Chapter 15 Identifying Failure & Repair Distributions

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1 Chape 5 Idefyg Falue & Repa Dsbuos Paamee Esmao maxmum lkelhood esmao C. Ebelg, Io o Relably & Maaably Chape 5 Egeeg, d ed. Wavelad Pess, Ic. Copygh 00

2 Maxmum Lkelhood Esmao (MLE) Fd esmaes fo he dsbuo paamees whch wll maxmze he pobably of obag he obseved sample mes. Max f( ) f( )... f( ) Chape 5

3 Why MLE s????. MLE s ae vaa:. MLE s ae Cosse: as, y = h( θ) he y = h( θ) θ θ 3. MLE s ae (bes) asympocally omal: σ σ ~ θ θ 4. Requed fo cea ess such as he Ch-Squae GOF es. 5. Has a uve appeal. 6. Ca accommodae cesoed daa Chape 5 3

4 MLE - Geomec Dsbuo Le X = a dscee adom vaable, he umbe of als ecessay o oba he fs falue. Assume he pobably of a falue emas a cosa p ad each al s depede, he: Pob{X=x} = f(x) = (-p) x- p, x =,,..., f ( x,x,,x ) = f( x )f( x ) f(x ) x,,x = (-p) x- p(- p ) x- p (- p ) x- p = p (- p ) x = ( ) Chape 5 4

5 Geomec Dsbuo Max 0 p ( ) = g(p) = p (- p) L = + l g( p) l p ( x ) l( p ) = 0 p + ( x ) = p NM = ( ) = 0 p = O QP x + (x -) = = = x Chape 5 5

6 Example 5.3 The followg daa was colleced o he umbe of poduco us whch esuled a falue whch sopped he poduco le: 5, 8,, 0, 7,,, 5. Theefoe, X = he umbe of poduco us ecessay o oba a falue. p = 8 = x 40 =. = (x-) Pob[X = x] = f(x) =.8 (.) Mea = /p = 40/8 = 5 P{X = 3} =.8 (0.) = 0.8 Chape 5 6

7 Lkelhood Fuco L( θ,..., θ ) = Π k f( θ,..., θk) = maxmze he log of he lkelhood fuco: l L( θ,..., θk) = 0 ; =,,...k θ fo Type I (gh cesoed) daa: L( θ,..., θ ) = Π k f( θ,..., θk) R( * ) = fo Type II, Chape 5 7

8 Expoeal MLE - Type II daa f( ) = λ -λ j e, j =,,... j P[T j> fo all j > ] = e - j L(,..., ) = e e R S T c -λ Π λ λ c - λ j= = λ exp λ - λ(-) Σ j= j h h - - U V W Chape 5 8

9 Expoeal MLE - Type II daa L= λ exp λ - λ(-) l d R S T Σ j= j λ λ j λ j= - (-) L = l - L = l - Σ j dλ λ j= U V W - (-) = 0 λ = Σ j= j + (-) = T Type I, use * Chape 5 9

10 Toal Tme o Tes - CFR = b o es = b falues k = b mulply cesos = falue me + = ceso me * = es me (Type I) = es m (Type II) MTTF = T / Complee: ; = Type I: Type II: = = = Type I mulply: Type II mulply: ( ) + ( ) = = + + * + Type I eplaceme: Type II eplaceme: * ( ) + k ( ) + k * Chape 5 0

11 Webull MLE - Type II Daa L F L( θ, ) = f ( ) R( ) = θ H G I θ K J NM = = e θ F H G I K J O L QP N M e θ F H G I K J l L l l θ ( ) l ( ) θ = = = + F H G I K J F H G I K J l L ( ) = + + θ θ θ = = 0 O QP θ l L ( ) l + = = + ( ) = = + l lθ + lθ = 0 Chape 5

12 Webull MLE - sgly cesoed g( ) = R S T = θ L NM = = l + ( ) = + ( ) +(-) s s O QP U V W s whee = s = R S T l = 0 fo complee daa * fo ype I daa fo ype II daa Chape 5

13 Newo-Raphso Mehod j+ = j g( ) j g'( ) j whee g(x) = d g(x) dx Chape 5 3

14 Nomal & Logomal MLEs complee daa σ NORMAL μ = x = ( -) s LOGNORMAL μ = = l = e MED μ ecall: s = ( - MTTF ) = - s = = ( l - μ ) Chape 5 4

15 Example 5.6 Ex. 5.8: 47., 84.8, 5.9,.5, 8., 99.6, 59.8, 38.8, 3.5, 53.4,0.4, 00.8, 30., 04.6, 6.5,., 86., 498.4, 77.0, 78.7,.3, 44.0, 5.3, 5.3,.8 L-S esmaes: med = 6 ad s =.63 μ = (l 44 + l l 498.4)/5 = 8.8/5 = 4.75 med = e = 5.93 s = [(l ) + (l ) + s = (l ) ]/5 =.3798 Chape 5 5

16 MLE wh Mulply Cesoed Daa pob of falue occug a me pob of falue occug afe me + L( θ) = f( ; θ) R( ; θ) ε F εc + F = se of dces fo falue mes C = se of dces fo cesoed mes (cludg sgly cesoed mes) Chape 5 6

