.0 [ − ln(.9) ]

1/2.6

Weibull distribution with β = 2.5 and θ = 9.0 yr.)

15-10

= 3.8 yr. (sample came from a

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

Waveland Press, Inc., Copyright © 2009

15.25 From the Excel Spreadsheet for Chapter 15: γ = 4.5755 and α = 528.836. (data was generated from a gamma distribution with γ = 4.5 and α = 525.) Using the course software, a 95% design life is 903.422 or the oil should be replaced every 9034 miles. The MTTF = 24,197 miles.

15.26 For failure mode A with Mode B times as censored times: Distribution is

Weibull

Intercept - a Slope - b Beta Theta Derived Values 95% design life 90% design life Mode Median Regression Statistics

-2.10663 0.4999162 0.4999162 67.62

R-squared

0.91964

0.178 0.75 0.0 32.486

For failure mode B with Mode A times as censored times: Distribution is Intercept - a Slope - b Mean (MTTF/MTTR) Standard Deviation Derived Values

Normal -2.80888 0.0190369 147.549 52.529

95% design life 90% design life Regression Statistics

61.138 80.049

R-squared

0.97048 .5

⎛ 1 ⎞ −⎜ ⎟ ⎝ 67.62 ⎠

⎛ 1 − 147.55 ⎞ = .8855; RB (1) = 1 − Φ ⎜ ⎟ = 1 − Φ ( −2.79 ) = .99736 ⎝ 52.53 ⎠ R (1) = RA (1) RB (1) = (.8855 )(.99736 ) = .8832 RA (1) = e

The Weibull distribution for Mode A has a DFR indicating that operational failures are less likely to occur the longer the line is in operation. If the line is not properly synchronized, then early or premature failures will occur soon after startup.

15-11

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

Waveland Press, Inc., Copyright © 2009

15.27 (a)

2 ⎛ t ⎞ − ( t /θ ) 2 e θ ⎜⎝ θ ⎟⎠ n 2 2 ⎛ t ⎞ − t /θ 2 2 n n − t /θ L(θ ) = ∏ ⎜ i ⎟ e ( i ) = 2 n ∏ ti e ( i ) θ i =1 i =1 θ ⎝ θ ⎠ f (t ) =

n

n

n

ln L(θ ) = n ln 2 − 2n ln θ + ∑ ln ti − ∑ ( ti / θ ) = n ln 2 − 2n ln θ + ∑ ln ti − i =1

2

i =1

i =1

d ln L(θ ) −2n 2 = + 3 ∑ ti2 = 0 θ θ i =1 dθ n

n

n

−2nθ 2 + 2∑ ti2 = 0 or θ =

∑t / n

i =1

i =1

2 i

2 2 ⎛ t ⎞ − t /θ 2 /θ (b) L(θ ) = ∏ ⎜ i ⎟ e ( i ) ⎡⎢ e −(tr ) ⎤⎥ ⎣ ⎦ i =1 θ ⎝ θ ⎠

r

r

ln L(θ ) = r ln 2 − 2n ln θ + ∑ ln ti − i =1

n−r

1

r

∑ t − (n − r )(t / θ ) θ 2

i =1

2 i

r

d ln L(θ ) −2r 2 2t = + 3 ∑ ti2 + 3r = 0 θ θ i =1 θ dθ r

r ⎛ r ⎞ −2rθ 2 + 2∑ ti2 + 2tr2 = 0 or θ = ⎜ ∑ ti2 + (n − r )tr2 ⎟ / r i =1 ⎝ i =1 ⎠

15-12

2

1

n

∑t

2

θ 2 i =1 i

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 '

= e −.2t ' |t = e −.2t 2

2

0

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

b) R( t ) = e −.2 t =.95

→

2

2.3 a) R( t ) = b) λ ( t ) =

z

z

100

t

−.2t 2 = ln(.95)

100

0 ≤ t ≤ 100

t

z

100

c) MTTF = R( t )dt = 0

z

100

0

0 ≤ t ≤ 100

2 100 2 (1−.01t )dt = t 100 0 −.005t 0 = 100−.005(100 ) = 50 days

z

d) σ 2 = 0 t 2 f ( t )dt − ( MTTF )2 =.01 0 t 2 dt − 502 =.033 t 3 100

100

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

→

2.4 a) R( t ) =

− ln(.95) =.506 yrs .2

t=

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

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

z

→

z

1000 t

.01tmedian =.5

LM N

z

c

h

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

b) MTTF = t ⋅ f ( t ) dt = 0

c) R( t ) = 1 −

0

→

− 502 = 8333 . (days )2

tmedian = 50 days

OP Q

100

3 1000 1 3 f ( t ′ )dt ′ = 9 t t ′ 2 dt ′ = t' 10 109 t

1 t3 3 3 = 9 1000 − t = 1 − 9 10 10

z

100

t3 =.99 109

z

0 ≤ t ≤ 1000 hrs 106 1 = 3 9 10 10

1000 3 1000 3 3 3 t dt = 9 t4 = 9 10004 − 0 = 750 hrs 9 0 0 10 10 ⋅ 4 10 ⋅ 4

→

t3 =.01 109

→

2.5 2-1

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 .0005 − d ( e − (.001t ) ) 1 1 1 ⋅ − (.001t )1/ 2 == (.001t )−1/ 2 (.001)e − (.001t ) ⋅ − (.001t )1/ 2 = b) λ ( t ) = dt 2 .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 − .001( 60 ) = = =.865 R(10) R(10) e− .001(10)

R(50 / 10) =

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

[ ln .859596] − 10 = 12.9 hrs t= 2

e

− .001( t +10)

2.6

z

a) R( t ) =

= .95e

z

10

t

− .001(10)

= .859596

10

→

.001

10

f ( t ′ )dt ′ = (.2−.02t ′ )dt ′ =.2 t ′ t −.01 t ′ 2 t

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 −.1 t 2 10

0

2 c) R( tmedian ) = 1−.2tmed +.01tmed =.5

→

10 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.

L.2t − .02t OP − 3.33 e) σ = z t f (t )dt − MTTF = z t (.2 −.02t )dt − 3.33 = M N3 4 Q 10

2

2

2

0

10

3

2

4

10

2

2

0

0

= (66.67 − 50) − 0 − 1109 . = 558 .

σ = σ = 558 . = 2.36 yrs

→

2

2.7 a) R(1) =

100 100 = =.826 2 121 (1 + 10)

L 200(t ′ + 10) OP = 0 − FG −100 IJ = 100 R( t ) = f ( t ′ )dt ′ = 200 ( t ′ + 10) dt ′ = M H (t + 10) K (t + 10) N −2 Q L (t + 10) OP = L −100 O = 0 − ( −10) = 10 yrs b) MTTF = R( t )dt = 100 ( t + 10) dt = 100M N −1 Q MN t + 10 PQ

z

z

∞

t

∞

−2 ∞

−3

2

t

z

∞

0

z

t

∞

−2

0

2-2

−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

t

t 1 1 f ( t ′ )dt ′ = dt ′ = t ′ = , b b 0 b t 0

R( t ) = 1 − F ( t ) = 1 −

t b−t = b 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

LM N

bb−t

0

0≤t ≤b

IFR

OP Q

b

t2 1 1 dt = bt − = b b 2 0 b

b − tmed 1 = b b 2

LMF b − b I − 0OP = 2b − b = b MNGH 2 JK PQ 2b 2 2

2

2

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 = b H K N 3b Q 4 3 4 12 12

b t2 dt − b 2

2

3

b

2

2

2

2

0

2.9

20 − t 1 t t = ; R( t ) = ; λ(t ) = 20 20 − t b 20 20 20 20 = 10 yrs; tmedian = = 10 yrs; σ = = 5.77 yrs MTTF = 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

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

Waveland Press, Inc., Copyright © 2009

F t I LF t I F −t IJ OP b) MTTF = R( t )dt = G1 − J dt = MG1 − J G H t K MNH t K H 3 K PQ O t − t LF t I M 1 − J − (1 − 0) P = = G 3 MH t K PQ 3 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

→

t = 2000(1 − .9 ) = 102.63 hrs

2.11

R( t / T0 ) =

R(T0 + t ) ( t + T0 + 1)−3/ 2 = R(T0 ) (T0 + 1)−3/ 2

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

− a t ′2 / 2

t 0

= e − at / 2 a > 0 and t ≥ 0 2

2 dR( t ) d ( e − at / 2 ) − at 2 / 2 f (t ) = − ( − at ) = ate − at / 2 =− −e dt dt 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

L (1+.05t ′ ) OP = 0 − d−(1+.05t ) i = (1+.05t ) R( t ) =.1 (1+.05t ′ ) dt ′ =.1M N −2(.05) Q

z

∞

t

−3

−2 ∞

−2

t

2-4

−2

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 =

LM (1+.05t ) OP = 0 − FG − 1 IJ = 20 H .05K N −.05 Q −1 ∞

0

z

z

∞ ∞ 1 1 MTTFafter = R(t / T0 )dt = R (t )dt = (1+.05t ) −2 dt − 2 1 T0 T 0 R (T0 ) (1+.05(1)) ∞

L (1+.05t ) OP = 0 − LM− (105 OP = 21 . ) = (105 . ) z (1+.05t ) dt = (105 . ) M N −(.05) Q N .05(1+.05(1)) Q 2

∞

−1

−2

∞

2

2

1

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 z −a

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

n ∫ ( x + c) dx =

t

dx ( x + c) n +1 = ln ( x + c ) , n ≠1 ; ∫ n +1 x+c

a ( t '+ a ) dt ' = (a) R (t ) = ∫ 2 t −1 ( t '+ a ) ∞

(b) R (t ) =

t

a

−1 ∞

= t

10 a ; 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 ) = 1 (e) λ (t ) = 2

a (t + a )

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) → ∞ ; does not exist 0

0

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2.18 (a ) f (t ) =

ka k

(t + a )

k +1

; (b) λ (t ) =

k → DFR t+a − k +1 ∞

ak (t + a ) (c) MTTF = ∫ dt = k 0 −k + 1 (t + a ) ak

∞

= 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

3t 3 3t 4 3 (c) MTTF = ∫ 3 dt = 3 = k 0 k 4k 0 4 k

(d ) 1 −

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 a

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

(b)

t=

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.

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.000064t 3 = .5 (c)

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

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

(e) MTTF =

∫ (1 − .000064t )dt = [t − .000016t ] = 25 − .000016(25) = 18.75 yr.

3

4 25 0

4

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) = ∞

1 1 (1 − .000125t 3 )dt = (5.3125) = 6.07 yr. ∫ R(10) 10 .875 ∞

225

225 dt = 2.24 MTTF = ∫ 2 −1( t + 15 ) 0 ( 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 ) = (1 + .5t + 10 / t ) exp[−(t + .25t 2 + 10 ln t − 1.25)] f (t ) = − dt

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CHAPTER 3 3.1 a)

R( t ) = e − λt = e − t / MTTF = e − t /1100 R( 200) = e −200/1100 =.834 b) R( t d ) = e − td /1100 =.90 td = −1100 ln(.90 ) = 115.90 hrs

c) R( t med ) = e − tmed /1100 =.5 tmed = −1100 ln(.5) = −1100( −.69315) = 762.46 hrs d)

R( t ) = 1 − Prob(both components fail) = 1 − (1 − Ri ( t ))2 = 1 − (1 − e − t /1100 )2 since components identical R (200) = 1 − (1 − e 200 /1100 ) 2 =.973

3.2 R(t / T0 ) = R(t ) because of memoryless property R (100 / 1000) = R(100) = e −.0004(100) =.96 P (T < 100) = F (100) = 1 − R (100) = 1−.96 =.04 R (1000 / 1000) = R (1000) = e −.0004(1000) =.67 P(T < 1000) = F(1000) = 1- R(1000) = 1-.67 =.33 3.3

λ 1 ( t ) =.0003 and λ 2 ( t ) = t / (5 ⋅105 ) 2

λ ( t ) = ∑ λ i ( t ) = λ 1 ( t ) + λ 2 ( t ) =.0003 + t / (5 ⋅105 ) i =1

R( t ) = e z

t

− λ ( t ′ ) dt ′ 0

=e z

t

− (.0003+ t ′ /( 510 ⋅ 5 ) dt ′ 0

= e − (.0003t + t /10 ) 2

6

R(100) = e − (.0003(100 )+100 /10 ) =.96 2

6

3.4 5

λ (t ) = ∑ λ i (t ) =.001+.005+.0007 +.0025+.00001 =.00921 i =1

a)

MTTF = 1 / λ = 1/.00921 = 108.58 days

σ 2 = 1 / λ2 and σ = 1 / λ = 108.58 days b) t.99 = −

1

λ

ln(.99) = 108.59 ln(.99) = 109 . days

3-1

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

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3.5

λ L =.001 upper bound: e − λ L t = e − .0001t ≥ R(t ) upper bound for t = 60: e −.0001( 60) =.942 ≥ R(60)

3.6 p .0028 = =.0056 where Δt = 1 / 2 day Δt 1 / 2 R(t) = e - λt

λ=

R(30) = e − .0056( 30) =.845 P(10 < T < 20) = R (10) − R(20) = e −.0056(10) − e −.0056( 20) =.9455−.894 =.0515 3.7

R( td ) = e− λtd = R(10) = e − λ⋅10 =.99 λ = − ln(.99) / 10

λ is the system's constant failure rate, MTTF = 1 / λ =

10 = 995 yrs and tmed =.69315 ⋅ MTTF = 690 yrs − ln.99

Since the components are identical, λ i = λ / n =

MTTFi = 1 / λ i =

200 = 19,900 yrs − ln(.99 )

− ln(.99) 10 ⋅ 20

and

t med i =.69315(19900) = 13793 yrs

3.8 R (3000) = 2e − λ ( 3000) − e −2 λ ( 3000) =.95 By trial and error: λ R(3000) .0001 .9328 .00005 .9806 .00008 .9545 .000085.9493 .000084.9504 .0000843 .9501 Therefore: MTTF=1/ λ =1/.0000843=11,862 hrs 3.9 a) R (t d ) = e − λ std R (5) = e − λ s ⋅5 =.95 λ s = − ln(.95) / 5 =.010287 → MTTFs = 1 / λ s = 97.478 yrs

λ s = λ 1 + λ 2 + λ 3 = λ 1 +.5λ 1 + 15 . λ 1 = 3λ 1 3-2

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

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λ 1 = λ s / 3 =.0102587 / 3 =.003419567 → MTTF1 = 1 / λ 1 = 292.43 yrs λ 2 =.5λ 1 =.001709784 → MTTF2 = 1 / λ 2 = 594.87 yrs . λ 1 =.005129351 → MTTF3 = 1 / λ 3 = 194.96 yrs λ 3 = 3λ 2 = 15 b) R (t ) = 1 − Prob(both power units fail) = 1- (1- e -.010287t ) 2 = 2e − .010287 t − e 2 (.010287) t R (5) = 2e −.010287(5) − e −2 (.010287)5 =.9975 15 .

MTTF = = 146.2 yrs

λs

3.10

e −.03125t (.03125t )n pn ( t ) = n! p0 (1 day) = p0 ( 24 hrs ) =

e−.03125⋅24 (.03125 ⋅ 24)0 =.4724, p1 =.3542, p2 =.1329 0!

2

∑ pi ( 24 hrs) =.9595 →

2 bulbs are sufficient

i=0

3.11 a) 2

2

1

1

λ = ∑ λ i = ∑1 / MTTFi = 1 / 5 + 1 / 5 =.4 F ( 2) = 1 − R( 2) = 1 − e −.4( 2 ) =.551 b) P(more than 1 failure in first 5 hrs)=1-P(no failures in first 5 hrs)-P(1 failure in first 5 hrs) = 1 − p0 (5) − p1 (5) = 1 − e − λt − e − λt ( λt ) = 1 − e −.4(5) − e −.4(5) (.4 ⋅ 5) =.595 c) No. Would expect batteries to have an increasing failure rate (IFR) due to wear-out.

3.12 a) R (t ) = 1 − (1 − e − λ 1t )(1 − e − λ 2 t ) = e − λ 1t + e − λ 2 t − e − ( λ 1 + λ 2 ) t ∞ 1 1 1 MTTF = R (t )dt = + − 0 λ1 λ 2 λ1 + λ 2 b)

z

R(1000) = e − .000356(1000) + e − .00156(1000) − e − (.000356 + .00156)1000 =.7634 MTTF =

1 1 1 + − = 2928.09 hrs .000356 .00156 .000356+.00156

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3.13 pn ( t ) =

e − λt ( λt ) n n!

where λt = (.00021)(10,000) = 2.1

P (third failure by 10000 hrs) = 1 − p0 − p1 − p2 = 1 − e

−2 .1

−

e −2.1 (2.1)1 e −2.1 ( 2.1) 2 − =.351 1!

2!

