Reliable lifetime modeling method, estimation method, vehicle, device, medium and product

Abstract

This invention discloses a reliable life modeling method, estimation method, vehicle, equipment, medium, and product, relating to the field of component reliability. The modeling method includes: calculating the first maintenance time of a batch of components in actual use; obtaining the fault state of the batch of components at the time of the first maintenance, and calculating the yield rate at the time of the first maintenance based on the fault state; calculating the second maintenance time based on the fault state, the first maintenance time, reliability requirements, and an initial reliable life model; obtaining the fault state of the batch of components at the time of the second maintenance, and calculating the yield rate at the time of the second maintenance based on the fault state; calculating the parameters of the reliability function based on the yield rates from multiple maintenances and the component reliability function, thereby obtaining the final reliable life model. This invention reduces maintenance time and costs, and avoids the problems of over-maintenance or under-maintenance.

Description

Technical Field

[0001] This invention belongs to the field of component reliability technology, and particularly relates to a component reliability life modeling method, a component maintenance cycle estimation method, rail vehicles, equipment, storage media and products. Background Technology

[0002] As the transportation equipment for urban rail transit, rail vehicles are subject to long-term aging due to factors such as friction, vibration, impact, and corrosion during use. When the aging of a component exceeds a certain level, malfunctions will occur, ranging from minor maintenance and repairs to serious threats to passenger safety and property. Especially as urban rail transit demands increasingly stringent safety performance from its vehicles, vehicle safety and reliability have become crucial for ensuring the safe and orderly operation of rail transit and achieving economic and social benefits. Therefore, the importance of rationally planning vehicle maintenance and repair schedules is growing, and determining appropriate maintenance intervals (i.e., maintenance cycles or maintenance intervals) is a key issue requiring further research.

[0003] Currently, maintenance intervals are defined based on maximum operating mileage and cumulative operating time. Regular preventative maintenance of rail vehicle components includes various levels of maintenance operations such as daily inspections, bi-weekly inspections, tri-monthly inspections, scheduled maintenance, overhauls, and major overhauls. The work is complex and redundant, consuming excessive operating time. While the intervals for daily, bi-weekly, and tri-monthly inspections are the same across different projects in various regions, the intervals for overhauls and major overhauls vary. However, all intervals comply with the "Regulations on the Operation and Maintenance Management of Urban Rail Transit Facilities and Equipment," which stipulate that overhaul intervals should not exceed 5 years or 800,000 kilometers of operating mileage, major overhaul intervals should not exceed 10 years or 1.6 million kilometers of operating mileage, and the overall service life generally should not exceed 30 years or 4.8 million kilometers of operating mileage.

[0004] For example, the overhaul interval for subway trains in City C1 is 5 years or 600,000 kilometers, and the major overhaul interval is 10 years or 1.2-1.5 million kilometers. For subway trains in City C2, overhaul and major overhaul are broken down into three-year inspections, minor repairs, and major overhauls. The three-year inspection interval is 3 years or 340,000-400,000 kilometers, the minor repair interval is 6 years or 620,000-750,000 kilometers, and the major overhaul interval is 12 years or 1.25-1.5 million kilometers. Subway trains in City C3 are divided into two types, A and B, with different overhaul and major overhaul intervals. Type A trains have an overhaul interval of 5 years or 500,000-600,000 kilometers and a major overhaul interval of 10 years or 1.0-1.2 million kilometers. Type B trains have an overhaul interval of 4 years or 750,000-800,000 kilometers and a major overhaul interval of 10 years or 1.6-1.7 million kilometers.

[0005] This method of determining maintenance cycles fails to consider the reliability of vehicle components and actual usage conditions, easily leading to either insufficient or excessive maintenance of these components. The former increases the probability of vehicle malfunctions during normal operation, potentially causing safety accidents, passenger evacuation, or decommissioning, posing both safety hazards and disrupting normal operations due to frequent maintenance needs, resulting in reduced vehicle uptime and utilization. The latter increases maintenance costs and time, similarly reducing uptime and utilization; to meet peak-hour operational demands, more rail vehicles must be deployed, further increasing operating costs. Summary of the Invention

[0006] The purpose of this invention is to provide a reliable life modeling method, estimation method, vehicle, equipment, medium and product to solve the problem that traditional maintenance cycle determination methods do not take into account the reliability of components and actual usage conditions, resulting in insufficient or excessive maintenance.

