CN105893696A - Multi-phase reliability increasing evaluation method of gradual investment of plurality of sets of products - Google Patents

Abstract

The invention provides a multi-phase reliability increasing evaluation method of gradual investment of a plurality of sets of products. The multi-phase reliability increasing evaluation method comprises the following steps: 1: calculating a failure distribution function of a final phase of the first set of product by utilizing reliability increasing testing data of the first set of product; 2: calculating equal initial failure time, inherited from the test of the first set of product, of the second set of product according to the failure distribution function of the final phase of the first set of product; 3: calculating a failure distribution function of a final phase of the second set of product according to the equal initial failure time of the second set of product and reliability increasing testing data of the second set of product; 4: repeating the step 2 and the step 3 and calculating failure distribution functions of final phases from the third set of product to the ith set of product in sequence; and 5: finally, calculating reliability indexes. According to the multi-phase reliability increasing evaluation method, the data obtained by a multi-phase reliability increasing test of the gradual investment of the plurality of sets of products is used, and the aims of directly calculating the reliability related indexes and carrying out reliability evaluation are achieved; and the multi-phase reliability increasing evaluation method is simple to calculate and convenient to use and has popularization and application values.

Description

A kind of multistage reliability evaluation method that multiple stage product gradually puts into

Technical field

The present invention proposes a kind of multistage reliability evaluation method that multiple stage product gradually puts into, integrated use condition Density function and pivot metering method, carry out multiple stage multistage reliability evaluation, and it can effectively solve multiple stage product gradually Input, every product, on the basis of inheriting previous product maturity state, persistently carry out the complicated propagation process of reliability growth Reliability assessment problem, it belongs to and is applicable to the correlative technology fields such as reliability growth data analysis, reliability assessment.

Background technology

Reliability growth technology, as an important component part of Reliability Engineering, has become as raising product reliable Property, save test period, reduce test number (TN) and reduce the effective way of research fund.

Conventional reliability evaluation method has Duane model, AMSAA model etc..Duane model, AMSAA model are joined The explicit physical meaning of number, form are succinct, it is simple to carry out tracking and the assessment of reliability growth process.But these models can only It is applicable to the reliability evaluation of separate unit product, or the reliability evaluation that multiple stage product is thrown in simultaneously.At present, actual Department limits according to developing needs, equipment time, appointed condition and research fund etc., it will usually in the multistage of a product After reliability growth terminates, research and develop a new product, the good technique state after product reliability increases before succession, then Carry on multistage reliability growth；Repeat said process, present the multistage reliability growth that multiple stage product gradually puts into Process.For the reliability growth process of this kind of complexity, at present, still there is no corresponding reliability evaluation method.

To this end, the present invention proposes a kind of multiple stage product gradually puts into multistage reliability evaluation method, scientific and reasonable The reliability level of corresponding product is evaluated on ground.

Summary of the invention

(1) purpose of the present invention: the present invention is directed in Practical Project, the multistage reliability that multiple stage product gradually puts into increases Growth process, integrated use conditional density function and pivot metering method, propose a kind of reliability evaluation method of practicality.This can The test data obtained by property growth test is as follows:

1st fault time of the 1st product is t₁₁, the 2nd fault time is t₁₂..., n-th₁Secondary fault time is2nd product is to dispose on the basis of improving and optimizating lifting in the 1st long term test of product, fault zero , its 1st fault time is t₂₁, the 2nd fault time is t₂₂..., n-th₂Secondary fault time is…；I-th product In long term test, the fault zero of the i-th-1 product and to carry out disposing on the basis of improving and optimizating lifting, its 1st time therefore Downtime is t_i1, the 2nd fault time is t_i2..., n-th_iSecondary fault time isThe multiple stage product proposed gradually puts into Multistage reliability evaluation method, can effectively solve the reliability assessment problem of above-mentioned complicated reliability growth process.

(2) technical scheme:

The multistage reliability evaluation method that the present invention a kind of multiple stage product gradually puts into, implementation step is as follows:

Step one: utilize the reliability growth test data of the 1st product: t₁₁, t₁₂...,Calculate the 1st product The failure distribution function of terminal stage

Its computational methods are as follows:

The likelihood function assuming parameter θ is L (θ), and θ ∈ Θ^*, Θ^*For Parameter Subspace, the distribution density of defined parameters θ Function is as follows:

$f\left(\mathrm{\θ}\right)=\frac{1}{C}L\left(\mathrm{\θ}\right),\mathrm{\θ}\∈{\mathrm{\Θ}}^{*}.---\left(1\right)$

WhereinFor norming constant.

