Method for carrying out life estimation according to product life distribution characteristics

Abstract

The invention discloses a method for carrying out life estimation according to product life distribution characteristics, and the method comprises the steps: constructing a ternary polynomial calculation model of correction coefficients K1, K2 and K3 about n, R and m based on historical data, and n is the number of tested products; r is a reliability requirement; m is a shape parameter of the product obeying Weibull distribution; and evaluating the service life of the product according to the following formula based on the constructed ternary polynomial calculation model: carrying out a service life test on n tested products at T, and if no fault occurs, determining the service life estimation value T1 of the product as T1 = E (t) / K1 (n, R, m), and if no fault occurs, determining the service life estimation value T1 of the product as T1 = E (t) / K1 (n, R, m); if r faults occur, 0 lt; rlt; if n is greater than or equal to E (t), the estimated life value T2 of the product is T2 = E (t) / K2 (n, R, m); if n is all failed, the life estimation value T3 of the product is T3 = E (t) / K3 (n, R, m); e (t) is the average value of the experiment time when the n tested products do not fail; k1 (n, R, m), K2 (n, R, m) and K3 (n, R, m) are ternary polynomial calculation models. According to the method, the product life can be quickly estimated, and the efficiency, stability and accuracy of product life estimation are improved.

Description

Technical Field

[0001] The present invention relates to the technical field of equipment life assessment, and particularly to a method for carrying out life estimation according to the characteristics of product life distribution. Background Art

[0002] Life is one of the important indicators of equipment. When verifying the life index of a product, due to the high cost of test pieces, long test cycles, and high test costs, usually only a small number of test pieces are selected to carry out life tests. Therefore, it is difficult to accurately estimate the product life. Currently, usually experts negotiate a correction coefficient and combine it with the product life experiment data to evaluate the product life. This method is time-consuming and laborious, with relatively low efficiency, and sometimes the given correction coefficient may be unstable or inaccurate.

[0003] In summary, when there is only a small amount of life data of the tested products, it is difficult to estimate their life. Usually, experts negotiate a correction coefficient based on experience and then give the product life estimate value; at the same time, the acquisition of the correction coefficient is relatively cumbersome and the results will vary due to people's subjective judgment. Summary of the Invention

[0004] The purpose of the present invention is to provide a method for carrying out life estimation according to the characteristics of product life distribution. The present invention can quickly estimate the product life and improve the efficiency, stability, and accuracy of product life estimation.

[0005] Technical Solution. A method for carrying out life estimation according to the characteristics of product life distribution, based on historical data, constructs a ternary polynomial calculation model of correction coefficients K 1 、K 2 、K 3 with respect to n, R, and m, where n is the number of tested products; R is the reliability requirement; m is the shape parameter of the Weibull distribution of the product; and the product life is evaluated according to the constructed ternary polynomial calculation model by the following formula:

[0006] For the life test of n tested products at time T, if no failures occur, the product life estimate value T 1 is:

[0007] T 1 = E(t) / K 1 (n, R, m) (1)

[0008] If r failures occur, 0 < r < n, then the product life estimate value T 2 is:

[0009] T 2 = E(t) / K 2 (n, R, m) (2)

[0010] If all n products fail, the estimated life of the product is T 3 for:

[0011] T 3 = E(t) / K 3 (n, R, m) (3)

[0012] E(t) is the average of the experimental time during which n tested products did not fail; K 1 (n,R,m),K 2 (n,R,m),K 3 (n,R,m) is a three-variable polynomial calculation model.

[0013] In the above-mentioned method of estimating life span based on the characteristics of product life span distribution,

[0014] K 1 (n,R,m)=-3.19-1.17n+22.50R+0.14m+1.35nR+0.33nm+0.58Rm-35.26R 2 -0.25

[0015] m 2 +0.29nRm-1.76nR 2 -4.53mR 2 -0.07nm 2 +0.68Rm 2 +27.14R 3 (4).

[0016] In the above-mentioned method of estimating life span based on the characteristics of product life span distribution,

[0017] K 2 (n,R,m)=-0.60-0.92n+12.89R-0.20m+0.78nR+0.35nm+1.53Rm-23.70R 2 -0.29

[0018] m 2 +0.14nRm-0.93nR 2 -5.30mR 2 -0.06nm 2 +0.75Rm 2 +22.38R 3 (5).

[0019] In the above-mentioned method of estimating life span based on the characteristics of product life span distribution,

[0020] K 3(n,R,m)=-7.10-1.89n+52.36R-1.97m+4.66nR+0.07nm+8.29Rm-99.54R 2 -0.41

[0021] m 2 -0.02nRm-3.09nR 2 -9.70mR 2 -0.01nm 2 +0.74Rm 2 +65.71R 3 (6).

