Product storage life evaluation method based on composite casual inspection success and failure data

Abstract

The invention provides a product storage life evaluation method based on composite sampling inspection success and failure data. An evaluation object is a complex mechanical and electrical product. Comprising the following steps: carrying out point estimation and confidence limit estimation of failure probability on special test data, and carrying out randomization processing on user use data containing fuzzy samples by adopting uniform distribution or truncated normal distribution; carrying out compatibility test on the special test data and the user use data; fusing the special test data meeting the compatibility test and the user use data by adopting a Bayesian method, and obtaining special stage failure probability estimation after updating of different storage periods; and further screening out degradation failure sample data based on the updated special failure probability estimation, and performing storage life evaluation on the degradation failure data by adopting a weighted least square method. The method combines the characteristics of two kinds of sampling data, and can solve the evaluation problems of small evaluation data volume, fuzzy samples and the like.

Description

Technical Field

[0001] The present invention relates to the technical field of product life assessment, and in particular to a product storage life assessment method based on composite random inspection success / failure data. Background Art

[0002] For the storage life assessment of complex electromechanical products, it is of great practical significance to conduct in-depth research on two types of sampling data. The first type is special test data, which is mainly the data generated by testing each batch of products through strict sampling tests at the special stage when the product leaves the factory or is just put into use. This type of data can reflect the failure information of the product in the special stage. However, since the sampling inspection in the special stage is usually strict and the scope is relatively limited, the amount of data is generally small. This brings certain challenges to the comprehensive and accurate analysis based on special test data. The second type is storage data, which is the data generated by regular sampling inspections of products in different storage stages and during actual use. This type of data can not only reflect the failure information caused by degradation of the product during the storage stage, but also implies the special failure information of the product. By analyzing the storage data, we can better understand the performance changes and failure modes of the product during long-term storage. However, the collection and collation of storage data also faces some problems, such as the possibility of fuzzy samples, that is, samples that are difficult to clearly judge whether they have failed, which brings great difficulties to data analysis and life assessment.

[0003] In view of the respective characteristics and existing problems of these two types of sampling data, in order to accurately evaluate the storage life of the product, it is necessary to combine the advantages of the two types of sampling data and adopt effective methods to solve problems such as small data volume and fuzzy samples. Summary of the invention

[0004] The technical problem to be solved by the present invention is to provide a product storage life evaluation method based on composite random inspection success / failure data in view of the deficiencies in the prior art.

[0005] To solve the above technical problems, the present invention includes:

[0006] A product storage life assessment method based on composite sampling success / failure data comprises the following steps:

[0007] Step S1: point estimation and confidence limit estimation of failure probability are performed on special test data and user usage data, wherein fuzzy sample data is randomized using uniform distribution or truncated normal distribution based on historical experience information;

[0008] Step S2: Perform compatibility check on the special test data and user usage data, and remove a small number of samples that do not meet the compatibility test;

[0009] Step S3: Using the Bayesian method to fuse the special test data and user usage data that meet the compatibility test, obtain the updated special stage failure probability estimates for different storage periods, and further screen out degradation failure sample data based on the updated special failure probability estimates, and then calculate the point estimate of the storage failure probability based on the obtained degradation failure sample data;

[0010] Step S4: Based on the storage failure probability data obtained in step S3, the product storage life is evaluated using the life distribution function, wherein the storage life parameters are estimated using the weighted least squares method with the sample capacity under different storage years as the weight.

[0011] Preferably, in step S1, point estimation and confidence limit estimation of failure probability of special test data are performed according to the value of sample size N to be sampled; when sample size N to be sampled is greater than a set value, binomial distribution is used to describe the number of special test sample failures; when sample size N to be sampled is not greater than a set value, hypergeometric distribution is used to describe the number of special test sample failures.

