Microsoft PowerPoint - Chap3 - Handout.ppt [Compatibility Mode]

37
C. Ebeling, Intro to Reliability & Maintainability Engineering, 2 nd ed. Waveland Press, Inc. Copyright © 2010 Chapter 3 The Constant Failure Rate Model Exponential Probability Distribution Chapter 3 1 Hurry, come here! Chapter 3 is starting now. You dont want to miss any of this.

description

See

Transcript of Microsoft PowerPoint - Chap3 - Handout.ppt [Compatibility Mode]

  • C. Ebeling, Intro to Reliability & Maintainability Engineering,

    2nd ed. Waveland Press, Inc. Copyright 2010

    Chapter 3

    The Constant Failure Rate Model

    Exponential Probability

    Distribution

    Chapter 3 1

    Hurry, come here!

    Chapter 3 is starting now.

    You dont want to miss any of this.

  • The Exponential Distribution

    Chapter 3 2

    0t , = (t)

    0t ,e =e = R(t)t-dt-

    t0 '

    Then

    Let

  • -The Reliability Function

    Chapter 3 3

  • -The CDF and PDF

    Chapter 3 4

    e-1 = F(t)t-

    e = dt

    F(t) d=f(t) t-

  • -Probability Density Function (PDF)

    Chapter 3 5

  • -The MTTF

    Chapter 3 6

    .368 = e = e = R(MTTF)1-

    MTTF

    -MTTFnote that

    or

    0|- t

    - t

    0

    1eMTTF = dt = = e

    -

    0

    1tMTTF te dt

    = =

  • -Variance & Standard Deviation

    Chapter 2 7

    definitional form:

    computational form:

    22

    0 = (t - MTTF f(t)dt)

    22 2

    0 = f(t)dt - (MTTF )t

  • -The Standard Deviation

    Chapter 3 8

    2

    = 0

    2

    t -1

    -te dt =

    12

    and MTTF

    = =1

  • -The Median Time to Failure

    Chapter 3 9

    .69315 = .5

    1- = tmed ln

    = .69315 MTTF

    R t e t( ) .= = 5

  • -The Design Life

    Chapter 3 10

    For a given reliability R, let

    lnthenR

    1= - Rt

    R- tRR( ) = = Rt e

  • -Example #1 - CFR model

    An electrical transmission line has been tested and found to have a CFR of .001 defects per foot. Find:

    a. R(100)

    b. R(1000)

    c. MTTF

    d. Standard Deviation

    e. tmedf. 90 percent design life

    Chapter 3 11

  • -Example #1 - solution

    Chapter 3 12

  • -Memoryless Property

    Chapter 3 13

    e

    e =

    )TR(

    )T+R(t = )T|R(t

    T-

    )T+(t-

    0

    00

    0

    0

    R(t) = e = e

    e e = t-

    T-

    T-t-

    0

    0

  • -Failure Modes

    R(t) =

    n

    =

    R (t)

    i

    i

    1

    Chapter 3 14

    Ri(t) is the reliability function for the ith failure mode, then, assuming independence among the failure modes,the system reliability, R(t) is found from

  • -More on Failure Modes

    iR (t) = - 0

    t i ( t ) d te

    Chapter 3 15

    Let

    R(t) =

    n

    i=1

    - 0t i ( t )dt 'e

    Then

    i

    i

    n

    =1

    (t) = (t) where

    01

    exp ( ') 'nt

    i

    i

    = t dt=

    t0

    - (t )d t= e

  • -Failure Modes and CFR

    Chapter 3 16

    in

    =1i

    ==(t)

    If a system consists of n independent,

    serially related components each with

    CFR, then

    and

    t0

    - d t - tR(t) = = e e

  • -Example - Failure Modes

    An engine tune-up kit consists of 3 parts each having CFRs (in failures per mile) of .000034, .000017, and .0000086.

    Find the MTTF, median time to failure, standard deviation, and reliability of the tune-up kit at 10,000 miles.

    Chapter 3 17

  • -Example - solution

    Chapter 3 18

  • -Parts Count Approach

    An integrated circuit board consists of the following components each having a CFR.

    Component a-Failure Rate(10-5) b- Quantity (a) x (b)

    Diodes, silicon .00041 10 .0041

    Resistors .014 25 .3500

    Capacitors .0015 12 .0180

    Transformer .0020 2 .0040

    Relays .0065 6 .0390

    Inductive devices .0004 12 .0048

    total .4199 x 10-5

    Chapter 3 19

    Therefore Rsys(t) = e- .000004199 t and MTTF = 1/.4199 x 105

  • -All Failure Modes are CFR

    MTTF = 1 =

    1

    1

    MTTF

    where MTTF = 1

    i=1

    n

    i

    i=1

    n

    i

    i

    i

    = 1

    Chapter 3 20

    = n MTTF = 1

    n 1

    1

    and

    If all components have identical failure rates, then:

  • -3.5 Poisson Process

    If a component having a constant failure rate is immediately repaired or replaced upon failing, the number

    of failures observed over a time period t has a Poisson

    distribution. The probability of observing n failures in time t

    is given by the Poisson probability mass function pn(t):

    Chapter 3 21

    ( ) ( )

    2

    0, 1, 2, !

