CN102663516B

CN102663516B - Model construction and evaluation method for service life and reliability of product under outfield circumstance

Info

Publication number: CN102663516B
Application number: CN201210085242.6A
Authority: CN
Inventors: 王立志; 姜同敏; 李晓阳; 王晓红
Original assignee: Beihang University
Current assignee: Beihang University
Priority date: 2012-03-28
Filing date: 2012-03-28
Publication date: 2015-02-11
Anticipated expiration: 2032-03-28
Also published as: CN102663516A

Abstract

The invention discloses a method for constructing and evaluating a product field life and reliability model. The specific steps are: step 1, establishing and determining a degradation model of the product; step 2, constructing a correction factor; step 3, establishing a Bayesian model; step Four, obtain the parameter value in the posterior distribution; Step five, evaluate life-span and reliability; The present invention has set up the relation between the laboratory information and the field information, clarified and quantified the difference between each stress environment, for the future The work of evaluation and result correction provides a strong basis. The present invention proposes a method of comprehensively utilizing two kinds of information (laboratory information and field information) to evaluate the life and reliability of the product in the field, thereby solving the problem of using accelerated degradation There are certain deviations in the direct evaluation of test information, and it is difficult to carry out evaluation work due to the scarcity of field information.

Description

Model construction and evaluation method of product field life and reliability

技术领域 technical field

本发明是一种综合运用加速退化试验数据及产品外场数据，对产品在外场情况下的寿命及可靠性开展评估的方法，属于寿命及可靠性评估技术领域。The invention is a method for evaluating the service life and reliability of a product in an external field by comprehensively using accelerated degradation test data and product field data, and belongs to the technical field of life and reliability evaluation.

背景技术 Background technique

产品的寿命及可靠性信息可以通过其失效数据及退化数据来体现，它们可以用来评估产品的寿命分布模型。而在产品外场的实际使用中，很难在短时间内获取足够的信息进行评估，此时加速试验技术应运而生。对于加速试验中传统的加速寿命试验而言，在评估长寿命高可靠性产品时，在有限的时间及经费条件下，很难获得足够的失效数据。此时应采用加速退化试验的方法，它能够在短时间内获得产品的性能退化数据，并外推得到产品在规定条件下的寿命及可靠性。Product life and reliability information can be reflected by its failure data and degradation data, which can be used to evaluate the product life distribution model. However, in the actual use of the product in the field, it is difficult to obtain enough information for evaluation in a short period of time. At this time, the accelerated test technology emerges as the times require. For the traditional accelerated life test in the accelerated test, it is difficult to obtain sufficient failure data under the condition of limited time and funds when evaluating long-life and high-reliability products. At this time, the accelerated degradation test method should be used, which can obtain the performance degradation data of the product in a short time, and extrapolate the life and reliability of the product under specified conditions.

然而，通过加速退化试验外推得到的结果仅代表产品在实验室条件下的寿命及可靠度，这种实验室条件同外场的使用条件并不完全相同。一方面，加速退化试验很难体现产品实际使用中的所有应力；另一方面，环境的噪声及测试水平也会随着各种情况的变化而变化。因此，通过加速退化试验获得的寿命及可靠性评估结果与外场使用中实际的寿命及可靠性有所不同，外场数据亦成为评估过程中最可靠的信息来源。However, the extrapolated results obtained through the accelerated degradation test only represent the life and reliability of the product under laboratory conditions, which are not exactly the same as those used in the field. On the one hand, it is difficult for the accelerated degradation test to reflect all the stresses in the actual use of the product; on the other hand, the noise of the environment and the test level will also change with the changes of various situations. Therefore, the life and reliability evaluation results obtained through accelerated degradation tests are different from the actual life and reliability in field use, and field data has become the most reliable source of information in the evaluation process.

由于外场数据非常的稀有，而实验室数据相对丰富，因此若能够建立外场及实验室信息之间的关系，并综合两种信息对产品在外场情况下的寿命及可靠性进行评估，将可以解决加速退化试验信息不能完全代表外场信息、外场信息稀少等问题，从而获得更准确的评估结果。Since field data is very rare and laboratory data is relatively abundant, if the relationship between field and laboratory information can be established, and the two kinds of information can be integrated to evaluate the life and reliability of the product in the field, it will be able to solve the problem. The accelerated degradation test information cannot fully represent the external field information, and the external field information is scarce, so as to obtain more accurate evaluation results.

发明内容 Contents of the invention

本发明的目的是为了解决上述问题，提出了一种能够综合利用实验室及外场信息对产品外场寿命及可靠性进行评估的模型构建及评估方法。从而解决了利用加速退化试验信息直接进行评估存在一定偏差、外场信息稀少难以开展评估工作等问题。The purpose of the present invention is to solve the above problems, and propose a model construction and evaluation method that can comprehensively utilize laboratory and field information to evaluate product field life and reliability. Therefore, it solves the problems that there is a certain deviation in the direct evaluation of the accelerated degradation test information, and it is difficult to carry out the evaluation work due to the scarcity of field information.

本发明的产品外场寿命及可靠性模型构建及评估方法，具体步骤为：The product field life and reliability model construction and evaluation method of the present invention, the specific steps are:

步骤一、建立并确定产品的退化模型；Step 1. Establish and determine the degradation model of the product;

步骤二、构建修正因子；Step 2, constructing the correction factor;

步骤三、建立贝叶斯模型；Step 3, establishing a Bayesian model;

步骤四、获取后验分布中的参数值；Step 4, obtaining parameter values in the posterior distribution;

步骤五、评估寿命及可靠性；Step 5. Evaluate life and reliability;

本发明的优点在于：The advantages of the present invention are:

(1)本发明建立了实验室信息同外场信息之间的关系，明确并量化了各应力环境之间的差异，为日后的评估、结果修正工作提供了有力的依据；(1) The present invention establishes the relationship between the laboratory information and the field information, clarifies and quantifies the differences between the stress environments, and provides a strong basis for future evaluation and result correction work;

(2)提出了综合利用两种信息(实验室信息和外场信息)对产品在外场情况下的寿命及可靠性进行评估的方法，从而解决了利用加速退化试验信息直接进行评估存在一定偏差、外场信息稀少难以开展评估工作等问题；(2) A method of comprehensively using two kinds of information (laboratory information and field information) to evaluate the life and reliability of products in the field is proposed, thus solving the problem of direct evaluation using accelerated degradation test information. Issues such as scarcity of information making it difficult to carry out assessments;

(3)本发明中，外场情况下的失效数据及退化数据都可同加速退化试验的数据进行综合，得到所需的评估结果。(3) In the present invention, the failure data and degradation data in the external field can be integrated with the data of the accelerated degradation test to obtain the required evaluation results.