[0012] Preferably, in step S1, the user usage data is divided into K groups by batches, and the number of samples in each group is N. i , i = 1, 2, ..., K; respectively count the number of failure samples in each group due to quality problems, non-quality problems and fuzzy problems, and obtain the failure probability point estimate of each group of samples based on the randomization assumption and total probability theory

[0013] Preferably, in step S2, the failure probability point estimate of each group of samples is determined by using a statistical test method. Whether it obeys the normal distribution with given mean and variance, and realizes data consistency test.

[0014] Preferably, the data compatibility test includes: (1) normal mean compatibility test, (2) normal variance compatibility test and (3) non-normal compatibility test;

[0015] The (1) normal mean consistency test includes:

[0016] (1-1) When the overall variance of the special test data and user usage data and When the distribution mean of the special test data and the user usage data is known, the Z test is used to determine whether there is a difference, and the null hypothesis H 0 : The two sample means are equal, calculate the Z statistic:

[0017]

[0018] Where: is the mean of the two test samples, n 1 、n 2 is the capacity of the two groups of samples;

[0019] Given a confidence level of 1-α, if |Z|>z α2 , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis;

[0020] (1-2) When the overall variance of the special test data and user usage data and When the values ​​are unknown but equal, use the T test to determine whether the distribution mean values ​​of the special test data and the user usage data are different, and establish the null hypothesis H 0 : The two sample means are equal, calculate the T statistic:

[0021]

[0022] Where: s 1 、s 2 is the standard deviation of the two groups of samples;

[0023] The degrees of freedom are expressed as:

[0024] f=n 1 +n 2 -2

[0025] Given a confidence level of 1-α, if T>t α / 2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis;

[0026] (1-3) When the overall variance of the special test data and user usage data and When they are unknown and not equal, Welch's T test is used to determine whether the distribution mean values ​​of the special test data and the user usage data are different, and the null hypothesis H is established. 0 : The two sample means are equal, calculate Welch's T statistic

[0027]

[0028] The degrees of freedom are expressed as:

[0029]

[0030] Given a confidence level of 1-α, if T>t α2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis;

[0031] (2) Normal variance compatibility test is to use F test to determine whether the population variances of two normal random variables are equal, and to establish the null hypothesis H 0 : The population variances of the two samples are equal, calculate the F statistic:

[0032]

[0033] The degrees of freedom are (n 1 -1,n 2 -1);

[0034] Given a confidence level of 1-α, if Then the null hypothesis is rejected, and the conclusion is that the population variances of the two samples are not equal, otherwise the null hypothesis is not rejected;

[0035] (3) Non-normality compatibility test is to use Kolmogorov-Smirnov test to determine whether the overall distribution of two random variables is consistent in the case of obvious non-normal distribution, and to establish the null hypothesis H 0 : The overall distribution of the two samples is equal, calculate the KS statistic:

[0036] D n,m =sup x |F 1,n (x)-F 2,m (x)|

[0037] Where: D n,m Indicates the maximum distance between the cumulative distribution functions of two samples, F 1,n (x) and F 2,m (x) represents the empirical distribution function of the two variables, n and m are the sizes of the two samples;

[0038] Given a confidence level of 1-α, if If the null hypothesis is rejected, the conclusion is that the population variances of the two samples are not equal; otherwise, the null hypothesis is not rejected.

[0039] Preferably, in step S3, the Bayesian method is used to fuse the special test data and the user usage data, specifically including: (1) selecting the special sampling test data as prior information to obtain the prior distribution of the special stage failure probability p; (2) selecting the user usage data as the observation information, and updating the posterior distribution of the special stage failure probability p based on the Bayesian theory; (3) calculating the special stage reliability of the batch product according to the posterior distribution of the special stage failure probability p.

