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

Model construction and evaluation method for service life and reliability of product under outfield circumstance Download PDF

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CN102663516B
CN102663516B CN201210085242.6A CN201210085242A CN102663516B CN 102663516 B CN102663516 B CN 102663516B CN 201210085242 A CN201210085242 A CN 201210085242A CN 102663516 B CN102663516 B CN 102663516B
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CN102663516A (en
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王立志
姜同敏
李晓阳
王晓红
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Beihang University
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Abstract

The invention discloses a model construction and evaluation method for service life and reliability of a product under the outfield circumstance. Specifically, the method comprises the following steps: step one, establishing and determining a degradation model of a product; step two, constructing a correction factor; step three, establishing a bayesian model; step four, obtaining a parameter value of posterior distribution; and step five, evaluating service life and reliability. According to the invention, a relation between laboratory information and outfield information is established and differences among all stress environments are determined and quantized, thereby providing a powerful basis for future work on evaluation and result correction; besides, a method in which two kinds of information, that is the laboratory information and the outfield information, is comprehensively utilized to evaluate service life and reliability of the product under the outfield circumstance is brought forward, so that problems that there is certain deviation due to direct evaluation with utilization of accelerated degradation testing information and it is difficult to do evaluation work due to sparse outfield information can be solved.

Description

Method for constructing and evaluating product outfield life and reliability model
Technical Field
The invention discloses a method for evaluating the service life and reliability of a product under an outfield condition by comprehensively utilizing accelerated degradation test data and outfield data of the product, and belongs to the technical field of service life and reliability evaluation.
Background
The life and reliability information of the product can be represented by the failure data and the degradation data of the product, and the life and reliability information can be used for evaluating a life distribution model of the product. In the practical use of the product outfield, it is difficult to obtain enough information in a short time for evaluation, and at this time, an accelerated test technology comes up. For the conventional accelerated life test in the accelerated test, it is difficult to obtain sufficient failure data under the limited time and expense conditions when evaluating a long-life and high-reliability product. In this case, an accelerated degradation test method is used, which can obtain the performance degradation data of the product in a short time and extrapolate the service life and reliability of the product under the specified conditions.
However, the extrapolation of the accelerated degradation test results is only representative of the life and reliability of the product under laboratory conditions, which are not exactly the same as the conditions under which the external field is used. On one hand, all the stresses in the actual use of the product are hardly reflected by an accelerated degradation test; on the other hand, the noise and test level of the environment also change with each situation. Therefore, the life and reliability evaluation results obtained by the accelerated degradation test are different from the actual life and reliability in the use of the external field, and the external field data also becomes the most reliable information source in the evaluation process.
Because the outfield data is rare and the laboratory data is rich, if the relationship between the outfield and the laboratory information can be established and the two kinds of information are integrated to evaluate the service life and the reliability of the product under the outfield condition, the problems that the accelerated degradation test information cannot completely represent the outfield information and the outfield information is rare and the like can be solved, and thus a more accurate evaluation result can be obtained.
Disclosure of Invention
The invention aims to solve the problems and provides a model construction and evaluation method capable of comprehensively utilizing laboratory and outfield information to evaluate the outfield service life and reliability of a product. Therefore, the problems that certain deviation exists when the accelerated degradation test information is used for directly evaluating, the external field information is rare and the evaluation work is difficult to carry out and the like are solved.
The invention relates to a method for constructing and evaluating a product outfield service life and reliability model, which comprises the following specific steps:
step one, establishing and determining a degradation model of a product;
step two, constructing a correction factor;
step three, establishing a Bayesian model;
step four, obtaining parameter values in posterior distribution;
evaluating the service life and reliability;
the invention has the advantages that:
(1) the invention establishes the relation between the laboratory information and the external field information, clearly and quantifies the difference between the stress environments, and provides a powerful basis for the evaluation and result correction work in the future;
(2) the method for evaluating the service life and reliability of the product under the condition of the external field by comprehensively utilizing two kinds of information (laboratory information and external field information) is provided, so that the problems that certain deviation exists when the accelerated degradation test information is used for directly evaluating, the external field information is rare and the evaluation work is difficult to carry out and the like are solved;
(3) in the invention, the failure data and the degradation data under the condition of the external field can be integrated with the data of the accelerated degradation test to obtain the required evaluation result.
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FIG. 1 is a flow chart of a method of the present invention;
FIG. 2 is a model established by WinBUGS in an embodiment of the present invention;
FIG. 3 shows the evaluation results of the embodiment of the present invention.
