CN106529693A - Method for determining and optimizing system maintenance rate - Google Patents
Method for determining and optimizing system maintenance rate Download PDFInfo
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Abstract
The invention discloses a method for determining and optimizing a system maintenance rate. The method is characterized in that the maintenance rate is adopted to ensure the system availability when a storage element and the element failure rate are affected by the working environment, the system is shown by using a dynamic fault tree, and the distribution of the element failure rate is determined by using the space fault tree theory to further determine the distribution of the element maintenance rate; three expected optimization states in operation of the system are provided, and the method is used to determine the element maintenance rates to reach the three states. The method comprises the following steps that when the system availability TA is set as 0.8, a work domain (namely, a working environment condition change scope) A<total>={t belongs to [0,100]day intersected c belongs to [0,50] DEG C}, the space fault tree is used to determine the system element fault rates lambdaE1E2 and lambdaE3E4, and then distribution of the element fault rates muE1E2 and muE3E4 under three system optimization targets are determined. The method can be used to determine and optimize the system maintenance rate.
Description
Technical field
The present invention relates to safety system engineering, more particularly to system maintenance rate determine and optimize.
Background technology
In practice, it is generally desirable to system is macroscopically reaching certain target, the such as reliability of system, availability, failure
Rate, total maintenance cost etc..The realization of system macro-goal, it is difficult to ensure from the general level of system, should be by composition system
The target that the primary element parameter of system is controlled to ensure system, these parameters include:The crash rate of element, maintenance rate, dimension
Accomplish this etc..The parameter as part failure rate is that element nature is determined, user cannot change；And maintenance rate is
Can adjust at work, you can some macro-goals of system are reached by the adjustment of maintenance rate.For part failure rate
It is generally a definite value, but actually many part failure rates is affected by working environment, are change crash rate,
Such as temperature, working time, air pressure etc..Another problem is that modern system can all take spare part strategy to ensure reliability.That
The how component maintenance rate in determination system after the problems referred to above are considered, so as to the realization for ensureing system macro-goal just becomes
Problem.
For the research for determining system maintenance rate mainly has at present:The deteriorating process optimum maintenance policy of Chen Jinyuan etc.:Base
In popularization cumulative shock model；The Chen Ji identical armament systems maintainability conflict based on genetic algorithm and extent function is planned
Study on Problems；Two parts, two mechanic's cold nonmaintained system repair and replacement policy of Wang Yanyu etc.；The failure interaction condition of Ge Yang etc.
Lower system maintenance interval decision-making etc..Dynamic fault tree is used for the description of the reliability containing spare part system is at present more, such as opened
CTCS-3 level ATP system fail-safe analyses of the Wen Tao etc. based on Dynamic fault tree；Zhang Zhi's English etc. based on Dynamic fault tree
Missile Pilot systems reliability analysis；The computer interlock system reliability based on Dynamic fault tree of Feng Xue etc. and performance point
Analysis research；The protection system dynamic reliability based on Dynamic fault tree with Monte-Carlo Simulation of Dai Zhihui etc. is assessed；Wu Wenbin
Deng the inertial navigation system safety analysiss based on improved Dynamic fault tree；The satellite system based on Dynamic fault tree of Zhang Xiaojie etc.
System fail-safe analysis.It is contemplated that element becomes in the case that crash rate and system contain spare part, to reach system macro-goal
Determine that the correlational study of component maintenance rate does not also occur.
Method proposes the change crash rate that element is represented using space fault tree theory, is represented using dynamic fault tree theory and is contained
There is contacting for spare part system structure and system macro-goal and component parameters, so as to be met the component maintenance rate point of requirement
Cloth.And expect to have obtained corresponding component maintenance rate distribution for systematic function in the case of three kinds in practice.