附图说明 Description of drawings

图1是本发明的方法流程图；Fig. 1 is method flowchart of the present invention;

图2是本发明的实施例中WinBUGS所建立的模型；Fig. 2 is the model established by WinBUGS in the embodiment of the present invention;

图3是本发明的实施例评估结果。Fig. 3 is the evaluation result of the embodiment of the present invention.

具体实施方式 Detailed ways

下面将结合附图和实施例对本发明作进一步的详细说明。The present invention will be further described in detail with reference to the accompanying drawings and embodiments.

本发明是一种综合利用实验室及外场信息对产品外场寿命及可靠性进行评估的模型构建及评估方法，流程图如图1所示，包括以下几个步骤：The present invention is a model construction and evaluation method that comprehensively utilizes laboratory and field information to evaluate product field life and reliability. The flow chart is shown in Figure 1, including the following steps:

本发明选择漂移布朗运动来描述产品的退化，对于漂移布朗运动模型：The present invention selects the drift Brownian motion to describe the degradation of the product, for the drift Brownian motion model:

Y(t)＝σB(t)+d(s)·t+y₀(1)Y(t)=σB(t)+d(s)·t+y ₀ (1)

其中：Y(t)为产品参数的退化过程；B(t)为均值为0，方差为时间t的标准布朗运动B(t)～N(0，t)；σ为扩散系数，不随应力和时间而改变，为常数；d(s)为漂移系数，即产品的性能退化率；y₀为产品性能的初始值；Among them: Y(t) is the degradation process of product parameters; B(t) is the standard Brownian motion B(t)～N(0,t) with the mean value of 0 and the variance of time t; σ is the diffusion coefficient, which does not vary with stress and Time changes, which is a constant; d(s) is the drift coefficient, that is, the performance degradation rate of the product; y ₀ is the initial value of the product performance;

1)加速模型；1) Acceleration model;

对基于漂移布朗运动退化模型而言，漂移系数d(s)为产品的性能退化率，它是一个与应力水平有关的函数，应用它可将加速模型与退化模型相结合。若假设产品性能退化率代表的加速模型为：For the degradation model based on drift Brownian motion, the drift coefficient d(s) is the performance degradation rate of the product, which is a function related to the stress level. It can be used to combine the acceleration model with the degradation model. If it is assumed that the acceleration model represented by the product performance degradation rate is:

其中，是应力s的某一已知函数。若得到加速模型中参数β₀、β₁的值，那么便可建立应力与退化数据之间的关系，确定加速模型，并得到漂移系数d(s)的值。in, is a known function of the stress s. If the values of parameters β ₀ and β ₁ in the acceleration model are obtained, then the relationship between stress and degradation data can be established, the acceleration model can be determined, and the value of the drift coefficient d(s) can be obtained.

2)贝叶斯总体分布及其数据形式；2) Bayesian population distribution and its data form;

由漂移布朗运动的性质可知，单位时间Δt的退化增量ΔY服从均值为d(s)·Δt，方差为σ²Δt的正态分布，即According to the nature of drifting Brownian motion, the degradation increment ΔY per unit time Δt obeys the normal distribution with mean d(s)·Δt and variance σ ² Δt, namely

ΔY～N(d(s)·Δt，σ²Δt)(3)ΔY～N(d(s)·Δt, σ ² Δt)(3)

为了便于贝叶斯方法的应用，本发明将ΔY作为后续运算的数据形式，式(3)作为贝叶斯方法中的总体分布；In order to facilitate the application of the Bayesian method, the present invention uses ΔY as the data form of subsequent operations, and formula (3) is used as the overall distribution in the Bayesian method;

3)评估模型；3) Evaluate the model;

如果设l为参数的失效阈值，即设Y(t)-l＜0时产品失效；那么产品失效的概率密度函数为：If l is the failure threshold of the parameter, that is, the product fails when Y(t)-l<0; then the probability density function of product failure is:

$h h ((t t)) = = {((\frac{{l l}^{22}}{{22 πt πt}^{33} {σ σ}^{22}}))}^{11 / / 22} exp exp {{- - \frac{{((11 - - d d ((s the s)) t t))}^{22}}{{22 tσ tσ}^{22}}}} - - - - - - ((44))$

产品的可靠度模型为：The product reliability model is:

$R R ((t t)) = = Φ Φ [[\frac{l l - - {y the y}_{00} - - d d ((s the s)) t t}{σ σ \sqrt{t t}}]] - - exp exp ((\frac{22 d d ((s the s)) ((l l - - {y the y}_{00}))}{{σ σ}^{22}})) Φ Φ [[- - \frac{l l - - {y the y}_{00} + + d d ((s the s)) t t}{σ σ \sqrt{t t}}]] - - - - - - ((55))$

其中R(t)为产品t时刻的可靠度，Ф为正态分布。Among them, R(t) is the reliability of the product at time t, and Ф is a normal distribution.

步骤二、构建修正因子；Step 2, constructing the correction factor;

由步骤一可知，d(s)和σ²是求解的关键，其中d(s)是与应力相关的函数，而σ²则反应了环境噪声及测试水平，他们的值会随着使用环境的改变而改变。因此构建两个修正因子k₁和k₂对d(s)和σ²进行修正。It can be known from step 1 that d(s) and σ ² are the key to the solution, where d(s) is a function related to stress, and σ ² reflects the environmental noise and test level, and their values will vary with the use environment. Change and change. Therefore, two correction factors k ₁ and k ₂ are constructed to correct d(s) and σ ² .