[0007] This invention solves the above-mentioned technical problems through the following technical solution: a method for modeling the reliable life of components, comprising:

[0008] Construct an initial reliable life model for the components;

[0009] Calculate the first maintenance time for a batch of components in actual use based on reliability requirements and the initial reliable life model;

[0010] Obtain the fault status of the batch of parts during the first overhaul, and calculate the yield rate during the first overhaul based on the fault status of the batch of parts during the first overhaul.

[0011] The second maintenance time is calculated based on the fault status of the batch of parts during the first maintenance, the time of the first maintenance, the reliability requirements, and the initial reliable life model.

[0012] Obtain the fault status of the batch of parts during the second overhaul, and calculate the yield rate during the second overhaul based on the fault status of the batch of parts during the second overhaul.

[0013] The calculation steps for repeated maintenance time and yield rate during maintenance are as follows: obtain the K maintenance times and yield rate during K maintenance times;

[0014] The parameters of the reliability function are calculated based on the yield rate during K maintenance cycles and the reliability function of the components, thereby obtaining the final reliable life model of the components.

[0015] Furthermore, an initial reliable life model for the component is constructed, including:

[0016] Construct a reliability function based on the service life distribution function of the components;

[0017] Acquire the time-of-failure data of multiple components during actual use;

[0018] The initial parameters of the reliability function are estimated based on the first failure time data of multiple components, thereby obtaining an initial reliable life model.

[0019] Furthermore, the specific formula for calculating the yield rate during the i-th maintenance is as follows:

[0020]

[0021] Where, η i M represents the yield rate during the i-th maintenance. Si M represents the number of normal parts during the i-th maintenance. Fi This represents the number of faulty parts during the i-th maintenance.

[0022] Furthermore, based on the fault status of the batch of components at the time of the i-th maintenance, the time of the i-th maintenance, the reliability requirements, and the initial reliable life model, the time of the (i+1)-th maintenance is calculated, including:

[0023] The maintenance interval for each component in this batch is calculated based on the i-th maintenance time, reliability requirements, and initial reliable life model.

[0024] Take the minimum maintenance interval among all components in this batch;

[0025] The time for the (i+1)th maintenance is calculated based on the minimum value and the time for the i-th maintenance.

[0026] Further, the step of calculating the maintenance interval for each component in the batch based on the i-th maintenance time, reliability requirements, and the initial reliable life model includes:

[0027] For each normal component, the first condition must be met during the (i+1)th maintenance:

[0028] P(t>τ i +Δτ ij |T>τ i )≥α;

[0029] Wherein, P(t>τ) i +Δτ ij |T>τ i This indicates that the reliable lifespan T of a normal component j exceeds the time τ of the i-th maintenance. i The probability that the failure time exceeds the time of the (i+1)th maintenance under certain conditions; α represents reliability; Δτ ij τ represents the i-th maintenance interval for component j. i+1 =τi +Δτ ij , τ i+1 This represents the (i+1)th maintenance time;

[0030] Calculate the i-th maintenance interval for each normal component based on the first condition and the initial reliable life model;

[0031] For each faulty component, the i-th maintenance interval is calculated based on the reliability requirements and the initial reliable life model.

[0032] Furthermore, the modeling method also includes calculating the maintenance time of the batch of components using the final reliable life model, specifically including:

[0033] The time for the (K+1)th maintenance is calculated based on the fault status of the batch of components during the Kth maintenance, the time of the Kth maintenance, the reliability requirements, and the final reliable life model.

[0034] Based on the same concept, this invention provides a method for estimating the maintenance cycle of components, including:

[0035] The final reliable life model of the component is constructed using the component reliable life modeling method described above;

[0036] The maintenance time of the component is calculated based on the ultimate reliable life model and the reliability requirements of the component.

[0037] Furthermore, the maintenance time of the component is calculated based on the ultimate reliable life model and the reliability requirements of the component, including:

[0038] The reliable life of the component is calculated based on the final reliable life model and the reliability requirements of the component, and the reliable life is used as the first maintenance time of the component.

[0039] Obtain the fault status of the components during the first overhaul;

[0040] The second maintenance time is calculated based on the fault status of the components described during the first maintenance, the time of the first maintenance, the reliability requirements, and the final reliable life model.

[0041] Repeat the steps of acquiring the fault status and calculating the maintenance time of the component to calculate the maintenance time for each component.

[0042] Based on the same concept, the present invention provides a rail vehicle, on which components are provided, and the maintenance time of the components is estimated using the component maintenance cycle estimation method described above.

