The quest to increase the performance of production systems that have become complex leads to the transfer to the maintenance function of the responsibility of guaranteeing the availability of such systems. Also, we will never stop saying that maintenance must integrate into all of the company’s initiatives, to affirm its role, which is to ensure greater availability and sustainability of the means of production. The objective of this paper is to evaluate the reliability and availability of a system without knowing the distribution law of the operating times. Among the methods for evaluating dependability criteria (Fault Trees, Petri Nets, etc.), we are interested in queues that have the advantage of taking into account functional dependencies, thus allowing a quantified optimization of maintenance. Indeed, queues make it possible to model parallel or sequential processes, implementing operations taking place at the same time or one after the other, meeting the needs of modeling production systems. The main result of this paper is the study of the influence of availability on the reliability of a multi-state production system.
Queuing theory was developed to provide models for predicting the behavior of systems responding to random requests. The first problems studied concerned the congestion of telephone traffic. Erlang will find that a telephone system can be modeled by arrivals of fish customers and exponential uptime [
The two queues schematized above are not the only types to be found in this area, but there are several. In a queuing process, if the client arrives and finds the server busy, it joins the queue associated with that server. At a certain point, the client is selected for service according to a rule known as a policy or service discipline. The expected service is then provided and the client exits the system.
In queue systems, a distinction is made between open systems where no restriction on the size of the queue is imposed; and closed systems where only a limited number of clients are allowed to stay in the queue.
In this paper, we are particularly interested in queuing networks, as they allow us to model and predict the behavior of the system in all its states. Thus, in any state of the system, reliability and availability can be assessed.
Markovian queues are those for which inter-arrivals and service times are exponential. Their Kendall notation will be of the form M/M/··· (M for Markovian...) [
This queue is characterized by a Poissonian arrival with birth rate λ and an exponential service life with death rate μ [
Let ρ = λ / μ , the queue can be considered as a process of birth and death, for which we have the relation (1).
λ n = λ , μ n { μ if n ≠ 0 0 if n = 0 (1)
The state probability (taking into account that ρ ≺ 1 for there to be a steady state) is given by relation (2).
{ π n = π 0 ρ n π 0 = 1 ∑ n = 0 + ∞ ρ n = 1 − ρ (2)
So π n = ( 1 − ρ ) ρ n (3)
All performance parameters are calculated for the case where the queue is stable ( λ ≺ μ , i . e ρ ≺ 1 ) and for the steady state of the queue.
The illustrative diagram is given by
Here d = λ because for all n ≥ 0 . Another way to look at it is to notice that the service is performed at a rate μ in every state where the system has at least one client [
d = P r o b a ( [ Queuenotempty ] ) μ = ∑ n = 1 + ∞ π n μ = [ 1 − π 0 ] μ = ρ μ = λ (4)
U = ∑ n = 1 + ∞ π n = 1 − π 0 = ρ = λ μ (5)
The average number of clients is calculated from the stationary probabilities of the following way:
L = ∑ n = 1 + ∞ n π n = ∑ n = 1 + ∞ n ( 1 − ρ ) ρ n = ρ ( 1 − ρ ) ∑ n = 0 + ∞ ( n + 1 ) ρ n = ρ ( 1 − ρ ) ( 1 + 2 ρ + 3 ρ 2 + ⋯ ) = ρ ( 1 − ρ ) d d p ( ρ + ρ 2 + ρ 3 + ⋯ ) = ρ ( 1 − ρ ) d d ρ ( 1 1 − ρ − 1 )
From where L = ρ 1 − ρ (6)
This parameter is obtained using Little’s law, given by the relation (7).
T = L d = 1 μ ( 1 − ρ ) (7)
This formula can be decomposed according to relation (8).
T = 1 μ + ρ μ ( 1 − ρ ) (8)
From the previous relationship, we deduce respectively the average time spent in the queue and the average number of customers in the queue by relationships (9) and (10).
T q = ρ μ ( 1 − ρ ) (9)
T = 1 μ + ρ μ ( 1 − ρ ) (10)
Consider a single server system identical to the M/M/1 queue except that the capacity of the queue is finite. We therefore always have the following assumptions: the process of arriving customers in the queue is a Poisson process with rate λ and a customer’s service time is an exponential random variable with rate μ . Let K be the capacity of the queue: this is the maximum number of clients that can be present in the system, either on standby or in service. When a customer arrives when there are already K customers present in the system, it is lost [
The process of birth and death modeling this type of queue is then defined by relation (11).