The content of the invention
1 space fault tree
Space fault tree (Space Fault Tree, SFT) was proposed in 2012, is had been achieved for some up to now and is ground
Study carefully achievement.Elementary event or thing among the basic theories of space fault tree thinks that system works in environment, due to constituting system
The property of reason element determines that the fault rate which works at different conditions is different.For example, such as two in electrical system
Pole pipe, its probability of malfunction is just with working time, operating temperature, have direct relation by electric current and voltage etc..If to this
System is analyzed, and the working time of each element and the temperature of work accommodation etc. may be all different, with system entirety
Working time and the change of ambient temperature, the probability of malfunction of system is also different.This phenomenon be it is in esse, but it is past
Toward being ignored, and think that probability of malfunction is invariable.
Using the change crash rate of element in SFT expression systems.The analysis theoretical and following in order to clearly describe SFT
Journey, lists the example of necessary SFT related contents and analyzed system.Simple electrical system is described, the system is by two
Pole pipe is constituted, and the rated operation of diode is affected by many factors, wherein importantly working time t and operating temperature c.
For the electrical system that affected by the two factors as object of study.Related definition is as follows:
1) space fault tree (or hyperspace accident tree):The probability of happening of elementary event is not fixed, is determined by n factor
Fixed, such accident tree is referred to as multidimensional accident tree, is represented with T.
2) influence factor of elementary event:Elementary event probability of happening is made to produce the factor of change.In this example, t is represented
Time factor, c represent temperature factor.
3) characteristic function (characteristic function) of the probability of happening of elementary event:Elementary event is affected in single influence factor
Under, with the probability of happening variation characteristic that the change of influence factor is showed.Which can be used with elementary function, piecewise function etc.Represent, i represents i-th element, x replaces influence factor.Such as the temporal characteristics function of i-th original paper of this exampleAnd temperature profile functionWherein, t is element use time, and λ is unit
Fault rate, A are divided into for temperature change.
4) the probability of happening spatial distribution of elementary event:Elementary event under the influence of n influence factor, with their change
The probability of happening change showed in hyperspace.N influence factor is sent out as separate independent variable, elementary event
Raw probability is used as functional value.Use Pi(x1, x2... xn) represent, i.e.,Wherein n for affect because
Plain number, in this example be
2 Dynamic fault trees
It is analysis space station and air-traffic control system that Dynamic fault tree method is the nineties in 20th century by professor J.B.Dugan
The method that reliability and a kind of analysis for proposing have the system reliability of dynamic random fault characteristic.Dynamic fault tree method is
Introduce dynamic logic gate on the basis of fault tree, such as cold reserve door, hot reserve door, order dependent door, preferentially with door etc., be used for
The cold reserve of sign system, hot reserve, can repair, the dynamic characteristic such as resource-sharing.
Here using Markov state transition matrix come process problem, including analytic method and matrix iteration.
When calculating the repaiied reliability index of single system, using analytic method；When calculating compared with complication system, especially solving system
Application matrix iterative method when reliability and failure frequency.Analytic method is, according to the characteristics of gate, to set up based on Markov shape
State shifts the equation group of figure, and with reference to the characteristics of gate, derivation can obtain the expression formula of reliability index, directly apply mechanically calculating reliable
Property refers to calibration method.Matrix iteration is the probability matrix and systematic state transfer matrix multiple iteration according to system, is asked
Solution given time system residing for different conditions probability, with reference to the characteristics of Different Logic door the normal working probability of compartment system,
Malfunction probability.Finally using whole dynamic fault tree model come each reliability index of assessment system.
λ is crash rate, and μ is maintenance rate.The function of each gate is respectively:With door:The incoming event X, Y of and if only if door
When all occurring, the outgoing event Z of door occurs；OR gate:As the incoming event X of door1, X2, X3At least 1 generation, the output thing of door
Part Z occurs；Cold reserve door:Primary input event X is in running order, and stand-by equipment Y is in cold standby state, only works as master/slave device
All failure when, the outgoing event Z of door occurs；Hot reserve door:Primary input event X is in running order, and stand-by equipment Y is in hot standby
With state, only when master/slave device all failures, the outgoing event Z of door occurs.