[0043] Based on the same concept, the present invention also provides an electronic device, including a memory, a processor, and a computer program / instructions stored in the memory, wherein the processor executes the computer program / instructions to implement the component reliability life modeling method or the component maintenance cycle estimation method as described above.

[0044] Based on the same concept, the present invention also provides a computer-readable storage medium having a computer program / instruction stored thereon, which, when executed by a processor, implements the component reliability life modeling method or component maintenance cycle estimation method as described above.

[0045] Based on the same concept, the present invention also provides a computer program product, including a computer program / instruction, which, when executed by a processor, implements the component reliability life modeling method or the component maintenance cycle estimation method as described above.

[0046] Beneficial effects

[0047] Compared with the prior art, the advantages of the present invention are as follows:

[0048] Based on the initial reliable life model, this invention calculates the next maintenance time by taking into account the fault state of the components during each maintenance, the previous maintenance time, and the reliability requirements. The parameters of the reliable life model are updated based on the yield rate during multiple maintenances to obtain the final reliable life model. This ensures that the components always operate within a high reliability range, thereby reducing maintenance time and costs and avoiding the problems of over-maintenance or under-maintenance.

[0049] This invention establishes a reliability function based on the service life data of components, and estimates parameters based on historical failure data and the reliability function to obtain an initial reliable life model, which ensures the accuracy of reliable life assessment and thus improves the accuracy of maintenance time estimation.

[0050] Based on the constructed reliable life model, this invention determines the inspection time for each step in a sequential manner, requiring less data and involving fast and simple calculations. It enables dynamic adjustment of component maintenance intervals, allowing components to maintain high reliability over a long period, thereby improving safety and reliability. Attached Figure Description

[0051] To more clearly illustrate the technical solution of the present invention, the accompanying drawings used in the description of the embodiments will be briefly introduced below. Obviously, the accompanying drawings described below are only one embodiment of the present invention. For those skilled in the art, other drawings can be obtained based on these drawings without creative effort.

[0052] Figure 1This is a flowchart of the component reliability life modeling method in an embodiment of the present invention. Detailed Implementation

[0053] The technical solutions of the present invention will be clearly and completely described below with reference to the accompanying drawings of the embodiments. Obviously, the described embodiments are only some embodiments of the present invention, and not all embodiments. Based on the embodiments of the present invention, all other embodiments obtained by those skilled in the art without creative effort are within the scope of protection of the present invention.

[0054] The technical solutions of this application will be described in detail below with specific embodiments. The following specific embodiments can be combined with each other, and the same or similar concepts or processes may not be described again in some embodiments.

[0055] Example 1

[0056] like Figure 1 As shown, the component reliability life modeling method provided by this embodiment of the invention includes the following steps:

[0057] Step 1: Construct an initial reliable life model for the components.

[0058] In a specific embodiment of the present invention, constructing an initial reliable life model for a certain component includes:

[0059] Step 1.1: Construct the reliability function based on the service life distribution function of the component.

[0060] For example, suppose the service life distribution function of a component is an exponential function, specifically:

[0061] D(t)=P(T≤t)=1-e -λt (1)

[0062] Where D(t) represents the service life distribution function, P(T≤t) represents the probability that the service life T is less than or equal to the failure time t, and λ represents the parameter to be determined. The reliability function of the component is:

[0063] R(t)=1-D(t)=e -λt (2)

[0064] Where R(t) represents the reliability of the component, and t represents time.

[0065] Step 1.2: Obtain the first failure time data of n components in actual use.

[0066] For example, suppose there are n components in use on a rail vehicle, and the first failure time data of these n components is acquired during actual use. In this embodiment, 100,000 kilometers is used as the unit for the first failure time data, that is, at what mileage does the rail vehicle travel before the component experiences its first failure. The first failure time data of the n components in actual use are shown in Table 1, where n = 30 in this embodiment.

[0067] Table 1. Time to first failure data for 30 components in actual use (unit: 100,000 km)

[0068] 3.578 1.277 2.896 4.123 2.868 1.260 1.082 0.301 0.211 0.088 3.666 4.895 5.023 2.369 3.213 4.125 1.865 2.789 3.652 4.256 4.123 5.201 3.842 5.314 4.665 3.645 5.936 4.251 2.689 4.197

[0069] Step 1.3: Estimate the initial parameters of the reliability function based on the first failure time data of n components, and then obtain the initial reliable life model.

[0070] Based on the first failure time data of n components, the maximum likelihood estimation method is used to obtain the estimated value of parameter λ: λ = 0.308. Therefore, the initial reliable life model is obtained as follows:

[0071] R(t) = e -0.308t (3)

[0072] Step 2: Calculate the first maintenance time for a batch of components in actual use based on reliability requirements and the initial reliable life model.