During actual reliability growth, generally according to testing, pinpoint the problems, take measures zero, continue test Process is carried out, and period can reduce product failure, improve the reliability level of product, and therefore, crash rate is monotone decreasing.In It is, if the crash rate of the 1st product is λ_1jGradually decrease:

$\mathrm{\∞}\≥{\mathrm{\λ}}_{11}\≥{\mathrm{\λ}}_{12}\≥...\≥{\mathrm{\λ}}_{1n_1}>0;$

It has been generally acknowledged that dead time interval obeys exponential, be then apparent from pivot amount

2t₁₁λ₁₁～χ²,

So, λ₁₁Distribution density function be

$f_{11}\left({\mathrm{\λ}}_{11}\right)=t_{11}{e}^{-t_{11}{\mathrm{\λ}}_{11}};$

For λ₁₂, pivot amount is 2 (t₁₂-t₁₁)λ₁₂～χ², and constrained: λ₁₂≤λ₁₁, then at given λ₁₁Under conditions of, λ₁₂Conditional Distribution Density Functions be

$f_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})=\frac{(t_{12}-t_{11})e^{-(t_{12}-t_{11}){\mathrm{\λ}}_{12}}}{1-e^{-(t_{12}-t_{11}){\mathrm{\λ}}_{11}}},({\mathrm{\λ}}_{12}\≤{\mathrm{\λ}}_{11}),$

Thus λ can be obtained₁₂Distribution density be

$f_{12}\left({\mathrm{\λ}}_{12}\right)={\∫}_{{\mathrm{\λ}}_{12}}^{+\mathrm{\∞}}{f}_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})f_{11}\left({\mathrm{\λ}}_{11}\right){\mathrm{d\λ}}_{11};$

For λ₁₃, likelihood functionThen at given λ₁₂Under conditions of, λ₁₃Condition distribution close Degree function is

$f_{13}\left({\mathrm{\λ}}_{13}\right|{\mathrm{\λ}}_{12})=\frac{e^{-(t_{13}-t_{12}){\mathrm{\λ}}_{13}}}{C_{12,{\mathrm{\λ}}_{12}}},({\mathrm{\λ}}_{13}\≤{\mathrm{\λ}}_{12}),$

WhereinTherefore, λ₁₃Distribution density be

$f_{13}\left({\mathrm{\λ}}_{13}\right)={\∫}_{{\mathrm{\λ}}_{13}}^{+\mathrm{\∞}}{\∫}_{{\mathrm{\λ}}_{12}}^{+\mathrm{\∞}}{f}_{13}\left({\mathrm{\λ}}_{13}\right|{\mathrm{\λ}}_{12})f_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})f_{11}\left({\mathrm{\λ}}_{11}\right){\mathrm{d\λ}}_{11}{\mathrm{d\λ}}_{12},$

By λ₁₁And λ₁₂Distribution density function substitute into, can obtain

$f({\mathrm{\λ}}_{11},{\mathrm{\λ}}_{12},{\mathrm{\λ}}_{13})=C_{13}{\mathrm{\λ}}_{11}{\mathrm{\λ}}_{12}{e}^{-t_{11}{\mathrm{\λ}}_{11}-(t_{12}-t_{11}){\mathrm{\λ}}_{12}-(t_{13}-t_{12}){\mathrm{\λ}}_{13}}{I}_{\{{\mathrm{\λ}}_{11}\≥{\mathrm{\λ}}_{12}\≥{\mathrm{\λ}}_{13}\}},$

WhereinFor norming constant, λ thus can be obtained₁₃Limit It is distributed as

$F_{13}\left({\mathrm{\λ}}_{13}\right)=C_{13}\{-{\mathrm{\λ}}_{13}^2{e}^{-t_{13}{\mathrm{\λ}}_{13}}-{\mathrm{\λ}}_{13}{e}^{-t_{13}{\mathrm{\λ}}_{13}}(\frac{2}{t_{13}}+\frac{2}{t_{12}}+\frac{1}{t_{11}})\}+1-e^{-t_{13}{\mathrm{\λ}}_{13}}$