Detailed Description
The present invention will be described in further detail with reference to the accompanying drawings and examples.
The invention relates to a model construction and evaluation method for evaluating the life and reliability of an outfield of a product by comprehensively utilizing information of a laboratory and the outfield, a flow chart is shown as a figure 1, and the method comprises the following steps:
step one, establishing and determining a degradation model of a product;
the invention selects drift Brownian motion to describe the degradation of the product, and for a drift Brownian motion model:
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 B (t) to N (0, t) with a mean of 0 and a variance of time t; σ is a diffusion coefficient, does not change with stress and time, and is a constant; 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;
for a drift-based brownian motion degradation model, the drift coefficient d(s) is the rate of performance degradation of the product, which is a function related to the stress level, and the acceleration model can be combined with the degradation model by applying the same. If an acceleration model represented by the product performance degradation rate is assumed to be:
wherein,is some known function of the stress s. If the parameter beta in the acceleration model is obtained0、β1Then a relationship between the stress and degradation data is established, an acceleration model is determined, and a value of the drift coefficient d(s) is obtained.
2) Bayesian population distribution and its data form;
from the nature of the drift Brownian motion, the degradation increment Δ Y per unit time Δ t follows a mean value of d(s). Δ t and a variance of σ2Normal distribution of Δ t, i.e.
ΔY~N(d(s)·Δt,σ2Δt)(3)
In order to facilitate the application of the Bayes method, the invention takes the delta Y as the data form of the subsequent operation, and the formula (3) is taken as the overall distribution in the Bayes 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:
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <msup> <mrow> <mn>2</mn> <mi>&pi;t</mi> </mrow> <mn>3</mn> </msup> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mn>1</mn> <mo>-</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <msup> <mrow> <mn>2</mn> <mi>t&sigma;</mi> </mrow> <mn>2</mn> </msup> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> </math>
the reliability model of the product is:
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mo>=</mo> <mi>&Phi;</mi> <mo>[</mo> <mfrac> <mrow> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>-</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> </mrow> <mrow> <mi>&sigma;</mi> <msqrt> <mi>t</mi> </msqrt> </mrow> </mfrac> <mo>]</mo> <mo>-</mo> <mi>exp</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>2</mn> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <mi>&Phi;</mi> <mo>[</mo> <mo>-</mo> <mfrac> <mrow> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>+</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> </mrow> <mrow> <mi>&sigma;</mi> <msqrt> <mi>t</mi> </msqrt> </mrow> </mfrac> <mo>]</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> </math>
where R (t) is the reliability of the product at time t, and phi is a normal distribution.
Step two, constructing a correction factor;
from step one, d(s) and σ2Is the key to the solution, where d(s) is a stress-related function and σ2Environmental noise and test levels are reflected and their values change as the usage environment changes. Thus constructing two correction factors k1And k2For d(s) and σ2And (6) correcting.
Stress is known as s1Drift coefficient d in the case of accelerated degradation testA(s1) Comprises the following steps:
wherein:denotes the stress s1Is a certain known function of.
Then the drift coefficient d in the case of an external fieldf(s) is:
df(s)=k1·dA(s1)(7)
although the diffusion coefficient σ is related only 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 change in tests and actual use of an external field. But may not be corrected without taking into account the characteristic change. If the diffusion coefficient in the case of accelerated degradation test is σADiffusion coefficient in the case of an external field is σfThen:
<math> <mrow> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> <mo>=</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> </math>
thereby completing the construction of the correction factor, i.e. correcting the factor k1And k2Introduced into the model.
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 testAExpressed as:
<math> <mrow> <msub> <mi>&Delta;y</mi> <mi>A</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>s</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>A</mi> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>&CenterDot;</mo> <msub> <mi>&Delta;t</mi> <mi>A</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein Δ yATo accelerate the increase in degradation in the degradation test, Δ tATo accelerate the time interval in the degradation test.
Degradation data y of the external fieldfExpressed as:
<math> <mrow> <msub> <mi>&Delta;y</mi> <mi>f</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <msub> <mi>&Delta;t</mi> <mi>f</mi> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> <mo>&CenterDot;</mo> <msub> <mi>&Delta;t</mi> <mi>f</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein Δ yfDelta t for the degradation increment in external field usefThe time interval in which the external field is in use.
In order to comprehensively use the accelerated degradation test data and the external field data, equations (9) and (10) are unified.