[0073] To ensure that components operate at a high level of reliability and that rail vehicles run safely, avoiding both over- and under-maintenance, a reliability α is set to 0.9 (i.e., the reliability requirement). Substituting 0.9 into formula (3), the reliability life T is calculated. α :

[0074]

[0075] With reliable life T α The time for the first overhaul of a batch of components in actual use, i.e., the first overhaul time τ1 = T α =0.342, indicating that the first maintenance check point was at mileage of 34,200 kilometers. It should be noted that this batch of parts is not from the same batch as the n parts in Table 1.

[0076] Step 3: Obtain the fault status of the batch of parts during the first overhaul, and calculate the yield rate during the first overhaul based on the fault status of the batch of parts during the first overhaul.

[0077] The batch of components was applied to a rail vehicle. When the rail vehicle traveled τ1 = 34200 km, the fault status of the batch of components was detected. Assume the number of components in the batch is 10, and the fault status of each component is shown in Table 2, where S indicates that the corresponding component is normal, and F indicates that the corresponding component is faulty.

[0078] Table 2 Fault Status of Batch Components

[0079]

[0080] As shown in Table 2, during the first overhaul, components #1 to #7 were in normal condition, while components #8 to #10 were in fault condition. That is, within the interval [0, τ1], there were 3 faulty components (#8, #9, #10) and 7 normal components (#1, #2, #3, #4, #5, #6, #7). In this embodiment, the specific formula for the yield rate during the i-th overhaul is:

[0081]

[0082] Where, η i M represents the yield rate during the i-th maintenance. Si M represents the number of normal parts during the i-th maintenance. Fi This represents the number of faulty parts during the i-th maintenance. According to formula (5), the yield rate during the first maintenance can be calculated.

[0083]

[0084] Step 4: Calculate the second maintenance time based on the fault status of the batch of parts during the first maintenance, the first maintenance time, reliability requirements, and the initial reliable life model.

[0085] In a specific embodiment of the present invention, the time for the (i+1)th maintenance is calculated based on the fault state of the batch of components at the time of the i-th maintenance, the time of the i-th maintenance, the reliability requirement, and the initial reliable life model, including:

[0086] Step 4.1: Calculate the maintenance interval for each component in this batch based on the maintenance time of the i-th maintenance, reliability requirements, and initial reliable life model; where the maintenance interval for each component refers to the interval between the maintenance time of the i-th maintenance and the maintenance time of the (i+1)-th maintenance.

[0087] Step 4.2: Take the minimum maintenance time interval among all parts in this batch;

[0088] Step 4.3: Calculate the (i+1)th maintenance time based on the minimum value in Step 4.2 and the maintenance time of the i-th maintenance.

[0089] For component j in this batch, in order to ensure that its reliability is not lower than α before the second overhaul, an interval Δτ is required. 1j A second overhaul was then conducted. The normally functioning components from the first overhaul (#1, #2, #3, #4, #5, #6, #7) were not disturbed. Therefore:

[0090] P(t>τ1+Δτ 1j |T>τ1)≥α,j=1,2,3,4,5,6,7 (6)

[0091] Wherein, P(t>τ1+Δτ) 1j |T>τ1) represents the probability that the failure time exceeds the second maintenance time given that the reliable life exceeds the first maintenance time, i.e., the probability that no failure will occur during the second maintenance; Δτ 1j This indicates the first maintenance interval of component j (i.e., the interval between the first maintenance time and the second maintenance time).

[0092] according to The solution is that the first maintenance interval for each normal component is: Δτ 11 =Δτ 12 =Δτ 13 =Δτ 14 =Δτ 15 =Δτ 16 =Δτ 17 =0.342. The faulty components (#8, #9, #10) from the initial overhaul need to be replaced or repaired, and then used as normal components. Since the reliability requirement is R(t)≥α, when α=0.9, the equation can be solved as follows:

[0093] R(Δτ 1j )=α,j=8,9,10 (7)

[0094] Solving for Δτ, we get: 1j =0.342,j=8,9,10.

[0095] To ensure that the reliability of each component is not lower than α before the second overhaul, the overhaul time interval Δτ1 = min between the first and second overhaul times. j {Δτ 1j} = 0.342, which means that the mileage between the first and second maintenance dates is 34,200 kilometers. Therefore, the second maintenance date is τ2 = τ1 + Δτ1 = 0.342 + 0.342 = 0.684, which means that the second maintenance date is when the rail vehicle has traveled 68,400 kilometers.

