CN104680005A - Non-parallel storage life test evaluation method based on accelerating factor feasible region selection - Google Patents

Abstract

The invention discloses a non-parallel storage life test evaluation method based on accelerating factor feasible region selection. The method comprises the following steps: I, calculating an accelerating factor and comprehensive failure time; II, discussing an accelerating factor feasible region; III, performing parameter selection by using an optimal linear unbiased estimation method; IV, calculating the point estimation and interval estimation of logarithm reliable life. Specific to the phenomenon of missing of storage data in non-parallel storage product life estimation, storage data and test data are integrated, and calculation is performed by making full use of data on the basis that the service life follows Weibull distribution, so that the accuracy of life model parameter estimation is ensured, an algorithm has low requirement on the initial value of a parameter, the algorithm iteration is rapid and simple, and the operability is high.

Description

Based on the non-parallel storage life test appraisal procedure that speedup factor feasible zone is selected

Technical field

The present invention relates to a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone, it is for Weibull distribution life model, based on the store data different by product and stress accelerated life test data, the feasible zone of speedup factor is discussed, and use the method for Best Linear Unbiased Estimate, the life-span of product is assessed.Be applicable to the fields such as life extension test life appraisal.

Background technology

For being worth expensive military hardware, after arriving storage period, needing to carry out life extension test and reappraising its storage life, for next step work program is prepared.Due to the restriction of sample number, the period of storage of carrying out the product of life extension test experienced before the test may not be identical, and we call non-parallel storing product these products.In the face of the product of high reliability long life, the method relying on traditional experiment to obtain complete data is not only unable to catch up with the speed of model change over a period of time, also causes very large burden, so life extension test mainly adopts accelerated life test to economy.

Current accelerated life test appraisal procedure mainly for be new product, and in actual applications, the life appraisal of non-parallel storing product carries out after having stored certain hour, in storage process, the environmental factor such as temperature, humidity, radiation can cause properties of product to be degenerated, and stores and be equivalent to one section of process of the test equally for product.Because the environmental stress in storage process is relatively little compared with proof stress, so often ignore store data, only consider accelerated life test data, but storage is a long-term process, if ignore storage information, there is certain error by causing between the result of assessment and actual value, so the life appraisal of non-parallel storing product simply can not use existing method to carry out, must consider that in storage process, environmental stress is on the impact of product.

Propose a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone based on this present invention, store data is combined with accelerated life test data, thus evaluates the life-span of product more accurately.

Summary of the invention

(1) object of the present invention: omit phenomenon for the storage information of non-parallel storing product when life appraisal, a kind of lifetime estimation method fully utilizing data is provided.Based on the store data of product and stress accelerated life test data, the feasible zone of speedup factor is discussed, and use the method for Best Linear Unbiased Estimate, the life-span of product is assessed.

(2) technical scheme:

The basic assumption that the present invention proposes is as follows:

Suppose that 1 life of product t obeys Weibull distribution, accumulative inefficacy function is

$F\left(t\right)=1-\exp (-{\left(\frac{t}{\mathrm{\η}}\right)}^m)---\left(1\right)$

Wherein, η is characteristics life, and m is form parameter.

Make y=lnt, be then converted into the extreme value distribution, its cumulative failure function is

$F\left(t\right)=1-\exp (-\exp \left(\frac{y-\mathrm{\μ}}{\mathrm{\σ}}\right))---\left(2\right)$

Wherein, location parameter μ=ln η, scale parameter σ=1/m.

Suppose that the failure mechanism of all products is the same under each stress level of 2 accelerated life tests, namely the scale parameter σ of life-span distribution is equal.

Suppose that 3 location parameters are the generalized linear functions of stress

Wherein S represents stress, represent with the relevant known function of stress.

Suppose that the remaining life-span of 4 products only depends on the failure probability accumulated, and stress level at that time, accumulation mode have nothing to do, this supposition is called as Nelson hypothesis.

Certain product of known storage certain hour carries out stress accelerated life test, at stress level S _j(j=1,2 ..., p), have n _jindividual sample carries out fixed failure number test, and storage stress is S ₀.Total q _jindividual sample fails, the out-of-service time is respectively period of storage before test is followed successively by truncated time is the period of storage of product of not losing efficacy is followed successively by

The method that the present invention proposes mainly comprises calculating speedup factor and comprehensive out-of-service time, discusses to speedup factor feasible zone, uses Best Linear Unbiased Estimate to carry out Selecting parameter, the point estimation calculating logarithm Q-percentile life and interval estimation.