[0022] In the aforementioned method of estimating the life span based on the characteristics of product life distribution, n=2-3; R=0.6-0.9; and m is calculated through the historical data of product life span.

[0023] In the aforementioned method of estimating life span based on the characteristics of product life distribution, when there is a lack of historical life span data, m defaults to 2.

[0024] In the above-mentioned method of estimating life according to the characteristics of product life distribution, in the process of constructing the ternary polynomial calculation model, when the highest index of n is 1, the highest index of m is 2, and the highest index of R is 3, the goodness of fit is 0.9982, 0.9976, and 0.9725 respectively.

[0025] The above-mentioned method of estimating the life span based on the characteristics of product life distribution is applied to the life span estimation of mechanical, electromechanical and other products.

[0026] Beneficial effects:

[0027] 1. Improved stability of life estimation. The life estimation value given by the present invention is stable. Compared with the expert evaluation method, the estimation result is not differentiated due to human subjective judgment.

[0028] 2. Improved efficiency of life estimation. It is possible to give an estimated value of product life without organizing experts to negotiate and determine parameters, thus improving the efficiency of life estimation.

[0029] 3. The accuracy of product life estimation is high. Compared with the life estimation value of the expert evaluation method, the error of the present invention is within 5%, and the results are close.

[0030] 4. To verify whether the product lifespan meets the expected lifespan, the present invention can help quickly determine when a lifespan test is required. BRIEF DESCRIPTION OF THE DRAWINGS

[0031] Figure 1 This is a graph showing the variation of the estimated lifespan with the number of tested products n;

[0032] Figure 2 It is a graph showing the change of the life estimation value with the reliability R;

[0033] Figure 3 It is a graph showing the change of the life estimation value with the shape parameter m;

[0034] Figure 4 It is a comparison of the life estimation value of the method of the present invention with the expert evaluation method. Specific implementation mode

[0035] Example 1. Currently, there are the following methods for evaluating the life of a product:

[0036] Perform a life test on n test products for time T. If no failures occur, the life estimation value T of the product 1 is:

[0037] T 1 = E(t) / K 1 (1)

[0038] If r failures occur (0 < r < n), the life estimation value T of the product 2 is:

[0039] T 2 = E(t) / K 2 (2)

[0040] If all n have failed, the life estimation value T of the product 3 is:

[0041] T 3 = E(t) / K 3 (3)

[0042] E(t) is the average value of the experimental time when no failures occur for n test products; K 1 、K 2 、K 3 are empirical correction coefficients.

[0043] Among them, the empirical correction coefficients are determined through consultation by organizing experts. This work is time-consuming and laborious, with low efficiency, and sometimes there may be inconsistent or inaccurate correction coefficients.

[0044] The correction coefficients K 1 、K 2 、K 3 are mainly determined by the main factors such as the number of test products and the reliability requirements. Sort out the correction coefficients K under different main factors in the historical data 1 、K 2 、K 3The values ​​are shown in Tables 1 to 6. In the table, n is the number of tested products; R is the reliability requirement; and m is the shape parameter of the product that follows the Weibull distribution.

[0045] Table 1 Correction coefficient K 1 (n=2)

[0046]

[0047] Table 2 Correction coefficient K 1 (n=3)

[0048]

[0049] Table 3 Correction coefficient K 2 (n=2)

[0050]

[0051] Table 4 Correction coefficient K 2 (n=3)

[0052]

[0053] Table 5 Correction coefficient K 3 (n=2)

[0054]

[0055] Table 6 Correction coefficient K 3 (n=3)

[0056]

[0057] The method of estimating life span of the present invention:

[0058] In order to determine the quantitative relationship between the correction coefficient and its main influencing factors, the correction coefficient K is constructed using historical data. 1 , K 2 , K 3 The three-variable polynomial calculation model of n, R, and m. When the highest index of n is 1, the highest index of m is 2, and the highest index of R is 3, the fitting effect is better, and the goodness of fit is 0.9982, 0.9976, and 0.9725 respectively, and the correction coefficient K is obtained. 1 , K 2 , K 3 The calculation expressions of are shown in equations (4) to (6).