[0096] Step 5: Obtain the fault status of the batch of parts during the second overhaul, and calculate the yield rate during the second overhaul based on the fault status of the batch of parts during the second overhaul.

[0097] When the rail vehicle has traveled τ2 = 68,400 kilometers, the fault status of this batch of parts is detected. As shown in Table 2, during the second overhaul, components #1, #3 to #7, and #10 were in normal condition, while components #2, #8, and #9 were in fault condition. Specifically, within the interval [0, τ2], there were 6 faulty components and 6 normal components. The number of faulty components (#2, #8, #9, #10) was 6 because: components #8 and #9 both failed in intervals [0, τ1] and [0, τ2], so they each need to be counted twice; component #10 failed in interval [0, τ1], so it needs to be counted once; component #2 failed in interval [0, τ2], so it needs to be counted once. The number of normal components (#1, #3, #4, #5, #6, #7) was 6 because component #10 failed in interval [0, τ1], therefore, component #10 needs to be removed from the normal components in interval [0, τ2]. According to formula (5), the yield rate during the second overhaul can be calculated.

[0098] Step 6: Repeat steps 4 and 5 to calculate the time of the third overhaul and the yield rate during the third overhaul.

[0099] First, the time for the third overhaul is calculated based on the fault status of the batch of parts during the second overhaul, the time of the second overhaul, the reliability requirements, and the initial reliable life model.

[0100] For component j in this batch, in order to ensure that its reliability is not lower than α before the third overhaul, an interval Δτ is required. 2j The third overhaul was then carried out. The normally functioning components from the second overhaul (#1, #3, #4, #5, #6, #7, #10) were not disturbed. Therefore:

[0101] P(t>τ2+Δτ 2j |T>τ2)≥α,j=1,3,4,5,6,7,10 (8)

[0102] Wherein, P(t>τ2+Δτ) 2j |T>τ2) represents the probability that the failure time exceeds the third maintenance time given that the reliable life exceeds the second maintenance time, i.e., the probability that no failure will occur during the third maintenance; Δτ 2j This indicates the second maintenance interval for component j (i.e., the interval between the second and third maintenance times).

[0103] according to The solution is that the second maintenance interval for each normal component is: Δτ 21 =Δτ 23 =Δτ 24 =Δτ 25 =Δτ 26 =Δτ 27 =Δτ 210 =0.342. The faulty components (#2, #8, #9) from the second overhaul need to be replaced or repaired. Since the reliability requirement is R(t)≥α, when α=0.9, the equation can be solved as follows:

[0104] R(Δτ 2j )=α,j=2,8,9 (9)

[0105] Solving for Δτ, we get: 2j =0.342,j=2,8,9.

[0106] To ensure that the reliability of each component is not lower than α before the third overhaul, the overhaul time interval Δτ2 = min between the second and third overhaul times. j {Δτ 2j} = 0.342, which means that the mileage between the second and third maintenance checkpoints is 34,200 kilometers. Therefore, the third maintenance checkpoint is τ3 = τ2 + Δτ2 = 0.684 + 0.342 = 1.026, which means that the third maintenance checkpoint is when the rail vehicle's mileage reaches 102,600 kilometers.

[0107] Secondly, obtain the fault status of the batch of parts during the third overhaul, and calculate the yield rate during the third overhaul based on the fault status of the batch of parts during the third overhaul.

[0108] When the rail vehicle travels τ3 = 102600 km, the fault status of this batch of parts is checked. Table 2 shows that during the third maintenance, parts #1 to #4 and #6 to #9 are in normal condition, while parts #5 and #10 are in fault condition. That is, in the interval [0, τ3], there are 8 faulty parts and 5 normal parts. The number of faulty parts (#2, #5, #8, #9, #10) is 8 because: parts #8 and #9 both fail in intervals [0, τ1] and [0, τ2], so both #8 and #9 need to be counted twice; part #10 fails in both intervals [0, τ1] and [0, τ3], so part #10 needs to be counted twice; #2 If a component fails in the interval [0,τ2], component #2 needs to be counted once; if component #5 fails in the interval [0,τ3], component #5 needs to be counted once; the number of normal components (#1, #3, #4, #6, #7) is 5 because: component #2 fails in the interval [0,τ2], component #9 fails in the intervals [0,τ1] and [0,τ2], and component #10 fails in the intervals [0,τ1] and [0,τ3]. Therefore, components #2, #9, and #10 need to be removed from the normal components in the interval [0,τ3]. According to formula (5), the yield rate during the third overhaul can be calculated.