Based on above-mentioned hypothesis and thinking, a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone provided by the invention, realizes as follows:

Step one: calculate speedup factor and comprehensive out-of-service time.

According to hypothesis 4, we can obtain: if product is at stress level S _ilower working time t _icumulative failure probability F _iequal at stress level S _jlower working time t _jcumulative failure probability F _j, then stress level S _jlower working time t _jstress level S can be turned to _ilower working time t _i.

If

F _i(t _i)＝F _j(t _j)

Wherein

Then

And obtain speedup factor thus

K _ijrepresent stress level S _jlower working time t _jbe converted into stress level S _ilower working time t _itime speedup factor.

By period of storage be converted into stress level S _junder equivalent time

$t_{\mathrm{ji}}^{S_0~S_j}=K_{j0}\·t_{\mathrm{ji}}^0$

Calculate the comprehensive out-of-service time

Step 2: speedup factor feasible zone is discussed.

Owing to containing unknown parameter b in speedup factor, according to the different values of parameter b, speedup factor can change, and causes the size of comprehensive out-of-service time to change, impacts the calculating of step 3, so need discuss to b, a point situation calculates.

With stress level S ₁for example, total n ₁individual sample carries out fixed time test, wherein q ₁individual sample fails, the out-of-service time is once period of storage is followed successively by truncated time is the period of storage of product of not losing efficacy is respectively the comprehensive out-of-service time is followed successively by

Will compare between two:

If t _1r> t _1s, and r, s=1,2 ..., n ₁and r ≠ s, then T _1r> T _1s;

If t _1r> t _1s, or t _1r< t _1s, then make T _1r=T _1s, obtain critical value b.

At stress level S ₁, one group of critical value can be obtained

Discuss respectively under each stress level, obtain the critical value of j group b, pressed order arrangement b from small to large ₁, b ₂..., b _f, wherein f=f ₁+ ... + f _j, obtain the interval of f+1 b.

Step 3: use Best Linear Unbiased Estimate method to carry out Selecting parameter.

With interval (b ₁, b ₂) be example

I, selection solve for parameter a, the initial value of b, σ wherein

The initial value suggestion of a, σ is selected: the result of calculation using Best Linear Unbiased Estimate when ignoring period of storage.

II, obtain speedup factor K _j0(j=1,2,3 ..., p).

By period of storage be converted into stress level S _junder equivalent time

$t_{\mathrm{ji}}^{S_0~S_j}=K_{j0}\&CenterDot;t_{\mathrm{ji}}^0$

Calculate the comprehensive out-of-service time

$T_{\mathrm{ji}}=t_{\mathrm{ji}}+t_{\mathrm{ji}}^{S_0~S_j}$

III, judge order statistic and Interval Statistic according to the size of comprehensive out-of-service time.

Will obtain according to order arrangement from small to large the i.e. front q of Weibull distribution ₁individual order statistic, obtains q thus ₁discuss in+1 interval.Will obtain according to order arrangement from small to large therefrom choose Interval Statistic.

With interval (Y ₁₁, Y ₁₂) be example, if in have s (s=1 ..., n ₁-q ₁+ 1) individual at interval (Y ₁₁, Y ₁₂) in, then choosing minimum is the 1st Interval Statistic.

To other q ₁individual interval is analyzed respectively, chooses all Interval Statistics.Finally obtain order statistic and the Interval Statistic of Weibull distribution.

Make y=lnY, obtain order statistic and the Interval Statistic of the extreme value distribution.Order obtain order statistic and the Interval Statistic of standard the extreme value distribution.

Estimates of parameters is obtained by the method for Best Linear Unbiased Estimate

IV, by estimated value with initial value contrast

If

$\left|\frac{{\hat{b}}_1-{\hat{b}}_0}{{\hat{b}}_0}\right|+\left|\frac{{\hat{a}}_1-{\hat{a}}_0}{{\hat{a}}_0}\right|+\left|\frac{{\widehat{\mathrm{\&sigma;}}}_1-{\widehat{\mathrm{\&sigma;}}}_0}{{\widehat{\mathrm{\&sigma;}}}_0}\right|<\mathrm{\&Delta;}$

Wherein Δ is the error amount of regulation.