In like manner, can obtainMarginal distribution functionThe i.e. invalid cost of the terminal stage of First product FunctionEspecially,

$F_{12}\left({\mathrm{\λ}}_{12}\right)=C_{12}\{-\frac{1}{t_{12}{t}_{11}}{\mathrm{\λ}}_{12}{e}^{-t_{12}{\mathrm{\λ}}_{12}}\}+1-e^{-t_{12}{\mathrm{\λ}}_{12}},$

Wherein:

$C_{12}=\frac{t_{12}^2{t}_{11}^2}{t_{12}+t_{11}};$

$\begin{array}{c}{F}_{14}\left({\mathrm{\λ}}_{14}\right)=1-\exp (-t_{14}{\mathrm{\λ}}_{14})-C_{14}\·\\ \left[\begin{array}{c}{\mathrm{\λ}}_{14}^3\exp (-t_{14}{\mathrm{\λ}}_{14})+(\frac{1}{t_{11}}+\frac{2}{t_{12}}+\frac{3}{t_{13}}+\frac{3}{t_{14}}){\mathrm{\λ}}_{14}^2\exp (-t_{14}{\mathrm{\λ}}_{14})\\ +(\frac{1}{t_{11}{t}_{12}}+\frac{1}{t_{11}{t}_{13}}+\frac{2}{t_{11}{t}_{14}}+\frac{2}{t_{12}^2}+\frac{2}{t_{12}{t}_{13}}+\frac{4}{t_{12}{t}_{14}}+\frac{6}{t_{13}^2}+\frac{6}{t_{13}{t}_{14}}+\frac{6}{t_{14}^2}){\mathrm{\λ}}_{14}\exp (-t_{14}{\mathrm{\λ}}_{14})\end{array}\right]\end{array};$

Wherein:

$\begin{array}{c}{C}_{14}=(\frac{1}{t_{11}{t}_{12}{t}_{13}}+\frac{1}{t_{11}{t}_{12}{t}_{14}}+\frac{1}{t_{11}{t}_{13}{t}_{14}}+\frac{2}{t_{11}{t}_{13}^2}+\frac{2}{t_{11}{t}_{14}^2}+\frac{2}{t_{12}^2{t}_{13}}+\frac{2}{t_{12}^2{t}_{14}}+\frac{2}{t_{12}{t}_{13}{t}_{14}}+\\ \frac{4}{t_{12}{t}_{13}^2}+\frac{4}{t_{12}{t}_{14}^2}+\frac{6}{t_{13}^3}+\frac{6}{t_{13}^2{t}_{14}}+\frac{6}{t_{13}{t}_{14}^2}+\frac{6}{t_{14}^3}{)}^{-1};\end{array}$

Step 2: according to the failure distribution function of the 1st product terminal stageCalculate the 2nd product from 1 product testing inherits the equivalent primary fault time obtained

Its computational methods are as follows:

Owing to the 2nd product is on the basis of the 1st long term test of product, fault zero optimize lifting, carry out portion Administration, therefore, the starting stage of the 2nd product inherits the state of the art of the 1st product terminal stage, i.e. the 2nd product Starting stage failure distribution function F₂₁(λ₂₁) it is the terminal stage failure distribution function of the 1st product

Remember that the 2nd product is in the primary fault timeUnder crash rate density function be

$f_{21}\left({\mathrm{\λ}}_{21}\right)=t_{21}^{\left(2\right)}{e}^{-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{21}},$

Corresponding distribution function is

$F_{21}\left({\mathrm{\λ}}_{21}\right)=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{21}),$

Choose successivelyM value: R₁, R₂..., R_m, then basisIt is calculated AccordinglyThen, utilizeCan obtain

$\left\{\begin{array}{c}{R}_1=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(1\right)})\\ R_2=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(2\right)})\\ .\\ .\\ .\\ R_m=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(m\right)})\end{array}\right.$

The then primary fault time of the 2nd productAvailable least-squares estimation draws