已知应力为s₁时，加速退化试验情况下的漂移系数d_A(s₁)为：When the known stress is s ₁ , the drift coefficient d _A (s ₁ ) in the case of accelerated degradation test is:

其中：表示应力s₁的某一已知函数。in: Represents some known function of stress _s1 .

那么外场情况下的漂移系数d_f(s)为：Then the drift coefficient d _f (s) in the case of external field is:

d_f(s)＝k₁·d_A(s₁)(7)d _f (s)=k ₁ ·d _A (s ₁ )(7)

虽然扩散系数σ只与产品本身有关，不随时间和应力而变，但由于试验和外场实际使用中产品本身的特性会发生改变，因此也需要修正。但在不考虑特性改变的情况下可不进行修正。若加速退化试验情况下的扩散系数为σ_A，外场情况下的扩散系数为σ_f，那么：Although the diffusion coefficient σ is only related to the product itself and does not change with time and stress, it also needs to be corrected because the characteristics of the product itself will change in the test and actual use in the field. However, corrections may not be made without considering changes in characteristics. If the diffusion coefficient in the case of accelerated degradation test is σ _A , and the diffusion coefficient in the case of external field is σ _f , then:

${σ σ}_{f f}^{22} = = {σ σ}_{A A}^{22} + + {k k}_{22} - - - - - - ((88))$

从而完成了构建修正因子，将修正因子k₁和k₂引入到模型之中。Thus, the construction of correction factors is completed, and the correction factors k ₁ and k ₂ are introduced into the model.

步骤三、建立贝叶斯模型；Step 3, establishing a Bayesian model;

1)建立加速退化试验数据及外场退化数据情况下的贝叶斯模型：1) Establish a Bayesian model in the case of accelerated degradation test data and external field degradation data:

由公式(3)可知，加速退化试验的数据y_A表示为：From the formula (3), it can be seen that the data y _A of the accelerated degradation test is expressed as:

${Δy Δy}_{A A} ~ ~ N N (({d d}_{A A} (({s the s}_{11})) \cdot &Center Dot; Δ Δ {t t}_{A A},, {σ σ}_{A A}^{22} \cdot &Center Dot; {Δt Δt}_{A A})) - - - - - - ((99))$

其中Δy_A为加速退化试验中的退化增量，Δt_A为加速退化试验中的时间间隔。Where Δy _A is the degradation increment in the accelerated degradation test, and Δt _A is the time interval in the accelerated degradation test.

外场的退化数据y_f表示为：The degradation data y _f of the external field is expressed as:

${Δy Δy}_{f f} ~ ~ N N (({d d}_{f f} ((s the s)) \cdot \cdot {Δt Δt}_{f f},, {σ σ}_{f f}^{22} \cdot &Center Dot; {Δt Δt}_{f f})) - - - - - - ((1010))$

其中Δy_f为外场使用中的退化增量，Δt_f为外场使用中的时间间隔。Among them, Δy _f is the degradation increment in the use of the external field, and Δt _f is the time interval in the use of the external field.

为了综合使用加速退化试验数据及外场数据，将公式(9)、(10)进行统一。In order to comprehensively use the accelerated degradation test data and field data, formulas (9) and (10) are unified.

由公式(7)、(8)便可得到：From formulas (7) and (8), we can get:

${Δy Δy}_{f f} ~ ~ N N (({k k}_{11} \cdot \cdot {d d}_{A A} ((s the s)) \cdot \cdot {Δt Δt}_{f f},, (({σ σ}_{A A}^{22} + + {k k}_{22})) \cdot &Center Dot; {Δt Δt}_{f f})) - - - - - - ((1111))$

若Δt_A≠Δt_f，取它们的最小公倍数Δt作为统一的时间间隔，那么：If Δt _A ≠Δt _f , take their least common multiple Δt as the unified time interval, then:

Δt＝p_AΔt_A (12)Δt = p _A Δt _A (12)

Δt＝p_fΔt_f (13)Δt = p _f Δt _f (13)

在得到倍数p_A、p_f后，统一的退化增量可以写为：After obtaining the multiples p _A and p _f , the unified degradation increment can be written as:

${Δy Δy}_{u u} = = (({y the y}_{A A 11} - - {y the y}_{A A 11 + + {p p}_{A A}},, {y the y}_{A A 22} - - {y the y}_{A A 22 + + {p p}_{A A}},, . . . . . .,, {y the y}_{Am Am - - {p p}_{A A}} - - {y the y}_{Am Am})) - - - - - - ((1414))$

或 ${Δy}_{u} = (y_{f 1} - y_{f 1 + p_{f}}, y_{f 2} - y_{f 2 + p_{f}}, . . ., y_{fm - p_{f}} - y_{fm})$ or ${Δy}_{u} = ({the y}_{f 1} - {the y}_{f 1 + p_{f}}, {the y}_{f 2} - {the y}_{f 2 + p_{f}}, . . ., {the y}_{fm - p_{f}} - {the y}_{fm})$

其中Δy_u为统一后的退化增量，y_Ai为加速退化试验的数据y_A中的第i个数据，为第i个数据同第i+p_A个数据的差，即p_AΔt_A(Δt)时间内的退化增量。y_fi为外场退化数据y_f中的第i个数据，为第i个数据同第i+p_f个数据的差，即p_fΔt_f(Δt)时间内的退化增量。Where Δy _u is the unified degradation increment, y _Ai is the i-th data in the accelerated degradation test data y _A , It is the difference between the i-th data and the i+p _A- th data, that is, the degradation increment within the time of p _A Δt _A (Δt). y _fi is the i-th data in the external field degradation data y _f , It is the difference between the i-th data and the i+p _f -th data, that is, the degradation increment within the time of p _f Δt _f (Δt).