[0109] Step 7: Calculate the parameters of the reliability function based on the yield rate and the reliability function of the components during the three maintenance cycles, and then obtain the final reliable life model of the components.

[0110] Using the yield rate during the three maintenance cycles as the reliability (as in formula (10)), and employing linear fitting, least squares, or maximum likelihood methods, we obtain λ = 1.1, and K = 3 in this embodiment.

[0111]

[0112] This leads to the final reliable lifetime model:

[0113] R(t) = e -1.1t (11)

[0114] Step 8: Calculate the time for the fourth overhaul based on the fault status of the batch of parts during the third overhaul, the time of the third overhaul, the reliability requirements, and the final reliable life model.

[0115] After the third overhaul, in order to ensure that its reliability is not lower than α before the fourth overhaul, an interval of Δτ is required. 3j If a third overhaul is performed, then:

[0116] ① For components #1, #3, #4, #6, and #7, which have been in normal use for τ3 hours, the reliability requirements must still be met before the next overhaul, i.e., the following must be satisfied:

[0117] P(t>τ3+Δτ 3j |T>τ3)≥α,j=1,3,4,6,7 (12)

[0118] Wherein, P(t>τ3+Δτ 3j |T>τ3) represents the probability that the failure time exceeds the fourth maintenance time given that the reliable life exceeds the third maintenance time, i.e., the probability that no failure will occur during the fourth maintenance; Δτ 3j This indicates the third maintenance interval for component j (i.e., the interval between the third and fourth maintenance times).

[0119] according to The solution is that the third maintenance interval for each normal component is: Δτ 31 =Δτ 33 =Δτ 34 =Δτ 36 =Δτ 37 =0.096.

[0120] ② For components #5 and #10, which were in a faulty state during the third overhaul, and became normal components after replacement or repair, R(Δτ) 3j )=α,j=5,10, the solution is: Δτ 3j =0.096,j=5,10.

[0121] ③ For components #2, #8, and #9, they were replaced after the second overhaul and were in normal condition during the third overhaul. Therefore, the usage time of components #2, #8, and #9 is Δτ2. To maintain reliability requirements before the next overhaul, they must satisfy the following equation:

[0122] P(T>Δτ2+Δτ 3j |T>Δτ2)≥α,i=2,8,9 (13)

[0123] Solving for Δτ 32 =Δτ 38 =Δτ 39 =0.096.

[0124] To ensure that the reliability of each component is not lower than α before the fourth overhaul, the interval between the third and fourth overhaul times is... This means that the mileage between the third and fourth maintenance dates is 9600 kilometers. Therefore, the fourth maintenance date is τ4 = τ3 + Δτ3 = 1.118, which means the fourth maintenance date is when the rail vehicle has traveled 111800 kilometers.

[0125] Step 9: Calculate the time of the kth maintenance based on the fault status of the batch of parts during the (k-1)th maintenance, the time of the (k-1)th maintenance, the reliability requirements, and the final reliable life model.

[0126] In a specific embodiment of the present invention, the time of the kth maintenance is calculated based on the fault state of the batch of components at the (k-1)th maintenance, the (k-1)th maintenance time, the reliability requirement, and the final reliable life model, including:

[0127] Step 9.1: Calculate the maintenance interval for each component in this batch based on the (k-1)th maintenance time, reliability requirements, and ultimate reliable life model; where the maintenance interval for each component refers to the interval between the (k-1)th maintenance time and the kth maintenance time.

[0128] Step 9.2: Take the minimum maintenance time interval among all parts in this batch;

[0129] Step 9.3: Calculate the time of the kth maintenance based on the minimum value in Step 9.2 and the time of the (k-1)th maintenance.

[0130] Example 2

[0131] The component maintenance cycle estimation method provided by this invention includes the following steps:

[0132] Step 1: Construct the final reliable life model of the component using the component reliable life modeling method in Embodiment 1 of this application;

[0133] Step 2: Calculate the maintenance time of the component based on the final reliable life model and the reliability requirements of the component. The number of components to be maintained can be 1 or multiple.

[0134] In a specific embodiment of the present invention, the maintenance time of a component is calculated based on the ultimate reliable life model and the reliability requirements of the component, including:

[0135] Step 2.1: Calculate the reliable life of the component based on the final reliable life model and the reliability requirements of the component, and use the reliable life as the time for the first overhaul of the component.