3.14 Yk is a random variable, the time of the kth failure. The expected value (mean) of Yk is k / λ . k 10 10 = = 294.11 hrs ≈ 12 days E (Yk ) = = λ 3.4 / 100 .034 3.15

λ i =.15 failures / yr ; t 0 = 2 months = 1 / 6 yr ; t = 10,000 hrs = 1.14 yrs R i (t ) = e − λ i ( t − t0 ) R (t ) = 1 − (1 − Ri (t )) 2 = 1 − (1 − e − λ ( t − t0 ) ) 2 = 2e − λ ( t − t0 ) − e 2 λ ( t − t0 ) R (114 . ) = 2e −.15(1.14 −1/ 6) − e −2 (.15)(1.14 −1/ 6) =.982

3.16

|R1 − 1 − e = 2e −e t>t R( t ) = S |T 1 t≤t MTTF = z R( t ) = z dt + z 2e −e dt = t + 2e z e dt − e z e dt L e OP − e LM e OP = t + 2 − 1 = t + 3 = t + 2e M λ 2λ 2λ N −λ Q N −2λ Q − λ ( t − t0 ) 2

− λ ( t − t0 )

−2 λ ( t − t0 )

0

0

∞ 0

∞ t0

t0 0

λt0

− λt

− λ ( t − t0 )

λt0 ∞ − λt

−2 λ ( t − t0 )

0

∞

2 λt0

−2 λt0

2 λt0 ∞ −2 λt t0

∞

0

0

t0

t0

0

t0

3.17 Assume exponential since constant failure rate, natural event, and equipment has no memory.

selective fading: λ sf = 4 / mo =

4 =.005556 / hr 24 × 30

.5 =.0006944 24 × 30 R( 24) = e− .005556( 24 ) +.0006944( 24 ) =.86

flat fading: λ ff =.5 / mo =

3.18 Failures on demand: λI = 0 , λO = 1/1000, tI = 14 hrs, tO = 10 hrs, p = 1/120 = .00833 λeff =

.001(10) .00833 + =.0007639 and td = - ln.90 / .0007639 = 137.9 hrs = 5.75 days 24 24

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3.19 f(t) = -dR(t) /dt = ba-bt (ln a) and λ(t) = f(t)/R(t) = b ln a , a constant

c h =e

Let λ = b ln a = ln ab . Then eλ = ab and R(t) = a b

−t

−λ t

3.20 Δt = 12/30 = .4 cartons per minute or 150 / hr; λ = p/ Δt = .005/.4 = .0125 failures/min

R(60) = e-.0125 (60) = .4724 3.21 With MTTF = 1772.46, the following bounds were obtained: time 100 200 500 1000 2000 5000 10000 3.22 MTTF(T0) =

lower bnd .9451 .8933 .7542 .5688 0 0 0

1 e − λ T0

z

∞

T0

R(t) ..9975 .9777 .9394 .7788 .3679 .00193 1.3888 x 10-11

e − λt ' dt ' =

ω .0001239 .000522884 .000562146

upper bnd 1 1 1 1 .7805 .0732 .00362

1 e − λT0 1 − λt ' ∞ e = = − λ T0 − T λ T0 λe 0 λ −λ e

3.23 t

o

o

λ k x k −1e− λ x

k k −1− r e − λ x k −1 r λ ( k − 1) ! x dx = ∑ ( −1) k − 1 − r ! −λ r +1 Γ ( k ) r =0 ( )( )

FY (t ) = ∫ fY (t ) dt = ∫ Since ∫

λ k x k −1e− λ x

t

Γ (k )

Γ (k )

dx

k −1 k −1 e − λ x ( k − 1) ! k −1 ( −1) λ k −1− r x k −1− r (λ x) (λ x) −λ x −λ x = − = = − e e ∑ ∑ ∑ r +1 i! ( k − 1)! r =0 ( −1) ( k − 1 − r )! i =0 r =0 ( k − 1 − r )! k −1− r

r

t

Then ∫ o

λ k x k −1e− λ x Γ (k )

t

i k −1 ⎡ − λ x k −1 ( λ x )i ⎤ λt ) ( − λt dx = ⎢ −e ∑ ⎥ = 1− e ∑ i ! ⎥⎦ i! i =0 i =0 ⎢⎣ 0

3.24 λ = 125/4,120,000 = 3.034 x 10-5

3-5

i

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

Waveland Press, Inc., Copyright © 2009 −

12,000(125)

(a) R(12, 000) = e 4,120,000 = .695 (b) t0.90 = −

12, 000(125) ( ln 0.90 ) = 3473 4,120, 000

(c) MTTF = 1/ λ = 32,960; tmed = .69315 ( 32,960 ) = 22,846 (d ) λs = 4λ = 1.2136 x 10−4 ; R (12, 000) = e

−1.2136 x 10−4 (12,000 )

= .2331

or (.695 ) = .2331 4

e − λst ( λs t ) − = e .00024272(12,360) [1 + 3 + 9 / 2] = .423 (e) λs = 8λ ; R2 (12,360 ) = ∑ i! i =0 2

i

3.25 The number of failures during time t is Poisson with mean .001t = 3.65. n 8 e −3.65 ( 3.65 ) = .9873 R8 (t ) = ∑ n! n =0 9 spares are needed Therefore: n 9 e −3.65 ( 3.65 ) = .9956 R9 (t ) = ∑ n! n=0

3.26 System times in operating hours failure rate = MTTF = Median = Std Deviation = R(8760) =

0.000003974 251,635.632 174,421.238 251,635.632 0.9658

Connectors as a group has the smallest reliability- R(8760) = .983.

3.27 R(t) = 1-(1-exp(-λt))^3 = 3exp(-λt) - 3exp(-2λt) + exp(-3λt) R(t) = 3exp(-.015t) - 3exp(-.03t) + exp(-.045t) R(100) = .5311 MTTF = 11/(6λ) = 11/.090 = 122 days Reliability without redundancy => R(100) = exp(-.015*100) = .2231 3.28 (a) R(t) = exp[- .002(t-100)] = .80 => t =100 - ln(.80)/.002 = 221 days Let X = a random variable, the number of failed bulbs; E[X] = np = 2 (b) From the binomial distribution with p = .2 and N = 10; Pr(x > 4) = .033.

3-6

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CHAPTER 4

4.1

FG t IJ β H K a) R( t ) = e θ

FG 100 IJ R(100 ) = e H 550 K =.912 1. 4

−

−

→ −

FG IJ

t. 99 β

H θ K =.99 b) B1 → R (t.99 ) = e t.99 = θ ( − ln(.99))1/ β = 550( − ln(.99))1/1.4 = 20.575 days . ) = 550 ⋅ Γ (171 . ) = 550⋅.910572 = 500.81 days c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 550 ⋅ Γ (1 + 1 / 14

{

d) σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

m

2

. ) } } = 550 {Γ(1 + 2 / 14. ) − Γ(1 / 14. ) } = 550 {Γ(2.43) − Γ(171 2

2

r

= 5502 1267032 . − (.910572)2 = 13246192 .

→

2

2

σ = σ 2 = 36395 . days

e) t.5 = t med = θ ( − ln(.5))1/ β = 550( − ln(.5))1/1.4 = 423.32 days f) t mod e = θ (1 − 1 / β )1/ β = 550(1 − 1 / 14 . )1/1.4 = 224.77 days g) t d = θ ( − ln R)1/ β = 550( − ln(.90))1/1.4 = 110.22 days 4.2

FG t IJ β H K a) R( t ) = e θ −

FG 100 IJ R(100) = e H 800 K =.750 .6

−

→

FG 300 IJ .6 R (100 + 200) e H 800 K b) R (100 / 200) = = =.887 F 200 IJ .6 R(200) −G e H 800 K −

L O 4.3 The power supply has a Weibull failure distribution with β=2.1 and θ = M∑ (1 / θ ) P N Q LF 1 IJ + FG 1 IJ + FG 1 IJ OP = 9328.98 hrs θ = MG MNH 12000K H 18500K H 21500K QP 3

−1/ β

β

i

i =1

2 .1

2 .1 −1/ 2 .1

2 .1

MTTF = θ ⋅ Γ(1 + 1 / β ) = 9328.98(.885633) = 8262.05 hrs

FG t IJ β R( t ) = e H θ K =.9 −

d

4.4

FG t −t IJ β H K a) R( t ) = e θ −

0

td = θ ⋅ ( − ln(.9 ))1/ 2.1 = 9328.98( − ln(.9 ))1/ 2.1 = 3194.79 hrs

→

d

FG 150−50 IJ R(150 ) = e H 8500 K =.9989 1. 54

→

−

b) MTTF = t 0 + θ ⋅ Γ (1 + 1 / β ) = 50 + 8500 ⋅ Γ (1 + 1 / 154 . ) = 50 + 8500⋅.9 = 7700 hrs c) tmed = t0 + θ ( − ln.50 )1/ β = 50 + 8500(.69315)1/1.54 = 6749.8 hrs

o

d) σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

l

q

2

. ) − .9 t t = 8500 oΓ(1 + 2 / 154

= 85002 Γ( 2.30 )−.81 = 25772442 e) td = t0 + θ ( − ln R )1/ β

→

2

→

2

σ = σ 2 = 5076.66 hrs

t.98 = 50 + 8500( − ln(.98))1/1.54 = 724.58 hrs 4-1

.

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

Waveland Press, Inc., Copyright © 2009

4.5 For 2 redundant components,

R( t ) = 1 − 1 − e− b t /θ g

FG IJ H K

18 . t 4.6 λ ( t ) =.07 + 10 10

2

β

.8

R(100) = 1 − 1 − e− b100/ 550g

1. 4

→

LM → R( t ) = e MN

LM ) = e MN

=.992

LM FG 1 IJ OP H 10 K QP R(1) = e MN =.9177 1. 8

→

− .07+

1. 8

− .07 t med +

For tmed , set R( tmed

FG t IJ OP H 10 K PQ

FG t IJ OP H 10 K PQ 1. 8

− .07 t +

2

med

=.50

FG t IJ =.6931 H 10 K 1.8

and solve for tmed : .07tmed +

FG t IJ H K 4.7 R( t ) = e θ

β

−n

→

→

med

tmed = 5.31 yrs, by trial and error

FG 100 IJ 1.3 R(100) = e H 10000 K =.95 −n

F 100 IJ = ln(.95) − nG H 10000K 1.3

n=

− ln(.95)

FG 100 IJ = 20.43 → 20 units H 10000 K 1.3

MTTF = θ ⋅ Γ (1 + 1 / β ) =

10000 ⋅ Γ (1.77) = 998177 . ⋅.923763 = 922.08 hrs 201/1.3

4.8 Failure time is Weibull with β =1.5 and θ =500.

FG 50 IJ a) R (50) = e H 500 K

1.5

−

=.96887

1/ β

b) td = θ ( − ln R ) → t.90 = 500( − ln(.90))2/ 3 = 11154 . hrs . hrs c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 500 ⋅ Γ (1 + 2 / 3) = 500⋅.903 = 4515

FG 100 IJ e H 500 K

1.5

−

d) R (50 / 50) = e)

R (50 + 50) = =.9438 R(50) .96887

FG t IJ +.002 H 500K 1/ 2

λ ( t ) =.003

LMFG t IJ +.002t OP H 500 K PQ R( t ) = e MN 1. 5

−

LM FG 50 IJ +.002( 50 )OP H 500 K PQ =.87667 R(50) = e MN 1. 5

→

−

4-2

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4.9 a) P

RS 200 − μ ≤ z ≤ 270 − μ UV =.90 = Pl-1.645 ≤ z ≤ 1.645q σ W T σ 200 − μ

σ σ =-

= −1645 . and

270 − μ

σ

200 - μ 1.645

= 1645 .

200 - μ 270 - μ =1.645 1.645

μ = 235 = MTTF b) R(210) = 1 − Φ

c) R( td ) = 1 − Φ

σ =-

200 - 235 . = 2128 .

1645

FG 210 − μ IJ = 1 − ΦFG 210 − 235IJ = 1 − Φ(−117 . ) = 1−.121 =.879 K H 2128 H σ K .

FG t − 235IJ =.99 H 2128 K . →

4.10 a) t R = tmed e sz1− R

→

FG 1 ln t IJ Hs t K

FG t − 235IJ =.01 H 2128 K . d

td = ( −2.32)2128 . + 235 = 185.63 hrs

R(10 + 200) = R( 200)

d) R(10 / 200 ) =

Φ

→

d

td − 235 = −2.32 2128 .

b) R( t ) = 1 − Φ

→

R( 210) .879 .879 = = =.926 200 − 235 1 − Φ( −164 . ) .9495 1− Φ 2128 .

FG H

IJ K

t.98 = 2236e.41( −2.05 ) = 964.82

med

R(5 weeks) = R(840 hrs) = 1 − Φ

FG 1 ln 840 IJ = 1 − Φ( −2.39) =.99158 H .41 2236K

Therefore probability of failure = 1 - .99158 = .00842 c) MTTF = t med

s2 e2

.412 = 2236e 2

= 2432.06 hrs

2 e s ( e s − 1) = 22362 e.41 ( e.41 − 1) = 1082754.22 d) σ 2 = t med 2236 t e) t mode = med2 = 2 = 1890.02 hrs s e e.41 2

2

2

2

4-3

→

σ = 1040.55 hrs

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

Waveland Press, Inc., Copyright © 2009

4.11

→ a) R( t ) = e b g b) B1 → R(t.99 ) =.99 − t /θ β

R(5000) = e− ( 5000/12000 ) =.8529 2 .1

t.99 = θ ( − ln(.99))1/ β = 12000( − ln(.99))1/ 2.1 = 1342.27 hrs B.1 → t.999 =.999 t.999 = 12000( − ln(.999))1/ 2.1 = 447.42 hrs

c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 12000 ⋅ Γ (148 . ) = 12000⋅.88574 = 10628.96 hrs

o

σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

m

2

. ) − Γ (148 . ) t t = 12000 oΓ(195 2

2

r

= 12000 2 .979881 − (.885747 )2 = 28127988.29

d)

→

σ = 5303.58 hrs

t.50 = t med = θ ( − ln(.5))1/ β = 12000( − ln(.5))1/ 2.1 = 10078.22 hrs t mod e = θ (1 − 1 / β )1/ β = 12000(1 − 1 / 2.1)1/ 2.1 = 8819.71 hrs

FG 8760 JI e) F (1 yr) = F (8760 hrs) = 1- R (8760) = 1 − e H 12000 K −

2 .1

=.4033

4.12 without burn - in: t.90 = 2450( − ln(.90))1/.86 = 178.96

R( t + 30) R( t + 30) R( t + 30) = F 30 IJ .86 = .9776 =.90 R(30) −G e H 2450 K R( t + 30 ) =.90(.9776) =.8798

with burn - in: R( t / 30) =

FG t +30 IJ H e 2450 K =.8798 . 86

→

t = 2450( − ln(.8798))1/.86 − 30 = 194.53 days

Burn-in testing increases design life by 194.53-178.96=15.57 days; may not be worth it.

4.13

L O a) The 4 belt system has a Weibull failure distribution with β=1.34 & θ = M∑ (1 / θ ) P N Q 4

−1/ β

β

i

.

i =1

b

g + b1 / 3400g + b1 / 8000g + b1 / 6100g

θ = 1 / 2500

1.34

1.34

F (72) = 1 − R (72) = 1 − e b

− 72 /1399 .757

1.34

g

1. 34

b)

F (72 / 4000) = 1 − R (72 / 4000) = 1 − β

where R (t / T0 ) = e − (( t + T0 )/θ ) + ( T0 /θ )

1.34 −1/1.34

= 1399.757 hrs

= 1−.981 =.019 1. 34 1. 34 R( 72 + 4000) − 4072 /1399 .757 g + b 4000 /1399.757 g = 1− e b =.094 R (4000)

β

4-4

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4.14

FG t IJ FtI −2 G J H θK a) R ( t ) = 2e − e HθK β

FG 5000 IJ F 5000 IJ −2 G H 12000 K Rp (5000) = 2e − e H 12000 K =.9784

β

−

p

2 .1

2 .1

−

→

b) MTTF = θ ⋅ Γ (1 + 1 / β ) ⋅ [2 − 2−1/ β ] = 12000 ⋅ Γ (148 . ) ⋅ [1281 . ] = 136171 . hrs

FG 8670 IJ c) F (1 yr) = F (8760 hrs) = 1 − R(8760) = 2e H 12000 K

2 .1

−

FG 8670 IJ − e H 12000 K

FG t IJ F t IJ −2 G H 12000 K d) Find t , such that 2e − e H 12000 K =.99 2 .1

a) R( td ) = 1 − Φ

Φ

=.163

2 .1

−

4.15

2 .1

−2

→

t = 4100 hrs

FG t − μ IJ = 1 − ΦFG t − 10IJ =.99 H θ K H 2.5 K d

d

FG t − μ JI =.01 H θ K d

t − 10 z = −164 . = d 2.5 td = 2.5( −12.325) + 10 = 4.19 days

→

FG 1 − 10IJ = 1 − Φ(−3.6) = 1−.00016 =.99984 H 2.5 K F 2 − 10IJ = 1 − Φ(−3.2) = 1−.00069 =.99931 ii) R(2) = 1 − ΦG H 2.5 K F 5 − 10IJ = 1 − Φ(−2) = 1−.02275 =.97725 iii) R (5) = 1 − ΦG H 2.5 K F 6 − 10IJ 1 − ΦG H 2.5 K = 1 − Φ(−16. ) = .94520 =.9672 R (1 + 5) R (6) c) R (1 / 5) = = = b) i) R (1) = 1 − Φ

R (5)

R (5)

4.16 a) R(1000 ) = 1 − Φ

.97725

F 1 lnF 1000I I GH t JK JK GH 125 .