数据中包括Δy_u，s，c，其中，Δy_u是退化增量，s是应力水平，c是状态参数(如果数据来自加速退化试验，c＝0，如果数据来自外场，c＝1)，那么统一后的贝叶斯模型分布函数为：The data includes Δy _u , s, c, where Δy _u is the degradation increment, s is the stress level, and c is the state parameter (if the data comes from an accelerated degradation test, c=0, if the data comes from an external field, c=1), Then the unified Bayesian model distribution function is:

${Δy Δy}_{u u} ~ ~ N N (({d d}_{A A} ((s the s)) \cdot &Center Dot; Δt Δt \cdot \cdot ((11 + + c c (({k k}_{11} - - 11)))),, (({σ σ}_{A A}^{22} + + {ck ck}_{22})) \cdot \cdot Δt Δt)) - - - - - - ((1515))$

β₀，β₁，k₁，k₂和σ_A ²是未知参数，它们的先验分布为：β ₀ , β ₁ , k ₁ , k ₂ and σ _A ² are unknown parameters, and their prior distributions are:

${β β}_{00} ~ ~ N N (({μ μ}_{{β β}_{00}},, {σ σ}_{{β β}_{00}}^{22})) - - - - - - ((1616))$

${β β}_{11} ~ ~ N N (({μ μ}_{{β β}_{11}},, {σ σ}_{{β β}_{11}}^{22})) - - - - - - ((1717))$

${k k}_{11} ~ ~ Γ Γ (({a a}_{{k k}_{11}},, {b b}_{{k k}_{11}})) - - - - - - ((1818))$

${k k}_{22} ~ ~ IGa IGa (({a a}_{{k k}_{22}},, {b b}_{{k k}_{22}})) - - - - - - ((1919))$

${σ σ}_{A A}^{22} ~ ~ IGa IGa ((a a,, b b)) - - - - - - ((2020))$

其中β₀，β₁服从正态分布，分别是分布中的参数。k₁服从伽马分布，是分布中的参数，σ_A ²，k₂服从倒伽马分布，a、b、分别是分布中的参数。Among them, β ₀ and β ₁ obey the normal distribution, are parameters in the distribution, respectively. k ₁ obeys the gamma distribution, is the parameter in the distribution, σ _A ² , k ₂ obeys the inverted gamma distribution, a, b, are parameters in the distribution, respectively.

那么后验分布为：Then the posterior distribution is:

$π π ((Θ Θ | | D D.))$

$= = π π (({β β}_{00},, {β β}_{11},, {σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22} | | Δy Δy,, s the s,, c c))$

$&Proportional; &Proportional; {Π Π}_{i i = = 11}^{n no} {((\frac{11}{(({σ σ}_{A A}^{22} + + {c c}_{i i} {k k}_{22})) \cdot &Center Dot; Δt Δt}))}^{11 / / 22} exp exp ((- - \frac{{(({Δy Δy}_{ui ui} - - {d d}_{A A} (({s the s}_{i i})) \cdot &Center Dot; Δt Δt \cdot &Center Dot; ((11 + + {c c}_{i i} (({k k}_{11} - - 11))))))}^{22}}{22 (({σ σ}_{A A}^{22} + + {c c}_{i i} {k k}_{22})) \cdot \cdot Δt Δt})) - - - - - - ((21 twenty one))$

$\cdot \cdot φ φ ((\frac{{β β}_{00} - - {μ μ}_{{β β}_{00}}}{{σ σ}_{{β β}_{00}}})) \cdot \cdot φ φ ((\frac{{β β}_{11} - - {μ μ}_{{β β}_{11}}}{{σ σ}_{{β β}_{11}}})) \cdot \cdot \frac{{b b}_{{k k}_{11}}^{{a a}_{{k k}_{11}}}}{Γ Γ (({a a}_{{k k}_{11}}))} {(({k k}_{11}))}^{{a a}_{{k k}_{11}} + + 11} {e e}^{{b b}_{{k k}_{11}} {k k}_{11}} \cdot \cdot \frac{{b b}_{{k k}_{22}}^{{a a}_{{k k}_{22}}}}{Γ Γ (({a a}_{{k k}_{22}}))} {((\frac{11}{{k k}_{22}}))}^{{a a}_{{k k}_{22}} + + 11} {e e}^{{b b}_{{k k}_{22}} {k k}_{22}} \cdot &Center Dot; \frac{{b b}^{a a}}{Γ Γ ((a a))} {((\frac{11}{{σ σ}_{A A}^{22}}))}^{a a + + 11} {e e}^{b b / / {σ σ}_{A A}^{22}}$

其中：Δy_ui是第i个退化增量数据，c_i是第i个数据的状态参数，s_i是作用在第i个数据的应力。Among them: Δy _ui is the i-th degradation incremental data, _ci is the state parameter of the i-th data, s _i is the stress acting on the i-th data.

2)建立加速退化试验数据及外场失效数据情况下的贝叶斯模型2) Establish a Bayesian model for accelerated degradation test data and field failure data

对于外场失效数据x＝(t_x1，t_x2，...，t_xn)，由公式(4)可知，其概率密度函数为：For the external field failure data x=(t _x1 , t _x2 ,..., t _xn ), it can be seen from the formula (4), its probability density function is:

$h h ((x x)) = = {((\frac{{l l}^{22}}{22 π π {x x}^{33} {σ σ}_{f f}^{22}}))}^{11 / / 22} exp exp {{- - \frac{{((l l - - {d d}_{f f} ((s the s)) x x))}^{22}}{22 x x {σ σ}_{f f}^{22}}}} - - - - - - ((22 twenty two))$

由(7)、(8)可知：From (7) and (8), it can be seen that:

$h h ((x x)) = = {((\frac{{l l}^{22}}{22 π π {x x}^{33} (({σ σ}_{A A}^{22} + + {k k}_{22}))}))}^{11 / / 22} exp exp {{- - \frac{{((l l - - {k k}_{11} \cdot \cdot {d d}_{A A} ((s the s)) x x))}^{22}}{22 x x (({σ σ}_{A A}^{22} + + {k k}_{22}))}}} - - - - - - ((23 twenty three))$