${\hat{t}}_{21}^{\left(2\right)}=\frac{{\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}\ln \left(\frac{1}{1-R_i}\right)-\frac{1}{m}{\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}{\Σ}_{i=1}^m\ln \left(\frac{1}{1-R_i}\right)}{{\Σ}_{i=1}^m{\left({\mathrm{\λ}}_{1n_1}^{\left(i\right)}\right)}^2-\frac{1}{m}{\left({\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}\right)}^2};$

Step 3: utilize the equivalent primary fault time of the 2nd productReliability growth test with the 2nd product Data, calculate the failure distribution function of the 2nd product terminal stage

Its computational methods are as follows:

The equivalent primary fault time according to the 2nd productWith the 2nd fault time that product is drawn: t₂₁, t₂₂...,Can obtain the equivalent reliability growth test data of the 2nd product:

$t_{21}^{\left(2\right)},t_{22}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{21},...,t_{2n_2}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{2n_2},t_{2n_2+1}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{2n_2},$

Then, the method using step one, calculate the failure distribution function of the 2nd product terminal stage

Step 4: repeat step 2 and three, calculates the 3rd product successively to the invalid cost of i-th product terminal stage Function

Step 5: final reliability index calculates；

Utilize the terminal stage failure distribution function of i-th product obtainedOrderCan obtainThe confidence upper limit that confidence level is γOrderProduct failure rate can be obtainedPoint estimationLife of productPoint EstimateFurther, according to above formula, for being the typical mission of t (unit: hour) task time, its task ReliabilityPoint estimationAnd under confidence level γ, its Task Reliability confidence Lower limit

By above step, the data that the multistage reliability growth test that multiple stage product can be used gradually to put into obtains, Carry out reliability assessment, reached can directly calculate reliability index of correlation, carried out the purpose of reliability assessment, solve test The complicated reliability growth process that product is few, the stage is few is difficult to carry out reliability assessment with existing reliability evaluation model Problem, it is ensured that crash rate monotone decline during reliability growth, meets engineering practice, calculates simple, side Just engineers and technicians use, and have stronger using value.

(3) advantage:

The present invention proposes a kind of multistage reliability evaluation method that multiple stage product gradually puts into, and its advantage is such as Under:

1. the present invention proposes a kind of multistage reliability evaluation method that multiple stage product gradually puts into, and effectively solves The reliability assessment problem of the above-mentioned complicated reliability growth process often occurred in reality.

Method the most proposed by the invention calculates simplicity, easily realizes, facilitates engineers and technicians to use, therefore has good Good using value.

Accompanying drawing explanation

Fig. 1 is the method for the invention flow chart.

In figure, symbol, code name are described as follows:

t_ij: the jth of i-th product that achieved reliability growth test obtains time fault time；

n_i: the number of faults occurred in the reliability growth test of i-th product that achieved reliability growth test obtains；

The primary fault time after equivalence is carried out with the i-th-1 product；

With the jth time fault time that the i-th-1 product carries out i-th product after equivalence；

Seek invalid cost method: propose in step one asks failure distribution function theoretical；

Equivalent method: step 2 proposes utilize previous product reliability to increase after premium properties and next The method of product equivalence.

Detailed description of the invention

The multistage reliability evaluation method that the present invention a kind of multiple stage product gradually puts into, its flow chart such as Fig. 1 institute Show.

Gradually put into reliability growth test data instance with certain type ground system multiple stage product, the present invention is done further Describe in detail.

Certain type ground system is carried out corresponding multiple stage product gradually throw according to test, fault, the strategy of test of making zero, continue Enter reliability test, obtain reliability test data as follows:

First ground system is tested 1120 hours altogether, breaks down altogether 2 times, this ground system when each fault occurs Accumulation test period is followed successively by: 180,285,1120.

Second ground system is to dispose under first ground system carries out the state of the art after fault effectively makes zero. Test 476 hours, break down 1 time, for hardware fault altogether.When fault occurs, the accumulation test period of this system is 95 hours.