[0040] Preferably, the fuzzy problem samples include special failure samples and storage failure samples in the storage process. According to the posterior estimation of the failure probability of the special stage obtained, the composition of the special failure samples in the fuzzy problem samples of each batch of products is obtained, and the degradation failure sample data is further screened to obtain Right now:

[0041]

[0042] Where: Indicates rounding down, r 3 The number of samples of fuzzy questions in the user usage data;

[0043] Then, based on the obtained degradation failure sample data Calculate a point estimate of the probability of storage failure

[0044]

[0045] Preferably, in step S4,

[0046] If the storage sampling test includes Z group storage years, the failure data is (t j ,p j ),j=1,2,...,Z;

[0047] The life distribution model adopts a two-parameter Weibull distribution, and its function form is:

[0048]

[0049] Where: m is the shape parameter, η is the scale parameter;

[0050] After linearization, we can get

[0051]

[0052] Let y j = lnt j , x j =lnln(1-p j ) -1 , μ=lnη,σ=m -1 ,but

[0053] y j =μ+σx j +ε j

[0054] Where: μ is the linearized fitting intercept; σ is the linearized fitting slope; ε j is the fitting error.

[0055] The weighted least squares method is used to estimate the parameters of the above formula, and the sample size is taken as the weight:

[0056]

[0057] Where: n jStorage period t j The sample size under ;

[0058] Then, the least squares estimates of parameters μ and σ are:

[0059]

[0060] According to the parameter relationship, the distribution parameter estimate can be obtained Then evaluate the product life.

[0061] The beneficial effects of the present invention are:

[0062] The present invention comprehensively considers special test data and user usage data, can more comprehensively reflect the actual situation of long-storage products, and improve the accuracy and reliability of evaluation. For user usage data containing fuzzy samples, uniform distribution or truncated normal distribution is used for randomization processing, which effectively handles uncertain information and increases the availability of data. Compatibility checks are performed on special test data and user usage data to ensure the rationality of data fusion and avoid erroneous evaluation results due to data incompatibility. The Bayesian method is used to fuse special test data and user usage data, which can make full use of prior information and sample information and improve the accuracy of failure probability estimation. The weighted least squares method is used to evaluate the product storage life, and weights can be performed according to the reliability of different data to further improve the accuracy of the evaluation. BRIEF DESCRIPTION OF THE DRAWINGS

[0063] Figure 1 is a flow chart of the evaluation method of the present invention;

[0064] Figure 2 This is the special failure probability update process for batch A;

[0065] Figure 3 Update the results of the failure probability of Batch B;

[0066] Figure 4 Update the results of the failure probability of Batch C;

[0067] Figure 5 Fitting curve for storage life. DETAILED DESCRIPTION

[0068] To facilitate understanding of the present invention, the present invention is further described in detail below in conjunction with the accompanying drawings and specific embodiments. It should be understood by those skilled in the art that the embodiments are only intended to help understand the present invention and should not be regarded as specific limitations of the present invention.

[0069] like Figure 1As shown, the present invention provides a product storage life evaluation method based on composite sampling success and failure data, the evaluation object is a complex electromechanical product, including the following steps:

[0070] Step S1: perform point estimation and confidence limit estimation of failure probability on the special test data according to the value of the sample size N to be sampled, and perform random processing on the user usage data containing fuzzy samples using uniform distribution or truncated normal distribution based on historical experience information;

[0071] Step S2: According to the random processing result of the user usage data containing fuzzy samples obtained in step S1, the compatibility test of the special test data and the user usage data is performed, and a small number of samples that do not meet the compatibility test are eliminated;

[0072] Step S3: Using the Bayesian method to fuse the special test data and user usage data that meet the compatibility test, obtain the updated special stage failure probability estimates for different storage periods, and further screen out degradation failure sample data based on the updated special failure probability estimates, and then calculate the point estimate of the storage failure probability based on the obtained degradation failure sample data;

[0073] Step S4: Based on the storage failure probability data obtained in step S3, the product storage life is evaluated using the life distribution function, wherein the storage life parameters are estimated using the weighted least squares method with the sample capacity under different storage years as the weight.