由(3)可知，加速退化试验中退化数据y＝(y₁，y₂，...，y_m)对于时间间隔Δt的退化增量Δy＝(Δy₁，Δy₂，…，Δy_m-1服从正态分布，其概率密度函数为：It can be known from (3) that in the accelerated degradation test, the degradation data y=(y ₁ , y ₂ ,...,y _m ) for the degradation increment Δy of the time interval Δt=(Δy ₁ , Δy ₂ ,...,Δy _{m- 1} obeys a normal distribution, and its probability density function is:

$f f ((Δy Δy)) = = \frac{11}{\sqrt{22 π π {σ σ}_{A A}^{22} Δt Δt}} exp exp ((- - \frac{{((Δy Δy - - {d d}_{A A} ((s the s)) Δt Δt))}^{22}}{22 {σ σ}_{A A}^{22} Δt Δt})) - - - - - - ((24 twenty four))$

它们的似然函数为：Their likelihood functions are:

$L L (({d d}_{A A} ((s the s)),, {σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22})) = = {Π Π}_{i i = = 11}^{n no} h h (({t t}_{xi xi})) {Π Π}_{j j = = n no + + 11}^{n no + + m m - - 11} f f (({Δy Δy}_{j j}))$

为了综合利用两组数据，并便于后续采用winBUGS软件求解参数，这里对模型的似然函数进行处理。假设一个模型的对数似然函数为w_i＝logf(z_i|θ)。那么模型的似然函数可以写作：In order to make comprehensive use of the two sets of data and facilitate the subsequent use of winBUGS software to solve the parameters, the likelihood function of the model is processed here. Assume that the log-likelihood function of a model is w _i =logf(z _i |θ). Then the likelihood function of the model can be written as:

$L L ((z z | | θ θ)) = = {Π Π}_{i i = = 11}^{n no} {e e}^{{w w}_{i i}} = = {Π Π}_{i i = = 11}^{n no} \frac{{e e}^{- - ((- - {w w}_{i i}))} {(({- - w w}_{i i}))}^{00}}{00!!} = = {Π Π}_{i i = = 11}^{n no} {f f}_{P P} ((00;; - - {w w}_{i i})) - - - - - - ((2525))$

由上可知，0！为0的阶乘，模型的似然函数被写作一组新随机变量的似然函数，它们服从泊松分布f_P(r；λ)，其均值λ等于-w_i，所有的新随机变量的观测值r都等于0。It can be seen from the above that 0! is the factorial of 0, the likelihood function of the model is written as the likelihood function of a group of new random variables, they obey the Poisson distribution f _P (r; λ), and its mean value λ is equal to -w _i , all observations of new random variables The values r are all equal to 0.

因此对于任意外场的失效数据t_xi及加速退化试验的数据Δy_j，(23)和(24)可写为：Therefore, for the failure data t _xi of any external field and the data Δy _j of the accelerated degradation test, (23) and (24) can be written as:

${w w}_{fi the fi} = = log log h h (({t t}_{xi xi} | | {β β}_{00},, {β β}_{11} {,, σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22}))$

${w w}_{Aj Aj} = = log log f f (({Δy Δy}_{j j} | | {β β}_{00},, {β β}_{11},, {σ σ}_{A A}^{22}))$

定义m_i为状态参数(当数据来自加速退化试验时，m_i＝0。当数据来自外场时，m_i＝1)，那么定义：Define m _i as the state parameter (when the data comes from the accelerated degradation test, m _i =0. When the data comes from the external field, m _i =1), then define:

w_i＝m_i·w_fi+(1-m_i)·w_Ai w _i ＝m _i ·w _fi +(1-m _i )·w _Ai

其中，w_i为统一w_fi和w_Aj的对数似然函数。where w _i is the logarithmic likelihood function that unifies w _fi and w _Aj .

运算中数据应包括z_i，s_i，m_i，其中，z_i代表失效数据或者退化增量数据y_i，s_i是应力水平。那么贝叶斯模型的似然函数为：The data in the operation should include z _i , s _i , m _i , where z _i represents the failure data or degradation increment data y _i , and s _i is the stress level. Then the likelihood function of the Bayesian model is:

$L L ((z z | | {β β}_{00},, {β β}_{11},, {σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22})) = = {Π Π}_{i i = = 11}^{n no} {e e}^{{w w}_{i i}} = = {Π Π}_{i i = = 11}^{n no} \frac{{e e}^{- - ((- - {w w}_{i i}))} {(({- - w w}_{i i}))}^{00}}{00!!} = = {Π Π}_{i i = = 11}^{n no} {f f}_{P P} ((00;; - - {w w}_{i i})) - - - - - - ((2626))$

其中β₀，β₁，k₁，k₂和σ_A ²是未知参数，它们的先验分布如(16)～(20)所示。因此后验分布为：Among them, β ₀ , β ₁ , k ₁ , k ₂ and σ _A ² are unknown parameters, and their prior distributions are shown in (16)～(20). So the posterior distribution is:

$π π ((Θ Θ | | D D.))$

$= = π π (({β β}_{00},, {β β}_{11},, {σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22} | | z z,, s the s,, m m))$

$&Proportional; &Proportional; L L ((z z | | {β β}_{00},, {β β}_{11},, {σ σ}_{A A}^{22},, {k k}_{11},, {k k}_{22})) - - - - - - ((2727))$

$\cdot &Center Dot; φ φ ((\frac{{β β}_{00} - - {μ μ}_{{β β}_{00}}}{{σ σ}_{{β β}_{00}}})) \cdot \cdot φ φ ((\frac{{β β}_{11} - - {μ μ}_{{β β}_{11}}}{{σ σ}_{{β β}_{11}}})) \cdot &Center Dot; \frac{{b b}_{{k k}_{11}}^{{a a}_{{k k}_{11}}}}{Γ Γ (({a a}_{{k k}_{11}}))} {(({k k}_{11}))}^{{a a}_{{k k}_{11}} + + 11} {e e}^{{b b}_{{k k}_{11}} {k k}_{11}} \cdot \cdot \frac{{b b}_{{k k}_{22}}^{{a a}_{{k k}_{22}}}}{Γ Γ (({a a}_{{k k}_{22}}))} {((\frac{11}{{k k}_{22}}))}^{{a a}_{{k k}_{22}} + + 11} {e e}^{{b b}_{{k k}_{22}} {k k}_{22}} \cdot \cdot \frac{{b b}^{a a}}{Γ Γ ((a a))} {((\frac{11}{{σ σ}_{A A}^{22}}))}^{a a + + 11} {e e}^{b b / / {σ σ}_{A A}^{22}}$