The multistage reliability evaluation method that the present invention a kind of multiple stage product gradually puts into, as it is shown in figure 1, it is implemented Step is as follows:

Step one: the fault data situation of first ground system is as follows:

0 ＜ 180 ＜ 285 ＜ 1120 (test dwell time T),

Then understand t₁₁=180, t₁₂=285, t₁₃=1120, above-mentioned data are substituted into:

$F_{13}\left({\mathrm{\λ}}_{13}\right)=C_{13}\{-{\mathrm{\λ}}_{13}^2{e}^{-t_{13}{\mathrm{\λ}}_{13}}-{\mathrm{\λ}}_{13}{e}^{-t_{13}{\mathrm{\λ}}_{13}}(\frac{2}{t_{13}}+\frac{2}{t_{12}}+\frac{1}{t_{11}})\}+1-e^{-t_{13}{\mathrm{\λ}}_{13}},$

WhereinIt is calculated: C₁₃=17563.4236, it may be assumed that

$F_{13}\left({\mathrm{\λ}}_{13}\right)=17563.4236\{-{\mathrm{\λ}}_{13}^2{e}^{-1120\mathrm{\λ}}-{\mathrm{\λ}}_{13}{e}^{-1120{\mathrm{\λ}}_{13}}(\frac{2}{1120}+\frac{2}{285}+\frac{1}{180})\}+1-e^{-1120{\mathrm{\λ}}_{13}};$

Step 2: owing to second ground system configures deployment on the basis of first ground system, it is assumed that the 2nd Platform product is in the primary fault timeUnder crash rate density function beCorresponding distribution function For:

$F_{21}\left({\mathrm{\λ}}_{21}\right)=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{21}),$

Choose F the most successively₁₃(λ₁₃) value { 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9} (can be according to essence Spend oneself select institute value number), utilize

$\left\{\begin{array}{c}{F}_{13}\left({\mathrm{\λ}}_{13}^{\left(1\right)}\right)=0.1\\ F_{13}\left({\mathrm{\λ}}_{13}^{\left(2\right)}\right)=0.2\\ .\\ .\\ .\\ F_{13}\left({\mathrm{\λ}}_{13}^{\left(9\right)}\right)=0.9\end{array}\right.$

Obtain Recycling

$\left\{\begin{array}{c}0.1=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(1\right)})\\ 0.2=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(2\right)})\\ .\\ .\\ .\\ 0.9=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(9\right)})\end{array}\right.\⇒\left\{\begin{array}{c}\ln \left(\frac{1}{1-0.1}\right)=t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(1\right)}\\ \ln \left(\frac{1}{1-0.2}\right)=t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(2\right)}\\ .\\ .\\ .\\ \ln \left(\frac{1}{1-0.3}\right)=t_{21}^{\left(2\right)}{\mathrm{\λ}}_{13}^{\left(9\right)}\end{array}\right.$

Can be obtained by least-squares estimation:

Therefore, the equivalence of second ground system obtained according to the reliability growth data of first ground system is initial Fault timeHour；

Step 3: utilize the equivalent primary fault time of the 2nd ground systemReliability with the 2nd ground system Growth test data, seek the failure distribution function of the 2nd ground system terminal stage

The equivalent primary fault time according to the 2nd ground systemDuring with the 2nd fault that ground system is drawn Between: t₂₁, t₂₂...,Can obtain the equivalent reliability growth test data of the 2nd ground system:

WillSubstitute into

$C_{23}={(\frac{2}{t_{23}^{{\left(2\right)}^2}}+\frac{2}{t_{23}^{\left(2\right)}{t}_{22}^{\left(2\right)}}+\frac{1}{t_{23}^{\left(2\right)}{t}_{21}^{\left(2\right)}}+\frac{2}{t_{22}^{{\left(2\right)}^2}}+\frac{1}{t_{22}^{\left(2\right)}{t}_{21}^{\left(2\right)}})}^{-1},$

Obtain: C₂₃=15468,

$F_{23}\left({\mathrm{\λ}}_{23}\right)=15468\{-{\mathrm{\λ}}_{23}^2{e}^{-1375.72{\mathrm{\λ}}_{23}}-{\mathrm{\λ}}_{23}{e}^{-1375.72{\mathrm{\λ}}_{23}}(\frac{2}{1375.72}+\frac{2}{994.72}+\frac{1}{994.72})\}+1-e^{-1375.72{\mathrm{\λ}}_{23}};$

Step 4: according to above formula, makes F₂₃(λ₂₃)=0.8, obtains in confidence level γ=0.8 time, the individually mistake of plane system Efficiency confidence upper limitLife-span confidence lower limitHour.Make F₂₃(λ₂₃)=0.5, can Obtain the crash rate point estimation of ground systemThe point estimation in life-spanHour.

