Florida International University - EIN 3235solution-manual-an-introducing-to-reliability-and-engineering

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CHAPTER 2 2.1 a) R R ( ) (. )( ) . ( ) (. )( ) 100 1 . 001 100 1 91 1000 1 001 1000 1 = 5 + = = + and = b) λ( ) ( ) ( ) ((. ) ) (. ) t dR t . (. ) (. dt R t d t dt t = t t − ... ⋅ = − + ⋅ + = + ⋅ − − − 1 001 1 1 001 1 001 001 1 001 1 1 1 2 + ) = + + = + . (. ) (. ) . . 001 001 1 001 1 001 2 001 1 t t t λ( ) t is decreasing because λ( ) t goes to zero as t goes to infinity. 2.2 a) R t e ( ) = −z0 0 tλ( ') ' ' ' t dt = e−.4ztt dt = e−. ' t t | = e−.2t 0 2 2 2 F R e ( / ) ( / ) 1 12 1 1 12 1 = − = − = −.2( / ) 1 12 2 .00139 b) R t e ( ) . t . ln(. ) t t ln(. ) yrs . . = = → − = → = . − = − 2 2 2 95 2 95 95 2 506 2.3 a) R t f t dt dt t t t t t t ( ) ( ) . . . ( ) . = 100 z z ′ ′ = 100 01 01 01 100 1 01 0 100 ′ = 100 ′ t = − = − ≤ ≤ b) λ( ) ( ) ( ) ( ) ( ) . . t dR t dt R t f t R t t = t − ⋅ = = − ≤ ≤ 1 01 1 01 0 100 c) MTTF R t dt t dt t t = = − = − = − = z z 0100( ) ( . ) . 0100 1 01 005 100 005 100 50 days 100 2 0 100 0 . ( )2 d)V 2 2 = − z0100t f t dt MTTF t dt t ( ) ( ) . 2 = 01 50 033 50 8333 z0100 2 − = − = 2 3 . 100 0 2 2 . ) (days V V = = 2 28 9 . days e) R t t ( ) . . . . median = − = → 1 01 5 01 5 median days tmedian = → tmedian = 50 2.4 a) R t f t dt t dt t t t t ( ) ( ) =z ′ ′ = z ′ ′ =L ' M N OPQ 1000 9 1000 2 9 3 100 3 10 1 10 = − = − ≤ ≤ 1 10 1000 1 10 0 1000 9 3 3 3 t 9 t c h hrs t F R ( ) ( ) ( / ) 100 1 100 1 1 100 10 10 10 1 10 3 9 6 = − = − − = = 9 3 b) MTTF t f t dt t dt t = ⋅ = = ⋅ = ⋅ z z 01000 ( ) 1039 01000 3 10 4 93 4 1000 0 10 4 93 1000 0 750 4 − = hrs c) R t ( ) . = − = → = → = = 1 t t . t . 10 99 10 01 10 215 443 3 9 3 9 3 7 hrs 2.5 2-1Ebeling, An Introduction to Reliability and Maintainability Engineering, 2nd ed. Waveland Press, Inc., Copyright © 2009 a) R( ) . 50 = e- .001(50) = 8 λ( ) ( ) (. ) (. ) . . (. ) (. ) / (. ) (. ) t d e dt e t e e t t t t t = − ⋅ == ⋅ = − − − − − 001 001 1 2 001 001 1/2 1/ 2 1/ 2 1/2 1 1 2 001 001 1 0005 001 b) λ( ) t is decreasing because λ( ) t goes to zero as goes to infinity. c) t R t T R T t ( ) + R R ( ) 50 10 60 + ( R T R R R e e ( / ) ( ) ( / ) ( ) ) ( ) . . ( ) 0 . (10) 0 0 001 60 001 50 10 10 10 = → = = = = 865 − − d) R t R t R e e t ( / ) ( ) ( ) . . ( ) . ( ) 10 10 10 95 001 10 001 10 = + = = − + − .001( 10) .001(10) >ln.859596@2 .95 .859596 10 12.9 hrs .001 t e e − + − = = → = − = t 2.6 R t f t dt t dt t t t t t t a) ( ) ( ) (. . ) . . =z10 ′ ′ = − z102 02 2 01 ′ ′ = ′ 10− ′2 10 = − − − = − + ≤ ≤ ( . ) ( . ) . . 2 2 1 01 1 2 01 0 10 t t t t t 2 2 yrs λ( ) ( ) ( ) . . . . . ( . ) ( . ) . . t f t R t t t t t t t = = − − + = − − = − 2 02 1 2 01 2 1 1 1 1 2 2 2 1 1 λ( ) . ( ) 0 2 10 = → and λ t = f so the hazard rate is, in fact, increasing. b) MTTF R t t = = − z z 010 ( ) ( . 010 1 2 + = − + . ) . . 01t dt t t 2 10 0 1 00333 2 10 0 t 3 100 = − + = 10 10 3 33 3 33 yrs . . R t t t t t c) ( ) . . . . . . median = − + = → − + = 1 2 01 5 01 2 5 0 med med 2 2 med med t ( ) median . ( . ) (. )(. ) (. ) . . (. ) = . , . r − − = r = 2 2 4 01 5 2 01 2 1414 2 01 17 07 2 93 2 t ( ) median is 2.93 years. (17.07 is outside range of values for t.) d) f(t) is linearly decreasing: f(0) = .2 and f(10)=0. Therefore, the mode occurs at t=0 yr. e) f t t ( mod 0 10 ≤ ≤ e ) [ ( )] = MAX f t V2 2 2 0 10 10 2 2 2 3 4 02 10 2 0 0 2 02 333 3 4 = − = − − = − 333 LMN Q P − O z t f t dt MTTF t t dt ( ) z (. . ) . . . t t . = − − − = → = = = ( . ) . . 66 67 50 0 1109 558 V V 2 558 2 36 . . yrs 2.7 R( ) ( ) 1 100 . 1 10 100 121 826 = 2 + a) = = R t f t dt t dt t t t t t t ( ) ( ) ( ) ( ) ( ) ( ) = ′ ′ = ′ + ′ = ′ + − LMN OPQ = − − + FGH IJK = + f − − f f z z 200 10 3 200 10 2 2 0 100 10 2 2 100 10 MTTF R t dt t dt t t = = + = + − LMN OPQ = − + LMN OPQ = − − = f − f − f f z z 0 ( ) ( ) 0 2 ( ) 1 ( ) 0 0 100 10 100 10 1 100 10 b) 0 10 10 yrs [Show More]

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