针对(21)或(27)中的后验分布，利用软件winBUGS或R对其进行求解，并得到未知参数β₀，β₁，k₁，k₂和σ_A ²的评估值。For the posterior distribution in (21) or (27), use the software winBUGS or R to solve it, and obtain the estimated values of unknown parameters β ₀ , β ₁ , k ₁ , k ₂ and σ _A ² .

步骤五、评估产品的寿命及可靠度Step 5. Evaluate the life and reliability of the product

得到参数β₀，β₁，k₁，k₂和σ_A ²的评估值后，便可得到外场情况下漂移系数d_f(s)及扩散系数σ_f的评估值，结合失效阈值l及产品性能初始值y₀的值，将它们带入公式(5)中便可获得产品在时刻t时的可靠度，以及在一定可靠度下产品的寿命值。After obtaining the evaluation values of parameters β ₀ , β ₁ , k ₁ , k ₂ and σ _A ² , the evaluation values of drift coefficient d _f (s) and diffusion coefficient σ _f under external field conditions can be obtained, combined with failure threshold l and product The value of the initial value of performance y ₀ , put them into the formula (5) to obtain the reliability of the product at time t, and the life value of the product under a certain reliability.

实施例：Example:

若对某光电产品实施温度步进应力加速退化试验，样本量为4，温度应力水平为4，温度分别为60℃、80℃、100℃、120℃；每个应力水平的试验时间分别为1250、750、500、500小时；产品的性能检测时间间隔Δt为5小时。同时，有一个产品在外场中使用，温度为25℃，搜集了它5000小时的数据，产品的性能检测时间间隔Δt为5小时。选择光功率作为其性能参数，光功率的初始值y₀为100，参数的失效阈值l为40。需要对产品外场条件下工作3年的可靠度以及可靠度为0.95时产品的寿命进行评估。If a temperature step stress accelerated degradation test is carried out on a photoelectric product, the sample size is 4, the temperature stress level is 4, and the temperature is 60°C, 80°C, 100°C, 120°C; the test time for each stress level is 1250 , 750, 500, 500 hours; the product performance testing time interval Δt is 5 hours. At the same time, a product is used in the field at a temperature of 25°C, and its data has been collected for 5,000 hours. The product's performance testing time interval Δt is 5 hours. The optical power is selected as its performance parameter, the initial value _y0 of the optical power is 100, and the failure threshold l of the parameter is 40. It is necessary to evaluate the reliability of the product under field conditions for 3 years and the life of the product when the reliability is 0.95.

步骤一、建立并确定相关模型Step 1. Establish and determine relevant models

由于对产品施加的应力为温度，因此选择阿伦尼乌兹(Arrhenius)模型作为加速模型，即：Since the stress applied to the product is temperature, the Arrhenius model is selected as the acceleration model, namely:

d(T)＝exp[A-B/T)]d(T)=exp[A-B/T)]

其中T为温度。退化模型为(1)。贝叶斯总体分布及其数据形式为(3)。可靠度模型为(5)。where T is the temperature. The degradation model is (1). The Bayesian population distribution and its data form are (3). The reliability model is (5).

步骤二、定义修正因子Step 2. Define the correction factor

引入两个修正因子k₁和k₂对d(T)和σ²进行修正。Introduce two correction factors k ₁ and k ₂ to correct d(T) and σ ² .

d_f(s)＝k₁·d(T)d _f (s)=k ₁ ·d(T)

若外场情况下的扩散系数为σ_f，那么：If the diffusion coefficient in the external field is σ _f , then:

${σ σ}_{f f}^{22} = = {σ σ}^{22} + + {k k}_{22}$

从而完成了修正因子的定义，将修正因子k₁和k₂引入到模型之中。Thus, the definition of the correction factor is completed, and the correction factors k ₁ and k ₂ are introduced into the model.

步骤三、建立贝叶斯模型Step 3: Build a Bayesian model

由公式(3)可知，将所有的退化数据转化为退化增量Δy_u的形式，Δt＝5小时。It can be seen from formula (3) that all degradation data are converted into the form of degradation increment Δy _u , Δt=5 hours.

运算所需的数据为Δy_u，T，c，The data required for the operation are Δy _u , T, c,

其中，T是温度应力水平，c是状态参数(如果数据来自加速退化试验，c＝0，如果数据来自外场，c＝1)，那么统一后的分布函数为：Among them, T is the temperature stress level, c is the state parameter (if the data comes from the accelerated degradation test, c=0, if the data comes from the external field, c=1), then the unified distribution function is:

Δy_u～N(d(s)·Δt·(1+c(k₁-1))，(σ²+ck₂)·Δt)Δy _u ～N(d(s)·Δt·(1+c(k ₁ -1)), (σ ² +ck ₂ )·Δt)

从而建立后验分布：This creates the posterior distribution:

π(Θ|D)π(Θ|D)

＝π(A，B，σ²，k₁，k₂|Δy_u，T，c)=π(A, B, σ ² , k ₁ , k ₂ |Δy _u , T, c)

步骤四、求解参数值Step 4. Solve for parameter values

利用软件winBUGS对步骤三得到的模型进行求解，建立的模型如图2所示，对其进行40000次运算，得到未知参数A，B，k₁，k₂和σ²的评估值，如表1所示。Use the software winBUGS to solve the model obtained in step 3. The established model is shown in Figure 2, and 40,000 operations are performed on it to obtain the evaluation values of unknown parameters A, B, k ₁ , k ₂ and σ ² , as shown in Table 1 shown.