.97725

→

tmed =

med

b) MTTF = t med

σ =t 2

2 med

s2 e2

1.252 = 7816.6e 2

s2

c) R (1000 / 1000) =

2 1.252

R (2000) = R(1000)

1000 1.25( −1.645 )

e

= 17077.39 cycles

e (e − 1) = 7816.6 e s2

.97725

1− Φ

= 7816.60 cycles

(high variability)

− 1) = 1,099,124,041 ; σ = 33,153 cycles

1.252

(e

FG 1 lnFG 2000 IJ IJ H 7816.6K K = 1 − Φ(−109 H 125 . . ) =.9075 .95

4-5

.95

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4.17 It is Weibull with β = 2 and θ = 1 × 105 : a) λ ( t ) = 2 × 10−10 t = at b where a = 2 × 10−10 and b = 1

b = β −1 → β = b +1 = 2 a=

FG IJ H K

β 1 β β 2 = 2 →θ = = = 1 × 105 β −1 −10 a θ θ θ 2 × 10

FG IJ HK b g

β −1

β t λ ( t ) = 2 × 10 t = where β = 2 and θ = 1 × 105 θ θ b) MTTFB = θ ⋅ Γ 1 + 1 / β = 100000 ⋅ Γ (15 . ) = 100000×.886227 = 88622.7 hrs MTTFA = 1 / λ = 1/.000011 = 909091 . hrs → Based only on MTTF , A preferred. −10

c) R( 2 / 3 yr ) = R(5840 hrs)

RB (5840 ) = e − ( 5840/1×10 ) =.997 5 2

; RA (5840 ) = e− (.000011)( 5840 ) =.938

Based only on warrantly costs, motor B is preferred. d)

e − (10×5840/1×10 ) RB (9 / 1) = =.713 ; RA (9 / 1) = e − (.000011)9( 5840 ) =.561 .9966 Motor B will reduce replacement costs.

5 2

Ft I Ft I −G J + G J HθK Hθ K = 1 ⋅t −t ln R (t ) − ln R (t ) 4.18 average failure rate = AFR = = β

1

1

2

t 2 − t1

t 2 − t1

FG t − μ IJ → Hσ K F 100 − μ IJ =.01 → ΦG H 12 K

4.19 R( t ) = 1 − Φ

β

2

R(100) = 1 − Φ

θβ

β

β

2

1

t 2 − t1

FG 100 − μ IJ =.99 H 12 K

100 − μ 12 μ = 100 + 12(2.32) = 127.84 hrs z = −2.32 =

4.20 a) tmed = MTTF ⋅ e − s / 2 = 3600e −2 = 487.21 hrs 2

FG 1 lnFG 100 IJ IJ = 1 − Φ(−.79) =.785 H 2 H 487.21K K F 1 F 200 IJ IJ 1 − ΦG lnG H 2 H 487.21K K =.67/.785 =.85 R( 200) = b) R (100 / 100) = R(100) = 1 − Φ

R(100)

.785

1− Φ

FG 1 lnFG 300 IJ IJ H 2 H 487.21K K =.888

R (300) = R (200) .67 d) This is a decreasing failure rate (DFR) so do not replace until a failure occurs. Operating past the maximum point on the hazard rate curve:

c) R (100 / 200) =

4-6

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4.21 a)

b

g

b

g

MTTF = θ ⋅ Γ 1 + 1 / β = 5.6419 ⋅ Γ 1 + 1 / 2 = 5.6419 ⋅ Γ (15 . ) = 5.6419⋅.886227 = 5 yrs b)

F (12 months) = F (1 year) = 1 − R(1) = 1 − e − (1/5.6429) =.031 2

c)

R (1 + 1) R ( 2) e − ( 2 /5.6419) .882 F (1 / 1) = 1 − R(1 / 1) = 1 − = 1− = 1− =.1 − =.09 R (1) R (1) .969 .969 2

d) Expected value of first year warranty: $500(.031)=$15.5 Expected value of second year warranty: $500(.09)=$45 Warranty costs: $25 and $40 respectively.

First year warranty should not be purchased. Second year warranty should be purchased. e) Value if both must be purchased: ($500)(1-R(2))=($500)(1-.88)=$60 Warranty costs: $25+$40=$65 Warranties should not be purchased.

4.22 Knotsew

Dell Fye

Ima Goode

(a) R(10,000 hours)

.961

.981

.948

(b) MTTF

91,142 (91,057)

97,454

126,666

(c) The mode

40,867

28,915

0

(d) The median time to failure

76,967

65,000

111,299

4-7

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(e) The 90 percent design life

20,040

20,448

19,500

(f) The reliability for the next 10,000 hours if it has survived the first 10,000 hours. (g) characterize the hazard rate function as DFR, CFR, or IFR

.9368

.92215

.947

IFR

IFR/DFR

IFR

4.23 Weibull distribution with beta = 1.2 and theta = 10,000 (a) R(1,000 hrs) .9388 (b) MTTF (in hrs) 9406 (c) The mode (in hours) 2247 (d) The median time to failure 7368 (e) The 90 percent design life 1533 (f) The conditional reliability to get to 1000 hours given it survives 500 hr. (g) characterize the hazard rate function as DFR, CFR, IFR, or combination (h) Average failure rate from 0 to 5,000 hrs

.965 IFR 8.71 x 10-5

4.24 Lognormal distribution with tmed = 10,000 and s = 1.2. (a) R(1,000 hrs) .9725 (b) MTTF (in hrs) 20,544 (c) The mode (in hours) 2369 (d) The median time to failure 10,000 (e) The 90 percent design life 2140 (f) The conditional reliability to get to 1000 hours given it survives 100 hr. (g) characterize the hazard rate function as DFR, CFR, IFR, or combination (h) Average failure rate from 0 to 5,000 hrs

.9725 IFR/DFR 6.62 x 10-5

4.25 (a) R(5000) = .8885, (b) t0.90 = 4773, (c) R(5000|5000) = R(10,000) / R(5000) = .4884 (d) 1 – [1 – R(5000|5000)]2 = .7383 4.26 gamma distribution with γ = 1.8 and α = 10,000 hr. (a) MTTF = 17,000 hr., (b) std. dev. = 13,038 hr., (c) tmed = 13,807 hr., (d) tmode = 7,000 (e) t0.95 = 2420 hr., (e) R(5,000) = .8534, (f) R(1,000|5,000) = R(6000) / R(5000) = .9511, (g) IFR

4-8

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4.27

2000 ⎞ ⎛ 1 (a ) R(2000) = 1 − Φ ⎜ ln ⎟ = 1 − Φ(−.851) = .80234 ⎝ 1.7 8500 ⎠ t ⎞ ⎛ 1 (b) 1 − Φ ⎜ ln ⎟ = .99 ⎝ 1.7 8500 ⎠ t 1 = −2.33 or t = 8500 e( −2.33)(1.7) = 161.87 ≈ 161 landings ln 1.7 8500 4.28 .7

(a ) R (50) = e

−.00045(50)

(b) R (50 | 50) =

⋅e

⎛ 50 ⎞ −⎜ ⎟ ⎝ 8939.43 ⎠

⎡ ⎛ 1 50 ⎞ ⎤ ⎢1 − Φ ⎜ .4 ln 100 ⎟ ⎥ = .9519 ⎝ ⎠⎦ ⎣

R(100) .87706 = = .9214 .9519 R(50)

(c) MTTFexternal = 1/ λ = 2222.22; MTTFquality = ( 8939.43) Γ (1 + 10 / 7 ) = 11,315.7 MTTFwear −out = 200e.4 /2 = 216.6 2

4.29 This is a gamma distribution with the shape parameter of 2.4. (a) Using trial and error and the Chapter 4 spreadsheet or the course software, α is found to be approximately 5865. (b) R(5,000) = .872443, (c) MTTF = 14,076, (c) tmed = 12,176.67, (d) 1 – [1 - . 872443]2 = .98373 4.30 (a) Manufacturer

R(72)

MTTF

tmed

RWT Space Division Rockweed-Smarton

.9118 .9276

.415 585 hr

339.4 443.7

(b) R(72|50) .8818 .9202

(c) R(t|50) = R(t+50)/R(50) = .95 ; solve (trial and error) for t ≈ 39.5 hr.

4.31 Rsys = R (5 / 6 | 50) ⋅ e

−.083( 5/6 )

⋅ e −.034(5/6) = (.998743)(.933171)(.972064 ) = .905962

4.32 ( t −μ ) (t − μ ) ⎤ ⎡⎛ 1 ⎞ (t − μ ) − e (t−αμ ) ⎤ 1 −e α ⎡ α α f '(t ) = d ⎢⎜ ⎟ e e − 1⎥ ⎥ / dt = 2 e ⎢e α ⎢⎣⎝ α ⎠ ⎥⎦ ⎢⎣ ⎥⎦

⎡ (t − μ ) ⎤ setting ⎢e α − 1⎥ = 0 and solving t = μ ⎢⎣ ⎥⎦

4-9

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4.33 t

t

2t ' t '2 t2 t2 F (t ) = ∫ 2 dt ' = 2 = 2 ; R(t ) = 1 − 2 b b 0 b b 0 2t 2 2t λ (t ) = b 2 = 2 2 IFR t b −t 1− 2 b b

b

2t 2 2t 3 2b3 2 MTTF = ∫ 2 dt = 2 = 2 = b 3b 0 3b 3 b 0 2 tmed = .5 or tmed = b .5 b2

4.34 2t t 2 2t t 2 ⎛ 2 2t ' ⎞ F (t ) = ∫ ⎜ − 2 ⎟ dt ' = − 2 ; R(t ) = 1 − + 2 b b ⎠ b b b b 0⎝ 2 2t − 2 2 ; IFR λ (t ) = b b 2 = 2t t (b − t ) 1− + 2 b b t

b

⎛ 2t t 2 ⎞ ⎛ t2 t3 ⎞ b b MTTF = ∫ ⎜1 − + 2 ⎟ dt = ⎜ t − + 2 ⎟ = b − b + = 3 3 b b ⎠ ⎝ b 3b ⎠ 0 0⎝ b

2 2tmed tmed − 2 = .5 solve for tmed using quadratic formula b b

4-10

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CHAPTER 4

4.1

FG t IJ β H K a) R( t ) = e θ

FG 100 IJ R(100 ) = e H 550 K =.912 1. 4

−

−

→ −

FG IJ

t. 99 β

H θ K =.99 b) B1 → R (t.99 ) = e t.99 = θ ( − ln(.99))1/ β = 550( − ln(.99))1/1.4 = 20.575 days . ) = 550 ⋅ Γ (171 . ) = 550⋅.910572 = 500.81 days c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 550 ⋅ Γ (1 + 1 / 14

{

d) σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

m

2

. ) } } = 550 {Γ(1 + 2 / 14. ) − Γ(1 / 14. ) } = 550 {Γ(2.43) − Γ(171 2

2

r

= 5502 1267032 . − (.910572)2 = 13246192 .

→

2

2

σ = σ 2 = 36395 . days

e) t.5 = t med = θ ( − ln(.5))1/ β = 550( − ln(.5))1/1.4 = 423.32 days f) t mod e = θ (1 − 1 / β )1/ β = 550(1 − 1 / 14 . )1/1.4 = 224.77 days g) t d = θ ( − ln R)1/ β = 550( − ln(.90))1/1.4 = 110.22 days 4.2

FG t IJ β H K a) R( t ) = e θ −

FG 100 IJ R(100) = e H 800 K =.750 .6

−

→

FG 300 IJ .6 R (100 + 200) e H 800 K b) R (100 / 200) = = =.887 F 200 IJ .6 R(200) −G e H 800 K −

L O 4.3 The power supply has a Weibull failure distribution with β=2.1 and θ = M∑ (1 / θ ) P N Q LF 1 IJ + FG 1 IJ + FG 1 IJ OP = 9328.98 hrs θ = MG MNH 12000K H 18500K H 21500K QP 3

−1/ β

β

i

i =1

2 .1

2 .1 −1/ 2 .1

2 .1

MTTF = θ ⋅ Γ(1 + 1 / β ) = 9328.98(.885633) = 8262.05 hrs

FG t IJ β R( t ) = e H θ K =.9 −

d

4.4

FG t −t IJ β H K a) R( t ) = e θ −

0

td = θ ⋅ ( − ln(.9 ))1/ 2.1 = 9328.98( − ln(.9 ))1/ 2.1 = 3194.79 hrs

→

d

FG 150−50 IJ R(150 ) = e H 8500 K =.9989 1. 54

→

−

b) MTTF = t 0 + θ ⋅ Γ (1 + 1 / β ) = 50 + 8500 ⋅ Γ (1 + 1 / 154 . ) = 50 + 8500⋅.9 = 7700 hrs c) tmed = t0 + θ ( − ln.50 )1/ β = 50 + 8500(.69315)1/1.54 = 6749.8 hrs

o

d) σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

l

q

2

. ) − .9 t t = 8500 oΓ(1 + 2 / 154

= 85002 Γ( 2.30 )−.81 = 25772442 e) td = t0 + θ ( − ln R )1/ β

→

2

→

2

σ = σ 2 = 5076.66 hrs

t.98 = 50 + 8500( − ln(.98))1/1.54 = 724.58 hrs 4-1

.

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4.5 For 2 redundant components,

R( t ) = 1 − 1 − e− b t /θ g

FG IJ H K

18 . t 4.6 λ ( t ) =.07 + 10 10

2

β

.8

R(100) = 1 − 1 − e− b100/ 550g

1. 4

→

LM → R( t ) = e MN

LM ) = e MN

=.992

LM FG 1 IJ OP H 10 K QP R(1) = e MN =.9177 1. 8

→

− .07+

1. 8

− .07 t med +

For tmed , set R( tmed

FG t IJ OP H 10 K PQ

FG t IJ OP H 10 K PQ 1. 8

− .07 t +

2

med

=.50

FG t IJ =.6931 H 10 K 1.8

and solve for tmed : .07tmed +

FG t IJ H K 4.7 R( t ) = e θ

β

−n

→

→

med

tmed = 5.31 yrs, by trial and error

FG 100 IJ 1.3 R(100) = e H 10000 K =.95 −n

F 100 IJ = ln(.95) − nG H 10000K 1.3

n=

− ln(.95)

FG 100 IJ = 20.43 → 20 units H 10000 K 1.3

MTTF = θ ⋅ Γ (1 + 1 / β ) =

10000 ⋅ Γ (1.77) = 998177 . ⋅.923763 = 922.08 hrs 201/1.3

4.8 Failure time is Weibull with β =1.5 and θ =500.

FG 50 IJ a) R (50) = e H 500 K

1.5

−

=.96887

1/ β

b) td = θ ( − ln R ) → t.90 = 500( − ln(.90))2/ 3 = 11154 . hrs . hrs c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 500 ⋅ Γ (1 + 2 / 3) = 500⋅.903 = 4515

FG 100 IJ e H 500 K

1.5

−

d) R (50 / 50) = e)

R (50 + 50) = =.9438 R(50) .96887

FG t IJ +.002 H 500K 1/ 2

λ ( t ) =.003

LMFG t IJ +.002t OP H 500 K PQ R( t ) = e MN 1. 5

−

LM FG 50 IJ +.002( 50 )OP H 500 K PQ =.87667 R(50) = e MN 1. 5

→

−

4-2

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4.9 a) P

RS 200 − μ ≤ z ≤ 270 − μ UV =.90 = Pl-1.645 ≤ z ≤ 1.645q σ W T σ 200 − μ

σ σ =-

= −1645 . and

270 − μ

σ

200 - μ 1.645

= 1645 .

200 - μ 270 - μ =1.645 1.645

μ = 235 = MTTF b) R(210) = 1 − Φ

c) R( td ) = 1 − Φ

σ =-

200 - 235 . = 2128 .

1645

FG 210 − μ IJ = 1 − ΦFG 210 − 235IJ = 1 − Φ(−117 . ) = 1−.121 =.879 K H 2128 H σ K .

FG t − 235IJ =.99 H 2128 K . →

4.10 a) t R = tmed e sz1− R

→

FG 1 ln t IJ Hs t K

FG t − 235IJ =.01 H 2128 K . d

td = ( −2.32)2128 . + 235 = 185.63 hrs

R(10 + 200) = R( 200)

d) R(10 / 200 ) =

Φ

→

d

td − 235 = −2.32 2128 .

b) R( t ) = 1 − Φ

→

R( 210) .879 .879 = = =.926 200 − 235 1 − Φ( −164 . ) .9495 1− Φ 2128 .

FG H

IJ K

t.98 = 2236e.41( −2.05 ) = 964.82

med

R(5 weeks) = R(840 hrs) = 1 − Φ

FG 1 ln 840 IJ = 1 − Φ( −2.39) =.99158 H .41 2236K

Therefore probability of failure = 1 - .99158 = .00842 c) MTTF = t med

s2 e2

.412 = 2236e 2

= 2432.06 hrs

2 e s ( e s − 1) = 22362 e.41 ( e.41 − 1) = 1082754.22 d) σ 2 = t med 2236 t e) t mode = med2 = 2 = 1890.02 hrs s e e.41 2

2

2

2

4-3

→

σ = 1040.55 hrs

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

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4.11

→ a) R( t ) = e b g b) B1 → R(t.99 ) =.99 − t /θ β

R(5000) = e− ( 5000/12000 ) =.8529 2 .1

t.99 = θ ( − ln(.99))1/ β = 12000( − ln(.99))1/ 2.1 = 1342.27 hrs B.1 → t.999 =.999 t.999 = 12000( − ln(.999))1/ 2.1 = 447.42 hrs

c) MTTF = θ ⋅ Γ (1 + 1 / β ) = 12000 ⋅ Γ (148 . ) = 12000⋅.88574 = 10628.96 hrs

o

σ 2 = θ 2 Γ (1 + 2 / β ) − Γ (1 + 1 / β )

m

2

. ) − Γ (148 . ) t t = 12000 oΓ(195 2

2

r

= 12000 2 .979881 − (.885747 )2 = 28127988.29

d)

→

σ = 5303.58 hrs

t.50 = t med = θ ( − ln(.5))1/ β = 12000( − ln(.5))1/ 2.1 = 10078.22 hrs t mod e = θ (1 − 1 / β )1/ β = 12000(1 − 1 / 2.1)1/ 2.1 = 8819.71 hrs

FG 8760 JI e) F (1 yr) = F (8760 hrs) = 1- R (8760) = 1 − e H 12000 K −

2 .1

=.4033

4.12 without burn - in: t.90 = 2450( − ln(.90))1/.86 = 178.96

R( t + 30) R( t + 30) R( t + 30) = F 30 IJ .86 = .9776 =.90 R(30) −G e H 2450 K R( t + 30 ) =.90(.9776) =.8798

with burn - in: R( t / 30) =

FG t +30 IJ H e 2450 K =.8798 . 86

→

t = 2450( − ln(.8798))1/.86 − 30 = 194.53 days

Burn-in testing increases design life by 194.53-178.96=15.57 days; may not be worth it.