The following equations (7) and (8) can be obtained:
<math> <mrow> <msub> <mi>&Delta;y</mi> <mi>f</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <msub> <mi>&Delta;t</mi> <mi>f</mi> </msub> <mo>,</mo> <mrow> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <msub> <mi>&Delta;t</mi> <mi>f</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> </math>
if Δ tA≠ΔtfTaking their least common multiple Δ t as the uniform time interval, then:
Δt=pAΔtA (12)
Δt=pfΔtf (13)
to obtain the multiple pA、pfThereafter, the uniform degradation delta can be written as:
<math> <mrow> <msub> <mi>&Delta;y</mi> <mi>u</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>1</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>2</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>Am</mi> <mo>-</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mi>Am</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mo>)</mo> </mrow> </mrow> </math>
or <math> <mrow> <msub> <mi>&Delta;y</mi> <mi>u</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>1</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>2</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>fm</mi> <mo>-</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mi>fm</mi> </msub> <mo>)</mo> </mrow> </mrow> </math>
Wherein Δ yuFor unified degradation increment, yAiData y for accelerated degradation testingAThe (d) th data of (1),for the ith data with the (i + p) th dataADifference of data, i.e. pAΔtA(Δ t) an increment of degradation over time. y isfiFor degradation of data y by external fieldfThe (d) th data of (1),for the ith data with the (i + p) th datafDifference of data, i.e. pfΔtf(Δ t) an increment of degradation over time.
Data included Δ yuS, c, wherein Δ yuIs the degradation increment, s is the stress level, c is the state parameter (if the data is from an accelerated degradation test, c is 0, if the data is from an external field, c is 1), then the unified bayesian model distribution function is:
<math> <mrow> <msub> <mi>&Delta;y</mi> <mi>u</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>&CenterDot;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>c</mi> <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> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>ck</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> </math>
β0,β1,k1,k2and σA 2Are unknown parameters whose prior distribution is:
<math> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>~</mo> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>,</mo> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> </math>
k 2 ~ IGa ( a k 2 , b k 2 ) - - - ( 19 )
<math> <mrow> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>~</mo> <mi>IGa</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein beta is0,β1Subject to a normal distribution of the signals,respectively, are parameters in the distribution. k is a radical of1Subject to the gamma distribution, is a parameter in the distribution, σA 2,k2Obeying an inverse gamma distribution, a, b,Respectively, are parameters in the distribution.
The posterior distribution is then:
<math> <mrow> <mi>&pi;</mi> <mrow> <mo>(</mo> <mi>&Theta;</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mo>=</mo> <mi>&pi;</mi> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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>&Delta;y</mi> <mo>,</mo> <mi>s</mi> <mo>,</mo> <mi>c</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mo>&Proportional;</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <mrow> <mrow> <mo>(</mo> <msubsup> <mi>&sigma;</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>&CenterDot;</mo> <mi>&Delta;t</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> <msub> <mi>&Delta;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>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>&CenterDot;</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>&sigma;</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>&CenterDot;</mo> <mi>&Delta;t</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>21</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</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>&Gamma;</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>&CenterDot;</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>&Gamma;</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>&CenterDot;</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>&sigma;</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>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mrow> </msup> </mrow> </math>
wherein: Δ yuiIs the i-th degradation incremental data, ciIs the state parameter of the ith data, siIs 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 ═ tx1,tx2,...,txn) From equation (4), the probability density function is:
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>&pi;</mi> <msup> <mi>x</mi> <mn>3</mn> </msup> <msup> <msub> <mi>&sigma;</mi> <mi>f</mi> </msub> <mn>2</mn> </msup> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>d</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>x</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mrow> <mn>2</mn> <mi>x</mi> </mrow> <msup> <msub> <mi>&sigma;</mi> <mi>f</mi> </msub> <mn>2</mn> </msup> </mrow> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>22</mn> <mo>)</mo> </mrow> </mrow> </math>
from (7) and (8):
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>&pi;</mi> <msup> <mi>x</mi> <mn>3</mn> </msup> <mrow> <mo>(</mo> <msup> <msub> <mi>&sigma;</mi> <mi>A</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>x</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>x</mi> <mrow> <mo>(</mo> <msup> <msub> <mi>&sigma;</mi> <mi>A</mi> </msub> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> </math>
