International Journal of Performability Engineering, Vol.10, No. 1, January 2014, pp.31-40. © RAMS Consultants Printed in India
Three-Parameter Weibull and Inverse Weibull Models Applied to a Combined Approach of a Sequential and an Accelerated Life Testing DANIEL I. DE SOUZA JR. 1,3, ASSED N. HADDAD2 and DANIELE R. FONSECA1 1 Fluminense Fed. University, Civil Engg. Dept., Grad. Program, Niterói, RJ, BRAZIL 2 Rio de Janeiro Federal University, Civil Engineering Dept., RJ, BRAZIL 3 North Fluminense State University, Industrial Engg Dept., Campos, RJ, BRAZIL (Received on August 28, 2013, revised on September 13, 2013) Abstract: In many situations where the amount of time available for testing is considerably less than the expected lifetime of the component, we often use accelerated life-testing. To translate test results obtained under accelerated conditions to normal using conditions, the “Maxwell Distribution Law “can be used. In this paper, a combined approach of a sequential life testing and an accelerated life testing is applied to a low alloy high-strength steel component. The underlying sampling distributions are assumed to be three-parameter Weibull and Inverse Weibull models and a linear acceleration is employed. To estimate the three parameters of both models, we use a maximum likelihood approach for censored failure data and apply truncation mechanisms developed by De Souza [1] for both models. An example illustrates the application of this procedure. Keywords: Low-Alloy high strength steel, three-parameter Weibull and inverse Weibull models, sequential life testing and accelerated life testing, truncation mechanisms, maximum likelihood estimator. 1.
Introduction
In a previous paper, De Souza and Rocha [2] have applied Eyring's accelerated model to times to breakdown of insulating fluid. The objective was to verify whether or not the accelerated model proposed by Eyring was able to translate results for the shape and scale parameters of an underlying two-parameter Inverse Weibull model, obtained under two accelerating using conditions, to expected normal using condition results for these two parameters. In this paper we will apply a combined approach of a sequential life testing and an accelerated life testing to a low alloy high-strength steel component. 2. Acceleration Mechanism The “Maxwell Distribution Law” is given by: (1) M TE = M tot × e − E KT In this equation, M TE represents the number of molecules at a particular absolute Kelvin temperature T (Kelvin = 273.16 plus the temperature in Centigrade) that passes a kinetic energy greater than E among the total number of molecules present, M tot ; E is the energy
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Daniel I. De Souza Jr. Assed N. Haddad and Daniele R. Fonseca
of activation of the reaction and K represents the gas constant (1.986 calories per mole). Equation (1) expresses the probability of a molecule having energy in excess of E. The accelerating factor AF 2/1 (or the ratio of the number of molecules at two different stress Kelvin temperatures T 2 and T 1 ; M TE (2)/M TE (1)) is given by:
M TE (2) e − E KT2 = , or yet; (2) M TE (1) e − E KT1 E 1 1 − (3) AF 2 / 1 = exp K T1 T2 Applying natural logarithm to both sides of Equation (3) and after some algebraic manipulation, we obtain: E 1 1 − (4) ln (AF 2 / 1 ) = K T1 T2 From Equation (4) we can estimate the term E/K by testing at two different stress temperatures and computing the acceleration factor on the basis of the fitted distributions. Then: E ln (AF2 / 1 ) = (5) K 1 1 − T1 T2
AF 2 / 1 =
The acceleration factor AF 2/1 will be given by the relationship θ 1 /θ 2 , with θ i representing a scale parameter or a percentile at a stress level corresponding to T i . Once the term E/K is determined, the acceleration factor AF 2/n to be applied at the normal stress temperature is obtained from Equation (3) by replacing the stress temperature T 1 with the temperature at normal condition of use T n . Then: E 1 1 − (6) AF 2 / n = exp K Tn T2 De Souza [3] has shown that under a linear acceleration assumption, if a threeparameter Weibull model represents the life distribution at one stress level, a threeparameter Weibull model also represents the life distribution at any other stress level. The shape parameter remains the same while the accelerated scale parameter and the accelerated minimum life are multiplied by the acceleration factor. The equal shape parameter is a necessary mathematical consequence to the other two assumptions, that is; assuming a linear acceleration model and assuming a three-parameter Weibull sampling distribution. If different stress levels yield data with very different shape parameters, then either the Weibull sampling distribution is the wrong model for the data or we do not have a linear acceleration condition. The same reasoning applies to the Inverse Weibull model. We will be assuming a linear acceleration condition. In general, the scale parameter and the minimum life can be estimated by using two different stress levels (temperature or cycles or miles, etc.), and their ratios will provide the desired value for the acceleration factors AF θ and AF ϕ . Then: θ (7) AF θ = n θa AF ϕ =