4.13

L O a) The 4 belt system has a Weibull failure distribution with β=1.34 & θ = M∑ (1 / θ ) P N Q 4

−1/ β

β

i

.

i =1

b

g + b1 / 3400g + b1 / 8000g + b1 / 6100g

θ = 1 / 2500

1.34

1.34

F (72) = 1 − R (72) = 1 − e b

− 72 /1399 .757

1.34

g

1. 34

b)

F (72 / 4000) = 1 − R (72 / 4000) = 1 − β

where R (t / T0 ) = e − (( t + T0 )/θ ) + ( T0 /θ )

1.34 −1/1.34

= 1399.757 hrs

= 1−.981 =.019 1. 34 1. 34 R( 72 + 4000) − 4072 /1399 .757 g + b 4000 /1399.757 g = 1− e b =.094 R (4000)

β

4-4

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4.14

FG t IJ FtI −2 G J H θK a) R ( t ) = 2e − e HθK β

FG 5000 IJ F 5000 IJ −2 G H 12000 K Rp (5000) = 2e − e H 12000 K =.9784

β

−

p

2 .1

2 .1

−

→

b) MTTF = θ ⋅ Γ (1 + 1 / β ) ⋅ [2 − 2−1/ β ] = 12000 ⋅ Γ (148 . ) ⋅ [1281 . ] = 136171 . hrs

FG 8670 IJ c) F (1 yr) = F (8760 hrs) = 1 − R(8760) = 2e H 12000 K

2 .1

−

FG 8670 IJ − e H 12000 K

FG t IJ F t IJ −2 G H 12000 K d) Find t , such that 2e − e H 12000 K =.99 2 .1

a) R( td ) = 1 − Φ

Φ

=.163

2 .1

−

4.15

2 .1

−2

→

t = 4100 hrs

FG t − μ IJ = 1 − ΦFG t − 10IJ =.99 H θ K H 2.5 K d

d

FG t − μ JI =.01 H θ K d

t − 10 z = −164 . = d 2.5 td = 2.5( −12.325) + 10 = 4.19 days

→

FG 1 − 10IJ = 1 − Φ(−3.6) = 1−.00016 =.99984 H 2.5 K F 2 − 10IJ = 1 − Φ(−3.2) = 1−.00069 =.99931 ii) R(2) = 1 − ΦG H 2.5 K F 5 − 10IJ = 1 − Φ(−2) = 1−.02275 =.97725 iii) R (5) = 1 − ΦG H 2.5 K F 6 − 10IJ 1 − ΦG H 2.5 K = 1 − Φ(−16. ) = .94520 =.9672 R (1 + 5) R (6) c) R (1 / 5) = = = b) i) R (1) = 1 − Φ

R (5)

R (5)

4.16 a) R(1000 ) = 1 − Φ

.97725

F 1 lnF 1000I I GH t JK JK GH 125 .

.97725

→

tmed =

med

b) MTTF = t med

σ =t 2

2 med

s2 e2

1.252 = 7816.6e 2

s2

c) R (1000 / 1000) =

2 1.252

R (2000) = R(1000)

1000 1.25( −1.645 )

e

= 17077.39 cycles

e (e − 1) = 7816.6 e s2

.97725

1− Φ

= 7816.60 cycles

(high variability)

− 1) = 1,099,124,041 ; σ = 33,153 cycles

1.252

(e

FG 1 lnFG 2000 IJ IJ H 7816.6K K = 1 − Φ(−109 H 125 . . ) =.9075 .95

4-5

.95

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4.17 It is Weibull with β = 2 and θ = 1 × 105 : a) λ ( t ) = 2 × 10−10 t = at b where a = 2 × 10−10 and b = 1

b = β −1 → β = b +1 = 2 a=

FG IJ H K

β 1 β β 2 = 2 →θ = = = 1 × 105 β −1 −10 a θ θ θ 2 × 10

FG IJ HK b g

β −1

β t λ ( t ) = 2 × 10 t = where β = 2 and θ = 1 × 105 θ θ b) MTTFB = θ ⋅ Γ 1 + 1 / β = 100000 ⋅ Γ (15 . ) = 100000×.886227 = 88622.7 hrs MTTFA = 1 / λ = 1/.000011 = 909091 . hrs → Based only on MTTF , A preferred. −10

c) R( 2 / 3 yr ) = R(5840 hrs)

RB (5840 ) = e − ( 5840/1×10 ) =.997 5 2

; RA (5840 ) = e− (.000011)( 5840 ) =.938

Based only on warrantly costs, motor B is preferred. d)

e − (10×5840/1×10 ) RB (9 / 1) = =.713 ; RA (9 / 1) = e − (.000011)9( 5840 ) =.561 .9966 Motor B will reduce replacement costs.

5 2

Ft I Ft I −G J + G J HθK Hθ K = 1 ⋅t −t ln R (t ) − ln R (t ) 4.18 average failure rate = AFR = = β

1

1

2

t 2 − t1

t 2 − t1

FG t − μ IJ → Hσ K F 100 − μ IJ =.01 → ΦG H 12 K

4.19 R( t ) = 1 − Φ

β

2

R(100) = 1 − Φ

θβ

β

β

2

1

t 2 − t1

FG 100 − μ IJ =.99 H 12 K

100 − μ 12 μ = 100 + 12(2.32) = 127.84 hrs z = −2.32 =

4.20 a) tmed = MTTF ⋅ e − s / 2 = 3600e −2 = 487.21 hrs 2

FG 1 lnFG 100 IJ IJ = 1 − Φ(−.79) =.785 H 2 H 487.21K K F 1 F 200 IJ IJ 1 − ΦG lnG H 2 H 487.21K K =.67/.785 =.85 R( 200) = b) R (100 / 100) = R(100) = 1 − Φ

R(100)

.785

1− Φ

FG 1 lnFG 300 IJ IJ H 2 H 487.21K K =.888

R (300) = R (200) .67 d) This is a decreasing failure rate (DFR) so do not replace until a failure occurs. Operating past the maximum point on the hazard rate curve:

c) R (100 / 200) =

4-6

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4.21 a)

b

g

b

g

MTTF = θ ⋅ Γ 1 + 1 / β = 5.6419 ⋅ Γ 1 + 1 / 2 = 5.6419 ⋅ Γ (15 . ) = 5.6419⋅.886227 = 5 yrs b)

F (12 months) = F (1 year) = 1 − R(1) = 1 − e − (1/5.6429) =.031 2

c)

R (1 + 1) R ( 2) e − ( 2 /5.6419) .882 F (1 / 1) = 1 − R(1 / 1) = 1 − = 1− = 1− =.1 − =.09 R (1) R (1) .969 .969 2

d) Expected value of first year warranty: $500(.031)=$15.5 Expected value of second year warranty: $500(.09)=$45 Warranty costs: $25 and $40 respectively.

First year warranty should not be purchased. Second year warranty should be purchased. e) Value if both must be purchased: ($500)(1-R(2))=($500)(1-.88)=$60 Warranty costs: $25+$40=$65 Warranties should not be purchased.

4.22 Knotsew

Dell Fye

Ima Goode

(a) R(10,000 hours)

.961

.981

.948

(b) MTTF

91,142 (91,057)

97,454

126,666

(c) The mode

40,867

28,915

0

(d) The median time to failure

76,967

65,000

111,299

4-7

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(e) The 90 percent design life

20,040

20,448

19,500

(f) The reliability for the next 10,000 hours if it has survived the first 10,000 hours. (g) characterize the hazard rate function as DFR, CFR, or IFR

.9368

.92215

.947

IFR

IFR/DFR

IFR

4.23 Weibull distribution with beta = 1.2 and theta = 10,000 (a) R(1,000 hrs) .9388 (b) MTTF (in hrs) 9406 (c) The mode (in hours) 2247 (d) The median time to failure 7368 (e) The 90 percent design life 1533 (f) The conditional reliability to get to 1000 hours given it survives 500 hr. (g) characterize the hazard rate function as DFR, CFR, IFR, or combination (h) Average failure rate from 0 to 5,000 hrs

.965 IFR 8.71 x 10-5

4.24 Lognormal distribution with tmed = 10,000 and s = 1.2. (a) R(1,000 hrs) .9725 (b) MTTF (in hrs) 20,544 (c) The mode (in hours) 2369 (d) The median time to failure 10,000 (e) The 90 percent design life 2140 (f) The conditional reliability to get to 1000 hours given it survives 100 hr. (g) characterize the hazard rate function as DFR, CFR, IFR, or combination (h) Average failure rate from 0 to 5,000 hrs

.9725 IFR/DFR 6.62 x 10-5

4.25 (a) R(5000) = .8885, (b) t0.90 = 4773, (c) R(5000|5000) = R(10,000) / R(5000) = .4884 (d) 1 – [1 – R(5000|5000)]2 = .7383 4.26 gamma distribution with γ = 1.8 and α = 10,000 hr. (a) MTTF = 17,000 hr., (b) std. dev. = 13,038 hr., (c) tmed = 13,807 hr., (d) tmode = 7,000 (e) t0.95 = 2420 hr., (e) R(5,000) = .8534, (f) R(1,000|5,000) = R(6000) / R(5000) = .9511, (g) IFR

4-8

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4.27

2000 ⎞ ⎛ 1 (a ) R(2000) = 1 − Φ ⎜ ln ⎟ = 1 − Φ(−.851) = .80234 ⎝ 1.7 8500 ⎠ t ⎞ ⎛ 1 (b) 1 − Φ ⎜ ln ⎟ = .99 ⎝ 1.7 8500 ⎠ t 1 = −2.33 or t = 8500 e( −2.33)(1.7) = 161.87 ≈ 161 landings ln 1.7 8500 4.28 .7

(a ) R (50) = e

−.00045(50)

(b) R (50 | 50) =

⋅e

⎛ 50 ⎞ −⎜ ⎟ ⎝ 8939.43 ⎠

⎡ ⎛ 1 50 ⎞ ⎤ ⎢1 − Φ ⎜ .4 ln 100 ⎟ ⎥ = .9519 ⎝ ⎠⎦ ⎣

R(100) .87706 = = .9214 .9519 R(50)

(c) MTTFexternal = 1/ λ = 2222.22; MTTFquality = ( 8939.43) Γ (1 + 10 / 7 ) = 11,315.7 MTTFwear −out = 200e.4 /2 = 216.6 2

4.29 This is a gamma distribution with the shape parameter of 2.4. (a) Using trial and error and the Chapter 4 spreadsheet or the course software, α is found to be approximately 5865. (b) R(5,000) = .872443, (c) MTTF = 14,076, (c) tmed = 12,176.67, (d) 1 – [1 - . 872443]2 = .98373 4.30 (a) Manufacturer

R(72)

MTTF

tmed

RWT Space Division Rockweed-Smarton

.9118 .9276

.415 585 hr

339.4 443.7

(b) R(72|50) .8818 .9202

(c) R(t|50) = R(t+50)/R(50) = .95 ; solve (trial and error) for t ≈ 39.5 hr.

4.31 Rsys = R (5 / 6 | 50) ⋅ e

−.083( 5/6 )

⋅ e −.034(5/6) = (.998743)(.933171)(.972064 ) = .905962

4.32 ( t −μ ) (t − μ ) ⎤ ⎡⎛ 1 ⎞ (t − μ ) − e (t−αμ ) ⎤ 1 −e α ⎡ α α f '(t ) = d ⎢⎜ ⎟ e e − 1⎥ ⎥ / dt = 2 e ⎢e α ⎢⎣⎝ α ⎠ ⎥⎦ ⎢⎣ ⎥⎦

⎡ (t − μ ) ⎤ setting ⎢e α − 1⎥ = 0 and solving t = μ ⎢⎣ ⎥⎦

4-9

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4.33 t

t

2t ' t '2 t2 t2 F (t ) = ∫ 2 dt ' = 2 = 2 ; R(t ) = 1 − 2 b b 0 b b 0 2t 2 2t λ (t ) = b 2 = 2 2 IFR t b −t 1− 2 b b

b

2t 2 2t 3 2b3 2 MTTF = ∫ 2 dt = 2 = 2 = b 3b 0 3b 3 b 0 2 tmed = .5 or tmed = b .5 b2

4.34 2t t 2 2t t 2 ⎛ 2 2t ' ⎞ F (t ) = ∫ ⎜ − 2 ⎟ dt ' = − 2 ; R(t ) = 1 − + 2 b b ⎠ b b b b 0⎝ 2 2t − 2 2 ; IFR λ (t ) = b b 2 = 2t t (b − t ) 1− + 2 b b t

b

⎛ 2t t 2 ⎞ ⎛ t2 t3 ⎞ b b MTTF = ∫ ⎜1 − + 2 ⎟ dt = ⎜ t − + 2 ⎟ = b − b + = 3 3 b b ⎠ ⎝ b 3b ⎠ 0 0⎝ b

2 2tmed tmed − 2 = .5 solve for tmed using quadratic formula b b

4-10

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CHAPTER 5

R(t ) =

5.1

z

∞

t

f (t ′)dt ′ =

z

∞

1 1 dt ′ = 2 t ( t + 1) t +1

L 1 OP a) R ( t ) = 1 − M1 − N t + 1Q L 1 OP =.999 b) 1 − M1 − N t + 1Q

'

L 1O R (.5) = 1 − M1 − P =.96296 N 15. Q

3

3

→

s

s

3

−1 1 = (1−.999)1/ 3 → t = 1- (1-.999)1/3 − 1 =.1111 yrs = 40.55 days t +1 1 c) R(.5) = =.667 .

15 3 3 x 3− x 2 3 .667 .333 Rs (.5) = ∑ = 3 .667 .333 + .667 =.44444+.296741 =.741185 x=2 x

1-

FG IJ b g b g b g b g b g HK 5.2 R = 1 − b1−.4g ⋅.97 =.95 b1−.4g = 1−.95/.97 → n = ln ln1−1.95−.4/.97 = 7.598 → 8 components n

n

5.3

d

(i) Rs (100) = 1 − 1 − e −100/1000 (ii) Rs (100) = e b

− 100 /10 , 000

5.4

b

a) Rtop = R ⋅ 1 − 1 − R

i = 2e 2

−100 /1000

− e −200 /1000 =.99094

g ⋅ e −.00005(100) =.99491 (higher reliability) 2

g ⋅ R = R 1 − b1 − Rg 2

2

2

Rbottom = R 3

and

Rs = 1 − (1 − Rtop )(1 − Rbottom ) ⋅ R

b g =.8019

b) Rtop =.9 2 1 − 1−.9

2

and

Rbottom =.9 3 =.729

Rs = 1 − (1−.8019)(1−.729) ⋅.9 =.85168

c

5.5 R( t ) = 1 − 1 − e −.1t

.93908e −5/ MTTF =.85

he 3

− t / MTTF

→

FG t IJ .8 −10FG t IJ Hθ K = e Hθ K 5.6 Rs ( t ) = e 10

−∑ i =1

i

i

→

c

R(5) = 1 − 1 − e−.1( 5 )

MTTF =

.8

3

5 . = 5017 .85 − ln .93908

IJ K

FG H

FG 1 IJ Rs (1) = e H θ K =.99 .8

→

he

−10

i

5-1

−5 / MTTF

=.85

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

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F 1I -10G J = ln(.99 ) → Hθ K

LM 1 ln(.99)OP N 10 Q

.8

i

5.7

m

a) Rs = 1 − (1−.9) 1 − (1 − (.1)(.2)).95

{

b) Rs = 1 − 1 − 1 − 1 − (.9 )(.9 )

−1/.8

θi = −

2

= 5588

rn1 − (1−.8) s =.985 3

}l

q

.97 1 − (.95)(.95) = 1 − (.065017 )(.0975) =.99366

c) Rs = 1 − (1 − R ⋅ R)(1 − R ) R = 1 − (1 − R − R 2 + R 3 ) R = R 2 (1 + R − R 2 ) =.99 Rs .8829 .9454 .9683 .9898

R .9 .95 .97 .99

5.8 a) Rs = 1 − (.1)(.1) 1 − 1−.8 1 − (.25)(.25) 1−.8 =.9405

hb g

c

b

gb

g

b) Rs = 1 − (1−.8) 3 1 − 1−.8(1 − (.3)(.3) 1 − (.9)(.9) =.9407 5.9 a) Rs = 1 − (1 − R3 )2 =.90

Rs (100) = e −100 λ =.881 b) Rs = 1 − (1 − R )2

3

→

(1 − R3 )2 =.10

→

λ=

=.90

Rs (100) = e −100 λ =.8142

→

ln(.881) =.001267 and MTTF = 1 / λ = 789.27 hrs −100

→ 1 − (1 − R )2 = 3 .90 →

R = 3 1 − .10 =.991

→

R = 1 − 1 − 3 .90 =.8142

λ=

ln(.8142 ) =.002055 and MTTF = 1 / λ = 486.62 hrs −100

→

n=

5.10 a) R(5) =

10 =.667 10 + 5

Rs (5) = 1 − (1−.677)n =.98

log(1−.98) = 356 . → 4 units log(1−.667)

b)