[0136] Using formula (11) from Example 1 as the final reliable lifetime model in this example, the formula for calculating reliable lifetime is:

[0137]

[0138] Among them, T α α represents the reliable lifespan of a component, and α represents the reliability of the component.

[0139] Step 2.2: Obtain the fault status of the components during the first overhaul.

[0140] Step 2.3: Calculate the second maintenance time based on the fault status of the components during the first maintenance, the time of the first maintenance, the reliability requirements, and the final reliable life model.

[0141] When the component is in normal condition, in order to ensure that its reliability is not lower than α before the second overhaul, the second overhaul needs to be carried out after an interval of Δτ1. Therefore, the component needs to meet the following requirements:

[0142] P(t>τ1+Δτ1|T>τ1)≥α (15)

[0143] Where P(t>τ1+Δτ1|T>τ1) represents the probability that the failure time of a component exceeds the second maintenance time when the reliable life T of the component exceeds the first maintenance time τ1; Δτ1 represents the first maintenance time interval of the component, τ2=τ1+Δτ1, and τ2 represents the second maintenance time.

[0144] When the component is in a faulty state, the first maintenance interval of the component is calculated based on the final reliable life model and reliability requirements. The specific formula is as follows:

[0145] R(Δτ1)=α (16)

[0146] Solving formula (16) will give the first maintenance interval when the component is in a faulty state.

[0147] When there is only one component, the second maintenance time τ2 of the component can be calculated based on the first maintenance interval and the first maintenance time τ1.

[0148] When there are multiple parts, take the minimum value of the first maintenance time interval of all parts, and calculate the second maintenance time τ2 of the parts based on this minimum value and the first maintenance time τ1.

[0149] Step 2.4: Repeat steps 2.2 and 2.3 to calculate the maintenance time for each component.

[0150] Example 3

[0151] Although not shown, embodiments of the present invention also provide a rail vehicle on which components are provided, the maintenance time of which is estimated using the component maintenance cycle estimation method in the embodiments of this application.

[0152] Although not shown, embodiments of the present invention also provide an electronic device, which includes: a memory, a processor, and a computer program / instructions stored in the memory, wherein the processor executes the computer program / instructions to implement the component reliability life modeling method or the component maintenance cycle estimation method in the embodiments of this application.

[0153] Although not shown, the electronic device includes a processor that can perform various appropriate operations and processes based on programs and / or data stored in read-only memory (ROM) or loaded from a storage portion into random access memory (RAM). The processor can be a multi-core processor or may contain multiple processors. In some embodiments, the processor may include a general-purpose main processor and one or more specialized coprocessors, such as a central processing unit, graphics processing unit (GPU), neural network processor (NPU), digital signal processor (DSP), etc. Various programs and data required for device operation are also stored in RAM. The processor, ROM, and RAM are interconnected via a bus. Input / output (I / O) interfaces are also connected to the bus.

[0154] The processor and memory described above are used together to execute programs / instructions stored in the memory. When the program / instructions are executed by the computer, they can implement the methods, steps, or functions described in the above embodiments.

[0155] Although not shown, embodiments of the present invention also provide a computer-readable storage medium storing a computer program / instruction thereon, which, when executed by a processor, implements the component reliability life modeling method or component maintenance cycle estimation method in the embodiments of this application.

[0156] Readable storage media include both permanent and non-permanent, removable and non-removable media that can store information using any method or technology. Information can be computer-readable instructions, data structures, program modules, or other data. Examples of computer storage media include, but are not limited to, phase-change memory (PRAM), static random access memory (SRAM), dynamic random access memory (DRAM), other types of random access memory (RAM), read-only memory (ROM), electrically erasable programmable read-only memory (EEPROM), flash memory or other memory technologies, CD-ROM, digital versatile optical disc (DVD) or other optical storage, magnetic tape, disk storage or other magnetic storage devices, or any other non-transferable medium that can be used to store information accessible by a computing device. As defined herein, computer-readable media does not include transient media, such as modulated data signals and carrier waves.

[0157] Although not shown, embodiments of the present invention also provide a computer program product, including: a computer program / instruction, which, when executed by a processor, implements the component reliability life modeling method or the component maintenance cycle estimation method in the embodiments of this application.

[0158] The above description only discloses specific embodiments of the present invention, but the scope of protection of the present invention is not limited thereto. Any changes or modifications that can be easily conceived by those skilled in the art within the scope of the technology disclosed in the present invention should be included within the scope of protection of the present invention.