In 3 systems, the maintenance rate of element determines
Studied Dynamic fault tree is given first, as shown in Figure 1.
3.1X1And X2Subtree is solved
X1Subtree module is the hot standby combination of element, and the crash rate of element is λ, and maintenance rate is μ. define its shape space:0 table of state
Show 2 modules all normal works, system worked well；During shape 1 represents 2 modules, in maintenance, another module is just for a failure
Often work, system worked well；State 2 represents that 2 modules all fail, and a module is being repaired, and another module is to be repaired, system
Failure.
According to probability combine hot reserve door the characteristics of, derive each reliability index expression formula such as formula (1)
In formula (1), availability A represents that system reaches available probability during steady operational status.Failure frequency be M (t) refer to [0,
T] the mean failure rate number of times of system in the time, the failure-frequency of system as in the unit interval.It is average on-stream time MUT, average
Downtime MDT.X2Solution procedure and X1It is identical.
The solution of 3.2T
System T connects lower floor's event by OR gate.Definition status space is:State 0 is that 2 modules are all normal, the normal work of system
Make；State 1 is X1Subtree module failure is being repaired, X2Subtree module is normal, T failures；State 2 is X2Subtree module failure repairing
Reason, X1Subtree module is normal, T failures. and its state transition diagram is as shown in Figure 2.
According to probability combine OR gate the characteristics of, derive each reliability index expression formula such as formula (2) shown in.
In formula, Λ=λ1+λ2。
Component maintenance rate distribution after 3.3 given system availabilities determines
If system availability TA=0.8, seek μE1E2And μE3E4, due to E1And E2It is X1Hot reserve event, use μE1E2Represent E1And E2
Maintenance rate (element X1Maintenance rate), μE3E4Define in the same manner.By A=T in formula (2)AAvailability understand,WhereinRepresent X1(the E in the case of with hot reserve1→E2) crash rate,Define in the same manner；Represent X1(the E in the case of with hot reserve1→E2) maintenance rate,Define in the same manner.Know againWherein MUTE1E2Represent E1And E2Average operation time, MDTE1E2Represent E1And E2
Average downtime, according to formula (1), obtainλ in formulaE1E2Represent E1And E2Maintenance
Rate (element X1Maintenance rate), λE3E4Define in the same manner.To solve the problems referred to above, variable k is introduced,In view of than
Symmetrically to facilitate research, if k ∈ 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,1/0.9,1/0.8,1/0.7,
1/0.6,1/0.5,1/0.4,1/0.3,1/0.2,1/0.1 }.Arrange said process and obtain formula (3).
If the excursion of the working field of element, i.e. working condition is AAlways={ t ∈ [0,100] day ∩ c ∈ [0,50] DEG C }, that is, be
The time of system work is 100 days, and operating temperature range is 0 to 50 DEG C.Obtain the failure probability point under the failure tree state of space
Cloth λE1E2And λE3E4So can be according to λE1E2And λE3E4With formula (3), the Dynamic fault tree is drawn in k=1, TAμ when=0.8E1E2
And μE3E4Scattergram.
It can be seen that meeting system availability TAUnder the conditions of=0.8, k=1, for system element X1Hot reserve event E1、E2's
Maintenance rate will meet maintenance rate distribution in working field；And X2Meet maintenance rate distribution, such system can reach TA=0.8
Macro-goal.
The determination of maintenance rate under 4 Different Optimization targets
Three kinds of in esse system optimization problems are proposed first, it is 1) in the case of research range (working field) determination, whole to be
System maintenance rate summation is minimum.The meaning of this actual demand is desirable to the maintenance frequency minimum in whole working field, so minimum
Maintenance frequency can reduce downtime, while maintenance cost can be reduced.2), in Practical Project, one can all be given typically
Acceptable minimal maintenance rate, just implements maintenance when the maintenance rate of element exceedes minimal maintenance rate, or maintainer is in
Armed state prepares maintenance.Generally wish system in whole working field, it is more little more than the region of minimal maintenance rate more
It is good, and then reduce maintainer waiting time raising maintenance efficiency.3) when the expense of different type element in maintenance system is different
When, between element, different maintenance rates can cause different system maintenance total costs in working field, certainly more low better.Certainly
Above-mentioned 3 points will ensure that system is maintained on certain availability basis, otherwise nonsensical.