as shown in (3), the degradation data y ═ y in the accelerated degradation test1,y2,...,ym) Degradation increment Δ y for time interval Δ t ═ Δ y1,Δy2,…,Δym-1Obeying a normal distribution, the probability density function is:
<math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mi>&Delta;y</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <msqrt> <mn>2</mn> <mi>&pi;</mi> <msup> <msub> <mi>&sigma;</mi> <mi>A</mi> </msub> <mn>2</mn> </msup> <mi>&Delta;t</mi> </msqrt> </mfrac> <mi>exp</mi> <mrow> <mo>(</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>&Delta;y</mi> <mo>-</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>&Delta;t</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <msup> <msub> <mi>&sigma;</mi> <mi>A</mi> </msub> <mn>2</mn> </msup> <mi>&Delta;t</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>24</mn> <mo>)</mo> </mrow> </mrow> </math>
their likelihood functions are:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>,</mo> <msup> <msub> <mi>&sigma;</mi> <mi>A</mi> </msub> <mn>2</mn> </msup> <mo>,</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>,</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mi>h</mi> <mrow> <mo>(</mo> <msub> <mi>t</mi> <mi>xi</mi> </msub> <mo>)</mo> </mrow> <munderover> <mi>&Pi;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mo>+</mo> <mi>m</mi> <mo>-</mo> <mn>1</mn> </mrow> </munderover> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>&Delta;y</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> </mrow> </math>
in order to comprehensively utilize the two groups of data and facilitate the subsequent solving of parameters by using winBUGS software, the likelihood function of the model is processed. Assume a model with a log-likelihood function of wi=logf(zi| θ). The likelihood function of the model can then be written as:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <mi>z</mi> <mo>|</mo> <mi>&theta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mi>e</mi> <msub> <mi>w</mi> <mi>i</mi> </msub> </msup> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <mrow> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <msup> <mrow> <mo>(</mo> <msub> <mrow> <mo>-</mo> <mi>w</mi> </mrow> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mn>0</mn> </msup> </mrow> <mrow> <mn>0</mn> <mo>!</mo> </mrow> </mfrac> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msub> <mi>f</mi> <mi>P</mi> </msub> <mrow> <mo>(</mo> <mn>0</mn> <mo>;</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>25</mn> <mo>)</mo> </mrow> </mrow> </math>
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 fP(r; lambda) with a mean value lambda equal to-wiAll observed values r of the new random variable are equal to 0.
Failure data t for any external fieldxiAnd data Δ y of accelerated degradation testjAnd (23) and (24) can be written as:
<math> <mrow> <msub> <mi>w</mi> <mi>fi</mi> </msub> <mo>=</mo> <mi>log</mi> <mi>h</mi> <mrow> <mo>(</mo> <msub> <mi>t</mi> <mi>xi</mi> </msub> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <msubsup> <mrow> <mo>,</mo> <mi>&sigma;</mi> </mrow> <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> </mrow> </math>
<math> <mrow> <msub> <mi>w</mi> <mi>Aj</mi> </msub> <mo>=</mo> <mi>log</mi> <mi>f</mi> <mrow> <mo>(</mo> <msub> <mi>&Delta;y</mi> <mi>j</mi> </msub> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </math>
definition miIs a state parameter (m when data comes from accelerated degradation testingi0. When the data comes from an external field, mi1), then define:
wi=mi·wfi+(1-mi)·wAi
wherein, wiTo unify wfiAnd wAjThe log-likelihood function of (a).
The data should include z in the operationi,si,miWherein z isiRepresenting failure data or degraded incremental data yi,siIs the stress level. Then the likelihood function of the bayesian model is:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <mi>z</mi> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mi>e</mi> <msub> <mi>w</mi> <mi>i</mi> </msub> </msup> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <mrow> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <msup> <mrow> <mo>(</mo> <msub> <mrow> <mo>-</mo> <mi>w</mi> </mrow> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mn>0</mn> </msup> </mrow> <mrow> <mn>0</mn> <mo>!</mo> </mrow> </mfrac> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msub> <mi>f</mi> <mi>P</mi> </msub> <mrow> <mo>(</mo> <mn>0</mn> <mo>;</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>26</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein beta is0,β1,k1,k2And σA 2Are unknown parameters and their prior distributions are shown in (16) to (20). The posterior distribution is therefore:
<math> <mrow> <mi>&pi;</mi> <mrow> <mo>(</mo> <mi>&Theta;</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mo>=</mo> <mi>&pi;</mi> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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> </mrow> </math>
<math> <mrow> <mo>&Proportional;</mo> <mi>L</mi> <mrow> <mo>(</mo> <mi>z</mi> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>27</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</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>&Gamma;</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>&CenterDot;</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>&Gamma;</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>&CenterDot;</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>&sigma;</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>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> </mrow> </msup> </mrow> </math>
step four, obtaining parameter values in posterior distribution;
aiming at the posterior distribution in (21) or (27), solving the posterior distribution by using software winBUGS or R, and obtaining an unknown parameter beta0,β1,k1,k2And σA 2Is measured.