Claims

1. A method for modeling the reliable life of components, characterized in that, The modeling method includes: Construct an initial reliable life model for the components; Calculate the first maintenance time for a batch of components in actual use based on reliability requirements and the initial reliable life model; Obtain the fault status of the batch of parts during the first overhaul, and calculate the yield rate during the first overhaul based on the fault status of the batch of parts during the first overhaul. The second maintenance time is calculated based on the fault status of the batch of parts during the first maintenance, the time of the first maintenance, the reliability requirements, and the initial reliable life model. Obtain the fault status of the batch of parts during the second overhaul, and calculate the yield rate during the second overhaul based on the fault status of the batch of parts during the second overhaul. The calculation steps for repeated maintenance time and yield rate during maintenance are as follows: obtain the K maintenance times and yield rate during K maintenance times; The parameters of the reliability function are calculated based on the yield rate during K maintenance cycles and the reliability function of the components, thereby obtaining the final reliable life model of the components. The calculation of the (i+1)th maintenance time is based on the fault status of the batch of components during the i-th maintenance, the i-th maintenance time, the reliability requirement, and the initial reliable life model. This includes: For each normal component, the first condition must be met during the (i+1)th maintenance: ； in, Indicates normal parts The reliable lifespan T exceeds the time of the i-th overhaul. The probability that the failure time exceeds the maintenance time of the (i+1)th maintenance under certain circumstances; Indicates reliability; This represents the i-th maintenance interval for the component. , This represents the (i+1)th maintenance time; Calculate the i-th maintenance interval for each normal component based on the first condition and the initial reliable life model; For each faulty component, calculate the i-th maintenance interval for each faulty component based on reliability requirements and the initial reliable life model; Take the minimum maintenance interval among all components in this batch; The time for the (i+1)th maintenance is calculated based on the minimum value and the time for the i-th maintenance.

2. The component reliability life modeling method according to claim 1, characterized in that, Constructing an initial reliable life model for the component includes: Construct a reliability function based on the service life distribution function of the components; Acquire the time-of-failure data of multiple components during actual use; The initial parameters of the reliability function are estimated based on the first failure time data of multiple components, thereby obtaining an initial reliable life model.

3. The component reliability life modeling method according to claim 1, characterized in that, The specific formula for calculating the yield rate during the i-th maintenance is as follows: ； in, This represents the yield rate during the i-th maintenance. This represents the number of normal parts during the i-th maintenance. This represents the number of faulty parts during the i-th maintenance.

4. The component reliability life modeling method according to any one of claims 1 to 3, characterized in that, The modeling method also includes calculating the maintenance time of the batch of components using the final reliable life model, specifically including: The time for the (K+1)th maintenance is calculated based on the fault status of the batch of components during the Kth maintenance, the time of the Kth maintenance, the reliability requirements, and the final reliable life model.

5. A method for estimating the maintenance cycle of components, characterized in that, The estimation method includes: The final reliable life model of the component is constructed using the component reliable life modeling method as described in any one of claims 1 to 4; The maintenance time of the component is calculated based on the ultimate reliable life model and the reliability requirements of the component.

6. The method for estimating the maintenance cycle of components according to claim 5, characterized in that, The maintenance time of the component is calculated based on the ultimate reliable life model and the reliability requirements of the component, including: The reliable life of the component is calculated based on the final reliable life model and the reliability requirements of the component, and the reliable life is used as the first maintenance time of the component. Obtain the fault status of the components during the first overhaul; The second maintenance time is calculated based on the fault status of the components described during the first maintenance, the time of the first maintenance, the reliability requirements, and the final reliable life model. Repeat the steps of acquiring the fault status and calculating the maintenance time of the component to calculate the maintenance time for each component.

7. A rail vehicle having components on it, characterized in that the maintenance time of the components is estimated using the component maintenance cycle estimation method as described in claim 5 or 6.

8. An electronic device, comprising a memory, a processor, and a computer program or instructions stored in the memory, characterized in that the processor executes the computer program or instructions to implement the component reliability life modeling method as described in any one of claims 1 to 4 or the component maintenance cycle estimation method as described in claim 5 or 6.

9. A computer-readable storage medium having a computer program or instructions stored thereon, characterized in that, when the computer program or instructions are executed by a processor, they implement the component reliability life modeling method as described in any one of claims 1 to 4 or the component maintenance cycle estimation method as described in claim 5 or 6.

10. A computer program product, comprising a computer program or instructions, characterized in that, when the computer program or instructions are executed by a processor, they implement the component reliability life modeling method as described in any one of claims 1 to 4 or the component maintenance cycle estimation method as described in claim 5 or 6.

Images