For Section of three system mentioned, due to TA=0.8, and λE1E2And λE3E4To know, then determine maintenance rate
It is critically depend on variable k.Due toWithI.e. in TAIn the case of fixation, k is represented
A kind of distributional effects, then different distribution (k is different) can bring differentWithSo by adjusting k
Under conditions of above three problem can be met, the μ in working field is determinedE1E2And μE3E4。
4.1 ask maintenance rate distribution during maintenance rate summation minimum
For first problem, A in working field is madeAlwaysThe maintenance rate minimum μ in={ t ∈ [0,100] day ∩ c ∈ [0,50] DEG C }min,
Its definition is as shown in formula (4).Result of calculation is as shown in table 1.
Table 1With the corresponding table of k value
Have shown in table 1, it is minimum as k=1Minimum, i.e.,Meet system availability TAIn the case of=0.8,
Maintenance rate in whole working area is minimum.
4.2 minimum zones for seeking specified maintenance rate
If the tolerable maintenance rate of systemThe μ in working field will so be determinedE1E2And μE3E4Respectively greater than 2 scope
K value when minimum.Method is the whole working field of traversal, when k is different to μE1E2And μE3E4Region more than 2 is counted, and counts knot
The corresponding k value of fruit reckling is required.Traversing result is as shown in table 2.
2 μ of tableE1E2And μE3E4Count value more than 2
As shown in table 2, during k=0.1, count minimum.ForFor other values (0.1～10) when, as a result identical, i.e. TA=0.8
In the case of, scope of the maintenance rate in whole working area more than 2 is minimum.
4.3 ask maintenance rate distribution during total maintenance cost minimum
In order to be described to element expense difference, definitionWithRespectively element X1And X2Maintenance cost, total maintenance cost
With definition as shown in formula (5).
IfI.e.It is from 0.1 to 10.So existUnder the change of k, one 19 can be obtained
× 19 CAlwaysMatrix.
Can be according to elementValue finds corresponding k value to determine μE1E2And μE3E4Distribution in working field.Example
Such as, if element X1And X2Maintenance cost be 20 yuan and 100 yuan, i.e.,During so k=0.3, system entirety
Maintenance cost is minimum, availability T of certain systemA=0.8.Seek μE1E2And μE3E4Distribution.
Using this method solving three kinds of common system availability demands.Maintenance rate when asking maintenance rate summation minimum
Maintenance rate distribution when being distributed, ask the minimum zone of specified maintenance rate, asking total maintenance cost minimum.If TA=0.8, by formula (4),
(5) and 4.2 section analysis in the case of above-mentioned three kinds component maintenance rate distribution μE1E2And μE3E4Settled accounts.That is element X1And X2
Maintenance rate meet respectively correspondence in the case of μE1E2And μE3E4, then the availability demand of system just can meet.
Description of the drawings
The Dynamic fault tree of Fig. 1 systems
The state transition diagram of Fig. 2 T
Specific embodiment
1 space fault tree
Space fault tree (SpaceFaultTree, SFT) was proposed in 2012, had been achieved for some researchs up to now
Achievement.Elementary event or physics among the basic theories of space fault tree thinks that system works in environment, due to constituting system
The property of element determines that the fault rate which works at different conditions is different.For example, such as two poles in electrical system
Pipe, its probability of malfunction is just with working time, operating temperature, have direct relation by electric current and voltage etc..If to this being
System is analyzed, and the working time of each element and the temperature of work accommodation etc. may be all different, with the work that system is overall
Make the change of time and ambient temperature, the probability of malfunction of system is also different.This phenomenon be it is in esse, but often
It is ignored, and thinks that probability of malfunction is invariable.
