Selection of the Optimal Operating Vector of Operational Systems by Mathematical Programming Methods
Abstract Generated abstract
This paper formulates the choice of operating intervals for long used complex systems subject to periodic preventive maintenance as a mathematical programming problem. It defines an objective function based on the mean useful output over a fixed operating period and, using known reliability characteristics, preventive maintenance time, productivity, and repair time, reduces the search for an optimal working vector to a stochastic linear programming problem under risk. The authors construct the constraint matrix and loss vector from assumed maintenance vectors and a worst operating regime, with special expressions for failure losses under an exponential reliability law. A numerical example illustrates how the resulting linear program yields an optimal allocation of working times and suggests how the approach may be extended to select among several total maintenance time resources.
Full Text
SELECTION OF THE OPTIMAL WORKING VECTOR OF OPERATED SYSTEMS USING MATHEMATICAL PROGRAMMING METHODS
V. A. KOCHETUROV, N. I. SABLIN
(Presented by the scientific seminar of NIIYaFEA)
In the course of operation, complex systems used over a long period are periodically stopped in order to carry out preventive maintenance.
The purpose of preventive maintenance is to carry out measures that contribute to improving the system’s serviceability, i.e., to improving its reliability characteristics.
Let us assume that:
1) the serviceability of the system is evaluated by the mean value of the main output effect produced by use of the system under consideration over a known time of its operation:
\[ \bar{y}=\frac{\sum_{i=1}^{n} c_i t_{pi}(p_i)}{T}, \tag{1} \]
where \(\bar{y}\) is the mean value of the main output effect over the operating time \(T\);
\(c_i\) is the productivity of the system on the \(i\)-th working section;
\(t_{pi}(p_i)\) is the time of fault-free operation of the system on the \(i\)-th working section, depending on the probability of fault-free operation of the system on the \(i\)-th section;
\(n\) is the number of working sections;
2) the reliability law of the system \(P(t>\tau)\) is known;
3) the total preventive-maintenance time \(T_{\mathrm{pr}}\) and \(n\), the frequency of its performance, are given;
4) the productivity vector of the system is known, \(C=\{c_1,c_2,\ldots,c_n\}\);
5) \(T\), the operating time of the system, is given;
6) \(t_y\), the mean time for eliminating one failure in the system, is given.
Under these conditions it is required, by methods of mathematical programming, to determine the optimal working vector:
\[ T_p^*=\{t_{p1}^*,t_{p2}^*,\ldots,t_{pn}^*\}, \]
which ensures the maximum of the objective function (1).
Obviously,
\[ T=T_p+T_{np}+T_y, \tag{2} \]
where
\[ T_{\mathrm{p}}=\sum_{i=1}^{n} t_{pi} \]
is the total operating time of fault-free operation of the system;
\[ T_{\mathrm{pr}}=\sum_{l=1}^{n} t_{\mathrm{pr}l} \]
is the total preventive-maintenance time;
\[ T_{\mathrm{u}}=\sum_{k=1}^{m} t_{\mathrm{u}k} \]
is the time spent on repairing the system in the event of failures during the operating periods of the system.
For a fixed time of preventive-maintenance work, the maximum of the objective function (1) is attained by maximizing \(T_{\mathrm{p}}\) or minimizing \(T_{\mathrm{u}}\), which is possible for some optimal operating vector.
We shall solve the stated problem by maximizing \(T_{\mathrm{p}}\).
Without loss of generality, let us pass from estimate (1) to an integral estimate of the operability of the system:
\[ L=\overline{T}_{\mathrm{u}}=\sum_{i=1}^{n} c_i t_{pi}(p_i). \tag{3} \]
Expression (3) is a linear function and, therefore, in view of the uncertainty of \(t_{pi}\), which depend on the known reliability characteristics of the system, the stated problem is solved by methods of stochastic linear programming and reduces to the case of risk [1] under the constraints:
\[ A T_{\mathrm{p}} \leq B;\qquad t_{pi} \geq 0, \tag{4} \]
where
\[ A= \begin{Vmatrix} a_{11}, & a_{12}, & \ldots, & a_{1n}\\ \cdot & \cdot & \cdot & \cdot & \cdot\\ a_{j1}, & a_{j2}, & \ldots & a_{jn}\\ \cdot & \cdot & \cdot & \cdot & \cdot\\ a_{k1}, & a_{k2}, & \ldots & a_{kn} \end{Vmatrix} \]
is the matrix of conditions,
\[ B=(b_1,b_2,\ldots,b_n)^T \]
is the vector of losses;
\(a_{ij}\ (j=1,k;\ i=1,n)\) are coefficients characterizing the share of losses on the \(i\)-th operating segment of the system.