Rs (5) = 1 − (1−.677)n e −.002( 5 ) =.98 → .323n = 1−.98e.01 → n =

log(.01015) = 417 . →5 log(.333)

The achieved reliability is: R s (5) = 1 − (1−.677 )5 e −.002( 5 ) =.9866

d id = 1 − n1 − e −e

id

i

od

id

5.11 Rs (t ) = 1 − 1 − e − λ 1t 1 − e − λ 2 t 1 − e − λ 3t = 1 − 1 − e − λ 1t 1 − e − λ 3t − e − λ 2 t + e − ( λ 2 + λ 3 ) t − λ 3t

− λ 2t

+ e − ( λ 2 + λ 3 ) t − e − λ 1t + e − ( λ 1 + λ 3 ) t + e − ( λ 1 + λ 2 ) t − e − ( λ 1 + λ 2 + λ 3 ) t

= e − λ 3t + e − λ 2t − e − ( λ 2 + λ 3 ) t + e − λ 1t − e − ( λ 1 + λ 3 ) t − e − ( λ 1 + λ 2 ) t + e − ( λ 1 + λ 2 + λ 3 ) t

5-2

s

it

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

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R(.1) = .8068 ∞ 1 1 1 MTTF = Rs (t ) dt = + + −

z

λ1

0

=

λ2

λ3

1

λ1 + λ 2

−

1

λ1 + λ 3

−

1

λ2 + λ3

+

1

λ1 + λ 2 + λ 3

1 1 1 1 1 1 1 + + − − − + =.2433 yrs = 88.8 days 5 10 15 15 20 25 30

5.12 With no redundancy R=.90.

With redundancy R= .95 1 − (1−.9) 2 .95 =.893. Redundancy does not improve reliability. .9

splitter

comparator

.9

5.13 Rs = (reliability with respect to open) - (probability of failing short)

b

g

b

= (1 − qo1qo 2 )(1 − qo5 )(1 − qo 3qo 4 ) − 1 − (1 − q s1 )(1 − q s 2 ) ⋅ q s5 ⋅ 1 − (1 − q s 3 )(1 − q s 4 ) = (1−.0004)(1−.05)(1−.0004) − (1 − (1−.15) 2 )(.2)(1 − (1−.15) 2 ) =.9338 5.14 3

High-level Network: R = (1 − q s2 ) 3 − 1 − (1 − qo ) 2 Low-level Network: R = 1 − qo3

2

− 1 − (1 − q s ) 3

2

= (1−.01) 3 − 1 − (.95) 2

= 1−.053

2

− 1 − (.9) 3

low level

2

3

=.96937

=.92631

high level

5.15 Rs = (reliability with respect to open) - (probability of failing short) = (1 − qo1 )(1 − qo 2 qo 3 )(1 − qo 4 ) − q s1 1 − (1 − q s 2 )(1 − q s 3 ) q s 4 5.16

FG FG IJ IJ = 1 − Φ( −210 . ) =.9821, H .7 H 435K K

FG 100 IJ 1 100 R1 (100) = e H 840 K =.9252, R2 (100) = 1 − Φ ln 1. 2

−

R3 (100) = e −.0001(100) =.99 a) Rs = R1 (100) ⋅ R2 (100) ⋅ R3 (100) = (.9252)(.9821)(.99) =.8996

n

b) Rs = 1 − 1 − (.9252)(.9821)

2

s(.99) =.9817

c ) Rs = 1 − (1−.9252) 2 1 − (1−.9821) 2 .99 =.9841

5-3

g

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5.17

R3 fails with probability of .05: R3 fails = (.9)2 (.99) =.8019 R3 works with probability of .95: R3 works =.99 Rs = R3 fails + R3 works =.05(.8019 )+.95(.99) =.980595 5.18 3

⎛ 3⎞

∑ ⎜⎝ x⎟⎠ R (1 − R)

a) R =

x

3− x

x=2

⎛ 3⎞ ⎛ 3⎞ = ⎜ ⎟ R 2 (1 − R) + ⎜ ⎟ R 3 = 3R 2 (1 − R ) + R 3 =3R 2 − 3R 3 + R 3 ⎝ 2⎠ ⎝ 3⎠

2 R − 3R2 + R = R( 2 R2 − 3 R + 1) = R( 2 R − 1)( R − 1) = 0 → R = 0, R =.5, or R = 1 3

R ≠ 0 and R ≠ 1. If R ≤.5, system reliability is less than a component's.

b) R =

4

⎛ 4⎞

∑ ⎜⎝ x⎟⎠ R (1 − R) x

4− x

x =3

⎛ 4⎞ ⎛ 4⎞ = ⎜ ⎟ R 3 (1 − R) + ⎜ ⎟ R 4 = 4 R 3 − 3R 4 + R 4 =4 R 3 − 3R 4 ⎝ 3⎠ ⎝ 4⎠

4 R − 3R4 − R = R( 4 R2 − 3R3 − 1) = 0 → 4 R2 − 3R3 − 1 = 0 since R ≠ 0 Solving numerically for the roots of 4 R 2 − 3R3 − 1 = 0 results in R=.767.

If R ≤.767 , system reliability is less than a single component.

3

5.19 There are 5 combinations of the 16 total which result in system success (non-failure): R1 R2 R3 R4 System Probability S S S S S .92(.95)(.99)=.761805 S S F S S .92(.05)(.99)=.040095 S F S S S .9(.1)(.95)(.99)=.084645 F S S S S .1(.9)(.95)(.99)=.084645 F F S S S .12(.95)(.99)=.009405 TOTAL: .980595 5.20

LM FG 3IJ R (1 − R) OP R = 3R (1 − R) + R R = R (3 − 2 R) R N H xK Q 3

Rs = ∑

x

3− x

2

3

v

x =2

2

v

v

To determine crossover reliability:

R = R2 (3 − 2 R ).95 → R - 2.85 R2 + 19 . R3 = R(1 - 2.85 R + 19 . R2 ) = 0 Solving (1 - 2.85 R + 19 . R 2 ) for R with the quadratic formula results in R=.9402 and R=.5598.

For .5598 ≤ R ≤.9402 , system reliability is MORE than a single component reliability.

5.21

FG nIJ (1 )b− Rg = 1 − ∑ FG nIJ (1)( −1) b− Rg = 1 − ∑ FG nIJ (−1) b− Rg H iK H iK H iK F nI F nI F nI R ( t ) = 1 − ∑ G J ( −1) e = 1 − 1 + ∑ G J ( −1) e = ∑ G J ( −1) e H iK H iK H iK n

Rs = 1 − (1 − R )n = 1 − ∑

n−i

i

i=0

n

s

i=0

i − λ it

n

i

i=0

n

i −1 − λ i t

i =1

i=0

n

i =1

5-4

n

i

i −1 − λ i t

i

i

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

Waveland Press, Inc., Copyright © 2009

z FGH niIJK (−1) e dt = ∑ FGH niIJK ( −1) z e dt = λ1 ∑ FGH niIJK ( −1i) ∞

n

n

i −1 − λ i t

MTTF = 0 ∑

i −1 ∞ 0

i =1

i =1

n

− λ it

i −1

i =1

5.22 a) reliability with respect to short - probability of failing open=

LM OPLM1 − ∏ 1 − q OP = ∏ 1 − q − ∏ 1 − (1 − q ) Q N QN b) reliability with respect to open - probability of a short= R = L1 − ∏ q O − L1 − ∏ (1 − q )O NM QP NM QP m

m

i =1

i =1

Rs = 1 − ∏ 1 − qoin m

s

n oi

m

n oi

i =1

m

n

si

i =1

m

oi

i =1

si

i =1

5.23

reliability with respect to open - probability of a short= L1 − ∏ q O − L1 − ∏ (1 − q )O MN PQ MN PQ n

m

i =1

5.24 n

n

i =1

i =1

FG H

m

oi

i =1

n

si

IJ K

n

λ s (t ) = ∑ λ i (t ) = ∑ ai t = ∑ ai t i =1

FG ∑ a IJ t F I H K − G ∑ a J t ' dt ' − − λ ( t ) dt ' =e H K =e 2 R(t ) = e

z

z

t

s

0

z

n

t

n

0 i =1

∞

MTTF = R(t )dt = 0

z

i =1

i

FG ∑ a IJ t H K ∞ −

0

n

i =1

e

2

i

2

2

i

=

π n

2∑ ai i =1

n

z

∞

NOTE: 0 e − a x dx = 2 2

1 π 2a

where, in this problem, a =

∑ ai

i =1

2

5.25 (a) Φ(XA XB XC ) = 1 - (1- XA) (1 - XB XC ) = 1 - [1 - XA - XB XC + XA XB XC ] = XA + XB XC - XA XB XC (b) Pr { Φ(XA XB XC ) = 1} = E [XA + XB XC - XA XB XC ] = RA + RB RC - RA RB RC = .9 + .82 - (.9) (.82 ) = .964 (c) Min Path Sets A B, C

Min Cut Sets A, B A,C

(d) Rl = [1 - (1-.9)(1-.8) ]2 = .9604 and Ru = 1 - (1-.9) (1 - .82 ) = .964

5-5

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

Waveland Press, Inc., Copyright © 2009

5.26 (a) Φ(XA XB XC ) = XA [1 - (1- XB) (1 - XC ) ]XD = XA XD ( XB + XC - XB XC ) = XA XD XB + XA XD XC - XA XD XB XC (b) Rsys = E[Φ(XA XB XC XD)] = E [XA XD XB + XA XD XC - XA XD XB XC ] = RA RD RB + RA RD RC - RA RD RBRC = 2 [ (.95) (.98) (.9) ] - (.95) (.92 )(.98) = .92169 (c) Min Path Sets A, B, D A, C, D

Min Cut Sets A D B, C

(d) Rl = (.95) (.98) [ 1-(1-.9)2 ] = .92169 and

Ru = 1 - [1 - (.95)(.9)(.98)]2 = .9737

5.27 (a) λsys = .27 x 10-3 ; Rsys(1000) = e-.27 = .7634; Rredund(1000) = 1 – (1-.7634)2 = .9440, MTTF = 1.5 / λ = 5555 (b) RA(1000) = e-.12 = .8869 ; RB(1000) = e-.15 = .8607; Rredund(1000) = [1 – (1-.8869)2 ] [1 – (1-.8607)2 ] e-.05 = .9208 Let x = exp(-.12) and y = exp(-.15) and ignoring for the moment the switching reliability function: R = [1 – (1- x)2] [1 – (1- y)2] = [1 – (1 – 2x + x2)] [1 – (1 – 2y + y2)] = (2x - x2) (2y - y2) = 4xy - 2xy2 – 2x2y + x2y2 or R = {4 exp(-.27) - 2 exp(-.42) - 2exp(-.39) + exp(-.54) } exp(-.05) = 4 exp(-.32) - 2 exp(.47) - 2 exp(-.44) + exp(-.59) Integrating to get the MTTF: 4/.32 - 2/.47 - 2/.44 + 1/.59 = 5.394 or 5,394 hrs. (c) R(1000) = p(2) + p(3) = 3 (.7634)2 (1-.7634) + (.7634)3 = .4136 + .4449 = .8585; MTTF = 1 / (.27 x 10-3 ) (.5 + .3333) = 3086 (eq. 5.10) 5.28

(a) R = 1 – (1 - .8187)2 = .9671 and MTTF =1.5 x 1/.2 = 7.5; (b) R = [1- (1- .9231)2 ] [1- (1- .8869)2 ]

= .9814

Letting x = exp(-.08) and y = exp(-.12): R = [1 – (1- x)2] [1 – (1- y)2] = [1 – (1 – 2x + x2)] [1 – (1 – 2y + y2)] = (2x - x2) (2y - y2) = 4xy - 2xy2 – 2x2y + x2y2 or R = 4 exp(-.20) - 2 exp(-.32) - 2exp(-.28) + exp(.40) Integrating to get the MTTF: 4/.2 - 2/.32- 2/.28 + 1/.4 = 9.107 (c) Rsubassemby = exp(-.2) = .8187 or Rsubassemby =exp(-.08) exp(-.12) = (.9231) (.8869)= .8187 4 ⎛4⎞ 1 ⎛1 1 1⎞ x 4− x R = ∑ ⎜ ⎟ (.8187 ) (1 − .8187 ) = .9794 and MTTF = ⎜ + + ⎟ = 5.42 .2 ⎝ 2 3 4 ⎠ x =2 ⎝ x ⎠

5.29 Rsystem = [1-(.98)(.97)] [1-(.95)(.94)] [1- (1- .94) 2] (.99) = .9812 5.30 If component g fails: Pfail = .10 and Rsys = (.98)(.97)(.94)(.98) = .8757 If component g does not fail: Rg = .9 and Rsys = {1 – [1-(.98)(.97)(.94)][1 – (.95)(.94)]} (.98) = .9688 Rsys = (.10) (.8757) + (0.90) (.9688) = 0.9595

5-6

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

Waveland Press, Inc., Copyright © 2009

5.31 R(1) = 1 – (1-exp(-λ) ) = .99; λ = .24268; MTTFunit = 4.12 3

∞

MTTF = ∫ ⎣⎡1 − (1 − e 0

∞

) ⎦⎤dt = ∫ ⎣⎡1 − (1 − e − λt )(1 − e − λt ) 2 ⎦⎤dt

− λt 3

0

∞

∞

0

0

= ∫ ⎣⎡1 − (1 − e − λt )(1 − 2e − λt + e −2 λt ) ⎦⎤dt = ∫ 1 − ⎣⎡1 − 3e − λt + 3e −2 λt − e −3λt ⎤⎦ dt ∞

= ∫ ( 3e − λt − 3e −2 λt + e −3λt ) dt = 0

3

λ

−

3 1 11 + = = 7.55 2λ 3λ 6λ

5.32 (a)

(b) Rs = {1 - (1-.9) [1 - .9 (1-.152 ) ]} (.98) (.95) = (.987975 ) (.98) (.95) = .9198 (c) Let Rp = reliability of processor. Then .987975 x Rp x .95 = .93 or Rp = .9908625 Therefore: R(t) = e- .0001 t = .9908 or t = -ln .9909 / .0001 = 91.4 minutes.

5.33 (a)

R=.8855

R=.8855

R=.8855

R=.8855

Rsystem = [1 – (1-.88552) ]2 = .974 – a form of low-level redundancy (b) K(2) out of N(4) system: R = p(2) + p(3) + p(4): R =

4

⎛ 4⎞

∑ ⎜ i ⎟ .8855 (1 − .8855) i=2

⎝ ⎠

i

4 −i

= .9945

5.34 Load sharing system with the two front engines in series.

λ1 = 2(.0039) = .0078, λ2 = .0054, λ1+ = 2(.02) = .04, λ2+ = .012 ; R(9) = .9890, MTTF = 135.23

5-7

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

Waveland Press, Inc., Copyright © 2009

5.35 (a) high level redundancy:

Router eng = e (

−2 .0054 ) (9)

= .9074; Rinner eng = e (

−2 .0039 ) (9)

= .9322

Rsys = 1 − (1 − .9074)(1 − .9322) = .9937

(

)(

)

Rsys = 1 − 1 − e −2(.0054)t 1 − e −2(.0054)t = e −2(.0054)t + e −2(.0054)t − e−2(.0054+.0039)t MTTF =

1 1 1 + − = 167.03 2 (.0054 ) 2 (.0039 ) 2 (.0054 + .0039 )

(b) Low level redundancy:

Rinner eng = e −(.0039)( 9) = .9655; Router eng = e −(.0054)( 9) = .9526 Rsys = ⎡⎣1 − (1 − .9655 )(1 − .9526 ) ⎤⎦ = .9967 2

MTTF = 363.33 (expand Rsys(t) and integrate) 5.36 Rhigh level = 1 – (1 - .74127)3 = .98268; Rlow level = Π [1 – (1-Ri)2] = .98221 5.37 Rlow level = [1 – (1-.9)2]9 = .9135; Rhigh level = 1 – (1 - .99)n = 1 – (1 - .38742)n For n = 5; Rhigh level = 1 – (1 - .38742)5 = .91374.

5.38 For two-engine aircraft, reliability = R2 ; For 4-engine aircraft, reliability = R4 + 4 R3 (1 - R) Find value for R in which R2 > R4 + 4 R3 (1 - R) or R2 - [R4 + 4 R3 (1 - R)] > 0 R2 - R4 - 4 R3 + 4 R4 = R2 - 4 R3 + 3 R4 = R2 (1 - 4R + 3R2) = R2 (1-R) (1-3R) > 0 Since 0<R<1, then 1-3R > 0 or R < 1/3 5.39 (a) Case I. Component fails: Probability = ‐ R = . : RI = R x R x R = . . . = .

Case II. Component does not fail: Probability = R = . : RII = ‐ [ ‐R x R ][ ‐ R ] = ‐ [ ‐ . . ][ ‐. ]=. . Note that component does not enter into the calculation since we would obtain the same value for RII regardless of whether component one fails or does not fail.

Rsys = ‐R RI + R RII = . +. . =.