Using the change crash rate of element in SFT expression systems.The analysis theoretical and following in order to clearly describe SFT
Journey, lists the example of necessary SFT related contents and analyzed system.Simple electrical system is described, the system is by two
Pole pipe is constituted, and the rated operation of diode is affected by many factors, wherein importantly working time t and operating temperature c.
For the electrical system that affected by the two factors as object of study.Related definition is as follows:
1) space fault tree (or hyperspace accident tree):The probability of happening of elementary event is not fixed, is determined by n factor
Fixed, such accident tree is referred to as multidimensional accident tree, is represented with T.
2) influence factor of elementary event:Elementary event probability of happening is made to produce the factor of change.In this example, t is represented
Time factor, c represent temperature factor.
3) characteristic function (characteristic function) of the probability of happening of elementary event:Elementary event is affected in single influence factor
Under, with the probability of happening variation characteristic that the change of influence factor is showed.Which can use P with elementary function, piecewise function etc.i x
X () represents, i represents i-th element, and x replaces influence factor.Such as the temporal characteristics function P of i-th original paper of this examplei t(t)=1-e-λt, and temperature profile functionWherein, t is element use time, and λ is cell failure rate, and A is warm
Degree change is divided into.
The probability of happening spatial distribution of elementary event:Elementary event exists with their change under the influence of n influence factor
The probability of happening change showed in hyperspace.N influence factor occurs as separate independent variable, elementary event
Probability is used as functional value.Use Pi(x1,x2,…xn) represent, i.e.,Wherein n is impact
Factor number, is P in this examplei(t, c)=1- (1-Pi t(t))(1-Pi c(c))。
2 Dynamic fault trees
It is analysis space station and air-traffic control system that Dynamic fault tree method is the nineties in 20th century by professor J.B.Dugan
The method that reliability and a kind of analysis for proposing have the system reliability of dynamic random fault characteristic.Dynamic fault tree method is
Introduce dynamic logic gate on the basis of fault tree, such as cold reserve door, hot reserve door, order dependent door, preferentially with door etc., be used for
The cold reserve of sign system, hot reserve, can repair, the dynamic characteristic such as resource-sharing.
Here using Markov state transition matrix come process problem, including analytic method and matrix iteration.
When calculating the repaiied reliability index of single system, using analytic method；When calculating compared with complication system, especially solving system
Application matrix iterative method when reliability and failure frequency.Analytic method is, according to the characteristics of gate, to set up based on Markov shape
State shifts the equation group of figure, and with reference to the characteristics of gate, derivation can obtain the expression formula of reliability index, directly apply mechanically calculating reliable
Property refers to calibration method.Matrix iteration is the probability matrix and systematic state transfer matrix multiple iteration according to system, is asked
Solution given time system residing for different conditions probability, with reference to the characteristics of Different Logic door the normal working probability of compartment system,
Malfunction probability.Finally using whole dynamic fault tree model come each reliability index of assessment system.
λ is crash rate, and μ is maintenance rate.The function of each gate is respectively:With door:The incoming event X, Y of and if only if door
When all occurring, the outgoing event Z of door occurs；OR gate:As the incoming event X of door1, X2, X3At least 1 generation, the output thing of door
Part Z occurs；Cold reserve door:Primary input event X is in running order, and stand-by equipment Y is in cold standby state, only works as master/slave device
All failure when, the outgoing event Z of door occurs；Hot reserve door:Primary input event X is in running order, and stand-by equipment Y is in hot standby
With state, only when master/slave device all failures, the outgoing event Z of door occurs.
In 3 systems, the maintenance rate of element determines
Studied Dynamic fault tree is given first, as shown in Figure 1.