For being task time the typical mission of 10 hours, the point estimation of its Task ReliabilityPutting Reliability γ=0.8 time, its Task Reliability confidence lower limit

In sum, The present invention gives a kind of multistage reliability evaluation method that multiple stage product gradually puts into. The method, after obtaining test data, is first according to test products order, calculates the invalid cost of the 1st product terminal stage Function；2nd product final state based on the 1st product is disposed, and utilizes its equivalent nature, calculates the 2nd product The equivalent primary fault time；The reliability growth data that test obtains are converted into equivalent fault time data, utilize step The method of one draws the failure distribution function of second product terminal stage；Gained failure distribution function is finally utilized to calculate respectively The point estimation of the crash rate of platform product, confidence upper limit, the reliability index such as the point estimation in life-span, confidence lower limit, complete reliability Growth Evaluation works.

The method cannot directly utilize existing model and carries out reliability evaluation for solving above-mentioned data cases Difficulty, and ensure that crash rate monotone decreasing characteristic during reliability growth, calculate simplicity, easily realize, convenient Engineers and technicians use, and have good using value.

Claims (4)

1. the multistage reliability evaluation method that a multiple stage product gradually puts into, it is characterised in that: implementation step is such as Under:

Step one: utilize the reliability growth test data of the 1st product:Calculate the 1st product The failure distribution function in whole stage

Step 2: according to the failure distribution function of the 1st product terminal stageCalculate the 2nd product from the 1st product The equivalent primary fault time obtained is inherited in product test

Step 3: utilize the equivalent primary fault time of the 2nd productWith the reliability growth test data of the 2nd product, Calculate the failure distribution function of the 2nd product terminal stage

Step 4: repeat step 2 and three, calculates the 3rd product successively to the failure distribution function of i-th product terminal stage

Step 5: final reliability index calculates；

Utilize the terminal stage failure distribution function of i-th product obtainedOrder Can obtainThe confidence upper limit that confidence level is γOrderProduct failure rate can be obtainedPoint estimationLife of productPoint estimationFurther, according to above formula, for task Time is the typical mission of t (unit: hour), its Task ReliabilityPoint estimation And under confidence level γ, its Task Reliability confidence lower limit

By above step, the data that the multistage reliability growth test using multiple stage product gradually to put into obtains, carrying out can Assess by property, reached can directly calculate reliability index of correlation, carried out the purpose of reliability assessment, solve test products Less, the complicated reliability growth process that the stage is few is difficult to carry out asking of reliability assessment with existing reliability evaluation model Topic, it is ensured that crash rate monotone decline during reliability growth, meets practical implementation situation.

The multistage reliability evaluation method that a kind of multiple stage product the most according to claim 1 gradually puts into, it is special Levy and be: described in step one " utilize the reliability growth test data of the 1st product: Calculate the failure distribution function of the 1st product terminal stage", its computational methods are as follows:

The likelihood function assuming parameter θ is L (θ), and θ ∈ Θ^*, Θ^*For Parameter Subspace, the distribution density function of defined parameters θ As follows:

$f\left(\mathrm{\θ}\right)=\frac{1}{C}L\left(\mathrm{\θ}\right),\mathrm{\θ}\∈\mathrm{\Θ}*.---\left(1\right)$

WhereinFor norming constant；

During actual reliability growth, generally according to testing, pinpoint the problems, take measures zero, continue the process of test Carrying out, period can reduce product failure, improve the reliability level of product, and therefore, crash rate is monotone decreasing；Then, if The crash rate of the 1st product is λ_1jGradually decrease:

$\mathrm{\∞}\≥{\mathrm{\λ}}_{11}\≥{\mathrm{\λ}}_{12}\≥...\≥{\mathrm{\λ}}_{1n_1}>0;$

It has been generally acknowledged that dead time interval obeys exponential, be then apparent from pivot amount

2t₁₁λ₁₁～χ²,

So, λ₁₁Distribution density function be

$f_{11}\left({\mathrm{\λ}}_{11}\right)=t_{11}{e}^{-t_{11}{\mathrm{\λ}}_{11}};$

For λ₁₂, pivot amount is 2 (t₁₂-t₁₁)λ₁₂～χ², and constrained: λ₁₂≤λ₁₁, then at given λ₁₁Under conditions of, λ₁₂'s Conditional Distribution Density Functions is