17 MLE Expoeal - mulply cesoed daa = - -λ ε F ε C ε F ε C -λ -λ L( ) = e e + λ λ λ λ e e l L( λ) = l λ- λ F ε dl L( λ ) = - dλ λ F ε C ε λ = F ε - λ C ε - + = 0 + C ε + + T Chape 5 7

18 MLE Webull mulply cesoed daa L F θ L( θ, ) = f ( ) R( ) = e F C F θ H G I θ K J NM F H G I K J QP C L F l L = l l + ( ) l H G I K J NM θ F θ = θ L F + H G I K J NM L θ θ θ P + F C θ θ O P Q L NM F H G I K J O QP = 0 O L NM O QP e θ F H G I K J C O QP F H G I K J θ L l l l θ θ θ θ θ F = + F H G I K J F H G I K J F H G I K J F H G I K J F H G I K J = C 0 Chape 5 8

19 MLE Webull - mulply cesoed daa l l = F ε all all ( - ) - solve umecally θ = L M N all O P Q mooocally ceasg RHS Chape 5 9

20 Example 5.7 Ffee us wee placed o es fo 500 hous. The followg falue mes ad ceso mes wee obseved po o cocludg he es: Fo he expoeal, T = (500) = 4498 ad he MLE fo he MTTF = T/ = 4498/8 = Fo he Webull, he 4 us whch had o faled by he ed of he es ae assged cesoed mes of 500 hous. The lef had sde of MLE Eq. equals Begg wh =. ad ceasg he gh had sde by.0 ul exceeds 5., esuls =.43. The θ = 49. Chape 5 0

21 Example 5.7 Gosh! Whch s he coec model? expoeal: R () = Webull: e / F H G I K J R ()= e R(00) =.837 R(00) =.90 Chape 5

22 Nomal Dsbuo - Cesoed Daa + L( μσ, ) = f( ) R( ) εf εc l L( μσ, ) = l e = πσ Maxmze usg a Numecal seach algohm (-μ ) - σ ( μ) σ + l e d' = πσ Chape 5

23 Mmum Exeme Value Dsbuo Complee o gh cesoed daa Lkelhood fuco: ( ) ( ) ( + s μ ) μ μ e α e α α L( αμ, ) = e e e = α = whee ( αμ) ( μ ) ( μ ) = = ( ) ( + ) s μ α α l L, = lα + e e α s fo complee daa = fo Type I daa * fo Type II daa Chape 5 3

24 Mmum Exeme Value Dsbuo l L( α, μ) l L( α, μ) = = α μ 0 / α + s / α e + ( ) se = α + = / + = / α s α e + ( ) e = + 0 / α + s / α μ = αl e + e = Chape 5 4

25 Mmum Exeme Value Dsbuo Complee daa Mehod of Momes: = / ; = / = = m m απ m = μ γα; m = + μ γα 6 solvg: ( ) ( 6 m ) m α = ; μ = m + γα; γ π Chape 5 5

26 Gamma Dsbuo Complee daa, lkelhood fuco: L (, =,..., γα, ) = = γ ( γ ) γ α ( γ) = α = / α γ α Γ( γ) l L= l l l Γ l L( αγ, ) = 0 ad solvg fo α : α = α γ = e Subsug fo α: Maxmze decly l L l l l = γ = ( γ ) = ( γ ) γ γ Γ( γ) Chape 5 6

27 Gamma Dsbuo Complee daa Mehod of Momes: m = ; m = + ( ) αγ γα γα m ; m m γ = α = m m m Chape 5 7

28 EXAMPLE 5.0 Twey us beleved o have a gamma dsbuo wee placed o a acceleaed lfe es wh falues days occug a he mes show: Usg Excel Solve, l l 69 Max L γ = γ γ γ 0l Γ γ γ 0γ ( ) ( ) ( ) γ =.893 α = The coespodg mehod of momes esmaos ae γ = ad α = Chape 5 8

29 Paamee Esmao fo Ieval Daa j, s he umbe of falues ha occu wh he eval (a j-, a j ) whee j =,,,k. Ay gh cesoed us ae coued he eval (a k, ). The lkelhood fuco ca be saed as k + j j j= L( θ) = R( a θ) R( a θ) j Chape 5 9

30 EXAMPLE 5. (Webull) Mohly falues of ffy us opeao wee ecoded ove a sx moh peod wh he followg esuls: Moh Uppe boud days Numbe of Falues L( θ, ) = e e j= aj aj k + θ θ j l L( θ, ) = l e e aj aj k + θ θ j j= Chape 5 30

31 EXAMPLE 5. (Webull) Usg Excel Solve, maxmze θ θ θ θ θ l L( θ, ) = l L( θ, ) = 0l e + l e e + 7 l e e θ θ θ θ θ θ θ + 4l e e + 3l e e + l e e + 3l e =.9486 ad θ = 5.6 Chape 5 3

32 Nex Tme Chape 6 Goodess-of-F Tesg The Ch-Squae es Tesg fo Ch-squaes C. Ebelg, Io o Relably & Maaably Chape 5 Egeeg, d ed. Wavelad Pess, Ic. Copygh 3 00

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