3.1X1And X2Subtree is solved
X1Subtree module is the hot standby combination of element, and the crash rate of element is λ, and maintenance rate is μ. define its shape space:0 table of state
Show 2 modules all normal works, system worked well；During shape 1 represents 2 modules, in maintenance, another module is just for a failure
Often work, system worked well；State 2 represents that 2 modules all fail, and a module is being repaired, and another module is to be repaired, system
Failure.
According to probability combine hot reserve door the characteristics of, derive each reliability index expression formula such as formula (1)
In formula (1), availability A represents that system reaches available probability during steady operational status.Failure frequency be M (t) refer to [0,
T] the mean failure rate number of times of system in the time, the failure-frequency of system as in the unit interval.It is average on-stream time MUT, average
Downtime MDT.X2Solution procedure and X1It is identical.
The solution of 3.2T
System T connects lower floor's event by OR gate.Definition status space is:State 0 is that 2 modules are all normal, the normal work of system
Make；State 1 is X1Subtree module failure is being repaired, X2Subtree module is normal, T failures；State 2 is X2Subtree module failure repairing
Reason, X1Subtree module is normal, T failures. and its state transition diagram is as shown in Figure 2.
According to probability combine OR gate the characteristics of, derive each reliability index expression formula such as formula (2) shown in.
In formula, Λ=λ1+λ2。
Component maintenance rate distribution after 3.3 given system availabilities determines
If system availability TA=0.8, seek μE1E2And μE3E4, due to E1And E2It is X1Hot reserve event, use μE1E2Represent E1And E2
Maintenance rate (element X1Maintenance rate), μE3E4Define in the same manner.By A=T in formula (2)AAvailability understand,WhereinRepresent X1(the E in the case of with hot reserve1→E2) crash rate,In the same manner
Definition；Represent X1(the E in the case of with hot reserve1→E2) maintenance rate,Define in the same manner.Know againWherein MUTE1E2Represent E1And E2Average operation time, MDTE1E2Represent
E1And E2Average downtime, according to formula (1), obtainλ in formulaE1E2Represent E1With
E2Maintenance rate (element X1Maintenance rate), λE3E4Define in the same manner.To solve the problems referred to above, variable k is introduced,In view of than it is symmetrical to facilitate research, if k ∈ 0.1,0.2,0.3,0.4,0.5,0.6,0.7,
0.8,0.9,1,1/0.9,1/0.8,1/0.7,1/0.6,1/0.5,1/0.4,1/0.3,1/0.2,1/0.1 }.Arrange said process
Obtain formula (3).
If the excursion of the working field of element, i.e. working condition is AAlways={ t ∈ [0,100] day ∩ c ∈ [0,50] DEG C }, that is, be
The time of system work is 100 days, and operating temperature range is 0 to 50 DEG C.Obtain the failure probability point under the failure tree state of space
Cloth λE1E2And λE3E4So can be according to λE1E2And μE3E4With formula (3), the Dynamic fault tree is drawn in k=1, TAμ when=0.8E1E2
And μE3E4Scattergram.
It can be seen that meeting system availability TAUnder the conditions of=0.8, k=1, for system element X1Hot reserve event E1、E2's
Maintenance rate will meet maintenance rate distribution in working field；And X2Meet maintenance rate distribution, such system can reach TA=0.8
Macro-goal.
The determination of maintenance rate under 4 Different Optimization targets
Three kinds of in esse system optimization problems are proposed first, it is 1) in the case of research range (working field) determination, whole to be
System maintenance rate summation is minimum.The meaning of this actual demand is desirable to the maintenance frequency minimum in whole working field, so minimum
Maintenance frequency can reduce downtime, while maintenance cost can be reduced.2), in Practical Project, one can all be given typically
Acceptable minimal maintenance rate, just implements maintenance when the maintenance rate of element exceedes minimal maintenance rate, or maintainer is in
Armed state prepares maintenance.Generally wish system in whole working field, it is more little more than the region of minimal maintenance rate more
It is good, and then reduce maintainer waiting time raising maintenance efficiency.3) when the expense of different type element in maintenance system is different
When, between element, different maintenance rates can cause different system maintenance total costs in working field, certainly more low better.Certainly
Above-mentioned 3 points will ensure that system is maintained on certain availability basis, otherwise nonsensical.