λ n = { λ if n ≺ K 0 if n = K , μ n = { μ if n ≠ 0 0 if n = 0 (11)
The integration of the recurring equation for performing calculations is given as follows:
π n = { π 0 ρ n if n ≤ K 0 if n ≻ K (12)
π 0 = { 1 ∑ 0 K ρ n = 1 − ρ 1 − ρ K + 1 if λ ≠ μ 1 K + 1 if λ = μ (13)
The system flow can be calculated in two equivalent ways: either by measuring the departure rate of clients leaving the server d S , or by measuring the actual arrival rate of clients accepted into the d e system. We would of course expect to obtain equality of these two rates.
The output rate of the server is equal to μ as soon as the queue is not empty:
d S = P r o b a ( [ Queuenotempty ] ) μ = ∑ n = 1 + ∞ π n μ = [ 1 − π 0 ] μ = ρ μ = ρ − ρ K + 1 1 − ρ K + 1 μ (14)
The effective input rate to the file is equal to λ from the moment a client arrives when the file is not full:
d e = P r o b a ( [ Queuenotfullatarrivaltimes ] ) λ = ∑ n = 0 K − 1 π n λ = [ 1 − π 0 K ] λ = 1 − ρ K 1 − ρ K + 1 λ (15)
Since ρ = λ μ , we have d e = d S = d , where d is the average throughput of the queue (entry or exit):
d = 1 − ρ K 1 − ρ K + 1 λ (16)
Note that, when K tends to infinity, we find the results of M/M/1, that is to say, provided that ρ ≺ 1 , which corresponds to the condition of stability of M/M/1.
U ( K ) = ∑ n = 1 K π n = 1 − π 0 = ρ − ρ K + 1 1 − ρ K + 1 = ρ 1 − ρ K 1 − ρ K + 1 (17)
Thus, in the case of a queue with limited capacity, the utilization rate is no longer equal to ρ . Indeed, the rate of use is always equal to the ratio of the average input rate to the average service rate (Little’s law): U = d μ .
Note that, as K → + ∞ , U ( K ) → ρ if ρ ≺ 1 , and towards one if ρ ≻ 1 [
The average number of customers in the system is given by relation (18).
L = ρ 1 − ρ 1 − ( K + 1 ) ρ K + K ⋅ ρ K + 1 1 − ρ K + 1 (18)
We calculate T by Little’s formula, but λ n not being constant, we must calculate λ ¯ , that is:
λ ¯ = ∑ n = 0 + ∞ P n with P n = 1 − ρ 1 − ρ K + 1 ρ n
Hence: T = L λ ¯ = ρ 1 − ρ ⋅ ( 1 − ( K + 1 ) ⋅ ρ K + K ⋅ ρ K + 1 ) ( ρ ≺ 1 ) (19)
| Symbols | Significations |
|---|---|
| P | System utilization rate |
| n | Average number of customers in the system (customers who are waiting and customers who are being served) |
| λ | Average customer arrival rate (failure rate) |
| μ | Average customer service throughput (repair rate) |
| T q | Average waiting time the queue |
| T | Average waiting time in the system |
| P 0 | Probability that there are zero units (customer) in the system |
| P n | Probability that there are n units (clients) in the system |
| π n ( t ) | Probability that the system is in state E n at time t |
Markov chains—or State Space Method (MEE)—were developed in the 1950s for the reliability analysis of repairable systems [
This method consists of representing the operation of a system by a set of components that can be in a finite number of operating and fault states.
A graphic support (the state graph) makes it possible to visualize the different states of a system which are represented by circles and interconnected by oriented arcs which correspond to the transitions (breakdowns and repairs) between states. For an n-component system, if each component has two states (working and failing), the maximum number of states is 2n.
To perform this analysis, you must first identify all the states of the system, classify them into operating states or failure states. Then, it is necessary to find out how to switch from one state to another during a malfunction or repair. With each transition from state E i to state E j , a transition rate L i j is associated which is defined in such a way that L i j ⋅ d t is equal to the probability of passing from E i to E j between two very close instants t and t + d t knowing that we are at time t in E i . Finally, the last step consists in calculating the probabilities of being in the different states during a system life period as well as of calculating the characteristics of dependability [
Using Markov graph modeling, time and stochastic dependencies are more widely taken into account than with classical methods.
In addition, they are easy to use [
We will assume that the system transits from state i to state j with a probability which depends only on states i and j. Such a system is said to be memoryless or even Markovian, which means that the probability does not depend on the states prior to i through which the system has passed during its history. Thus, the future state of the studied system depends only on its present state and not on its past states.