5-8

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

Waveland Press, Inc., Copyright © 2009

(b)

R1

R2

R3

R4

R5

System

Prob

Prob

count

0.8

0.9

0.8

0.9

0.9

1

S

S

S

S

S

S

0.46656

0

1

S

S

S

S

F

S

0.05184

0

2

S

S

S

F

S

S

0.05184

0

3

S

S

F

S

S

S

0.11664

0

4

S

F

S

S

S

S

0.05184

0

5

F

S

S

S

S

S

0.11664

0

1

S

S

S

F

F

S

0.00576

0

2

S

S

F

F

S

F

0.01296

0.01296

3

S

F

F

S

S

S

0.01296

0

4

F

F

S

S

S

S

0.01296

0

5

S

S

F

S

F

F

0.01296

0.01296

6

S

F

S

S

F

F

0.00576

0.00576

7

S

F

S

F

S

F

0.00576

0.00576

8

F

S

S

S

F

S

0.01296

0

9

F

S

S

F

S

F

0.01296

0.01296

10

F

S

F

S

S

S

0.02916

0

1

S

S

F

F

F

F

0.00144

0.00144

2

S

F

F

F

S

F

0.00144

0.00144

3

F

F

F

S

S

S

0.00324

0

4

S

F

F

S

F

F

0.00144

0.00144

5

S

F

S

F

F

F

0.00064

0.00064

6

F

S

S

F

F

F

0.00144

0.00144

7

F

S

F

F

S

F

0.00324

0.00324

8

F

S

F

S

F

F

0.00324

0.00324

failure

9

F

F

S

S

F

F

0.00144

0.00144

10

F

F

S

F

S

F

0.00144

0.00144

1

S

F

F

F

F

F

0.00016

0.00016

2

F

S

F

F

F

F

0.00036

0.00036

3

F

F

S

F

F

F

0.00016

0.00016

4

F

F

F

S

F

F

0.00036

0.00036

5

F

F

F

F

S

F

0.00036

0.00036

1

F

F

F

F

F

F

0.00004

0.00004

32

totals

5-9

1

0.0676

Rsys =

0.9324

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 6

6.1 Load Sharing System:

R (t ) = e − 2λt + t 1 2 3 4 5

MTTF =

[

]

+ 2λ .2 e −λ t − e − 2 λt = e −.2t + e −.3t − e −.2t = e −.2t − 2 e −.3t − e −.2t + .2 − .3 2λ − λ

LM N

[

]

[

R(t) .9745 .9133 .8333 .7456 .6574

OP Q

1 .2 1 1 = 8.33 yrs + − .2 −.1 .3 .2

6.2 Standby System: λ1=1/MTTF1=1/1000=.001, λ2 =1/333=.003, λ−2 =1/2000=.0005 a) R( t ) =.90 = e − λ1t + t 275 280 290 291

− − . 003 t λ1 e − λ 2t − e − ( λ1 + λ 2 )t = e −.001t −.667 e − e −.0015t − λ1 + λ 2 − λ 2

R(t) 0.9087 0.9060 0.9005 0.8999

design life=291 hours λ1 1 1 .001 b) MTTF = + = + = 1222 hrs λ 1 λ 2 λ 1 + λ−2 .001 .003 .001+.0005

d

b

i

g

6.3 Standby System: a) R( t ) = e − λ1t +

λ1 .001 e− λ t − e − ( λ +λ ) t = e−.001t + e −.0005t − e− (.001+0 ) t − λ1 + λ 2 − λ 2 .001−.005 2

1

− 2

R(72) = e-.072 −.25 e −.36 − e−.072 =.9887 b) MTTFsystem =

1

1

λ1

+

λ1 1 λ1 1 1 = + = + = 2000 − λ2 (λ1 + λ2 ) λ1 .005(λ1 + 0) λ1 .005

1 = 2000 − λ1 .005

1 O 1 L λ = M2000 − = P N .005 Q 1800 −1

→

1

6-1

]

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

Waveland Press, Inc., Copyright © 2009

6.4 Standby System with Switching Failures: a) R( t ) = e − λ1t +

− (1 − p )λ 1 (.995).001 −.0005t − e − (.001+0 ) t e − λ 2t − e − ( λ1 +λ 2 ) t = e−.001t + e − .001−.005 λ1 + λ 2 − λ 2

R(72) = e-.072 − (.995).25 e −.36 − e−.072 =.9884 b) MTTFsystem =

=

1

λ1 1

λ1

LM N

LM N

OP Q

OP Q

+

1 1 1 (1 − p )λ 1 1 1 (1 − p )λ 1 = + − − − − λ1 + λ 2 − λ 2 λ 2 λ1 + λ 2 λ1 λ1 − λ 2 λ 2 λ1

+

(1 − p )λ 1 λ 1 − λ 2 1 (1 − p ) 1 .995 1 = + = + = + 199 = 2000 λ1 − λ 2 λ 2λ 1 λ1 λ2 λ 1 .005 λ 1

LM N

λ1 = ( 2000 − 199)−1 =.000555

OP Q

→

MTTF = 1 / λ 1 = 1801

6.5 Three-state Device (eqn 6-41):

R(t R ) = e − ( λ 1 + λ 2 ) t R = e − (.00034 + .000021) t R = e − .000361t R =.99 P( t ) =

→

λ2 .000021 1 − e − ( λ1 + λ 2 ) t = 1 − e −.000361(1000 ) =.01763 .000361 λ1 + λ 2

tR =

ln(.99) = 27.84 hrs −.000361

6.6 state 1 2 3 4 5

unit1 on-line failed on-line failed failed

unit2 on-line on-line failed failed failed

unit3 standby standby standby on-line failed

dP1 ( t ) = −2λP1 ( t ) → P1 ( t ) = e −2 λt dt dP2 ( t ) = λP1 ( t ) − λP2 ( t ) = λe −2 λt − λP2 ( t ) dt dP2 ( t ) + λP2 ( t ) = λe −2 λt dt

z

z

P2 ( t )eλt = e −2 λt eλt dt + C = e − λt dt + C =

e − λt +C −λ

e −2 λt e −2 λt 1 − λt 1 − λt + Ce − λt = + e = e 1 − e − λt λ −λ −λ λ 1 1 since P2 ( 0) = 0 = + C, C = −λ λ 1 − λt dP3 ( t ) = λP1 ( t ) − λP3 ( t ) = e 1 − e − λt since state 3 is equivalent to state 2 dt λ P2 ( t ) =

6-2

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

Waveland Press, Inc., Copyright © 2009

dP4 ( t ) = λP2 ( t ) + λP3 ( t ) − λP4 ( t ) = 2e − λt 1 − e − λt − λP4 ( t ) dt

LM e OP + C N λQ L e OP + Ce = 2e LMt + e − 1 OP P ( t ) = 2e M t + N λQ N λ λQ

z

P4 ( t )eλt = 2e − λt 1 − e − λt eλt dt + C = 2 t + − λt

− λt

− λt

− λt

− λt

− λt

4

since P4 ( 0) = 0 =

2

λ

+C→C = −

2

λ

R( t ) = P1 ( t ) + P2 ( t ) + P3 ( t ) + P4 ( t ) = e −2 λt + 2te − λt ∞

MTTF = ∫ R(t )dt = 0

∞ 1 1 2 + 2∫ te − λt dt = + 2 0 2λ 2λ λ

6.7 Load Sharing System with λ1=λ2=λ=.05 and λ+1 = λ+2 =.2. + 2λ R( t ) = e −2 λt + e − λ t − e −2 λt + 2λ − λ 2(.05) −2(.05)( 5 ) R(5) = e e −.2( 5) − e2(.05)5 =.6065+.2387 =.8452 + 2(.05)−.2

MTTF =

LM N

OP Q

LM N

OP Q

1 2λ 1 1 1 2(.05) 1 1 = + − = 10 + ( −1) 5 − 10 = 15 hrs + − + + 2λ 2 λ − λ λ 2λ 2(.05) 2(.05)−.2 .2 .1

6.8 Single 1000 gpm pump: MTTF = 700 hrs and t d = − MTTF ln R = −700 ln(.80) = 156.2 hrs Load Sharing System: MTTF = 600 hrs MTTF=700 is preferred MTTF. + λ 2 2 / 800 R( t ) = e −2 λt + e − λ t − e −2 λt = e −2 t / 800 + e − t / 200 − e −2 t / 800 + 2 / 800 − 1 / 200 2λ − λ = e − t / 400 − (1) e − t / 200 − e − t / 400 =.80 By trial and error:

R(t) .8452 .8149 .8026 .8014 .8001

t 200 225 235 236 237

tR = 237 hrs is preferred design life.

6.9 Standby System with λ 1 = 1 / 700 =.00142857, λ 2 =.005, and λ-2 =.001:

R(t ) = e − λ 1t +

λ1 .001428 e − λ t − e − ( λ + λ ) t = e −.001428t + e −.005t − e − (.001428 + .001) t − λ1 + λ 2 − λ 2 .001428+.001−.005 2

1

− 2

= e −.001428t −.5552 e −.005t − e −.002428t =.80 By trial and error: td=295.5 hrs

MTTF =

1

λ1

+

.001428 1 λ1 = + = 817.6 hrs − λ 2 (λ 1 − λ 2 ) .001428 .005(.001428+.001)

This system is preferred based on both design time and MTTF. 6.10 6-3

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

Waveland Press, Inc., Copyright © 2009

a) R (100) = 1 − (1 − e −.0034(100) )(1 − e −.0105(100) ) = .8126 b) Standby System with Switching Failure ( λ 1 =.0034, λ 2 =.0105, and λ−2 =.0005 ): − (1 − p) λ 1 R (t ) = e − λ 1t + e − λ 2 t − e − ( λ 1 + λ 2 ) t = e −.0034 t + ( −.4378) e −.0105t − e − (.0034 + .0005) t − λ1 + λ 2 − λ 2 R (100) =.855 c) Load Sharing System (eqns 6-16,17 & 18):

λ1 =.0034, λ 2 =.0105, λ+1 =.0051, λ+2 =.01575 R(t ) = e − ( λ 1 + λ 2 ) t +

+ + λ1 λ2 e −λ 2 − e −( λ1 +λ 2 )t + e −λ1 − e −( λ1 +λ 2 )t + + λ1 + λ 2 − λ 2 λ1 + λ 2 − λ1

= e − (.0139) t + ( −1838 . ) e −.01575t − e −.0139 t + 1193 . e − .0051t − e −.0139 t

R(100) =.2491+.077 +.4192 =.745 MTTF(a) = 1/.0034 + 1/.0105 - 1/[.0034+.0105] = 317.4

MTTF(b ) =

1

λ1

+

(1 − p )λ1 ⎡ 1 1 ⎤ = 364.7 ⎢ − − λ1 + λ2 − λ2 ⎣ λ2 λ1 + λ2− ⎥⎦

MTTF(c) =

LM N

LM N

OP Q

OP Q

1 1 1 1 1 λ1 λ2 =235.57 + − + − + + + + λ1 + λ 2 λ1 + λ 2 − λ 2 λ 2 λ1 + λ 2 λ1 + λ 2 − λ1 λ1 λ1 + λ 2

6.11 Three Component Standby System:

LM N

R( t ) = e − λt 1 + λt +

OP 2 Q

λ2t 2

LM N

R(10 ) = e −.02(10) 1+.2 +

→

OP Q

.22 =.9988 2

6.12 Let: λ a =active unit’s failure rate, λ s =standby units failure rate while in use ( λ−s =0) State 1 2 3 4

Unit 1

Unit 2

active failed failed failed

standby active failed failed

Unit 3 standby standby active failed

Solving these 3 differential equations with initial conditions: P1 ( 0) = 1, P2 ( 0) = P3 ( 0) = 0 dP1 ( t ) dP2 ( t ) dP3 ( t ) = − λ a P1 ( t ) , = λ a P1 ( t ) − λ s P2 ( t ) , = λ s P2 ( t ) − λ s P3 ( t ) dt dt dt results in: P1 ( t ) = e − λ a t

FG λ IJ e Hλ −λ K F λ λ IJ te P (t ) = G Hλ −λ K P2 ( t ) =

a

a

− e− λ at

− λ st

+

s

a

s

3

a

− λ st

s

FG λ λ IJ e H (λ − λ ) K a

s

2

a

s

6-4

− λ at

− e − λ st

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

Waveland Press, Inc., Copyright © 2009

R( t ) = P1 ( t ) + P2 ( t ) + P3 ( t ) = e− λ at +

z

FG λ IJ e Hλ −λ K a

a

− λ st

− e− λ at +

FG λ λ IJ te Hλ −λ K a

s

a

s

− λ st

s

2λ a 2λ s − 2λ a MTTF = R( t )dt = + + 0 λ a λ s ( λ a − λ s ) ( λ a − λ s )2 ∞

1

6.13

z

z

MTTF = 0 R( t )dt = 0 e− b λ1 + λ 2 g t + ∞

θ

FG λ λ IJ e H (λ − λ ) K a

z

+ 2

s

− λ at

2

a

note:

λ1 ∞ e− λ t − e−b λ +λ gt + 0 λ1 + λ 2 − λ 2 1

z

− e − λ st

s

te − λ st dt =

1

λ2s

2

LM N

+ 1

2

1

+ 1

2

+ 1

1

1

2

2

6.14 λ 1 = 3 / 8, λ 2 = 2 / 8, λ+1 = 6 / 8, λ+2 = 3 / 8, λ 1 + λ 2 = 5 / 8 MTTF =

LM N

OP Q

LM N

OP Q

1 3/ 8 1 1 2/8 1 1 = 3.733 hrs + − + − 5/ 8 2 / 8 3/ 8 5/ 8 −1 / 8 6 / 8 5 / 8

R( t ) = P1 ( t ) + P2 ( t ) + P3 ( t ) = e− ( 5/ 8 )t + 3 / 2 e − ( 3/ 8 ) t − e − ( 5/ 8 ) t + 2 e − ( 6/ 8 ) t − e − ( 5/ 8 ) t R(1) =.8891 6.15

OP Q

1 1 1 λ2 λ1 ∞ e− λ t − e− b λ +λ gt = + + − + 0 + + λ1 + λ 2 − λ1 λ1 + λ 2 λ1 + λ 2 − λ 2 λ 2 λ1 + λ 2

z LM 1 − 1 OP λ + λ + λ − λ Nλ λ + λ Q

+

+

R(8) =.07983

and

k −1 1 k −1 λi i ! k −1 λi k ( λt )i ∞ dt = ∑ 0 e− λt t i dt = ∑ = ∑ = i +1 i=0 λ i=0 i ! λ i=0 i ! i=0 i ! λ n! Γ( n + 1) ∞ where 0 x n e− ax dx = = n +1 for n positive integer n +1 a a

z

z

∞

∞

z

k −1

MTTF = 0 Rk ( t ) = 0 e− λt ∑

z

6.16 λ1 1

dP1 ( t ) = − λ 1 P1 ( t ) dt

→

λ2 2

P1 ( t ) = e− λ1t

6-5

3

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dP2 ( t ) = λ 1 P1 ( t ) − λ 2 P2 ( t ) dt dP2 ( t ) + λ 2 P2 ( t ) = λ 1 P1 ( t ) = λ 1e− λ1t dt

z

P2eλ 2t = λ 1e− ( λ1 −λ 2 ) t dt = P2 ( t ) =

z

λ1

−( λ 1 − λ 2 )

z

∞

λ1

−( λ 1 − λ 2 )

e− λ1t + Ce− λ 2t =

z

∞

e − ( λ1 − λ 2 ) t + C

λ1 λ1 e− λ t − e− λ t , P2 ( 0) = 0 → C = ( λ1 − λ 2 ) ( λ1 − λ 2 ) 2

∞

MTTF = 0 R( t )dt = 0 P1 ( t )dt + 0 P2 ( t )dt = =

1

λ1

+

LM N

OP Q

1

λ1

+

1

LM N

λ1 1 1 − λ1 − λ 2 λ 2 λ1

OP Q

λ1 λ1 − λ 2 λ 1 1 1 = + 1 = + λ 1 − λ 2 λ 1λ 2 λ 1 λ 1λ 2 λ 1 λ 2

6.17 State Description 2 units operating, one in standby 1 unit failed, 2 units operating 2 units failed, 1 operating (system has failed)

State 1 2 3

2λ 1

2λ 2

dP1 (t ) = −2λ P1 (t ) dt

3

dP2 (t ) = 2λ P1 (t ) − 2λ P2 (t ) dt

R(1000)

(a) .966

(b) .946

MTTF

7037

6703

6.18

and

R = P1 ( t ) + P2 ( t )

(a) λ = .07 + .08 = .15 and λ+ = .27: R = .9664 using eq. 6.19 and MTTF = 7037 using eq. 6.20. (b) λ = λ1 = λ2 = .27 and λ- = λ-2 = .01 + .02 = .03, and p = .10: R = .9446 using eq. 6.33 and MTTF = 6703 from 1/λ + (1-p) / (λ + λ-) (eq. 6.30 adjusted for p) 6.19 (a) λ = .08 + .12 = .2, λ- = .01 + .04 = .05; use Eq. 6.29 for R(1) and Eq. 6.30 for MTTF R(1) = .9761, MTTF = 8.57 (b) λ+ = .08 + .12 = .2, λ = .05 + .09 = .14; use Eq. 6.19 for R(1) and Eq. 6.20 for MTTF R(1) = .9133, MTTF = 4.167 6.20 λ1 = .01; λ1+ = .02 load-sharing: λ2 = .02, λ2+ = .04 ; standby λ1 = .01, λ2- = .005, λ2 = .04 Load-sharing: R(8) = .978, MTTF = 75hr Standby: R(8) =.976, MTTF = 70 hr P{ Layze dig} = P2 (8) = .123

6-6

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

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6.21 (a) R = [1-(1-.9)(1-.95)][1-(1-(.96)(.93))(1-(.92)(.93))](.75) = (.9796)(.75) = .7347 (b) RF(1000) = exp(-1000λ) = .75; λ = .0002877 (1) λ/4 = .0000719; R (1000) = exp[-.0719] = .9306 R2/3 = .93063 + 3(.9306)2 (1-.9306) = .9862; RFAILS = (.9796)(.9862) = .9661 (2) λ = .9(.0002977); λ+ = .0002877; RF (1000) = .9426; RFAILS = (.9796)(.9426) = .9234 (3) RF(1000) = .9217; RFAILS = (.9796)(.9217) = .9029 6.22 Alternative

Distribution

Rsystem(1 day)

MTTFsystem (in days)

Load sharing system with two different units.