Then obtain a, the estimated value of b, σ

${\hat{b}=\hat{b}}_1,\hat{a}={\hat{a}}_1,\widehat{\mathrm{\&sigma;}}={\widehat{\mathrm{\&sigma;}}}_1$

Otherwise, order

${\hat{b}}_0={\hat{b}}_1,{\hat{a}}_0={\hat{a}}_1,{\widehat{\mathrm{\&sigma;}}}_0={\widehat{\mathrm{\&sigma;}}}_1$

Proceed to II.

Step 4: the point estimation and the interval estimation that calculate logarithm Q-percentile life.

Given fiduciary level R, degree of confidence γ

The point estimation of logarithm Q-percentile life

${\hat{y}}_R=\hat{a}+\hat{b}x+\widehat{\mathrm{\&sigma;}}\ln \ln \frac{1}{R}---\left(5\right)$

Logarithm Q-percentile life interval estimation

${\hat{y}}_{R_L}=\hat{a}+{\hat{b}x+\mathrm{\&mu;}}_R\widehat{\mathrm{\&sigma;}}-k\widehat{\mathrm{\&sigma;}}---\left(6\right)$

${\hat{y}}_{R_U}=\hat{a}+\hat{b}x+{\mathrm{\&mu;}}_R\widehat{\mathrm{\&sigma;}}-k\widehat{\mathrm{\&sigma;}}---\left(7\right)$

Wherein

$k=u_{\mathrm{\&gamma;}}\frac{u_{\mathrm{\&gamma;}}(c_{13}+c_{23}x+c_{33}{u}_R)-\sqrt{w}}{c_{33}{u}_{\mathrm{\&gamma;}}-1}$

$w={u_{\mathrm{\&gamma;}}}^2{(c_{13}+c_{23}x)}^2+(1-c_{33}{u_{\mathrm{\&gamma;}}}^2)(c_{11}+c_{22}{x}^2+2c_{12}x)+c_{33}{u_R}^2+2c_{13}{u}_R+2c_{23}{u}_{R}x$

Wherein, " the Best Linear Unbiased Estimate method " mentioned in step 3, the concrete practice is as follows:

At stress S _junder, total sample number is n _j, for q before the extreme value distribution _jindividual order statistic, $y_{j1}^{*}\&le;...\&le;y_{{\mathrm{jm}}_j}^{*}(m_j\&Element;\{1,...,q_1+1\left\}\right)$ For the m of this sample _jindividual Interval Statistic.

Order

$Q=\underset{j=1}{\overset{p}{\mathrm{\&Sigma;}}}\underset{k,l=1}{\overset{q_j+m_j}{\mathrm{\&Sigma;}}}(y_{\mathrm{ik}}-a-{\mathrm{bx}}_j-\mathrm{\&sigma;}{u}_{\mathrm{jk}})g_{\mathrm{jkl}}(y_{\mathrm{jl}}-a-b{x}_j-\mathrm{\&sigma;}{u}_{\mathrm{jl}})---\left(8\right)$

Ask local derviation can obtain the estimated value of parameter a, b, σ.

The covariance matrix of parameter estimation amount is

$\mathrm{cov}(\hat{a},\hat{b},\widehat{\mathrm{\&sigma;}})={\mathrm{\&sigma;}}^{2}C---\left(9\right)$

$C={\left[\begin{array}{ccc}\underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}\\ \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jil}}{x_j}^2& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j{\mathrm{\&mu;}}_{\mathrm{ji}}\\ \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j{\mathrm{\&mu;}}_{\mathrm{ji}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}{\mathrm{\&mu;}}_{\mathrm{ji}}\end{array}\right]}^{-1}$