Step five, evaluating the service life and reliability of the product
To obtainParameter beta0,β1,k1,k2And σA 2After the evaluation value, the drift coefficient d under the external field condition can be obtainedf(s) and diffusion coefficient σfThe 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 the time t and the life value of the product with a certain reliability can be obtained by substituting the values into the formula (5).
Example (b):
if a temperature stepping stress accelerated degradation test is carried out on a certain photoelectric product, the sample size is 4, the temperature stress level is 4, and the temperatures are 60 ℃, 80 ℃, 100 ℃ and 120 ℃ respectively; the test time for each stress level was 1250, 750, 500 hours, respectively; the performance test interval Δ t of the product was 5 hours. Meanwhile, a product is used in an external field, the temperature is 25 ℃, 5000-hour data of the product is collected, and the performance detection time interval delta t of the product is 5 hours. Selecting optical power as its performance parameter, the initial value y of optical power0The failure threshold l for the parameter is 100 and 40. An assessment of the reliability of a product operating for 3 years at outfield conditions and the lifetime of the product at a reliability of 0.95 is required.
Step one, establishing and determining a relevant model
Since the stress applied to the product is temperature, an Arrhenius (Arrhenius) model was chosen as the acceleration model, namely:
d(T)=exp[A-B/T)]
wherein T is the temperature. The degradation model is (1). The Bayesian population distribution and its data form are (3). The reliability model is (5).
Step two, defining correction factor
Introducing two correction factors k1And k2For d (T) and σ2And (6) correcting.
ThenDrift coefficient d in the case of an external fieldf(s) is:
df(s)=k1·d(T)
if the diffusion coefficient in the case of an external field is σfThen:
<math> <mrow> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> <mo>=</mo> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> </mrow> </math>
thereby completing the definition of the correction factor k1And k2Introduced into the model.
Step three, establishing a Bayesian model
As can be seen from equation (3), all the degradation data is converted into degradation increments Δ yuΔ t is 5 hours.
Data required for the operation is Δ yu,T,c,
Where T is the temperature stress level and c is the state parameter (if the data is from an accelerated degradation test, c is 0, if the data is from an external field, c is 1), then the unified distribution function is:
Δyu~N(d(s)·Δt·(1+c(k1-1)),(σ2+ck2)·Δt)
thus establishing a posterior distribution:
π(Θ|D)
=π(A,B,σ2,k1,k2|Δyu,T,c)
step four, solving parameter values
Solving the model obtained in the third step by using software winBUGS, and buildingThe vertical model is shown in FIG. 2, and 40000 times of calculation are carried out to obtain unknown parameters A, B, k1,k2And σ2The evaluation values of (2) are shown in table 1.
TABLE 1 evaluation of parameters
Parameter(s) A B k1 k2 σ2
Evaluation value 9.21 4818.3 1.5089 6.08e-4 6.97e-4
Step five, evaluating the service life and reliability of the product
After obtaining the evaluation value of the parameter, the drift coefficient d under the condition of external field can be obtainedf(s) and diffusion coefficient σfThe evaluation value of (a) is combined with the failure threshold value l and the initial value y of the product performance0By substituting them into equation (5) the life of the product andreliability, as shown in fig. 3.
From this, it is known that the reliability of the product at the external field condition for 3 years is 0.99, and the life of the product at the reliability of 0.95 is about 3.5 years. If the data from the accelerated degradation test were used directly for evaluation without modification by the methods herein, the results were obtained as follows: the reliability of 3 years of operation is 1, and the lifetime of 0.95 is about 6 years. Therefore, the correction modeling of the external field condition is necessary, and the evaluation precision can be improved.