Standby system with 3 identical units and no failures in standby Active redundant system with 2 identical units A 2 out of 3 system having identical units

Exponential with λ1 =.25, λ2 = .3 λ1+ = .4 and λ2+ = .5 Exponential with λ = .9

.9116

4.091

.9371

3.333

Weibull with β = 1.5 and θ = 2

.9113

Exponential with λ = .2

.9133

2.475 (See equation 4.18)

4.167

6.23 Using equation 6.31: 1 ⎛ λt ⎞ (a ) λ = λA + λB = .233; R(t ) = e − λt ∑ ⎜ ⎟ = e −.233(2) [1 + .466 /1] = .9199 i =0 ⎝ i ! ⎠ MTTF = 2 / λ = 8.58

(b) RA (2) = e −.087(2) [1 + .174 /1] = .9865; RB (2) = e −.146(2) [1 + .292 /1] = .9865 R (2) = (.9865 )(.9865 ) = .9518

[1 + .087t ][1 + .146t ] = e−(.233)(t ) ⎡⎣1 + .087t + .146t + .0127t 2 ⎤⎦ ∞ 1 .087 .146 2 (.0127 ) − .233 ( t ) + + + = 10.59 MTTF = e ( ) ⎡1 + .087t + .146t + .0127t 2 ⎤ dt = R (t ) = RA (t ) RB (t ) = e (

− .087 +.146 ) ( t )

∫

⎦

⎣

0

∞

since ∫ x n e − ax dx = 0

.233 .2332

.2332

.2333

n! for a > 0, n positive integer a n +1

6.24 For this system, kλt is the expected number of failures during time t of k identical and active units in series. This becomes the mean of a Poisson process and N −k

( k λt )

i =0

i!

Rsys (t ) = e − k λt ∑

6-7

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R (t) is the equivalent to finding the probability of no more than N-k failures (the number of sys

standby or replacement units) during time t. To find the MTTF, ∞

∞

N −k

∫ R (t ) dt = ∫ e ∑ − k λt

sys

0

N −k

=∑

( kλ )

i =0

0

1

=

( k λt ) dt = N −k ( k λ ) ∞ te− kλt dt i!

∑ i! ∫ i =0

N −k

1 1 MTTF 1 = ∑ ∑ k λ i =0 i ! k i =0 i !

i! k λ where MTTF = 1/ λ i =0

2

2

0

N −k

6-8

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CHAPTER 7

7.1

x1 = 45 and x2 = 120 / 160 =.75 λ ( t / x ) = e−9.48 +.01759 x1 + 7.017 x2 = e−9.48 +.01759( 45 )+ 7.017(.75 ) =.03253 × 10−3 MTTF = 1 / λ = 30740.85 −3

R(1000 ) = e − (1000)(.03253×10 ) =.96799 x1 = 30 and x2 = 120 / 160 =.75

λ (t / x ) = e −9.48 + .01759 x1 + 7.017 x2 = e −9.48+ .01759( 30) + 7.017(.75) = e −3.6896 =.02498 × 10 −3 −3

R (1000) = e − (1000)(.02498×10 ) =.97533 x1 = 45 and x2 = 120 / 200 =.60

λ (t / x ) = e −9.48 +.01759 x1 + 7.017 x2 = e −9.48 +.01759( 45) + 7.017(.6) = e −4.4783 =.01135 × 10 −3 −3

R (1000) = e − (1000)(.01135×10 ) =.98871 Increasing the capacitor rating improves the reliability the most because λ is decreased. .98871−.96799 The percentage increase is: =.0214 → 2.14% .96799 7.2

u( x ) = 12.31−.0157( 200)−.35( 7.3) = 6.615

FG ln t − u( x )IJ → R(20) = 1 − ΦFG ln(20) − 6.615IJ = 1 − Φ( −2.53) =.9943 H s K H 143 K .

7.3 L O 1 a) R = z f ( x )dx = e dx = M− e P = − e − ( −1) =.86466 z 500 N Q R( t ) = 1 − Φ

1000 0

1000 0

−

x 500

−

x 500

1000

−2

0

IJ FG IJ K H K F 1 t IJ = ΦFG 1 ln 1000IJ = Φ(2.31) =.98956 c) R = P (ln X ≤ ln 1000) = ΦG ln H s t K H .3 500 K b) R = P ( X < 1000) = Φ

FG H

1000 − μ

1000 − 500 =Φ = Φ(3.03) =.99878 165

θ

med

7.4 a)

R=

z

∞

250

z z FGH IJK

250 − 1 ∞ − 2600 e dx = 0 − ( − e 2600 ) =.90832 2600 250 x

f ( y )dy =

z

β y b) R = P (Y ≥ s) = f ( y )dy = s s θ θ ∞

∞

β −1

−

e

FG y IJ β HθK

FG y IJ F 250 IJ −G dy = e H θ K = e H 2600 K =.85762 β

.8

−

c) R = P(Y ≥ 250) = 1 − P(Y ≤ 250) = 1 − P(ln Y ≤ ln 250) = 1 − Φ

= 1 − Φ( −2.60) =.99534

7-1

FG 1 ln 250 IJ H .9 2600K

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7.5 Using equation 7.15: .95(25) 1 1 R= = =.95 → μ y = = 475 μx 25 .05 1+ 1+

μy

μy

Or:

R( y) = R=

z

∞

0

z

x

x

y

− − 1 y − 25 e dx = − e 25 0y = 1 − e 25 25 0

R ( y ) f ( y )dy =

z

0

∞

F1 − e Iλe dy = λe dy − λe FGH GH JK −

y 25

z

− λy

∞

z

− λy

∞

−

0

0

IJ K dy

1 +λ y 25

LM −e FGH IJK OP λ .05 − λM F = 1− =.95 → λ = → MTTF = 475 I P 1 GH JK .95 × 25 F I MN PQ GH 25 + λJK −

= − e− λy 0

∞

∞

1 +λ y 25

1 +λ 25

0

7.6

L x OP =.85738 R = P ( X ≤ 950) = dx = M 10 N10 Q 7.7 F μ − μ I F 275 − 180 I Using equation 7.5: R = ΦG GH θ + θ JJK = ΦGH 25 + 30 JK = Φ(2.43) =.99245

z

950 3x 2

3

9

9

0

950

0

7.8

y

x

2 y

2 x

z

R = P ( X < Y ) = P ( X < 20) =.25 R (t ) = e − (1− R )αt = e − (.00674)(.5) t

20

0

2

2

a) R (10) = e − (.00674)(.5)(10) =.96686

and

0

= − e −.25( 20) + 1 =.99326

b) R (20) = e − (.00674)(.5)(20) =.93482

7.9 from equation 7.22,

z

20

e − .25 x dx = − e −.25 x

∞

R = P(Y ≥ X ) = P(Y ≥ 120 ) = 120 f ( y )dy = 1 − Φ R( t ) = R + (1 − R )e − at =.96246+.03754e −.25t

FG 120 − 200 IJ = 1 − Φ( −178 . ) =.96246 H 45 K

→

R(5) =.9732

7.10

1 dx = z .02( y − 15) zz z z 2 1 1L .02 O dx = M.5 x − x − 15g P =.4733 = z .5−.02( x − 15) b 3 2 2N Q

R( y ) = 15 x fY ( y ) f X ( x )dydx = 15 x .04( y − 15)dy y 20

17

17 15

20

17

17 15

3

2

15

7-2

2 20 x

1 dx 2

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7.11

OP Q

LM N

1200

x x − 1 1200 − 300 300 R = P( X ≤ Y ) = P( X ≤ 1200 ) = e dx = − e 300 0 0

z

= −e

−

1200 300

− ( −1) =.9817

R( t ) =.981687 =.8786 7.12 a) R = P (Y ≥ X ) = P (Y ≥ 464) = b) R( y) =

z

y

0

f X ( x )dx = 2

z

y

0

z

L y OP dy = M 10 N10 Q

1000 3 y 2

3

9

9

464

1000

= 464

10003 − 464 3 =.9001 109

(.001−.000001x )dx = 2 .001x −.0000005x 2

z R( y ) f ( y )dy =2z (.001y −.0000005 y )FGH 310y IJK dy 6 L .001 y .0000005 y O 6 = 2.5 × 10 − 10 =.9 = − M P 10 N 4 5 Q 10

y 0

= 2 .001y −.0000005 y 2

2

R= 0

1000 0

1000

Y

2

9

5 1000

4

8

9

8

9

0

c) R( td ) = e− (1− R )αtd = e − (1−.8786 )(.01) td =.99

→

td =

− ln(.99 ) = 10.050 yrs (1−.9 )(.01)

7.13

FG x IJ F 100 IJ −G R = P( X ≤ Y ) = P( X ≤ 100) = 1 − e H θ K = 1 − e H 50 K =.98168 β

2

−

R( t ) = e − (1− R )αt = e − (1−.98168 )2 t = e −.03664 t = e − λt → λ =.03664 and MTTF = 1 / λ = 27.293 days 7.14 13,500 − 10,000 R = P ( X ≤ Y ) = P ( X ≤ 13,500) = Φ = Φ(35 . ) =.99977 1000

FG H

R (100) = (.99977)100 =.97726 7.15 R = P( X ≤ Y ) =

z

25

0

f X ( x )dx =

z

IJ K

LM N

OP LM Q N 25

FG H

−100 −100 200 −100 = − dx = 3 2 2 0 ( x + 10) ( x + 10) 0 35 102 25

IJ OP =.91837 KQ

7.16

R( t ) = e − (1− R )αt = e − (1−.992 )(1) t = e −.008 t R( td ) = e−.008 td =.95

→

td =

→

R( 25) = e −.008( 25 ) =.81873

− ln.95 = 6.41 yrs .008

7.17 Using equation 7.17: med y 100 ln ln med x 20 R=Φ =Φ .62 +.82 s y2 + sx2

F F GG GH GG H

IJ I K JJ JJ K

7.18 R = P ( X ≤ 2400) =

F FG IJ I GG H K JJ = Φ(160 . ) =.9452 JJK GGH

z

2400

0

f X ( x )dx =

z

2400

0

LM N

OP Q

2400

x x2 dx = 3125 . × 106 2(3125 . × 106 ) 0

7-3

=.9216

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

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7.19 Flat Fading: R = 1 − Φ

FG 100 − 75IJ = 1 − Φb1g =.15866 H 25 K

for random loads: R( t ) = e − (1− R )αt = e − (.84134 )(.0006944 ) t

→

R( 24) =.98608

→

R( 24) =.96703

FG 1 ln 5 × 10 IJ = FFG 1 ln 5 × 10 IJ = Fb.672g =.74857 H s t K H .76 3 × 10 K 12

Selective Fading: R = F

12 2

med

where s is determined from m = t med es /2 2

for random loads: R( t ) = e − (1− R )αt = e − (.25143)(.005555 ) t

Rtotal =.98608×.96703 =.95357 7.20 a) Fc =

π 2 EI L2

FG H

=

b) μ = θ ⋅ Γ 1 +

π 2 (16 . × 106 )(5.359) 1202

FG 5877 IJ = 5877 lbs → R = 1 − e H 2500 K

1. 76

−

IJ = 2500ΓFG1 + 1 IJ = 2225 → F = 3μ = 6675 H 176 . K βK 1

c

FG 6675 IJ R = 1 − e H 2500 K

1. 76

−

=.9964

7.21

F ln 1260 I R = ΦGG 835 JJ = Φb2.71g =.9969 GH .15 JK R( t ) = R t / Δt =.99693 12 t

(t in yrs) →

L 7.4(125) OP 7.22 v = M N 240(.03) (.02) Q 1.1

2 .11

1.84

R(5) =.99693 60 =.8315

1/ 4 .76

. ft / sec = 3131

7.23

⎛ 1 ⎛ 350 ⎞ ⎞ ln ⎜ R = Φ⎜ ⎟ ⎟ = Φ ( 3.24 ) = .9994 ⎝ .33 ⎝ 120 ⎠ ⎠ R (90) = e − (1−.9994)(.5)(90) = .9734 ⎛

⎞ ⎟ = Φ (1.6368 ) = .949 2 2 ⎝ .3592 + 1.1774 ⎠

7.24 (a) From table 7.2: R = Φ ⎜

ln(300 / 40)

7-4

=.9889

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

Waveland Press, Inc., Copyright © 2009

⎛ ln(300 / 40) ⎞ R = Φ⎜ ⎟ = Φ (1.711) = .956 (b) ⎝ 1.1774 ⎠ R (30) = .95615 = .51 7.25 R =Φ

FG ln 40 / 34 IJ = Φ FG .1625IJ = Φ b.108g =.5438 H .9 + 12. K H 15. K 2

2

α = 6.2/31 = .2 per day = .008333 per hr.

R(t) = exp(- [1-.5438] x .008333 x t) R(56 hr.) = exp(-.0038 x 56) = exp(-.2129) = .8082 MTTF = 1/.0038 = 263.16 hrs = 10.96 or 11 days 7.26 Stress (X)

Strength (Y)

Frequency of Load

a.

Constant = 600 psi

b.

Exponential with mean = 120 volts

Weibull with beta = 2.1 and theta = 2500 psi Exponential with mean = 2500 volts

Random (Poisson) averaging once every 300 days Periodic – every other day

c.

lognormal with median = 600 lbs. s = .38

constant = 1200 lbs.

Periodic – every 40 days

Static Reliability R .951

Dynamic reliability R(100 days) .984

.9542

.096

.966

.933

a. R = exp[-(600/2500)2.1 = .951; R(100) = exp[-(1-.951)/3] = .984 b. R = 2500 / (2500 + 120) = .9542; R(100) = .954250 = .096

⎛ 1 ⎛ 1200 ⎞ ⎞ 2 ln ⎜ ⎟ ⎟ = Φ (1.824 ) = .966; R(100) = .966 = .933 ⎝ .38 ⎝ 600 ⎠ ⎠

c. R = Φ ⎜

7.27 (a) RGMABX = 1 - exp(-25/5) = .99326; R(500) = exp[ - (1-.99326) 500/30] = .8938

⎛ ln ( 30 / 5 ) ⎞ 5 ⎟ = Φ ( 2.1728 ) = .985 ; R(500) = .985 = .9272 2 2 ⎝ .8 + .2 ⎠

RDELP1 = Φ ⎜

RChass2(500) = exp(-500/20,000) = .9753 System reliability = (.8938) (.9277) (.9753) = .8087 (b) Low level redundancy: [1 – (1-.8938)2 ] [1 – (1-.9277)2 ] .9753 = .9592 High level redundancy: {1 - [1 – (.8938) (.9277) ]2 } .9753 = .9468

7.28 (a)

7-5

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

Waveland Press, Inc., Copyright © 2009 600 y − x / 200 600 ⎡ e ⎤ 1 1 R = ∫ ⎢∫ dx ⎥ dy = ∫ (1 − e − y / 200 ) dy 200 300 300 300 ⎣ 0 300 ⎦ 600

y e − y / 200 = + 200 = 2 + .0331914 − 1 − .1487534 = .8844 300 300 300 Or for the more for the discriminating student:

R = 1.0 (1 − e

−300 / 200

− x / 200

∞

− x / 200

e ⎡ 600 1 ⎤ e dy ⎥ dx + 0 ∫ dx ) + ∫ ⎢∫ 300 x 300 ⎦ 200 200 ⎣ 600 600

x ⎤ e − x / 200 ⎡ dx = .7769 + .1075 = .8844 − 2 300 ⎢ ⎣ 300 ⎥⎦ 200

= .7769 + ∫

600

b. R(30) = .884430 = .025; c. R(30) = Exp[-(1-.8844) 30] = .0312; d. MTTF = 1/ (1-.8844) = 8.65 days 7.29 b

b

y2 2y = 1 − 1/ 4 = 3 / 4 a. R = ∫ 2 dy = 2 b b b/2 b/2 5

b. R(110) = (3/4) = .237 c. R(110) = exp[-(1-3/4) (1/20) (110)] = .2528 b ⎡ y 1 ⎤ 2y 2 y3 2 ⎡ y ⎤ 2y = = = = R dx dy dy ⎢ ⎥ ∫0 ∫0 b b2 ∫0 ⎢⎣ b ⎥⎦ b2 d.

3b3 0 3 ⎣ ⎦ b

b

7.30

5, 000 = 4780 e.09/2 ⎛ 1 10, 000 ⎞ (a) R = Φ ⎜ ln ⎟ = .993 4780 ⎠ ⎝ .3 (b) R(20) = .99340 = .757

tmed =

7.31

σ 2 = (4780) 2 e.09 ( e.09 − 1) = 2,354,369.5; σ = 1534.4

⎛ 10, 000 − 5, 000 ⎞ (a) R = Φ ⎜ ⎟ = Φ ( 3.26 ) = .99944 1534.4 ⎝ ⎠ 40 (b) R(20) = .99944 = .9778 Two distributions having the same mean and variance can generate significantly different results.