[0074] In step S1, the point estimation and confidence limit estimation of the failure probability of the special test data are performed according to the value of the sample size N to be inspected, which specifically includes the following contents:

[0075] The number of samples to be inspected for special test data is N, including the number of failures D; the total number of special inspection samples is n, and the number of failure samples is r 0 , the number of successful samples is nr 0 ;

[0076] (1) When the number of samples to be inspected N is greater than the set value, the number of failed samples X in the special test is described by the binomial distribution, that is,

[0077]

[0078] Then the point estimate of the failure probability p of the special test data is:

[0079]

[0080] In the formula: n is the total number of special inspection samples, and r is the number of failure samples in special inspection.

[0081] Given a confidence level of 1-α, the one-sided lower confidence limit pL Determined by the following formula:

[0082]

[0083] One-sided upper confidence limit p for the parameter U Determined by the following formula:

[0084]

[0085] Two-sided confidence interval for the parameter (p L ,p U ) is determined by the following formula:

[0086]

[0087] (2) When the number of samples to be inspected N is not greater than the set value, the number of failed samples X in the special test is described by the hypergeometric distribution, that is:

[0088]

[0089] Then the point estimate of the failure probability p of the special test data is:

[0090]

[0091] Given a confidence level of 1-α, the one-sided confidence lower limit p of the parameter L Determined by the following formula:

[0092]

[0093] In the formula: the symbol ! represents the factorial function; according to the value of N, n, r and α, p can be obtained from the above formula L value.

[0094] In step S1, the random processing method of invalid samples of user usage data containing fuzzy samples is:

[0095] Assume that the total number of samples in the user usage data is M, and the number of quality problem samples is r 1 , the number of samples without quality problems is r 2 , the number of fuzzy problem samples is r 3 , then the number of successful samples is Mr 1 -r 2 -r 3 ; Consider the number of failure samples r as a random variable, and its distribution is described by uniform distribution or truncated normal distribution, then r 1 ≤r≤r 1 +r 3 .

[0096] If the probability density function of the number of failure samples r is f(r), for continuous uniform distribution, we have:

[0097]

[0098] If the discrete characteristics of the sample are considered, discrete uniform distribution is adopted:

[0099]

[0100] For the truncated normal distribution, we have:

[0101]

[0102] Where: φ(·) represents the standard normal probability density function, and Φ(·) represents the standard normal cumulative distribution function.

[0103] For user usage data, it is divided into K groups by batch, and the number of samples in each group is N i ,i=1,2,…,K. Count the number of failed samples in each group due to quality problems, non-quality problems and fuzzy problems respectively, and obtain the failure probability point estimate of each group of samples based on the randomization assumption and total probability theory.

[0104] The point estimate of the failure probability p for each group of samples is calculated by the following formula:

[0105]

[0106] If the number of failure samples r follows a continuous uniform distribution, then:

[0107]

[0108] If the number of failure samples r follows a discrete uniform distribution, then:

[0109]

[0110] If the number of failures follows a truncated normal distribution, then:

[0111]

[0112] Where: μ is the mean of the parent distribution of the truncated normal distribution, σ is the standard deviation of the parent distribution of the truncated normal distribution.

[0113] In step S2, in order to meet the requirements of data fusion, the compatibility test method of the special test data and the user data is as follows:

[0114] Use statistical test methods to determine the failure probability point estimate of each group of samples Whether it obeys the normal distribution with given mean and variance, and realizes data consistency test.

[0115] The problem of whether the special test data and user usage data are compatible belongs to the problem of whether the two types of data come from the same population. Therefore, the compatibility test can be transformed into a test to determine whether the population distribution of the two types of data is statistically consistent. If the special quality status of each batch of products is consistent, then the failure probability value obtained from the K group of user usage data is It should obey the normal distribution, and its mean and variance can be determined based on the point estimate and interval estimate of the p-value obtained from special sampling tests.

[0116] Based on this, the compatibility problem between special sampling data and user usage data can be transformed into Whether it obeys the normal distribution with given mean and variance, and the statistical test method is used to realize data compatibility test.