    [ ]

    nt

    n

    e tp t n

    nE n t

    = =

    = =

    K

  • The Poisson and Exponential

    Chapter 3 22

    ( ) ( ) ( )0

    00!

    t

    te tp t e R t

    = = =

  • -The Gamma Distribution

    LetTi = time between failure i 1 and failure i having an

    exponential distribution with parameter .

    Yk = the time of the kth failure.

    The sum of k independent exponential random variables has a

    gamma distribution with parameters and k.

    If k is an integer (i.e. the Erlang distribution),

    Chapter 3 23

    1

    k

    k i

    i

    Y T=

    =

    ( )( ) ( )

    1

    ( ) for , , 0

    1 !, positive int.

    k k t

    Y

    t ef t k t

    k

    n n n

    =

    =

  • -The Gamma Distribution

    The CDF is the probability that the kth failure will occur by

    time t.

    The mean value for Yk is k/, and the variance is k/2.

    The mode is (k 1)/(why?).

    Chapter 3 24

    { } ( ) ( ) ( ) ( )

    ( ) ( ) ( )

    0 1 1

    0 11

    0

    Pr 1 ...

    1 ... 10! 1! !

    it can be obtained by integration on Gamma function too.

    kk Y k

    k ik

    t t t

    i

    Y t F t p t p t p t

    t t te e e

    k i

    =

    = = =

    =

  • ?

    Related Failure Distributions

    Chapter 3 25

    Random Variable Distribution Parameter(s)

    T, time to failure Exponential Yk, time of the k

    th failure Gamma (Erlang) , kN, number of failures in time t Poisson

  • EXAMPLE 3.9

    A specially designed welding machine has a nonrepairable motor

    with a constant failure rate of 0.05 failures per year. The company

    has purchased two spare motors. If the design life of the welding

    machine is 10 yr, what is the probability that the two spares will be

    adequate?

    Chapter 3 26

  • EXAMPLE 3.9

    Let Y3 be the time of the third failure. Y3 has a gamma

    distribution with k = 3 and = 0.05.

    The expected, or mean, time to obtain 3 failures is 3/0.05 =

    60 yr.

    The probability that the third failure will occur within 10 yr is

    Chapter 3 27

  • -Redundancy2 identical components with CFR

    ( )2

    1 Pr one fails

    2- t

    R(t) =

    1-(1- )e

    =

    Chapter 3 28

    = 1-(1-2e +e )- t -2 t

    = 2e - e- t -2 t

  • Redundancy - Hazard Rate Function

    [ ]1 - t -2 t- t -2 t

    R(t)f(t) 2 - 2e e(t) = =

    R(t) R(t) 2 -e e

    =

    Chapter 3 29

    = (1 - e )

    (1-.5 e )

    - t

    - t

    Not a CFR process!

  • Chapter 3 30

  • Redundancy, CFR & MTTF

    Chapter 3 31

    = 2-1

    2 =

    3

    2 =

    1.5

    - t -2 t

    0 0MTTF = R(t)dt = (2 - )dte e

  • -Two-parameter Exponential

    Chapter 3 32

  • More 2-parameter Exponential

    Chapter 3 33

    R t e

    t t t

    t tR

    med

    t t

    med

    R

    med o( ) .

    ln . .

    ln

    ( )= =

    = +

    = +

    = +

    5

    05 0 693150 0

    0

    Note that the variance does not change and the mode occurs at t0.

  • Example CFR Model

    A certain type of surge protector is observed to fail at the constant rate of .0005 failures per day. Find:

    a. The MTTF, median, and the standard deviation.

    b. Find the reliability during the first year; the second year; the second year given it has survived the first year.

    c. The 90% design life.

    Chapter 3 34

  • Example - solution

    Chapter 3 35

  • Bonus Round - Return Period

    An asteroid 2/3 of a mile across smashing into the Earth at a high speed which would cause a blast with the power of thousands of hydrogen bombs could happen within the next 300,000 years.

    - Dave Morrison (NASA)

    Chapter 3 36

  • The End

    Chapter 3 37

    Our next assignment

    will be in Chapter 4.

    First, however, you will

    want work the

    problems identified for

    chapter 3.