Wherein, $C={\left(g_{\mathrm{jkl}}\right)}_{(q_j+m_j)\&times;(q_j+m_j)}=V^{-1}={\left(v_{\mathrm{jkl}}\right)}_{(q_j+m_j)\&times;(q_j+m_j)},$ U _jk(k=1,2 ..., q _j) be standard the extreme value distribution size of population be n _jthe average of a kth order statistic of sample, v _jkl(k, l=1,2 ..., q _j) for standard the extreme value distribution size of population be the kth of the sample of nj and the covariance of l order statistic; u _jk(k=q _j+ 1 ..., q _j+ m _j) for standard the extreme value distribution size of population be n _jthe average of kth+1 order statistic of the sample of+1, v _jkl(k, l=q _j+ 1, q _j+ m _j, k≤l) for standard the extreme value distribution size of population be n _jthe kth+1 of sample of+2 and the covariance of l+2 order statistic, v _jkl(k, l=q _j+ 1, q _j+ m _j, k > l) for standard the extreme value distribution size of population be n _jthe l+1 of sample of+2 and the covariance of kth+2 order statistics, v _jkl(k=1,2 ..., q _j; L=q _j+ 1, q _j+ m _j) for standard the extreme value distribution size of population be n _jthe kth of sample of+1 and the covariance of l+1 order statistic, v _jkl(l=1,2 ..., q _j; K=q _j+ 1, q _j+ m _j) for standard the extreme value distribution size of population be n _jthe kth+1 of sample of+1 and the covariance of l+1 order statistic, these averages and covariance obtain by formulae discovery or table look-up.

(3) advantage and effect: the present invention is a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone, and its advantage is:

1. the present invention is directed to the phenomenon of omitting store data in non-parallel storing product life appraisal, on the basis of Weibull Distributed Units, by speedup factor, store data and test figure is comprehensive, make full use of data and calculate, ensure that the accuracy of life model parameter estimation.

2. algorithm of the present invention requires lower to the initial value of parameter, and algorithm iteration is simple, workable fast.

Accompanying drawing explanation

Fig. 1 the method for the invention process flow diagram

Embodiment

Below in conjunction with embodiment, the present invention is described in further details.

Previous work of the present invention:

Certain storage gyro Weibull Distributed Units, carry out Censoring accelerated life test, assess its life-span, period of storage is in table 1.

Storage temperature is 20 DEG C, and test temperature is respectively 45 DEG C, 57 DEG C, 69 DEG C, 80 DEG C.

Acceleration model is Arrhenius relationship:

η＝Aexp(E/kT) (10)

Wherein A is normal number, and E is activation energy, k=1.38 × 10 ^-23j/K is Boltzmann constant.

Taking the logarithm in both sides, can obtain

lnη＝a+bx

Wherein a=lnA, therefore, the logarithm of characteristics life is the linear function of inverse temperature.

Test figure is as table 1.

Table 1 test figure

Note: d represents sky.

The present invention is based on the non-parallel storage life test appraisal procedure that speedup factor feasible zone is selected, as shown in Figure 1, concrete implementation step is as follows:

Step one: calculate speedup factor and comprehensive out-of-service time.

Speedup factor

K _ij＝exp(b(1/T _i-1/T _j))

By period of storage be converted into stress level S _junder equivalent time

$t_{\mathrm{ji}}^{S_0~S_j}=K_{j0}\&CenterDot;t_{\mathrm{ji}}^0$

Calculate the comprehensive out-of-service time

$T_{\mathrm{ji}}=t_{\mathrm{ji}}+t_{\mathrm{ji}}^{S_0~S_j}=t_{\mathrm{ji}}+K_{j0}{\&CenterDot;t}_{\mathrm{ji}}^0=t_{\mathrm{ji}}+\exp \left(b\left(b(1/T_j-1/T_0)\right)\right)\&CenterDot;t_{\mathrm{ji}}^0$

Step 2: speedup factor feasible zone is discussed.

The difference along with speedup factor can be found by 3 sample datas of battery of tests in observation table 1, the comprehensive out-of-service time size of 1 and 2 two sample can change, make the comprehensive out-of-service time of sample 1 and 2 equal, obtain b=3415, during b>3415, comprehensive out-of-service time 1<2.

Observe other group test figure in table 1, can find that the size of the comprehensive out-of-service time of sample 1 and 2,1 and 3 in the 3rd group of test also can change along with the difference of speedup factor.Make the comprehensive out-of-service time of sample 1 and 2 in the 3rd group equal, obtain b=8505; Make the out-of-service time of sample 1 and 3 equal, obtain b=5184.During b>8505, comprehensive out-of-service time 1<2<3; 5184<b<8505, comprehensive out-of-service time 2<1<3; B<5184, comprehensive out-of-service time 2<3<1.

To sum up obtain four intervals (-∞, 3415) of b, (3415,5184), (5184,8505), (8505 ,+∞)

Step 3: use Best Linear Unbiased Estimate to carry out Selecting parameter.

Determine the position of order statistic and Interval Statistic under different b value, as shown in table 2.