[0059] K 1 (n,R,m)=-3.19-1.17n+22.50R+0.14m+1.35nR+0.33nm+0.58Rm-35.26R 2 -0.25

[0060] m 2 +0.29nRm-1.76nR 2 -4.53mR 2 -0.07nm 2 +0.68Rm 2 +27.14R 3 (4)

[0061] K 2 (n,R,m)=-0.60-0.92n+12.89R-0.20m+0.78nR+0.35nm+1.53Rm-23.70R 2 -0.29

[0062] m 2 +0.14nRm-0.93nR 2 -5.30mR 2 -0.06nm 2 +0.75Rm 2 +22.38R 3 (5)

[0063] K 3 (n,R,m)=-7.10-1.89n+52.36R-1.97m+4.66nR+0.07nm+8.29Rm-99.54R 2 -0.41

[0064] m 2 -0.02nRm-3.09nR 2 -9.70mR 2 -0.01nm 2 +0.74Rm 2 +65.71R 3 (6)

[0065] Substituting equations (5) to (6) into equations (1) to (3), we can obtain the estimated product life value T under different conditions. 1 , T 2 , T 3 The calculation expression of the product number n is generally 2 to 3; the reliability R is generally 0.6 to 0.9; the shape parameter m can be calculated from the historical life data of the product. If there is no historical life data, the default value is 2. This formula is more suitable for the life estimation of mechanical, electromechanical and other products.

[0066] The estimated lifespan T of the method of the present invention is 1 , T 2 , T 3 As the number of tested products n changes, Figure 1As shown, under the same conditions, T 3 The life limit of all tested products is measured, so the estimated life value is the lowest, that is, T 1 >T 2 >T 3 The more products are tested, the lower the chance of accidental occurrence, and therefore the larger the estimated life of the product. Figure 1 The displayed situation is consistent with the actual trend.

[0067] Correction factor T 1 , T 2 , T 3 As R changes Figure 2 As shown in the figure, under the same conditions, the higher the reliability requirement, the more conservative the product life estimate. Figure 2 The displayed situation is consistent with the actual trend.

[0068] Correction factor T 1 , T 2 , T 3 As m changes, Figure 3 As shown, m is the shape parameter of the Weibull distribution of the product. The larger the m, the better the product stability. Therefore, under the same conditions, the larger the estimated lifespan, Figure 3 The displayed situation is consistent with the actual trend.

[0069] The lifespan estimation method of the present invention is used to compare the lifespan estimation value with the lifespan estimation value of the expert evaluation method. The results are as follows: Figure 4 As shown, we can see that the two estimated values ​​are relatively close, and the average error is within 5%, indicating that this method can be used for life estimation.

[0070] This method can be used to quickly estimate the time required for life testing. If the expected life of a product is T, there are n test products, the reliability requirement is R, and the shape parameter is m, in order to verify whether the product meets the expected life, the minimum time required for life testing is T. test =K 1 (n,R,m)×T.

[0071] If the life test time T test If no product failure occurs within the specified period, the estimated product life will meet the expected life, otherwise the estimated product life will be less than the expected life.

[0072] Example 2. There are two products undergoing life tests. One product fails after working for 5000 hours, and the other fails after working for 6000 hours. The reliability R is required to be 0.9, and the shape parameter m is 2. Then the estimated life value T of the product is 0 for:

[0073] T 0 =T 3=E(T) / K 3 (n,R,m)=E(T) / (-7.10-1.89n+52.36R-1.97m+4.66nR+0.07nm+8.2

[0074] 9Rm-99.54R 2 -0.41m 2 -0.02nRm-3.09nR 2 -9.70mR 2 -0.01nm 2 +0.74Rm 2 +65.71R 3 )=1628.90h.

[0075] Three products were tested for a lifespan of 5000 hours. One product failed after working for 3000 hours, one product failed after working for 3250 hours, and one product did not fail. The reliability R requirement is 0.8, and the shape parameter m is 2.5. The estimated lifespan of the product is T 0 for:

[0076] T 0 =T 2 =E(T) / K 2 (n,R,m)=E(T) / (-0.60-0.92n+12.89R-0.20m+0.78nR+0.35nm+1.5

[0077] 3Rm-23.70R 2 -0.29m 2 +0.14nRm-0.93nR 2 -5.30mR 2 -0.06nm 2 +0.75Rm 2 +22.38R 3 )=2301.62h.

[0078] The life test of two tested products was carried out for 3250 hours. Both products did not fail. The reliability R requirement is 0.7. The shape parameter m is 2. The estimated life value of the product is T 0 for:

[0079] T 0 =T 1 =E(T) / K 1 (n,R,m)=E(T) / (-3.19-1.17n+22.50R+0.14m+1.35nR+0.33nm+0.5

[0080] 8Rm-35.26R 2 -0.25m2 +0.29nRm-1.76nR 2 -4.53mR 2 -0.07nm 2 +0.68Rm 2 +27.14R 3 )=2164.38h.