Claims (1)

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:
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>&pi;</mi> <msup> <mi>t</mi> <mn>3</mn> </msup> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>t</mi> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> </mrow> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>4</mn> <mo>)</mo> </mrow> </mrow> </math>
the reliability model of the product is:
<math> <mrow> <mi>R</mi> <mrow> <mo>(</mo> <mi>t</mi> <mo>)</mo> </mrow> <mo>=</mo> <mi>&Phi;</mi> <mo>[</mo> <mfrac> <mrow> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>-</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> </mrow> <mrow> <mi>&sigma;</mi> <msqrt> <mi>t</mi> </msqrt> </mrow> </mfrac> <mo>]</mo> <mo>-</mo> <mi>exp</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <mn>2</mn> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>)</mo> </mrow> </mrow> <msup> <mi>&sigma;</mi> <mn>2</mn> </msup> </mfrac> <mo>)</mo> </mrow> <mi>&Phi;</mi> <mo>[</mo> <mo>-</mo> <mfrac> <mrow> <mi>l</mi> <mo>-</mo> <msub> <mi>y</mi> <mn>0</mn> </msub> <mo>+</mo> <mi>d</mi> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>t</mi> </mrow> <mrow> <mi>&sigma;</mi> <msqrt> <mi>t</mi> </msqrt> </mrow> </mfrac> <mo>]</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>5</mn> <mo>)</mo> </mrow> </mrow> </math>
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 testA(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 fieldf(s) is:
df(s)=k1·dA(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 σAIn the case of an external fieldHas a diffusion coefficient of σfAnd then:
<math> <mrow> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> <mo>=</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>8</mn> <mo>)</mo> </mrow> </mrow> </math>
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 testAExpressed as:
<math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mn>4</mn> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <msub> <mi>s</mi> <mn>1</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>A</mi> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>A</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>9</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein: Δ yATo accelerate the increase in degradation in the degradation test, Δ tATime intervals in accelerated degradation tests;
degradation data y of the external fieldfExpressed as:
<math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>f</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>f</mi> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>f</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>10</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein Δ yfDelta t for the degradation increment in external field usefTime intervals in which the external field is in use;
unifying the formulas (9) and (10); the following equations (7) and (8) can be obtained:
<math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>f</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>f</mi> </msub> <mo>,</mo> <mrow> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;</mi> <msub> <mi>t</mi> <mi>f</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>11</mn> <mo>)</mo> </mrow> </mrow> </math>
if Δ tA≠ΔtfTaking their least common multiple Δ t as the uniform time interval, then:
Δt=pAΔtA (12)
Δt=pfΔtf (13)
to obtain the multiple pA、pfThereafter, the uniform degradation delta can be written as:
<math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>u</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>1</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>A</mi> <mn>2</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>Am</mi> <mo>-</mo> <msub> <mi>p</mi> <mi>A</mi> </msub> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mi>Am</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>14</mn> <mo>)</mo> </mrow> </mrow> </math>
or <math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>u</mi> </msub> <mo>=</mo> <mrow> <mo>(</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>1</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>1</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>2</mn> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mrow> <mi>f</mi> <mn>2</mn> <mo>+</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>,</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>,</mo> <msub> <mi>y</mi> <mrow> <mi>fm</mi> <mo>-</mo> <msub> <mi>p</mi> <mi>f</mi> </msub> </mrow> </msub> <mo>-</mo> <msub> <mi>y</mi> <mi>fm</mi> </msub> <mo>)</mo> </mrow> </mrow> </math>
Wherein Δ yuFor unified degradation increment, yAiData y for accelerated degradation testingAThe (d) th data of (1),for the ith data with the (i + p) th dataADifference of data, i.e. pAΔtA(Δ t) an incremental degradation over time; y isfiFor degradation of data y by external fieldfThe (d) th data of (1),for the ith data with the (i + p) th datafDifference of data, i.e. pfΔtf(Δ t) an incremental degradation over time;
data included Δ yuS, c, wherein Δ yuIs 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:
<math> <mrow> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>u</mi> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>&CenterDot;</mo> <mrow> <mo>(</mo> <mn>1</mn> <mo>+</mo> <mi>c</mi> <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> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>ck</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>15</mn> <mo>)</mo> </mrow> </mrow> </math>
β01,k1,k2andare unknown parameters, a priori of whichThe distribution is as follows:
<math> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>16</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>~</mo> <mi>N</mi> <mrow> <mo>(</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>17</mn> <mo>)</mo> </mrow> </mrow> </math>
<math> <mrow> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>~</mo> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <msub> <mi>a</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>,</mo> <msub> <mi>b</mi> <msub> <mi>k</mi> <mn>1</mn> </msub> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>18</mn> <mo>)</mo> </mrow> </mrow> </math>
k 2 ~ IGa ( a k 2 , b k 2 ) - - - ( 19 )
<math> <mrow> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>~</mo> <mi>IGa</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>,</mo> <mi>b</mi> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>20</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein: beta is a01Subject 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>&pi;</mi> <mrow> <mo>(</mo> <mi>&Theta;</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <mi>&pi;</mi> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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>&Delta;y</mi> <mo>,</mo> <mi>s</mi> <mo>,</mo> <mi>c</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>&Proportional;</mo> <munderover> <mi>&Pi;</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>&sigma;</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>&CenterDot;</mo> <mi>&Delta;t</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>&Delta;</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>&CenterDot;</mo> <mi>&Delta;t</mi> <mo>&CenterDot;</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>&sigma;</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>&CenterDot;</mo> <mi>&Delta;t</mi> </mrow> </mfrac> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</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>&Gamma;</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>&CenterDot;</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>&Gamma;</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>&CenterDot;</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>&sigma;</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>&sigma;</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 yuiIs the i-th degradation incremental data, ciIs the state parameter of the ith data, siIs 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 ═ tx1,tx2,…,txn) From equation (4), the probability density function is:
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>&pi;</mi> <msup> <mi>x</mi> <mn>3</mn> </msup> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>d</mi> <mi>f</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>x</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>x</mi> <msubsup> <mi>&sigma;</mi> <mi>f</mi> <mn>2</mn> </msubsup> </mrow> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>22</mn> <mo>)</mo> </mrow> </mrow> </math>
from (7) and (8):
<math> <mrow> <mi>h</mi> <mrow> <mo>(</mo> <mi>x</mi> <mo>)</mo> </mrow> <mo>=</mo> <msup> <mrow> <mo>(</mo> <mfrac> <msup> <mi>l</mi> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>&pi;</mi> <msup> <mi>x</mi> <mn>3</mn> </msup> <mrow> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>2</mn> </mrow> </msup> <mi>exp</mi> <mo>{</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>l</mi> <mo>-</mo> <msub> <mi>k</mi> <mn>1</mn> </msub> <mo>&CenterDot;</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>x</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <mi>x</mi> <mrow> <mo>(</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>+</mo> <msub> <mi>k</mi> <mn>2</mn> </msub> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>}</mo> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>23</mn> <mo>)</mo> </mrow> </mrow> </math>
as shown in (3), the degradation data y ═ y in the accelerated degradation test1,y2,…,ym) Degradation increment Δ y for time interval Δ t ═ Δ y1,Δy2,…,Δym-1) Obeying a normal distribution, the probability density function is:
<math> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mi>&Delta;y</mi> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mn>1</mn> <msqrt> <mn>2</mn> <mi>&pi;</mi> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mi>&Delta;t</mi> </msqrt> </mfrac> <mi>exp</mi> <mrow> <mo>(</mo> <mo>-</mo> <mfrac> <msup> <mrow> <mo>(</mo> <mi>&Delta;y</mi> <mo>-</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mi>&Delta;t</mi> <mo>)</mo> </mrow> <mn>2</mn> </msup> <mrow> <mn>2</mn> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mi>&Delta;t</mi> </mrow> </mfrac> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>24</mn> <mo>)</mo> </mrow> </mrow> </math>