[0117] (1) Normal mean consistency test

[0118] (1-1) When the overall variance of the special test data and user usage data and When the distribution mean of the special test data and the user usage data is known, the Z test is used to determine whether there is a difference, and the null hypothesis H 0 : The two sample means are equal, calculate the Z statistic:

[0119]

[0120] Where: is the mean of the two test samples, n 1 、n 2 is the capacity of the two groups of samples;

[0121] Given a confidence level of 1-α, if |Z|>z α2 , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis.

[0122] (1-2) When the overall variance of the special test data and user usage data and When the values ​​are unknown but equal, use the T test to determine whether the distribution mean values ​​of the special test data and the user usage data are different, and establish the null hypothesis H 0 : The two sample means are equal, calculate the T statistic:

[0123]

[0124] Where: s 1 、s 2 is the standard deviation of the two groups of samples;

[0125] The degrees of freedom are expressed as:

[0126] f=n 1+n 2 -2

[0127] Given a confidence level of 1-α, if T>t α / 2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis.

[0128] (1-3) When the overall variance of the special test data and user usage data and When they are unknown and not equal, Welch's T test is used to determine whether the distribution mean values ​​of the special test data and the user usage data are different, and the null hypothesis H is established. 0 : The two sample means are equal, calculate Welch's T statistic:

[0129]

[0130] The degrees of freedom are expressed as:

[0131]

[0132] Given a confidence level of 1-α, if T>t α2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis.

[0133] (2) Normal variance consistency test

[0134] Use the F test to determine whether the population variances of two normal random variables are equal and establish the null hypothesis H 0 : The population variances of the two samples are equal, calculate the F statistic:

[0135]

[0136] The degrees of freedom are (n 1 -1,n 2 -1);

[0137] Given a confidence level of 1-α, if If the null hypothesis is rejected, the conclusion is that the population variances of the two samples are not equal; otherwise, the null hypothesis is not rejected.

[0138] (3) Non-normality consistency test

[0139] For cases with obvious non-normal distribution, the Kolmogorov-Smirnov test is used to determine whether the overall distributions of the two random variables are consistent and to establish the null hypothesis H 0 : The overall distribution of the two samples is equal, calculate the KS statistic:

[0140] D n,m =sup x |F1,n (x)-F 2,m (x)|

[0141] Where: D n,m Indicates the maximum distance between the cumulative distribution functions of two samples, F 1,n (x) and F 2,m (x) represents the empirical distribution function of the two variables, n and m are the sizes of the two samples;

[0142] Given a confidence level of 1-α, if If the null hypothesis is rejected, the conclusion is that the population variances of the two samples are not equal; otherwise, the null hypothesis is not rejected.

[0143] A small number of samples that do not meet the compatibility test are eliminated, and the remaining batches of samples are used for the next step of information fusion.

[0144] In step S3, the Bayesian method is used to fuse the special test data and the user usage data, and the storage sample composition is determined based on the fused failure probability. The specific steps are as follows:

[0145] Estimation of failure probability of special test sampling based on Bayesian fusion:

[0146] (1) Determination of the prior distribution of failure probability: Select the special test data as the prior information and give the prior distribution of the failure probability p (taking the truncated normal distribution as an example):

[0147]

[0148] Where: represents the mean estimate of the failure probability p of the special test data, represents the variance estimate of the failure probability p.

[0149] (2) Update of the Posterior Distribution of Failure Probability

[0150] Select user usage data as observation information and update the posterior distribution of failure probability p based on Bayesian theory:

[0151]

[0152] Where: Θ represents the actual data set used.

[0153] (3) According to the posterior distribution of failure probability p, calculate the special stage reliability of batch products:

[0154]

[0155] Where: E(·) represents the expected function.

[0156] Through the above steps, the special test data and user usage data can be integrated to evaluate the special failure probability of batch products.