These numbers are arranged in the matrix P = ( P i , j ) ( 0 ≤ i , j ≤ n ) . This matrix is stochastic because the (stochastic) vector in row i contains the probabilities of all
possible transitions starting from state i and therefore their sum is equal to one. The sequence of stochastic vectors π ( t ) for t = 0 , 1,2, ⋯ satisfies the matrix recurrence formula given by Relation (20).
( π 0 ( t + 1 ) , π 1 ( t + 1 ) , ⋯ , π n ( t + 1 ) ) = ( π 0 ( t ) , π 1 ( t ) , ⋯ , π n ( t ) ) ( P 0 , 0 P 0 , 1 ⋯ P 0 , n P 1 , 0 P 1 , 1 ⋯ P 1 , n ⋮ ⋮ ⋱ ⋮ P n , 0 P n , 1 ⋯ P n , n ) (20)
The more compact form is given by Equation (21).
π ( t + 1 ) = π ( t ) P (21)
The recurrence formula π ( t + 1 ) = π ( t ) P applied at t = 0 , 1,2, ⋯ , gives:
π ( 1 ) = π ( 0 ) P π ( 2 ) = π ( 1 ) P = π ( 0 ) P 2 π ( 3 ) = π ( 2 ) P = π ( 0 ) P 3 ⋮
At any instant t (integer), the stochastic vector is therefore given by Equation (22).
π ( t ) = π ( 0 ) P t (22)
If the eigenvalue 1 of the stochastic matrix P of a homogeneous Markov chain is simple and dominant (any other eigenvalue has a modulus strictly less than 1) then the sequence ( P n ) n ∈ N converges to a strictly positive matrix P ∞ of the form given by expression (23), with P 0 + P 1 + ⋯ + P n = 1 [
P ∞ = ( P 0 P 1 ⋯ P n P 0 P 1 ⋯ P n ⋮ ⋮ ⋱ ⋮ P 0 P 1 ⋯ P n ) (23)
Moreover, any sequence ( π n ) n ∈ N defined by: π 0 = ( P ( X 0 = 0 ) P ( X 0 = 1 ) ⋯ P ( X 0 = n ) ) and ∀ n ∈ N , π n + 1 = π n × P , π ∞ = ( P 0 P 1 ⋯ P n ) converges to the vector π ∞ as defined by relation (24).
π ∞ = ( P 0 P 1 P 2 ⋯ P n ) (24)
π ∞ is the unique distribution of probabilities verifying: π × P = π .
A homogeneous Markov chain with a set of states E can be represented by the directed graph G such that:
· its vertices are the states E,
· the arc from vertex i to vertex j exists if, P i , j ≻ 0 ,
· the evaluation of the arc i → j is given by the probability of transition P i , j .
A Markov chain can then be seen as a random walk on Graph G: we draw randomly x 0 realization X 0 according to the π 0 law, then of x 0 we draw x 1 (realization of X 1 ) according to the transition probabilities of the arcs coming from x 0 , etc [
Either a production system, so the history of operating hours or failures during a year of operation reveals the times between failures (TBF) and the times to repair (TTR), given in
We suppose that our production system can be found in three states: the state (E0) where there is no failure with a probability of 1 − λ , the state (E1) where there is a partial failure with a probability of 1 − ( μ + λ ) and the state (E2) where we observe a total failure of the system with a probability of 1 − μ . The probabilities of transitions between the different states are defined in the matrix of transitions given in relation (25).
P = ( 1 − λ P 0 λ λ ( 1 − P 0 ) P 1 μ 1 − ( μ + λ ) P 1 λ μ ( 1 − P 2 ) μ P 2 1 − μ ) (25)
We make the following working hypotheses:
· service disciplines first come, first served (FIFO),
· arrivals according to a Poisson process,
| Months of operating | TBF (h) | TTR (h) |
|---|---|---|
| January | 510 | 150 |
| February | 680 | 32 |
| March | 312 | 23 |
| April | 540 | 40 |
| May | 618 | 57 |
| June | 514 | 42 |
| July | 613 | 40 |
| August | 223 | 38 |
| September | 146 | - |
| October | 233 | 21 |
| November | 424 | 5 |
| December | 620 | 26 |
| Total | 444 | 5433 |
· exponential service time,
· either the M/M/1 queue.
From the previous matrix, we deduce the graph of the availability states, by
It is assumed that the production system is reliable if and only if the probabilities of repairs resulting from transitions from state (E2) are zero (as long as we opt instead for palliative maintenance in this state).
Hence the reliability graph in
The processing of data from the history of operating hours makes it possible to calculate the mean time between failure (MTBF), the mean time to repair (MTTR), the failure rate (λ) and the repair rate (μ).