their likelihood functions are:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <msub> <mi>d</mi> <mi>A</mi> </msub> <mrow> <mo>(</mo> <mi>s</mi> <mo>)</mo> </mrow> <mo>,</mo> <msubsup> <mi>&sigma;</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> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mi>h</mi> <mrow> <mo>(</mo> <msub> <mi>t</mi> <mi>xi</mi> </msub> <mo>)</mo> </mrow> <munderover> <mi>&Pi;</mi> <mrow> <mi>j</mi> <mo>=</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> <mrow> <mi>n</mi> <mo>+</mo> <mi>m</mi> </mrow> </munderover> <mi>f</mi> <mrow> <mo>(</mo> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>j</mi> </msub> <mo>)</mo> </mrow> </mrow> </math>
assume a model with a log-likelihood function of wi=logf(yi| θ); then the likelihood function of the model is:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <mi>y</mi> <mo>|</mo> <mi>&theta;</mi> <mo>)</mo> </mrow> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mi>e</mi> <msub> <mi>w</mi> <mi>i</mi> </msub> </msup> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <mrow> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mn>0</mn> </msup> </mrow> <mrow> <mn>0</mn> <mo>!</mo> </mrow> </mfrac> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msub> <mi>f</mi> <mi>P</mi> </msub> <mrow> <mo>(</mo> <mn>0</mn> <mo>;</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>25</mn> <mo>)</mo> </mrow> </mrow> </math>
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 fP(r; lambda) with a mean value lambda equal to-wiAll observed values r of the new random variable are equal to 0;
failure data t for any external fieldxiAnd data Δ y of accelerated degradation testjAnd (23) and (24) can be written as:
<math> <mrow> <msub> <mi>w</mi> <mi>fi</mi> </msub> <mo>=</mo> <mi>log</mi> <mi>h</mi> <mrow> <mo>(</mo> <msub> <mi>t</mi> <mi>xi</mi> </msub> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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> </mrow> </math>
<math> <mrow> <msub> <mi>w</mi> <mi>Aj</mi> </msub> <mo>=</mo> <mi>log</mi> <mi>f</mi> <mrow> <mo>(</mo> <mi>&Delta;</mi> <msub> <mi>y</mi> <mi>j</mi> </msub> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</mi> <mi>A</mi> <mn>2</mn> </msubsup> <mo>)</mo> </mrow> </mrow> </math>
let miFor the state parameter, when the data is from an accelerated degradation test, mi0; when the data comes from an external field, mi1, then:
wi=mi·wfi+(1-mi)·wAj
wherein, wiTo unify wfiAnd wAjA log-likelihood function of;
the data in operation includes zi,si,miWherein z isiRepresenting failure data or degraded incremental data Δ yi,siIs the stress level; then the likelihood function of the bayesian model is:
<math> <mrow> <mi>L</mi> <mrow> <mo>(</mo> <mi>z</mi> <mo>|</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msup> <mi>e</mi> <msub> <mi>w</mi> <mi>i</mi> </msub> </msup> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <mfrac> <mrow> <msup> <mi>e</mi> <mrow> <mo>-</mo> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> </mrow> </msup> <msup> <mrow> <mo>(</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mn>0</mn> </msup> </mrow> <mrow> <mn>0</mn> <mo>!</mo> </mrow> </mfrac> <mo>=</mo> <munderover> <mi>&Pi;</mi> <mrow> <mi>i</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi>n</mi> </munderover> <msub> <mi>f</mi> <mi>P</mi> </msub> <mrow> <mo>(</mo> <mn>0</mn> <mo>;</mo> <mo>-</mo> <msub> <mi>w</mi> <mi>i</mi> </msub> <mo>)</mo> </mrow> <mo>-</mo> <mo>-</mo> <mo>-</mo> <mrow> <mo>(</mo> <mn>26</mn> <mo>)</mo> </mrow> </mrow> </math>
wherein beta is01,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>&pi;</mi> <mrow> <mo>(</mo> <mi>&Theta;</mi> <mo>|</mo> <mi>D</mi> <mo>)</mo> </mrow> </mtd> </mtr> <mtr> <mtd> <mo>=</mo> <mi>&pi;</mi> <mrow> <mo>(</mo> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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>&beta;</mi> <mn>0</mn> </msub> <mo>,</mo> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>,</mo> <msubsup> <mi>&sigma;</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>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>0</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</mo> <mi>&phi;</mi> <mrow> <mo>(</mo> <mfrac> <mrow> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> <mo>-</mo> <msub> <mi>&mu;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> <mo></mo> </mrow> <msub> <mi>&sigma;</mi> <msub> <mi>&beta;</mi> <mn>1</mn> </msub> </msub> </mfrac> <mo>)</mo> </mrow> <mo>&CenterDot;</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>&Gamma;</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>&CenterDot;</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>&Gamma;</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>&CenterDot;</mo> <mfrac> <msup> <mi>b</mi> <mi>a</mi> </msup> <mrow> <mi>&Gamma;</mi> <mrow> <mo>(</mo> <mi>a</mi> <mo>)</mo> </mrow> </mrow> </mfrac> <msup> <mrow> <mo>(</mo> <mfrac> <mn>1</mn> <msubsup> <mi>&sigma;</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>&sigma;</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 beta01,k1,k2Andthe evaluation value of (1);
evaluating the service life and reliability of the product;
obtaining a parameter beta01,k1,k2Andafter the evaluation value, the drift coefficient d under the external field condition is obtainedf(s) and diffusion coefficient σfThe 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).
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