$f_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})=\frac{(t_2-t_{11})e^{-(t_{12}-t_{11}){\mathrm{\λ}}_{12}}}{1-e^{-(t_{12}-t_{11}){\mathrm{\λ}}_{11}}}({\mathrm{\λ}}_{12}\≤{\mathrm{\λ}}_{11}),$

Thus obtain λ₁₂Distribution density be

$f_{12}\left({\mathrm{\λ}}_{12}\right)={\∫}_{{\mathrm{\λ}}_{12}}^{+\mathrm{\∞}}{f}_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})f_{11}\left({\mathrm{\λ}}_{11}\right){\mathrm{d\λ}}_{11};$

For λ₁₃, likelihood functionThen at given λ₁₂Under conditions of, λ₁₃Condition distribution density letter Number is

$f_{13}\left({\mathrm{\λ}}_{13}\right|{\mathrm{\λ}}_{12})=\frac{e^{-(t_{13}-t_{12}){\mathrm{\λ}}_{13}}}{C_{12,{\mathrm{\λ}}_{12}}}({\mathrm{\λ}}_{13}\≤{\mathrm{\λ}}_{12}),$

WhereinTherefore, λ₁₃Distribution density be

$f_{13}\left({\mathrm{\λ}}_{13}\right)={\∫}_{{\mathrm{\λ}}_{13}}^{+\mathrm{\∞}}{\∫}_{{\mathrm{\λ}}_{12}}^{+\mathrm{\∞}}{f}_{13}\left({\mathrm{\λ}}_{13}\right|{\mathrm{\λ}}_{12})f_{12}\left({\mathrm{\λ}}_{12}\right|{\mathrm{\λ}}_{11})f_{11}\left({\mathrm{\λ}}_{11}\right){\mathrm{d\λ}}_{11}{\mathrm{d\λ}}_{12},$

By λ₁₁And λ₁₂Distribution density function substitute into,

$f({\mathrm{\λ}}_{11},{\mathrm{\λ}}_{12},{\mathrm{\λ}}_{13})=C_{13}{\mathrm{\λ}}_{11}{\mathrm{\λ}}_{12}{e}^{-t_{11}{\mathrm{\λ}}_{11}-(t_{12}-t_{11}){\mathrm{\λ}}_{12}-(t_{13}-t_{12}){\mathrm{\λ}}_{13}}{I}_{\{{\mathrm{\λ}}_{11}\≥{\mathrm{\λ}}_{12}\≥{\mathrm{\λ}}_{13}\}},$

WhereinFor norming constant, thus obtain λ₁₃Limit be distributed as

$F_{13}\left({\mathrm{\λ}}_{13}\right)=C_{13}\{-{\mathrm{\λ}}_{13}^2{e}^{-t_{13}{\mathrm{\λ}}_{13}}-{\mathrm{\λ}}_{13}{e}^{-t_{13}{\mathrm{\λ}}_{13}}(\frac{2}{t_{13}}+\frac{2}{t_{12}}+\frac{1}{t_{11}})\}+1-e^{-t_{13}{\mathrm{\λ}}_{13}}$

In like manner, can obtainMarginal distribution functionThe i.e. failure distribution function of the terminal stage of First productEspecially,

$F_{12}\left({\mathrm{\λ}}_{12}\right)=C_{12}\{-\frac{1}{t_{12}{t}_{11}}{\mathrm{\λ}}_{12}{e}^{-t_{12}{\mathrm{\λ}}_{12}}\}+1-e^{-t_{12}{\mathrm{\λ}}_{12}},$

Wherein:

$C_{12}=\frac{t_{12}^2{t}_{11}^2}{t_{12}+t_{11}};$

$\begin{array}{c}{F}_{14}\left({\mathrm{\λ}}_{14}\right)=1-\exp (-t_{14}{\mathrm{\λ}}_{14})-C_{14}\·\\ \left[\begin{array}{c}{\mathrm{\λ}}_{14}^3\exp (-t_{14}{\mathrm{\λ}}_{14})+(\frac{1}{t_{11}}+\frac{2}{t_{12}}+\frac{3}{t_{13}}+\frac{3}{t_{14}}){\mathrm{\λ}}_{14}^2\exp (-t_{14}{\mathrm{\λ}}_{14})\\ +(\frac{1}{t_{11}{t}_{12}}+\frac{1}{t_{11}{t}_{13}}+\frac{2}{t_{11}{t}_{14}}+\frac{2}{t_{12}^2}+\frac{2}{t_{12}{t}_{13}}+\frac{4}{t_{12}{t}_{14}}+\frac{6}{t_{13}^2}+\frac{6}{t_{13}{t}_{14}}+\frac{6}{t_{14}^2}){\mathrm{\λ}}_{14}\exp (-t_{14}{\mathrm{\λ}}_{14})\end{array}\right]\end{array}$

Wherein:

$\begin{array}{c}{C}_{14}=(\frac{1}{t_{11}{t}_{12}{t}_{13}}+\frac{1}{t_{11}{t}_{12}{t}_{14}}+\frac{1}{t_{11}{t}_{13}{t}_{14}}+\frac{2}{t_{11}{t}_{13}^2}+\frac{2}{t_{11}{t}_{14}^2}+\frac{2}{t_{12}^2{t}_{13}}+\frac{2}{t_{12}^2{t}_{13}}+\frac{2}{t_{12}{t}_{13}{t}_{14}}+\\ 4t12t132+4t12t142+6t133+6t132t14+6t13t142+6t143)\\ -1\end{array}.$

The multistage reliability evaluation method that a kind of multiple stage product the most according to claim 1 gradually puts into, it is special Levy and be: " the failure distribution function according to the 1st product terminal stage described in step 2Calculate the 2nd Platform product inherits the equivalent primary fault time obtained from the 1st product testing", its computational methods are as follows:

Owing to the 2nd product is on the basis of the 1st long term test of product, fault zero optimize and promote, carry out disposing, Therefore, the starting stage of the 2nd product inherits the state of the art of the 1st product terminal stage, the i.e. initial rank of the 2nd product Segment fault distribution function F₂₁(λ₂₁) it is the terminal stage failure distribution function of the 1st product

Remember that the 2nd product is in the primary fault timeUnder crash rate density function be

$f_{21}\left({\mathrm{\λ}}_{21}\right)=t_{21}^{\left(2\right)}{e}^{-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{21}},$

Corresponding distribution function is

$F_{21}\left({\mathrm{\λ}}_{21}\right)=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{21}),$

Choose successivelyM value: R₁, R₂..., R_m, then basisIt is calculated corresponding 'sThen, utilize?

$\left\{\begin{array}{c}{R}_1=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(1\right)})\\ R_2=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(2\right)})\\ .\\ .\\ .\\ R_m=1-\exp (-t_{21}^{\left(2\right)}{\mathrm{\λ}}_{1n_1}^{\left(m\right)})\end{array}\right.$

The then primary fault time of the 2nd productCan draw with least-squares estimation

${\hat{t}}_{21}^{\left(2\right)}=\frac{{\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}\ln \left(\frac{1}{1-R_i}\right)-\frac{1}{m}{\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}{\Σ}_{i=1}^m\ln \left(\frac{1}{1-R_i}\right)}{{\Σ}_{i=1}^m{\left({\mathrm{\λ}}_{1n_1}^{\left(i\right)}\right)}^2-\frac{1}{m}{\left({\Σ}_{i=1}^m{\mathrm{\λ}}_{1n_1}^{\left(i\right)}\right)}^2}.$

The multistage reliability evaluation method that a kind of multiple stage product the most according to claim 1 gradually puts into, it is special Levy and be: described in step 3, " utilize the equivalent primary fault time of the 2nd productReliability with the 2nd product Growth test data, calculate the failure distribution function of the 2nd product terminal stageIts calculating side Method is as follows:

The equivalent primary fault time according to the 2nd productWith the 2nd fault time that product is drawn: Can obtain the equivalent reliability growth test data of the 2nd product:

$t_{21}^{\left(2\right)},t_{22}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{21},\mathrm{...},t_{2n_2}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{2n_2},t_{2,n_2+1}^{\left(2\right)}=t_{21}^{\left(2\right)}+t_{2n_2},$

Then, the method using step one, calculate the failure distribution function of the 2nd product terminal stage