[0157] Point estimate of storage failure probability:

[0158] Since the failure samples in the storage process include special failure samples and storage failure samples, and the fuzzy problem samples are actually composed of the above two samples, after obtaining the posterior estimate of the failure probability in the special stage, the composition of special failure samples in the fuzzy problem samples of each batch of products is obtained, and the degradation failure sample data is further screened, that is,

[0159]

[0160] Where: Indicates rounding down.

[0161] Therefore, the point estimate for the probability of storage failure is:

[0162]

[0163] Based on this, the failure information composition of user usage data can be obtained and the product storage life can be further evaluated.

[0164] In step S4, since the sample size of a single random inspection is different, it is generally believed that the larger the sample size of the random inspection batch, the higher the credibility of the estimated failure probability. Therefore, the present invention provides a weighted least squares method based on sample size weights for parameter estimation.

[0165] If the storage sampling test includes Z group storage years, the failure data is (t j ,p j ),j=1,2,...,Z. The life distribution model adopts the two-parameter Weibull distribution, and its function form is:

[0166]

[0167] Where: m is the shape parameter, η is the scale parameter;

[0168] After linearization, we can get

[0169]

[0170] Let y j = lnt j , x j =ln ln(1-p j ) -1 , μ=lnη,σ=m -1 ,but

[0171] y j =μ+σxj +ε j

[0172] Where: μ is the linearized fitting intercept; σ is the linearized fitting slope; ε j is the fitting error.

[0173] The weighted least squares method is used to estimate the parameters of the above formula. For the periodic sampling data, since the sampling sample sizes of different storage years are inconsistent, the estimation errors of failure probability under different storage years are different. It is generally believed that the sample size is inversely proportional to the estimation error of failure probability. Therefore, the weighted least squares method is used for parameter estimation, and the sample size is taken as the weight:

[0174]

[0175] Where: n j Storage period t j The sample size under ;

[0176] Then, the least squares estimates of parameters μ and σ are:

[0177]

[0178] Where:

[0179] According to the parameter relationship, the distribution parameter estimate can be obtained Then the product storage life and storage reliability are estimated.

[0180] Case application:

[0181] Both the special test and storage test of a certain type of product can collect sampling data. The special test records the total number of samples and the number of failed samples, among which the failed samples are basically caused by production quality problems. The storage test records the total number of samples, the number of quality problem samples, the number of storage problem samples, and the number of fuzzy problem samples under each storage period. Both the special test and the storage test include 3 batches, namely batches A, B, and C, as shown in Table 1 and Table 2.

[0182] For the special test data, the number of samples to be inspected in each batch is 300, which is a large number, so the binomial distribution is used for description; for the user usage data, the uniform distribution is used to characterize the randomness of the failure samples. Taking the confidence level 1-α=0.95, the compatibility test of the special test data and the user usage data is carried out, and the results are shown in Table 3.

[0183] Table 1 Special test data

[0184] Sample group number Number of samples for random inspection Number of failed samples Sample group number Number of samples for random inspection Number of failed samples A0-1 10 0 B0-3 10 1 A0-2 10 1 C0-1 10 0 A0-3 10 1 C0-2 10 1 A0-4 10 1 C0-3 10 1 B0-1 10 0 C0-4 10 0 B0-2 10 1

[0185] Table 2 User usage data

[0186] Sample group number Storage life Total number of samples Number of samples with quality problems Number of stored problem samples Number of fuzzy problem samples A1 5 50 3 2 0 A2-1 6 50 2 3 2 A2-2 6 50 5 3 2 A3 7 50 2 5 3 A4 8 50 4 5 0 B1 5 50 3 2 2 B2 6 50 1 2 2 B3 7 50 4 5 1 B4 8 50 2 11 2 C1 5 50 2 4 0 C2 7 50 1 3 0 C3 8 50 3 8 3 C4-1 10 50 1 15 0 C4-2 10 50 3 13 3 C4-3 10 50 4 17 2

[0187] Table 3 Compatibility test results

[0188]

[0189] From the above table, we can see that the special test data and user usage data of batches A, B, and C all meet the compatibility test. Therefore, the two parts of information can be fused based on the Bayes fusion method to obtain a more accurate special failure probability estimate. The update process of the special failure probability of each batch is as follows: Figure 2 , Figure 3 and Figure 4 shown.