For Section of three system mentioned, due to TA=0.8, and λE1E2And λE3E4To know, then determine maintenance rate
It is critically depend on variable k.Due toWithI.e. in TAFixed situation
Under, k represents a kind of distributional effects, then different distribution (k is different) can bring differentWithSo
Under conditions of above three problem can be met by regulation k, the μ in working field is determinedE1E2And μE3E4。
4.1 ask maintenance rate distribution during maintenance rate summation minimum
For first problem, A in working field is madeAlwaysThe maintenance rate minimum μ in={ t ∈ [0,100] day ∩ c ∈ [0,50] DEG C }min,
Its definition is as shown in formula (4).Result of calculation is as shown in table 1.
Table 1With the corresponding table of k value
Have shown in table 1, it is minimum as k=1Minimum, i.e.,Meet system availability TAIt is in the case of=0.8, whole
Maintenance rate in individual working area is minimum.
4.2 minimum zones for seeking specified maintenance rate
If the tolerable maintenance rate of systemThe μ in working field will so be determinedE1E2And μE3E4Respectively greater than 2 scope
K value when minimum.Method is the whole working field of traversal, when k is different to μE1E2And μE3E4Region more than 2 is counted, and counts knot
The corresponding k value of fruit reckling is required.Traversing result is as shown in table 2.
2 μ of tableE1E2And μE3E4Count value more than 2
As shown in table 2, during k=0.1, count minimum.ForFor other values (0.1～10) when, as a result identical, i.e. TA=0.8
In the case of, scope of the maintenance rate in whole working area more than 2 is minimum.
4.3 ask maintenance rate distribution during total maintenance cost minimum
In order to be described to element expense difference, definitionWithRespectively element X1And X2Maintenance cost, total maintenance cost
With definition as shown in formula (5).
IfI.e.It is from 0.1 to 10.So existUnder the change of k, one 19 can be obtained
× 19 CAlwaysMatrix.
Can be according to elementValue finds corresponding k value to determine μE1E2And μE3E4Distribution in working field.Example
Such as, if element X1And X2Maintenance cost be 20 yuan and 100 yuan, i.e.,During so k=03, the overall dimension of system
Repair expense minimum, availability T of certain systemA=0.8.Seek μE1E2And μE3E4Distribution.
Using this method solving three kinds of common system availability demands.Maintenance rate when asking maintenance rate summation minimum
Maintenance rate distribution when being distributed, ask the minimum zone of specified maintenance rate, asking total maintenance cost minimum.If TA=0.8, by formula (4),
(5) and 4.2 section analysis in the case of above-mentioned three kinds component maintenance rate distribution μE1E2And μE3E4Settled accounts.That is element X1And X2
Maintenance rate meet respectively correspondence in the case of μE1E2And μE3E4, then the availability demand of system just can meet.
Claims (10)
1. a kind of system maintenance rate determines and optimization method, it is characterised in that contain storage element and part failure rate to understand
When being affected by working environment, it is to ensure the maintenance rate taken of system availability, the system is used into dynamic fault tree representation, is made
Determine that its part failure rate is distributed with space fault tree theory, and then determine the distribution of component maintenance rate；In proposing system operation,
Expect the three kinds of Optimal States for reaching, and carried out really to reaching component maintenance rate distribution during these three states using the method
It is fixed；Which comprises the steps：In initialization system availability TA=0.8, working field (working environment excursion) AAlways={ t
∈ [0,100] day ∩ c ∈ [0,50] DEG C }, and system element fault rate λ is determined by space fault treeE1E2And λE3E4Afterwards, it is determined that
Element failure rate μ under three kinds of system optimization aimsE1E2And μE3E4Distribution；The present invention can be used for system maintenance rate and determine and optimize.