M T B F = ∑ T B F N = 452.75 h → λ = 1 M T B F = 0.00221 h − 1
M T T R = ∑ T T R N = 37 h → μ = 1 M T T R = 0.02702 h − 1
By definition, the utilization rate is the probability that the server in the queue is busy.
From relation (5), U = ρ = λ μ = 0.08179 .
From relation (6), L = ρ 1 − ρ = 0.08907 .
From relation 10, we deduce the average number of customers in the queue, i.e. L q = ρ 2 1 − ρ = 0.00728 .
This parameter is obtained using Little’s law of relation (7).
T = 1 μ ( 1 − ρ ) = 40.30627
From relation (9), we deduce the average time spent in the queue, namely:
T q = ρ μ ( 1 − ρ ) = 3.29664
In steady state, we have:
{ P 0 λ = μ P 1 P 1 λ = μ P 2 ⇒ { P 1 = λ μ P 0 = ρ P 0 P 2 = λ μ P 1 = ρ P 1 = ρ 2 P 0 ⇒ P n = ρ n P 0 ( n ∈ N )
We find the same general term of probability as that found in the literature [
Now, P 0 = 1 − ρ = 0.91821 , so P 1 = 0.07510 and P 2 = 0.00614 .
We can see that P 0 + P 1 + P 2 = 1 , so the production system can admit 0, 1 or 2 customers.
We have: P = ( 1 − λ P 0 λ λ ( 1 − P 0 ) P 1 μ 1 − ( μ + λ ) P 1 λ μ ( 1 − P 2 ) μ P 2 1 − μ ) = ( 0.99779 0.00202 0.00018 0.00202 0.97077 0.00016 0.02685 0.00016 0.97298 )
We set the following initial conditions: ( π 0 , π 1 , π 2 ) ( 1 ) = ( 1 , 0 , 0 ) .
The probability of the system in its states ( ( E 0 ) , ( E 1 ) , ( E 2 ) ) after two years of operation is given as follows:
( π 0 , π 1 , π 2 ) ( 2 ) = ( 1 , 0 , 0 ) ( 0.99779 0.00202 0.00018 0.00202 0.97077 0.00016 0.02685 0.00016 0.97298 ) = ( 0.99779 , 0.00202 , 0.00018 )
From the recurrence relation given by Equation (21), we deduce by conjecture that:
( π 0 , π 1 , π 2 ) ( n ) = ( π 0 , π 1 , π 2 ) ( 1 ) P n − 1 ( n ∈ N ) (26)
So, after four years, the probability of the system in its states is given as follows:
( π 0 , π 1 , π 2 ) ( 4 ) = ( π 0 , π 1 , π 2 ) ( 1 ) P 3 = ( 1 , 0 , 0 ) ( 0.99779 0.00202 0.00018 0.00202 0.97077 0.00016 0.02685 0.00016 0.97298 ) 3 = ( 1 , 0 , 0 ) ( 0.99341 0.00587 0.00052 0.00588 0.91486 0.00045 0.07821 0.00061 0.92112 ) = ( 0 .99340 , 0.00587 , 0.00052 )
( π 0 , π 1 , π 2 ) ( 10 ) = ( π 0 , π 1 , π 2 ) ( 1 ) P 9 = ( 1 , 0 , 0 ) ( 0.99779 0.00202 0.00018 0.00202 0.97077 0.00016 0.02685 0.00016 0.97298 ) 9 = ( 1 , 0 , 0 ) ( 0.98048 0.00160 0.00143 0.01616 0.76571 0.00114 0.21512 0.00167 0.78165 ) = ( 0.98048 , 0.00160 , 0.00143 ) = ( 0.99340 , 0.00587 , 0.00052 )
When n → + ∞ , Equation (26) becomes:
( π 0 , π 1 , π 2 ) + ∞ = ( π 0 , π 1 , π 2 ) ( 1 ) P + ∞
However, according to the expression (23),
P ∞ = ( P 0 P 1 P 2 P 0 P 1 P 2 P 0 P 1 P 2 ) = ( 0.91821 0.07510 0.00614 0.91821 0.07510 0.00614 0.91821 0.07510 0.00614 )
( π 0 , π 1 , π 2 ) + ∞ = ( 1 , 0 , 0 ) ( 0.91821 0.07510 0.00614 0.91821 0.07510 0.00614 0.91821 0.07510 0.00614 ) = ( 0.91821 , 0.07510 , 0.00614 )
We deduce that the production system reaches the operating limit (maximum number of years of operation) with an availability of 91.82%, when it is in state (E0); with an availability of 7.51%, when it is in state (E1) and with an availability of 0%, when it is in state (E2).