[0190] As the batches are updated, the distribution of the failure probability of the special item becomes more concentrated and less dispersed, that is, it has a smaller estimated variance. Based on this, the fuzzy sample composition of each batch and each group of data can be determined, and the storage failure probability under each storage period can be further updated. In contrast, the parameter estimation and storage life evaluation are performed based on the weighted least squares method without considering the fuzzy sample. The fitting results are as follows: Figure 5 The parameter estimation results and the life assessment results when the reliability is 0.80, 0.90 and 0.95 are shown in Table 4. It can be seen that the life assessment result without considering the fuzzy sample is greater than the life assessment result of the fuzzy sample based on Bayes fusion processing of the present invention. Therefore, only using the determined sample for life assessment ignores the fuzzy sample information, which may lead to hidden risks.

[0191] Table 4 Storage evaluation results

[0192]

[0193] Although embodiments of the present invention have been shown and described, it will be appreciated by those skilled in the art that various changes, modifications, substitutions and variations may be made to the embodiments without departing from the principles and spirit of the present invention, and that the scope of the present invention is defined by the appended claims and their equivalents.

Claims

1. A product storage life assessment method based on composite sampling success and failure data, characterized in that: The following steps are involved: Step S1: point estimation and confidence limit estimation of failure probability are performed on special test data and user usage data, wherein fuzzy sample data is randomized using uniform distribution or truncated normal distribution based on historical experience information; Step S2: Perform compatibility check on the special test data and user usage data, and remove a small number of samples that do not meet the compatibility test; Step S3: Using the Bayesian method to fuse the special test data and user usage data that meet the compatibility test, obtain the updated special stage failure probability estimates for different storage periods, and further screen out degradation failure sample data based on the updated special failure probability estimates, and then calculate the point estimate of the storage failure probability based on the obtained degradation failure sample data; Step S4: Based on the storage failure probability data obtained in step S3, the product storage life is evaluated using the life distribution function, wherein the storage life parameters are estimated using the weighted least squares method with the sample capacity under different storage years as the weight.

2. A product storage life assessment method based on composite sampling success / failure data according to claim 1, characterized in that: In step S1, point estimation and confidence limit estimation of failure probability of special test data are performed according to the value of the sample size N to be sampled. When the sample size N to be sampled is greater than the set value, binomial distribution is used to describe the number of special test sample failures. When the sample size N to be sampled is not greater than the set value, hypergeometric distribution is used to describe the number of special test sample failures.

3. A product storage life assessment method based on composite sampling success / failure data according to claim 1, characterized in that: In step S1, the user usage data is divided into K groups by batch, and the number of samples in each group is N. i , i = 1, 2, ..., K; respectively count the number of failure samples in each group due to quality problems, non-quality problems and fuzzy problems, and obtain the failure probability point estimate of each group of samples based on the randomization assumption and total probability theory 4. A product storage life assessment method based on composite sampling success / failure data according to claim 3, characterized in that: In step S2, the failure probability point estimate of each group of samples is determined by using a statistical test method. Whether it obeys the normal distribution with given mean and variance, and realizes data consistency test.