2. in system, the maintenance rate determination of element includes：X1And X2Subtree is solved, the solution of T, element dimension after given system availability
Repair rate distribution to determine.
3.X1Subtree module is the hot standby combination of element, and the crash rate of element is λ, and maintenance rate is μ；Define its shape space:State 0
Represent 2 modules all normal works, system worked well；During shape 1 represents 2 modules, a failure is in maintenance, another module
Normal work, system worked well；State 2 represents that 2 modules all fail, and a module is being repaired, and another module is to be repaired, is
System failure；According to probability combine hot reserve door the characteristics of, derive each reliability index expression formula such as formula (1),
Availability A represents that system reaches available probability during steady operational status, and failure frequency is that M (t) was referred in [0, the t] time
The mean failure rate number of times of system, the failure-frequency of system as in the unit interval, average on-stream time MUT, average downtime
MDT, X2Solution procedure and X1It is identical.
4. system T connects lower floor's event by OR gate, and definition status space is:State 0 is that 2 modules are all normal, and system is normal
Work；State 1 is X1Subtree module failure is being repaired, X2Subtree module is normal, T failures；State 2 is X2Subtree module failure exists
Repair, X1Subtree module is normal, T failures；According to probability combine OR gate the characteristics of, derive each reliability index expression formula such as
Shown in formula (2),
In formula, Λ=λ1+λ2。
5. system availability T is setA=0.8, seek μE1E2And μE3E4, due to E1And E2It is X1Hot reserve event, use μE1E2Represent E1With
E2Maintenance rate (element X1Maintenance rate), μE3E4Define in the same manner；By A=T in formula (2)AAvailability understand,WhereinRepresent X1(the E in the case of with hot reserve1→E2) crash rate,Define in the same manner；Represent X1(the E in the case of with hot reserve1→E2) maintenance rate,Define in the same manner；Know againWherein MUTE1E2Represent E1And E2Average operation time, MDTE1E2Represent E1And E2
Average downtime, according to formula (1), obtainλ in formulaE1E2Represent E1And E2Dimension
Repair rate (element X1Maintenance rate), λE3E4Define in the same manner；Variable k is introduced, said process is arranged and is obtained formula (3),
6.In view of than it is symmetrical to facilitate research, if
k∈{0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,1/0.9,1/0.8,1/0.7,1/0.6,1/0.5,
1/0.4,1/0.3,1/0.2,1/0.1}。
7. a kind of system maintenance rate according to claim 1 determines and optimization method, it is characterised in that Different Optimization target
The determination of lower maintenance rate includes：Maintenance rate distribution when asking maintenance rate summation minimum, seeks the minimum zone of specified maintenance rate, asks
Maintenance rate distribution during total maintenance cost minimum.
8. it is according to claim 7 to ask maintenance rate during maintenance rate summation minimum to be distributed, it is characterised in that to make in working field
AAlwaysThe maintenance rate minimum μ in={ t ∈ [0,100] day ∩ c ∈ [0,50] DEG C }min, which is defined as shown in formula (4),
It is minimum as k=1Minimum, i.e.,Meet system availability TAIn the case of=0.8, in whole working area
Maintenance rate it is minimum.
9. the minimum zone for seeking specified maintenance rate according to claim 7, it is characterised in that set the tolerable maintenance of system
RateThe μ in working field will so be determinedE1E2And μE3E4K value when respectively greater than 2 scope is minimum；Method is whole for traversal
Individual working field, when k is different to μE1E2And μE3E4Region more than 2 is counted, and the corresponding k value of count results reckling is
It is required.
10. it is according to claim 7 to ask maintenance rate during total maintenance cost minimum to be distributed, it is characterised in that to define CX1And CX2
Respectively element X1And X2Maintenance cost, total maintenance cost defined as shown in formula (5),
IfI.e.It is from 0.1 to 10, thenUnder the change of k, can obtain one 19 ×
19 CAlwaysMatrix.
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