[0081] The errors of the above three life estimation results and the expert evaluation method estimation results are 4.84%, 0.04% and 0.11% respectively, all within 5%, indicating that this method can be used for life estimation.

[0082] When the above method is used for life estimation, there is no need to organize experts to negotiate and determine parameters, and an estimated value of the product life is given, which improves the efficiency of life estimation. In addition, the estimated value of the life given by this method will not lead to differences in the estimation results due to human subjective judgment.

[0083] The expected life of a product is 3000h. There are 3 test products. The reliability R requirement is 0.7. The shape parameter m is 3. To verify whether the product meets the expected life, the product needs to be tested for at least T test for:

[0084] T test =K 1 (n,R,m)×T=(-3.19-1.17n+22.50R+0.14m+1.35nR+0.33nm+0.58Rm-35.2

[0085] 6R 2 -0.25m 2 +0.29nRm-1.76nR 2 -4.53mR 2 -0.07nm 2 +0.68Rm 2 +27.14R 3 )×T=3457.50h.

[0086] If the product does not fail within the life test time of 3457.50h, the estimated product life value meets the expected life of 3000h; otherwise, the estimated product life value may be less than the expected life, affecting product evaluation and delivery.

[0087] In the above, the method of the present invention can quickly determine the time when the product needs to undergo a life test to verify whether the product life meets the expected life.

Claims

1. A method for estimating product life based on product life distribution characteristics, characterized in that: Based on historical data, a ternary polynomial calculation model of correction coefficients K1, K2, and K3 with respect to n, R, and m is constructed, where n is the number of tested products; R is the reliability requirement; m is the shape parameter of the Weibull distribution of the products; the life of the products is evaluated according to the following formula based on the constructed ternary polynomial calculation model: For the life test of n tested products at time T, if no failures occur, the life estimate value T1 of the products is: T1 = E(t) / K1(n, R, m) (1) If r failures occur, 0 < r < n, the life estimate value T2 of the products is: T2 = E(t) / K2(n, R, m) (2) If all n have failed, the life estimate value T3 of the products is: T3 = E(t) / K3(n, R, m) (3) E(t) is the average value of the experimental time when no failures occur for n tested products; K1(n, R, m), K2(n, R, m), and K3(n, R, m) are ternary polynomial calculation models.

2. The method for life estimation according to the characteristics of product life distribution as described in claim 1, characterized in that K1(n,R,m)=-3.19-1.17n+22.50R+0.14m+1.35nR+0.33nm+0.58Rm-35.26R 2 -0.25 m 2 +0.29nRm-1.76nR 2 -4.53mR 2 -0.07nm 2 +0.68Rm 2 +27.14R 3 (4)。 3. The method for life estimation according to the characteristics of product life distribution as described in claim 1, characterized in that <h2 style=";text-align:left;direction:ltr">K2(n,R,m) = -0.60 - 0.92n + 12.89R - 0.20m + 0.78nR + 0.35nm + 1.53Rm - 23.70R<h2 style=";text-align:left;direction:ltr"> 2 <h2 style=";text-align:left;direction:ltr"> -0.29 m 2 +0.14nRm-0.93nR 2 -5.30mR 2 -0.06nm 2 +0.75Rm 2 +22.38R 3 (5)。 4. The method for life estimation according to the characteristics of product life distribution as described in claim 1, characterized in that <h2 style=";text-align:left;direction:ltr">K3(n,R,m) = -7.10 - 1.89n + 52.36R - 1.97m + 4.66nR + 0.07nm + 8.29Rm - 99.54R<h2 style=";text-align:left;direction:ltr"> 2 <h2 style=";text-align:left;direction:ltr"> -0.41 m 2 -0.02nRm-3.09nR 2 -9.70mR 2 -0.01nm 2 +0.74Rm 2 +65.71R 3 (6)。 5. The method for estimating product life according to product life distribution characteristics according to claim 1, characterized in that: n = 2 - 3; R = 0.6 - 0.9; m is calculated from the historical product life data.

6. The method for estimating product life according to product life distribution characteristics according to claim 5, characterized in that: When there is a lack of historical life data, m takes 2.

7. The method for estimating product life according to product life distribution characteristics according to claim 1, characterized in that: During the process of constructing the ternary polynomial calculation model, when the highest exponent of n is 1, the highest exponent of m is 2, and the highest exponent of R is 3, the goodness of fit is 0.9982, 0.9976, and 0.9725 respectively.

8. The method for estimating product life according to product life distribution characteristics according to claim 1, characterized in that: It is applied to the life estimation of mechanical and electro-mechanical products.

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