The position of table 2 order statistic and Interval Statistic

According to the analysis result in table 2, calculating is divided into 4 groups by us, and result of calculation is as table 3.

Table 3 result of calculation

Step 4: the point estimation and the interval estimation that calculate logarithm Q-percentile life, result of calculation is as table 4.

Table 4 fiduciary level is 0.9, and degree of confidence is the interval estimation result of 0.7

	Logarithm Q-percentile life	Q-percentile life
			Point estimation	8.3287	4141.0307d(11.345a)
Confidence upper limit	8.3320	4154.7187d(11.383a)
			Confidence lower limit	8.3253	4126.9751d(11.307a)

Note: d represents sky, a represents year.

So based on all processes above, assess the storage life of product is 13.345 years.

Result shows, adopts the inventive method can realize the life appraisal of non-parallel storing product, and is consistent with actual, reach the object of expection.

In sum, The present invention gives a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone.The method, for Weibull distribution life model, based on the store data different by product and stress accelerated life test data, is discussed to life model parameter, and is used the method for Best Linear Unbiased Estimate, assess the life-span of product.The concrete steps of the method are: first calculate speedup factor and comprehensive out-of-service time according to acceleration model, then speedup factor feasible zone is discussed, Best Linear Unbiased Estimate is used to calculate respectively according to different situations, the estimated value of Confirming model parameter, according to point estimation and the interval estimation of given fiduciary level and confidence calculations logarithm Q-percentile life.The present invention is equally applicable to the lognormal distribution etc. that can be converted into logarithm position-yardstick race, and the life appraisal of Based on Censored Data accelerated life test, has very strong operability.

Claims (2)

1., based on the non-parallel storage life test appraisal procedure that speedup factor feasible zone is selected, suppose as follows:

Suppose that 1 life of product t obeys Weibull distribution, accumulative inefficacy function is

$F\left(t\right)=1-\exp \left({-\left(\frac{t}{\mathrm{\&eta;}}\right)}^m\right)---\left(1\right)$

Wherein, η is characteristics life, and m is form parameter;

Make y=lnt, be then converted into the extreme value distribution, its cumulative failure function is

$F\left(t\right)=1-\exp (-\exp \left(\frac{y-\mathrm{\&mu;}}{\mathrm{\&sigma;}}\right))---\left(2\right)$

Wherein, location parameter μ=ln η, scale parameter σ=1/m;

Suppose that the failure mechanism of all products is the same under each stress level of 2 accelerated life tests, namely the scale parameter σ of life-span distribution is equal;

Suppose that 3 location parameters are the generalized linear functions of stress

Wherein S represents stress, represent with the relevant known function of stress;

Suppose that the remaining life-span of 4 products only depends on the failure probability accumulated, and stress level at that time, accumulation mode have nothing to do, this supposition is called as Nelson hypothesis;

Certain product of known storage certain hour carries out stress accelerated life test, at stress level S _j(j=1,2 ..., p), have n _jindividual sample carries out fixed failure number test, and storage stress is S ₀; Total q _jindividual sample fails, the out-of-service time is respectively period of storage before test is followed successively by truncated time is the period of storage of product of not losing efficacy is followed successively by

It is characterized in that: the concrete implementation step of this appraisal procedure is as follows:

Step one: calculate speedup factor and comprehensive out-of-service time

According to hypothesis 4, we obtain: if product is at stress level S _ilower working time t _icumulative failure probability F _iequal at stress level S _jlower working time t _jcumulative failure probability F _j, then stress level S _jlower working time t _jstress level S can be turned to _ilower working time t _i;

If

F _i(t _i)＝F _j(t _j)

Wherein

Then

And obtain speedup factor thus

K _ijrepresent stress level S _jlower working time t _jbe converted into stress level S _ilower working time t _itime speedup factor;

By period of storage be converted into stress level S _junder equivalent time

$t_{\mathrm{ji}}^{S_0~S_j}=K_{j0}\&CenterDot;t_{\mathrm{ji}}^0$

Calculate the comprehensive out-of-service time

Step 2: speedup factor feasible zone is discussed

Owing to containing unknown parameter b in speedup factor, according to the different values of parameter b, speedup factor can change, and causes the size of comprehensive out-of-service time to change, impacts the calculating of step 3, so need discuss to b, a point situation calculates;