5. A product storage life assessment method based on composite sampling success / failure data according to claim 4, characterized in that: The data consistency test includes: (1) normal mean consistency test, (2) normal variance consistency test and (3) non-normal consistency test; The (1) normal mean consistency test includes: (1-1) When the overall variance of the special test data and user usage data and When the value is known, use the Z test to determine whether the distribution mean of the special test data and the user usage data is different, establish the null hypothesis H0: the two sample means are equal, and calculate the Z statistic: Where: is the mean of the two test samples, n1 and n2 are the capacity of the two groups of samples; Given a confidence level of 1-α, if |Z|>z α2 , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis; (1-2) When the overall variance of the special test data and user usage data and When they are unknown but equal, use the T test to determine whether the distribution mean of the special test data and the user usage data is different, establish the null hypothesis H0: the two sample means are equal, and calculate the T statistic: In the formula: s1, s2 are the standard deviations of the two groups of samples; The degrees of freedom are expressed as: f=n1+n2-2 Given a confidence level of 1-α, if T>t α / 2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis; (1-3) When the overall variance of the special test data and user usage data and When the two samples are unknown and not equal, use Welch's T test to determine whether the distribution mean of the special test data and the user usage data is different, establish the null hypothesis H0: the two sample means are equal, and calculate Welch's T statistic The degrees of freedom are expressed as: Given a confidence level of 1-α, if T>t α2,f , then reject the null hypothesis and conclude that the two sample means are not equal, otherwise do not reject the null hypothesis; (2) Normal variance compatibility test uses the F test to determine whether the population variances of two normal random variables are equal, establishes the null hypothesis H0: the population variances of the two samples are equal, and calculates the F statistic: The degrees of freedom are (n1-1,n2-1); Given a confidence level of 1-α, if Then the null hypothesis is rejected, and the conclusion is that the population variances of the two samples are not equal, otherwise the null hypothesis is not rejected; (3) Non-normal compatibility test is used to determine whether the population distributions of two random variables are consistent in the case of obvious non-normal distribution. The null hypothesis H0 is established: the population distributions of the two samples are equal, and the KS statistic is calculated: D n,m =sup x |F 1,n (x)-F 2,m (x)| Where: D n,m Indicates the maximum distance between the cumulative distribution functions of two samples, F 1,n (x) and F 2,m (x) represents the empirical distribution function of the two variables, n and m are the sizes of the two samples; Given a confidence level of 1-α, if If the null hypothesis is rejected, the conclusion is that the population variances of the two samples are not equal; otherwise, the null hypothesis is not rejected.

6. A product storage life assessment method based on composite sampling success / failure data according to claim 1, characterized in that: In step S3, the Bayesian method is used to fuse the special test data and the user usage data, specifically including: (1) selecting the special sampling test data as prior information to obtain the prior distribution of the special stage failure probability p; (2) selecting the user usage data as the observation information, and updating the posterior distribution of the special stage failure probability p based on the Bayesian theory; (3) calculating the special stage reliability of the batch product according to the posterior distribution of the special stage failure probability p.

7. A product storage life assessment method based on composite sampling success / failure data according to claim 6, characterized in that: The fuzzy problem samples include special failure samples in the storage process and storage failure samples. Based on the posterior estimation of the failure probability of the special stage, the composition of special failure samples in the fuzzy problem samples of each batch of products is obtained, and the degradation failure sample data is further screened to obtain Right now: Where: represents rounding down, r3 is the number of fuzzy question samples in user usage data; Then, based on the obtained degradation failure sample data Calculate a point estimate of the probability of storage failure 8. The method for evaluating product storage life based on composite sampling success / failure data according to claim 1, characterized in that: In the step S4, The storage sampling test includes Z group storage years, and the failure data is (t j ,p j ),j=1,2,...,Z; The life distribution model adopts a two-parameter Weibull distribution, and its function form is: Where: m is the shape parameter, η is the scale parameter; After linearization, we can get Let y j = lnt j , x j = lnln(1 - p j ) -1 , μ = lnη, σ = m -1 , then y j =μ+σx j +e j Where: μ is the linearized fitting intercept; σ is the linearized fitting slope; ε j is the fitting error. The weighted least squares method is used to estimate the parameters of the above formula, and the sample size is taken as the weight: Where: n j Storage period t j The sample size under ; Then, the least squares estimates of parameters μ and σ are: According to the parameter relationship, the distribution parameter estimate can be obtained Then evaluate the product life.

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