7.32 Random stress, constant strength, periodic loads Static reliability: R = Pr{X< 1000} = Φ[ (1000 – 400) / 140] = Φ[ 4.285714] = 0.999990886 Dynamic reliability: R(8) = .999990886 (8 x 180) = 0.986961

7-6

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7.33 Random stress, constant strength, random loads Static reliability: R = Pr{X < 750} = 1 – exp[-750/250] = 0.950213 R(t) = Exp[-(1-0.950213)(5)t]; R(1/12) = Exp[-(1-0.950213)(5)/12] = 0.979469 ∞

MTTF = ∫ e −(1− R )α t dt = 0

1 1 = = 4.0 yr (1 − R ) α (1 − .950213) 5

7.34 constant stress, random strength, periodic loads Static reliability: R = Pr{Y>100} = 1 - Φ[ (1/.5) ln( 100/400) ] = 1 - Φ[-1.2] = .8857 R4 = .88574 = .6154 7.35 Static reliability: R = 1 / (1 – uy/ux ) = 1 / (1 – 400/200) = 2/3 R(t) = Exp[- (1 – 2/3)( .5) t] ; R(1) = Exp[- (1 – 2/3) (.5)] = .8465; MTTF = 6 yr 7.36 (a) s = pD/2t; E(s) = (25 x 290)/(2 x 1/16) = 58,000 psi and STD(s) = (25 x 50)/(2 x 1/16) = 10,000 psi R = Pr{s < 86,200} = Φ[(86,200 - 58,000)/10,000] = Φ[2.82] = .9976

⎛

⎞

⎝

39002 + 108 ⎠

(b) R = Φ ⎜ 86, 200 − 58, 000 ⎟ = Φ ( 2.63) = .99573 (c)

(.5)(25)(290) ⎞ ⎛ ⎜ 86, 200 − ⎟ t Since Φ ( 3.09 ) = .999, ⎜ ⎟ = 3.09 2 2 2 ⎜ 3900 + [(.5)(25)(50)] / t ⎟ ⎝ ⎠ (.5)(25)(290) = 3.09 39002 + [(.5)(25)(50)]2 / t 2 86, 200 − t squaring both sides, multiplying through by t 2 , and using the quadratic formula results in t = .0663 7.37 (a) h = 90 x 10-4 x 25 x 34 / (82 x 450) = 2.073 x 10-4 mm per application. Time to failure = 2.5 / (15 x 2.073 x 10-4) = .0804 x 104 = 804 days. (b) For platinum: h = 130 x 10-4 x 25 x 24 / (100 x 450) = 1.733 x 10-4 mm per application. Time to failure = 2.5 / (15 x 1.733 x 10-4) = 962 days.

7.38 Since F = ma = 20 kg x 9.81 m/sec2 = 196.2 Newtons, then s = 196.2 N / [π x (.8 x 10-3 m)2 ] / 4 = 390.33 MPa. And R = 1 – Fy(s) = exp[-390.33/807] = .6165 7.39 Let Y = a RV, the stress on a single member and X = a RV, the load placed on the platform.

X is normal with μx = 18,000 and σx = 3600.

Then Y = [3600/4 + X/4] / 9 with μy = 100 + μx /36 = 600 psi and σy = σx / 36 = 100; therefore

⎛ 2100 − 600 ⎞ R = Φ⎜ ⎟ = Φ ( 2.94 ) = .99836 2 2 + 500 100 ⎝ ⎠ 7.40 (a) This is case 1, k = 600 psi. R = 1 – exp[-(600/190)2.15] = .99992859 Rn = .99 or n log R = log .99; n = log .99 / log R = 1407.37 landings or 469 days

7-7

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

Waveland Press, Inc., Copyright © 2009

(b) R(90|469) = .99992859559 x 3/ .99 = .998; F(499) = 1 - . 998 = .002

7.41 ∞

R = ∫ (e 0

− x/ μ

∞

γ −1 − x /α γ −1 − x (1/α +1/ μ ) ) αx γ Γe(γ ) dx = α1γ ∫ x eΓ (γ ) dx 0 γ

⎛ αμ ⎞ γ ∞ γ −1 − x / k ⎜ α + μ ⎟ ⎛ μ ⎞γ k x e αμ ⎟ =⎜ dx = ⎜ Letting k = , R= γ ∫ γ ⎟ α +μ α 0 k Γ (γ ) ⎜ α ⎟ ⎝ μ +α ⎠ ⎜ ⎟ ⎝ ⎠ For the case where α = 60, γ = 1.3, and R = .99, μ = 140, 1.3 20 (.991/1.3 ) ⎛ μ ⎞ μ 1/1.3 1/1.3 = = = + = = 2577 .99, .99 , 20 99 or μ μ μ ( ) ⎜ ⎟ 1 − .991/1.3 μ + 20 ⎝ μ + 20 ⎠

7.42 Find α first and then compute R. The mode is the most frequent or likely value.

40 = 35 + .5772157α ; α = 8.66; R = F (70) = exp[−e−( 70−35) /8.66 ] = .9826 7.43 The stress distribution is Weibull with parameters β = 2 and θx and the strength distribution is Weibull with parameters β = 2 and θy. Therefore from Eq. 7.13, ∞ ∞ 2 − ⎡( y /θ )2 + ( y /θ y )2 ⎤ 2 − ( y /θ y ) 2 1 ⎦ R = ∫ ⎡1 − e − ( y /θ x ) ⎤ 2 ye dy = 1 − 2 ∫ e ⎣ x ( 2 y ) dy ⎣ ⎦θ θ y

0

y 0

⎛ 1 ⎛ θ x2 + θ y2 ⎞ 1 ⎞ let u = ( y / θ x ) + ( y / θ y ) ; then du = 2 y ⎜ 2 + 2 ⎟ dy = 2 y ⎜ 2 2 ⎟ dy ⎜θ ⎟ ⎜ θ θ ⎟ ⎝ x θy ⎠ ⎝ x y ⎠ 2

θ x2θ y2

2

θ y2 θ x2 = R = 1− 2 2 e du = 1 − 2 θ y (θ x + θ y2 ) ∫0 (θ x + θ y2 ) θ x2 + θ y2 ∞

−u

7-8

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

Waveland Press, Inc., Copyright © 2009

CHAPTER 9

LM N

OP Q

t

t '2 t '3 t3 − 1 dt ' = = 9.1 H ( t ) = 333 3(333) 1 999

z

t 1

z

z

→

LM N

H ( 6) =

63 − 1 =.215 999

OP Q

10

t3 t4 104 − 1 = = 7.507 hrs MTTR = t ⋅ h( t )dt = dt = 1 1 333 4(333) 1 1332 10

10

t3 −1 H ( tmed ) =.5 = med 999 9.2 a)

R( t ) = 1 − Φ

b

→

g = 7.940 hrs

tmed = 999(.5) + 1

FG 1 ln t IJ = 1 − ΦFG 1 ln 5 IJ =.90 Hs t K H .7 t K med

1 5 = −128 ln . .7 tmed

1/ 3

Φ

→

med

→

2

tmed =

5 .7 ( −1.28 )

e

FG 1 ln 5 IJ =.10 H .7 t K med

= 12.249 yrs

2

MTTF = t med e s / 2 = 12.249e.7 / 2 = 15.650 yrs b)

tmed =

MTTF e

s2 / 2

=

2 . hrs = 1213 e1/ 2

H ( 4) = Φ

→

FG 1 ln 4 IJ = ΦFG ln 4 IJ =.883 . K H s t K H 1213 med

.

1213 tmode = s2 = 1 =.446 hrs e e t med

9.3

H (t ) = Φ

FG 1 ln t IJ = ΦFG 1 ln 10 IJ =.90 . t K H s t K H 121 med

→ tmed =

med

10 1.21(1.28 )

e

. yrs = 213

. e1.21 / 2 = 4.43 hrs MTTR = tmed e s / 2 = 213 2

9.4 tmed =

MTTF e

s2 / 2

2

=

10,000 2

e2 / 2

= 1353353 . hrs

With PM: T = 100 hrs , n = 5 , and t = 550 hours , and R ( t ) = R(T )n R( t − nT )

LM N

m

FG H

1 100 R (550) = R(100)5 R(50) = 1 − Φ ln m .

2 1353353

IJ OP5 LM1 − ΦFG 1 ln 50 IJ OP K Q N H 2 1353353 . KQ

= 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

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

Waveland Press, Inc., Copyright © 2009

9.5

FG t IJ a) R( t ) = de − λt ie H θ K

FG 5 IJ − .008 ( 5 ) R(5) = e e H 10 K = .9606 .7504 =.7208

β

−

c

→

h

1. 8

b

−

g

b) Considering reliability at the end of maintenance interval such that 5=nT:

d

Rm (5) = e −

e

FG IJ H K

− .008( 5 )

5 T 1.8 T 10

FG T IJ e i H 10 K

1.8

−n

−

FG IJ H K

5 T T 10

1.8

b

g

→

101.8 ln.9890 =.0853 yr → n = 5 / T = 58.6 ≈ 59 T= − 5

= .9606 e

=.95

.95 =.9890 .9606

=

5 ⋅ T 1.8 = ln.9890 − 1.8 10 ⋅ T 101.8 ln.9890 T =− 5 .8

.8

FG T IJ − .008( 5 ) c) R( t ) = de ie H 10 K b1−.005gn 1.8

−n

b

g

n

The 1−.005 term greatly reduces R(t) for large n so a .95 reliability is not possible.

9.6

z R(t )dt = z (1−.001t )dt = t − (.001t ) / 2 z (1−.001t )dt = 95 = 950 hrs b) MTTF = a) MTTF = 0

1000

1000 0

2

1000 0

= 500 hrs

100 0

1 − R(100 ) .1 c) R( 225) = 1−.001( 225) =.775 M

b

g

RM ( 225) = R(100)2 (.99)2 R( 225 − 200) =.92 (.99)2 1−.001( 25) =.774 d) RM ( 225) = R(50 ) (.99) R( 225 − 200) =.95 (.99) 1−.001( 25) =.763 No improvement.

4

4

4

4

9.7 MTTR = 2, s =.2

t

a) tmed = MTTRe −.5 s = 2e −.5(.2 ) = 196 . hrs and tmod e = med = s2 2

b) H ( t ) =.95 = Φ

FG 1 ln t IJ H .2 196 . K

2

e

→

196 . = 188 . hrs e.04

t = 196 . e.2(1.645 ) = 2.72 hrs

. hr ) = Φ c) P( H < 100 min ) = P( H < 167

. I FG 1 ln 167 J = Φ( −.80) =.212 H .2 196 . K

9.8

λ ( t ) = at b =.0004521t .8 → a =.0004521 and b =.8 β = b + 1 = 18 . and θ = 1.8 β / a = 1.8 18 . /.0004521 = 100

FG 20 IJ H K A: R( 20 ) = e 100 =.9463 1. 8

−

FG 1 IJ H K B: R( 20) = e 100 =.995 1. 8

−20

C: Rsys ( 20) = 1 − (1 − R( 20 ))2 = 1 − (1−.9463)2 =.9971 An active redundant system results in the greatest reliability.

9-2

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

Waveland Press, Inc., Copyright © 2009

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

LM.01t −.1t + t OP z R(t )dt = z (.01t −.2t + 1)dt = N 3 Q = 4.871 yrs b) MTTF = .5

3

2

.5 0

T 0

1 − R(T )

2

0

1 − .01(.5)2 −.2(.5) + 1

.0975

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 =1

i i

i

5

∑q f i =1

=

i i

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

F 50000IJ . (b) For power law intensity # failures= f = G H θ K β

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] = 13396.8 / 3304 = 4.05 MTTR is not very sensitive to the maintenance concept but the number of failures is.

9.11 Replace (renewal process): R (300) = 1 − Φ

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

(.4 x 10−8 )(2.8) ∫

700

400

t1.8 dt =

(.4 x 10−8 )(2.8) 2.8 700 ⎡⎣t ⎤⎦ = .2928757 400 2.8

Therefore Pr{N(700) - N(400) = 0} = e-.2928757 = .7461 Conclusion: Replace the part.

LMt − t OP tI F − 1 dt J G R( t )dt H b K = N 2b Q = b LMT − T OP = b − T 9.12 MTTF = = T 1 − R(T ) 2b Q 2 TN F TI 1 − G1 − J H bK b

z

T

0

z

T

0

2

T

2

0

9-3

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

Waveland Press, Inc., Copyright © 2009

9.13

LM− 2 e + 1 e OP − 1 c4e − e 2 − e e dt c h λ 2λ z R(t )dt = z Q = 2λ =N MTTF = 1 − 2e + e 1 − R(T ) 1 − c2e − e h c1 − e h T

− λt

T 0

=

9.14

−

−2 λt

− λT

c

− λt

hc

− λT

−2 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

−2 λt

h

−2 λT

− λT

−2 λT

h

−3

0

− λT 2

h as T → ∞, MTTF → 3 λ M

2 as T → 0, 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 −r 1 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 ( 2λ + r )2 − 4λ2 λ + r + x1 λ + r + x2 − ( 2λ + r ) ± ; A= ; B= 2 2 x1 − x2 x2 − x1 λ +r+s λ + r + x1 x1t λ + r + x2 x2t A B = + → P1 ( t ) = e e − z1 = x1 − x2 x1 − x2 ( s − x1 )( s − x2 ) s − x1 s − x2 x1 , x2 =

s+λ 1 −λ 0 λ λ A B = = + → P2 ( t ) = z2 = 2 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

c

z

∞

MTTF = R (t )dt = 0

9.15 H ( t ) =

t−a b−a

MTTR =

z

b a

LM N

−1 2λ + r + x1 2λ + r + x2 − x1 − x2 x1 x2

LM N

OP Q

b

OP Q

t t2 b2 − a 2 ( b + a )( b − a ) ( b + a ) = = = dt = b−a 2( b − a ) a 2( b − a ) 2( b − a ) 2

9-4

h

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

Waveland Press, Inc., Copyright © 2009

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 = −(b − T ) + b = T − T =z G H b K N 2b Q 2b 2b 2b 2

T0

2

T0 0

E TOV

2 0

2

0

0

0

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

LM N

t 100 − t ; R( t ) = ; 100 100

T = 5 days

5N ≤ t ≤ 5( N + 1) → Rm (17) =.8402275 ( N = 3)

OP Q

5

−(100 − t ) 952 1002 100 − t − + dt 2(100 ) 0 2(100) 2(100) 100 MTTFM = = = = 97.5 hrs R . .05 ( ) 1− 95 1− 5 Without maintenance: R (17) =.83 and MTTF = 50 hrs

z

5 0

2

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 i 2 3 i =1

∑

FG H

IJ K

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

− 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 =

9.20 K1 = λ 1 + λ 2 + r = 1 + 2 + 2 = 5; K 2 = λ 1 ⋅ λ 2 = 2; r = 2

− K1 ± K12 − 4 K 2 −5 ± 25 − 8 x1 , x2 = = = −.4384, − 4.561 2 2 (5−.4384)e −.4384 t − (5 − 4.561)e −4.561t = 11065 .

R( t ) = e −.4384 t −.1065e−4.561t −.4384 + 4.561 R(1) =.7125

9-5

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

Waveland Press, Inc., Copyright © 2009

9.21 r =.10; λ =.023; eqn 9.18

−(3(.023) + .10) ± .0232 + 6(.023).10 + .102 x1 , x2 = = −.006511, −.162489 2 ( −.006511)e−.162489 t − ( −.162489 )e−.006511t = −.0417e−.162489 t + 10417 .

R( t ) = e −.006511t −.006511+.162489 R( 72) =.65185 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 = = − 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

x1 , x2 =

9.23 NHPP:

(a) m(0,1000) =

z

1000

0

e

−7 .5 + .003t

OP Q

OP Q

1000

e −7.5+ .003t dt = .003 0

= 352 .

100

e −7.5+ .003t (b) m(0,100) = = .0645 ; R(100) = e-.0645 = .9375 .003 0 (c) Since ρ(t) is increasing, the automobile is deteriorating.

9.24 The CDF for the cycle time is G(t) = 1 −

Therefore Pr{T<1} = G(1) = 1 −

e − .1t −.1e − t .

1−.1

e −.1 −.1e −1 =.0355 .9

9.25 For this distribution, MTTF = 3.026 and σ = 3.435.

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

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

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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 = 5.74 2

⎛ 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 ⎞⎟ (d ) mB (5) = .00025(5) 2.4 = .01189; RM (10) = ⎢e ⎝ 22.5 ⎠ ⎢⎣

1.7

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

(

9.29 (a) tmed = 5 / exp .688 / 2 2

1 6.25 ⎞ ln ) = 4; H (6.25) = Φ ⎜⎝⎛ .668 ⎟ = .7479 4 ⎠

1 t⎞ t ⎛ 1 ln ⎟ = .95; z = 1.645 = ln ; t = 12 hrs .668 4 ⎝ .668 4 ⎠

(b) H (t ) = Φ ⎜

12 work hrs or noon the next day 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 9.31 R(t) = exp[-(t/5000)1.45] ; R(30) = .9994 ; Rm(1825) = .999460 R(25) = .9642 9.32

d 2 ρ (t ) = ab(b − 1)(b − 2)t b −3 2 dt 2 d ρ (t ) = b 2 e a +bt log-linear: 2 dt 2 d ρ (t ) = −ab 2 e − bt bounded: 2 dt power law:

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.

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Waveland Press, Inc., Copyright © 2009

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.

t

∫

⎧λ t

for t ≤ c

9.34 Since E[ N (t )] = ρ (t ') dt ' = ⎨

b ⎩λt + a (t − c) for t > c Then E[N(5,000)] = .001 (5000) + .0001 (5000 – 1000)1.4 = 5 + 11.0 = 16.0 0

9-8