With stress level S ₁for example, total n ₁individual sample carries out fixed time test, wherein q ₁individual sample fails, the out-of-service time is once period of storage is followed successively by truncated time is the period of storage of product of not losing efficacy is respectively the comprehensive out-of-service time is followed successively by

Will compare between two:

If t _1r> t _1s, and r, s=1,2 ..., n ₁and r ≠ s, then T _1r> T _1s;

If t _1r> t _1s, or t _1r< t _1s, then make T _1r=T _1s, obtain critical value b;

At stress level S ₁, obtain one group of critical value

Discuss respectively under each stress level, obtain the critical value of j group b, pressed order arrangement b from small to large ₁, b ₂..., b _f, wherein f=f ₁+ ... + f _j, obtain the interval of f+1 b;

Step 3: use Best Linear Unbiased Estimate method to carry out Selecting parameter

With interval (b ₁, b ₂) be example

I, selection solve for parameter a, the initial value of b, σ wherein

The initial value suggestion of a, σ is selected: the result of calculation using Best Linear Unbiased Estimate when ignoring period of storage;

II, obtain speedup factor K _j0(j=1,2,3 ..., p);

By period of storage be converted into stress level S _junder equivalent time

$t_{\mathrm{ji}}^{S_0~S_j}=K_{j0}\&CenterDot;t_{\mathrm{ji}}^0$

Calculate the comprehensive out-of-service time

$T_{\mathrm{ji}}=t_{\mathrm{ji}}+t_{\mathrm{ji}}^{S_0~S_j}$

III, judge order statistic and Interval Statistic according to the size of comprehensive out-of-service time;

Will obtain according to order arrangement from small to large the i.e. front q of Weibull distribution ₁individual order statistic, obtains q thus ₁discuss in+1 interval; Will obtain according to order arrangement from small to large therefrom choose Interval Statistic;

With interval (Y ₁₁, Y ₁₂) be example, if in have s (s=1 ..., n ₁-q ₁+ 1) individual at interval (Y ₁₁, Y ₁₂) in, then choosing minimum is the 1st Interval Statistic;

To other q ₁individual interval is analyzed respectively, chooses all Interval Statistics, finally obtains order statistic and the Interval Statistic of Weibull distribution;

Make y=lnY, obtain order statistic and the Interval Statistic of the extreme value distribution, order obtain order statistic and the Interval Statistic of standard the extreme value distribution;

Estimates of parameters is obtained by the method for Best Linear Unbiased Estimate

IV, by estimated value with initial value contrast

If

$\left|\frac{{\hat{b}}_1-{\hat{b}}_0}{{\hat{b}}_0}\right|+\left|\frac{{\hat{a}}_1-{\hat{a}}_0}{{\hat{a}}_0}\right|+\left|\frac{{\widehat{\mathrm{\&sigma;}}}_1-{\widehat{\mathrm{\&sigma;}}}_0}{{\widehat{\mathrm{\&sigma;}}}_0}\right|<\mathrm{\&Delta;}$

Wherein Δ is the error amount of regulation,

Then obtain a, the estimated value of b, σ

$\hat{b}={\hat{b}}_1,\hat{a}={\hat{a}}_1,\widehat{\mathrm{\&sigma;}}={\widehat{\mathrm{\&sigma;}}}_1$

Otherwise, order

${\hat{b}}_0={\hat{b}}_1,{\hat{a}}_0={\hat{a}}_1,{\widehat{\mathrm{\&sigma;}}}_0={\widehat{\mathrm{\&sigma;}}}_1$

Proceed to II;

Step 4: the point estimation and the interval estimation that calculate logarithm Q-percentile life

Given fiduciary level R, degree of confidence γ

The point estimation of logarithm Q-percentile life

${\hat{y}}_R=\hat{a}+\hat{b}x+\widehat{\mathrm{\&sigma;}}1n1n\frac{1}{R}---\left(5\right)$

Logarithm Q-percentile life interval estimation

${\hat{y}}_{R_L}=\hat{a}+\hat{b}x+{\mathrm{\&mu;}}_R\widehat{\mathrm{\&sigma;}}-k\widehat{\mathrm{\&sigma;}}---\left(6\right)$

${\hat{y}}_{R_U}=\hat{a}+\hat{b}x+{\mathrm{\&mu;}}_R\widehat{\mathrm{\&sigma;}}-k\widehat{\mathrm{\&sigma;}}---\left(7\right)$