On the other hand, we also note that during the life cycle of the production system, availability decreases in state (E0). On the other hand, when it is in state (E2), availability is zero, which I translate by the fact of its inactivity in this state.
According to the equations of state of the system, in a steady state, we have:
P 0 + P 1 + P 2 = 1 ⇒ P 0 + ρ P 0 + ρ 2 P 0 = 1
⇒ P 0 = 1 1 + ρ + ρ 2 = 0.91871 ; P 1 = 0.07514 and P 2 = 0 .00614 .
In this case, the new matrix of system state transitions, in agreement with
P = ( 1 − λ P 0 λ λ ( 1 − P 0 ) P 1 μ 1 − ( μ + λ ) P 1 λ 0 0 0 ) = ( 0.99779 0.00203 0.00017 0.00203 0.97077 0.00016 0 0 0 )
We set the following initial conditions:
( π 0 , π 1 , π 2 ) ( 1 ) = ( 1 , 0 , 0 )
The probability of the system in its states after two years of operation is given as follows:
( π 0 , π 1 , π 2 ) ( 2 ) = ( 1 , 0 , 0 ) ( 0.99779 0.00203 0.00017 0.00203 0.97077 0.00016 0 0 0 ) = ( 0.99779 , 0.00203 , 0.00017 )
From relation 26, we deduce after four years that the probability of the system in its states is given as follows:
( π 0 , π 1 , π 2 ) ( 4 ) = ( π 0 , π 1 , π 2 ) ( 1 ) P 3 = ( 1 , 0 , 0 ) ( 0.99779 0.00203 0.00017 0.00203 0.97077 0.00016 0 0 0 ) 3 = ( 1 , 0 , 0 ) ( 0.99339 0.00590 0.00016 0.00590 0.91486 0.00015 0 0 0 ) = ( 0.99339 , 0.00590 , 0.00016 )
( π 0 , π 1 , π 2 ) ( 10 ) = ( π 0 , π 1 , π 2 ) ( 1 ) P 9 = ( 1 , 0 , 0 ) ( 0.99779 0.00203 0.00017 0.00203 0.97077 0.00016 0 0 0 ) 9 = ( 1 , 0 , 0 ) ( 0.98040 0.01612 0.00015 0.01612 0.76580 0.00012 0 0 0 ) = ( 0.98040 , 0.01612 , 0.00015 )
We see that the last row of the matrix P is zero, therefore the matrix P is no longer stochastic, therefore the relation (23) is no longer applicable in a steady state.
On the other hand, in a steady state, we have: π × P = π , because π n + 1 = π n , with π = ( π 0 , π 1 , π 2 ) and ∑ i = 1 3 π i = 1 .
So: { 0.00203 π 0 + 0.97077 π 1 = π 1 0.00017 π 0 + 0.00016 π 1 = π 2 π 0 + π 1 + π 2 = 1 ⇒ ( π 0 , π 1 , π 2 ) = ( 0.93490 , 0.06492 , 0.00016 ) .
The production system reaches the operating limit with a reliability of 93.49%, when it is in state (E0); with 6.49% reliability, when in the (E1) state and with 0% reliability, when in the (E2) state.
To have better availability and adequate reliability, the following operations must be carried out:
· Set up a group of executives and technicians specializing in the field of production chain operation;
· Develop a preventive maintenance program corresponding to the needs of the production line;
· Work in close collaboration with all other departments of the company;
· Ensure rigorous control of maintenance operations.
Having reached the end of the writing of this paper, whose general objective was to evaluate the availability and the reliability of the networks of Markov queues, we showed how to take advantage of the knowledge on the theory of the chains of Markov, in order to help in the analysis of waiting phenomena. We have presented an application on a production line where the arrival process is Poissonian. This approach allowed us to show the influence of availability on reliability. Based on the results obtained, proposals for improving maintenance were made.
The authors would like to thank the anonymous reviewers for their time and effort. Their constructive comments and helpful suggestions helped us to clarify the main paper’s research contributions and improve its quality.
The author(s) received no financial support for the research, authorship, and/or publication of this article.
The authors declare no conflicts of interest regarding the publication of this paper.
Djami, A.B.N., Nzié, W., Ousman, B., Anyi, J.N. and Nguelcheu, U.N. (2024) A Maintenance Service Improvement Approach Based on Queue Networks: Application. Open Journal of Applied Sciences, 14, 425-440. https://doi.org/10.4236/ojapps.2024.142030