ϕn ϕa
(8)
Three-Parameter Weibull and Inverse Weibull Models Applied to a Combined Approach of a Sequential and an Accelerated Life Testing
33
3. Maximum Likelihood Estimation a) For the Three-Parameter Inverse Weibull Model The likelihood function for a three-parameter Inverse Weibull sampling model for censored Type II data (failure censored) is given by: r n −r ;t>0 (9) L (β; θ ; ϕ) = k! ∏ f (t i ) [R (t r )] i =1 The density function f(t i ) and the reliability function R(t r ) are given respectively by: β β +1 β θ θ exp − (10) f (t i ) = ti − ϕ θ ti − ϕ β θ R (t r ) = exp − tr − ϕ β +1
(11) r
− ∑ (θ t i − ϕ)β r 1 r r β × e i =1 L (β; θ ; ϕ) = k! β θ ∏ i =1 (t i − ϕ) The log-likelihood function L = ln [L (β ; θ ; ϕ)] is given by:
− (θ t r − ϕ)β e
n −r
(12)
β
β
r r θ θ (13) – (n − r ) L = ln (k!) + r ln (β) + r β ln (θ) - (β + 1) × ∑ ln(t i − ϕ) – ∑ tr − ϕ i =1 i =1 t i − ϕ To find the value of θ and β that maximize the log-likelihood function, we take the derivatives of L with respect to θ, β and ϕ, and make them equal to zero. We obtain: r 1 rβ dL = – βθ β−1 ∑ θ − ϕ t dθ i i =1
β
– (n − r )β θ
β
1 t − ϕ = 0 r
β −1
(14) β
β
r r θ θ θ dL r – (n − r ) × × ln = + r ln (θ) – ∑ ln (t i − ϕ) – ∑ dβ β t −ϕ ti − ϕ tr − ϕ i =1 i =1 i θ = 0 ln tr − ϕ
β +1 β +1 r r 1 1 1 dL β =0 + (n − r ) = (β + 1) ∑ – β θ × ∑ t i − ϕ t ( ) − ϕ t − ϕ dϕ r i i =1 i =1 From Equation (14), we obtain: 1β
(15)
(16)
β r 1 β 1 + (n − r ) (17) θ = r ∑ ti − ϕ tr − ϕ i 1 = It may be noticed that when β = 1, the Equation (17) reduces to the maximum likelihood estimator for the inverse two-parameter exponential distribution. Using (17) for
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Daniel I. De Souza Jr. Assed N. Haddad and Daniele R. Fonseca
θ in Equations (15) and (16) and applying some algebra, Equations (15) and (16) reduce to: r r – ∑ ln (t i − ϕ) + β i =1
r
(β + 1) ∑
β β r 1 1 ln(t i − ϕ) + (n − r ) ln (t r − ϕ) r × ∑ ti − ϕ tr − ϕ i =1
1 – i =1 (t i − ϕ)
β
r
1 1 + (n − r ) − ϕ tr − ϕ i =1 i
∑ t
β
β +1 β +1 r 1 1 β r ∑ + (n − r ) i =1 t i − ϕ tr − ϕ β
1 1 ∑ t − ϕ + (n − r ) t − ϕ r i =1 i r
β
=0
=0
(18)
(19)
Equations (18) and (19) can be solved iteratively. b) For the Three-Parameter Weibull Model Applying the same approach used in the Inverse Weibull model case, we obtain the maximum likelihood equations for the Weibull distribution: 1β
r (20) θ = ∑ (t i − ϕ)β + (n − r )(t r − ϕ)β r i =1 r r × ∑ (t i − ϕ)β ln (t i − ϕ) + (n − r ) (t r − ϕ)β ln (t r − ϕ) r i =1 r = 0 (21) + ∑ ln (t i − ϕ) – r β i =1 ∑ (t i − ϕ)β + (n − r )(t r − ϕ)β i =1
r r 1 + – ∑ (t i − ϕ)β + (n − r )(t r − ϕ)β r (β − 1) ∑ ( t i =1 i − ϕ) i =1
r β × ∑ (t i − ϕ) β−1 + (n − r )(t r − ϕ) β−1 = 0 i =1
(22)
The problem is reduced to solving Equations (21) and (22) iteratively. It may be noticed that when β = 1, (20) reduces to the MLE for the two-parameter exponential model. 4.
The Sequential Life Testing
The hypothesis testing situations were given in [1] and [4]: 1. For the scale parameter θ: H 0 : θ ≥ θ 0 ; H 1 : θ < θ 0 The probability of accepting H 0 will be set at (1-α) if θ = θ 0 . Now, if θ = θ 1 where θ 1 < θ 0 , then the probability of accepting H 0 will be set at a low level γ. 2. For the shape parameter β: H 0 : β ≥ β 0 ; H 1 : β < β 0 The probability of accepting H 0 will be set at (1-α) if β = β 0 . Now, if β = β 1 where β 1 < β 0 , then the probability of accepting H 0 will be also set at a low level γ. 3. For the location parameter ϕ: H 0 : ϕ ≥ ϕ 0 ; H 1 : ϕ < ϕ 0
Three-Parameter Weibull and Inverse Weibull Models Applied to a Combined Approach of a Sequential and an Accelerated Life Testing
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Again, the probability of accepting H 0 will be set at (1-α) if ϕ = ϕ 0 . Now, if ϕ = ϕ 1 where ϕ < ϕ 0 , then the probability of accepting H 0 will be once more set at a low level γ. The sequential probability ratio (SPR) will be given by SPR = L 1,1,1,n / L 0,0,0,n . a) In the Three-parameter Inverse Weibull Case, The SPR will be n n t −ϕ 0 ∏ i i =1 t −ϕ 1 i
β β θ 0 θ 1 n 0 1 exp − ∑ − β β β +1 0 1 i =1 1 ϕ − t − ϕ t i 1 0 i b) In the Three-parameter Weibull Case,
β θ 1 β SPR= 1 × 1 β0 β 0 θ 0
(
)
(
)
β
0
+1
(
(
)
)
We will have:
β β 0 1 n t −φ − φ t 0 i i 1 − − exp ∑ ∏ β0 β1 β −1 i =1 i =1 0 θ θ t −φ 1 0 0 i The continue region becomes A < SPR < B, where A = γ /(1-α) and also B = (1-γ)/α. We will accept the null hypothesis H 0 if SPR ≥ B and we will reject H 0 if SPR ≤ A. Now, if A