表1参数评估值Table 1 Parameter Evaluation Values

参数 parameters A A B B k₁ k ₁ k₂ k ₂ σ² ^σ2 评估值 The assessed value 9.21 9.21 4818.3 4818.3 1.5089 1.5089 6.08e-4 6.08e-4 6.97e-4 6.97e-4

得到参数的评估值后，便可得到外场情况下漂移系数d_f(s)及扩散系数σ_f的评估值，结合失效阈值l及产品性能初始值y₀的值，将它们带入公式(5)中便可获得产品的寿命及可靠度，如图3所示。After obtaining the evaluation values of the parameters, the evaluation values of the drift coefficient d _f (s) and the diffusion coefficient σ _f in the external field can be obtained, combined with the values of the failure threshold l and the initial value of product performance y ₀ , they can be brought into the formula (5 ) to obtain the life and reliability of the product, as shown in Figure 3.

由此可知产品在外场条件下工作3年的可靠度为0.99，可靠度为0.95时的寿命约为3.5年。若不采用本文的方法进行修正，而直接用加速退化试验的数据进行评估，那么所得到的结果为：工作3年的可靠度为1，可靠度为0.95时的寿命约为6年。由此可知，对外场情况的修正建模是十分必要的，并可以提高评估的精度。It can be seen from this that the reliability of the product working in the field for 3 years is 0.99, and the service life is about 3.5 years when the reliability is 0.95. If the method in this paper is not used for correction, but the data of the accelerated degradation test is directly used for evaluation, then the obtained results are: the reliability of 3 years of work is 1, and the life expectancy is about 6 years when the reliability is 0.95. It can be seen that the revised modeling of the external field situation is very necessary and can improve the accuracy of the evaluation.

Claims

1. A method for constructing and evaluating a product outfield service life and reliability model is characterized by comprising the following steps:

step one, establishing and determining a degradation model of a product;

the product degradation is described by adopting drift Brownian motion, and a drift Brownian motion model is as follows:

Y(t)＝σB(t)+d(s)·t+y₀ (1)

wherein: y (t) is the degradation process of the product parameters; b (t) is the standard Brownian motion with mean 0 and variance at time t, B (t) N (0, t); sigma isA diffusion coefficient; d(s) is a drift coefficient, namely the performance degradation rate of the product; y is₀Is an initial value of product performance;

1) accelerating the model;

the drift coefficient d(s) is the performance degradation rate of the product, and an acceleration model represented by the performance degradation rate of the product is as follows:

wherein,is some known function of stress s; beta is a₀、β₁Two parameters are represented;

2) bayesian population distribution and its data form;

the degradation increment Δ Y per unit time Δ t obeys a mean value d(s). Δ t and a variance σ²Normal distribution of Δ t, i.e.

ΔY～N(d(s)·Δt,σ²Δt) (3)

Taking the delta Y as a data form of subsequent operation, and taking an expression (3) as overall distribution in a Bayesian method;

3) evaluating the model;

if l is set as the failure threshold of the parameter, the product fails when Y (t) -l is set to be less than 0; the probability density function for product failure is then:

the reliability model of the product is:

wherein: r (t) is the reliability of the product at the time t, and phi is normal distribution;

step two, constructing a correction factor;

construction of two correction factors k₁And k₂For d(s) and σ²Correcting;

stress is known as s₁Drift coefficient d in the case of accelerated degradation test_A(s₁) Comprises the following steps:

wherein:denotes the stress s₁A certain known function of;

then the drift coefficient d in the case of an external field_f(s) is:

d_f(s)＝k₁·d_A(s₁) (7)

although the diffusion coefficient σ is related only to the product itself and does not change with time and stress, since the characteristics of the product itself are changed in tests and actual use of an external field, correction is also required, but correction may not be performed without considering the characteristic change; if the diffusion coefficient in the case of accelerated degradation test is σ_AIn the case of an external fieldHas a diffusion coefficient of σ_fAnd then:

step three, establishing a Bayesian model;

1) establishing a Bayesian model under the conditions of accelerated degradation test data and external field degradation data:

as can be seen from equation (3), data y of the accelerated degradation test_AExpressed as:

wherein: Δ y_ATo accelerate the increase in degradation in the degradation test, Δ t_ATime intervals in accelerated degradation tests;

degradation data y of the external field_fExpressed as:

wherein Δ y_fDelta t for the degradation increment in external field use_fTime intervals in which the external field is in use;

unifying the formulas (9) and (10); the following equations (7) and (8) can be obtained:

if Δ t_A≠Δt_fTaking their least common multiple Δ t as the uniform time interval, then:

Δt＝p_AΔt_A (12)

Δt＝p_fΔt_f (13)

to obtain the multiple p_A、p_fThereafter, the uniform degradation delta can be written as:

or

Wherein Δ y_uFor unified degradation increment, y_AiData y for accelerated degradation testing_AThe (d) th data of (1),for the ith data with the (i + p) th data_ADifference of data, i.e. p_AΔt_A(Δ t) an incremental degradation over time; y is_fiFor degradation of data y by external field_fThe (d) th data of (1),for the ith data with the (i + p) th data_fDifference of data, i.e. p_fΔt_f(Δ t) an incremental degradation over time;

data included Δ y_uS, c, wherein Δ y_uIs the degradation increment, s is the stress level, c is the state parameter, if the data comes from the accelerated degradation test, c is 0, if the data comes from the external field, c is 1, then the unified bayesian model distribution function is:

β₀,β₁,k₁,k₂andare unknown parameters, a priori of whichThe distribution is as follows:

k_{2} ~ IGa (a_{k_{2}}, b_{k_{2}}) - - - (19)

wherein: beta is a₀,β₁Subject to a normal distribution of the signals,are the parameters in the distribution, respectively; k is a radical of₁Subject to the gamma distribution,is a parameter in the distribution that is,k₂obeying an inverse gamma distribution, a, b,Are the parameters in the distribution, respectively;

the posterior distribution is then:

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>π</mi> <mrow> <mo>(</mo> <mi>Θ</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <mi>π</mi> <mrow> <mo>(</mo> <msub> <mi>β</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>β</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>,</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>|</mo> <mi>Δy</mi> <mo>,</mo> <mi>s</mi> <mo>,</mo> <mi>c</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>&Proportional;</mo> <munderover> <mi>Π</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <mrow> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>c</mi> <mi>i</mi> </msub> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>·</mo> <mi>Δt</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mrow> <mo>(</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>Δ</mi> <msub> <mi>y</mi> <mi>ui</mi> </msub> <mo>-</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>s</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>·</mo> <mi>Δt</mi> <mo>·</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <msub> <mi>c</mi> <mi>i</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>-</mo> <mn>1</mn> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mrow> <mo>(</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>c</mi> <mi>i</mi> </msub> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>·</mo> <mi>Δt</mi> </mrow> </mfrac> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>·</mo> <mi>φ</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>β</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>μ</mi> <msub> <mi>β</mi> <mn>0</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>σ</mi> <msub> <mi>β</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>·</mo> <mi>φ</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>β</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>μ</mi> <msub> <mi>β</mi> <mn>1</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>σ</mi> <msub> <mi>β</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>·</mo> <mfrac> <msup> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mrow> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <msub> <mi>k</mi> <mn>1</mn> </msub> </mrow> </msup> <mo>·</mo> <mfrac> <msup> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>k</mi> <mn>2</mn> </msub> </mfrac> <mo>)</mo> </mrow> <mrow> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <msub> <mi>k</mi> <mn>2</mn> </msub> </mrow> </msup> <mo>·</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mfrac> <mo>)</mo> </mrow> <mrow> <mi>a</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <mi>b</mi> <mo>/</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mrow> </msup> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>21</mn> <mo>)</mo> </mrow> </mrow> </math>

wherein: Δ y_uiIs the i-th degradation incremental data, c_iIs the state parameter of the ith data, s_iIs the stress acting on the ith data;

2) establishing a Bayesian model under the conditions of accelerated degradation test data and outfield failure data:

for outfield failure data x ═ t_x1,t_x2,…,t_xn) From equation (4), the probability density function is:

from (7) and (8):

as shown in (3), the degradation data y ═ y in the accelerated degradation test₁,y₂,…,y_m) Degradation increment Δ y for time interval Δ t ═ Δ y₁,Δy₂,…,Δy_m-1) Obeying a normal distribution, the probability density function is:

their likelihood functions are:

assume a model with a log-likelihood function of w_i＝logf(y_i| θ); then the likelihood function of the model is:

from the above, 0! For a factorization of 0, the likelihood function of the model is written as a likelihood function of a set of new random variables subject to a Poisson distribution f_P(r; lambda) with a mean value lambda equal to-w_iAll observed values r of the new random variable are equal to 0;

failure data t for any external field_xiAnd data Δ y of accelerated degradation test_jAnd (23) and (24) can be written as:

let m_iFor the state parameter, when the data is from an accelerated degradation test, m_i0; when the data comes from an external field, m_i1, then:

w_i＝m_i·w_fi+(1-m_i)·w_Aj

wherein, w_iTo unify w_fiAnd w_AjA log-likelihood function of;

the data in operation includes z_i,s_i,m_iWherein z is_iRepresenting failure data or degraded incremental data Δ y_i,s_iIs the stress level; then the likelihood function of the bayesian model is:

wherein beta is₀,β₁,k₁,k₂Andare unknown parameters, and the prior distribution of the unknown parameters is shown as (16) to (20); the posterior distribution is therefore:

<math> <mrow> <mfenced open='' close=''> <mtable> <mtr> <mtd> <mi>π</mi> <mrow> <mo>(</mo> <mi>Θ</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <mi>π</mi> <mrow> <mo>(</mo> <msub> <mi>β</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>β</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>,</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>|</mo> <mi>z</mi> <mo>,</mo> <mi>s</mi> <mo>,</mo> <mi>m</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>&Proportional;</mo> <mi>L</mi> <mrow> <mo>(</mo> <mi>z</mi> <mo>|</mo> <msub> <mi>β</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>β</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>,</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>·</mo> <mi>φ</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>β</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>μ</mi> <msub> <mi>β</mi> <mn>0</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>σ</mi> <msub> <mi>β</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>·</mo> <mi>φ</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>β</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>μ</mi> <msub> <mi>β</mi> <mn>1</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>σ</mi> <msub> <mi>β</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>·</mo> <mfrac> <msup> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mrow> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <msub> <mi>k</mi> <mn>1</mn> </msub> </mrow> </msup> <mo>·</mo> <mfrac> <msup> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msub> <mi>k</mi> <mn>2</mn> </msub> </mfrac> <mo>)</mo> </mrow> <mrow> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>2</mn> </msub> </msub> <msub> <mi>k</mi> <mn>2</mn> </msub> </mrow> </msup> <mo>·</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>Γ</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mfrac> <mo>)</mo> </mrow> <mrow> <mi>a</mi> <mo>+</mo> <mn>1</mn> </mrow> </msup> <msup> <mi>e</mi> <mrow> <mi>b</mi> <mo>/</mo> <msubsup> <mi>σ</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mrow> </msup> </mtd> </mtr> </mtable> </mfenced> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>27</mn> <mo>)</mo> </mrow> </mrow> </math>

step four, obtaining parameter values in posterior distribution;

solving the posterior distribution in (21) or (27) to obtain an unknown parameter beta₀,β₁,k₁,k₂Andthe evaluation value of (1);

evaluating the service life and reliability of the product;

obtaining a parameter beta₀,β₁,k₁,k₂Andafter the evaluation value, the drift coefficient d under the external field condition is obtained_f(s) and diffusion coefficient σ_fThe evaluation value of (a) is combined with the failure threshold value l and the initial value y of the product performance₀The reliability of the product at time t and the lifetime of the product at reliability can be obtained by substituting them into equation (5).