Wherein

$\begin{array}{c}k=u_{\mathrm{\&gamma;}}\frac{u_{\mathrm{\&gamma;}}(c_{13}+c_{23}x+c_{33}{u}_R)-\sqrt{w}}{c_{33}{u}_{\mathrm{\&gamma;}}-1}\\ w={u_{\mathrm{\&gamma;}}}^2{(c_{13}+c_{23}x)}^2+(1-c_{33}{u_{\mathrm{\&gamma;}}}^2)(c_{11}+c_{22}{x}^2+{2c}_{12}x)+c_{33}{u_R}^2+{2c}_{13}{u}_R+{2c}_{23}{u}_{R}x\end{array}.$

2. a kind of non-parallel storage life test appraisal procedure selected based on speedup factor feasible zone according to claim 1, it is characterized in that: " Best Linear Unbiased Estimate method " described in step 3, the concrete practice is as follows:

At stress S _junder, total sample number is n _j, the front q of the extreme value distribution _jindividual order statistic,

$y_{j1}^{*}\&le;\&CenterDot;\&CenterDot;\&CenterDot;\&le;y_{{\mathrm{jm}}_j}^{*}(m_j\&Element;\{1,\&CenterDot;\&CenterDot;\&CenterDot;,q_1+1\left\}\right)$ For the m of this sample _jindividual Interval Statistic;

Order

$Q=\underset{j=1}{\overset{p}{\mathrm{\&Sigma;}}}\underset{k,l=1}{\overset{q_j+m_j}{\mathrm{\&Sigma;}}}(y_{\mathrm{jk}}-a-{\mathrm{bx}}_j-{\mathrm{\&sigma;u}}_{\mathrm{jk}})g_{\mathrm{jkl}}(y_{\mathrm{jl}}-a-{\mathrm{bx}}_j-{\mathrm{\&sigma;u}}_{\mathrm{jl}})---\left(8\right)$

Local derviation is asked namely to obtain the estimated value of parameter a, b, σ;

The covariance matrix of parameter estimation amount is

$\mathrm{cov}(\hat{a},\hat{b},\widehat{\mathrm{\&sigma;}})={\mathrm{\&sigma;}}^{2}C---\left(9\right)$

$C={\left[\begin{array}{ccc}\underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}\\ \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jil}}{x_j}^2& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j{\mathrm{\&mu;}}_{\mathrm{ji}}\\ \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{x}_j{\mathrm{\&mu;}}_{\mathrm{ji}}& \underset{j,i,k}{\mathrm{\&Sigma;}}{g}_{\mathrm{jik}}{\mathrm{\&mu;}}_{\mathrm{ji}}{\mathrm{\&mu;}}_{\mathrm{jk}}\end{array}\right]}^{-1}$

Wherein, $C={\left(g_{\mathrm{jkl}}\right)}_{(q_j+m_j)\&times;(q_j+m_j)}=V^{-1}={\left(v_{\mathrm{jkl}}\right)}_{(q_j+m_j)\&times;(q_j+m_j)},$ U _jk(k=1,2 ..., q _j) be standard the extreme value distribution size of population be n _jthe average of a kth order statistic of sample, v _jkl(k, l=1,2 ..., q _j) for standard the extreme value distribution size of population be n _jthe kth of sample and the covariance of l order statistic; u _jk(k=q _j+ 1 ..., q _j+ m _j) for standard the extreme value distribution size of population be n _jthe average of kth+1 order statistic of the sample of+1, v _jkl(k, l=q _j+ 1, q _j+ m _j, k≤l) for standard the extreme value distribution size of population be n _jthe kth+1 of sample of+2 and the covariance of l+2 order statistic, v _jkl(k, l=q _j+ 1, q _j+ m _j, k > l) for standard the extreme value distribution size of population be n _jthe l+1 of sample of+2 and the covariance of kth+2 order statistics, v _jkl(k=1,2 ..., q _j; L=q _j+ 1, q _j+ m _j) for standard the extreme value distribution size of population be n _jthe kth of sample of+1 and the covariance of l+1 order statistic, v _jkl(l=1,2 ..., q _j; K=q _j+ 1, q _j+ m _j) for standard the extreme value distribution size of population be n _jthe kth+1 of sample of+1 and the covariance of l+1 order statistic, these averages and covariance obtain by formulae discovery and tabling look-up.