The matrix of conditions \(A\) characterizes the losses of output effect due to the unreliability of the system, taking account of preventive-maintenance work; the vector of losses \(B\) is equal to the mean admissible losses due to the unreliability of the system over the operating period \(T\) for deterministic vectors of preventive maintenance.
Since the preventive-maintenance vector \(T_{\mathrm{pr}}\) basically determines the reliability of the system on the operating vector \(T_{\mathrm{p}}\) of the system operating period, we determine a set of vectors \(\{T_{\mathrm{pr}}^j\}\ (j=\overline{1,k})\), and, on the basis of this set, shall determine the matrix \(A\) and the vector of losses \(B\); in this case, to each \(j\)-th row of the matrix there will correspond a deterministic vector from the set \(\{T_{\mathrm{pr}}^j\}\).
Since after each \(i\)-th stage the system is put into preventive maintenance, the following functional relationship is obvious:
\[ a_{ij}=f(t_{\mathrm{pr}i})=f[p_i(t_{\mathrm{pr}i})], \tag{5} \]
where \(p_i(t_{\mathrm{pr}i})\) is the reliability of the system on the \(i\)-th operating segment.
Next, the worst mode of system use should be selected.
We shall consider the worst operating regime of the system to be the case when
\[ \begin{cases} t_{p1}=t_{p2}=\ldots=t_{pn}=t_p=\dfrac{T_p}{n},\\[6pt] t_{np1}=t_{np2}=\ldots=t_{npn}=t_{np}=\dfrac{T_{np}}{n}. \end{cases} \tag{6} \]
The choice of such a worst regime is practically confirmed, since it is quite rare to encounter an operated system with identical durations of working periods and preventive maintenance; based on a priori information about the system, the worst operating regime may also be different, which does not change the generality of the problem.
Having chosen the worst operating regime of the system, let us determine the coefficients of the constraint matrix \(a_{ij}\) from the relation:
\[ a_{ij}=\frac{l_i t_y}{t_p}, \tag{7} \]
where \(l_i\) is the number of failures over \(t_{pi}\) (on the \(i\)-th working segment);
\(t_y\) is the average time for eliminating a failure for the given system;
\(t_p\) is the working period of the system corresponding to the worst operating regime.
The number of failures \(l_i\) depends on the reliability \(p_i\) of the system on the \(i\)-th working segment, which in turn depends on the duration of preventive maintenance on the \((i-1)\)-th preventive-maintenance segment \(t_{np(i-1)}\). We shall take into account the degree of influence of \(t_{npi}\) on the reliability of the system on the \((i+1)\)-th working segment \(t_{p(i+1)}\) according to the following approximate dependence:
\[ p_{i+1}\simeq \left(1+\kappa\frac{t_{npi}}{T_{np}}\right)p_i \tag{8} \]
\[ (\text{when } p_{i+1}\geq 1 \quad p_{i+1}\equiv 1), \]
where \(p_{i+1}\) is the reliability of the system after preventive maintenance;
\(p_i\) is the reliability of the system before preventive maintenance, i.e., after completion of the \(i\)-th working segment;
\(\kappa\geq 1\) is a constant, determined experimentally.
In the case of the exponential reliability law this dependence can be obtained exactly [see the appendix].
\(l_i\) is a random variable, but when information on the reliability of the system is available it is determined with sufficient reliability. Under the exponential reliability law, for example, \(l_i\) is determined from the following relation:
\[ l_i=-\frac{T_p}{t_{pi}}\ln p_i. \tag{9} \]
The loss vector \(\mathbf{B}=\{b_1,b_2,\ldots,b_k\}^{T}\) will be considered equal to the total losses of output effect for the worst working vector \(\mathbf{T}_p\); the components of \(\mathbf{B}\) are determined in this case by the relation:
\[ b_j=t_p\cdot \sum_{i=1}^{n} a_{ij}. \tag{10} \]
Having determined the matrix \(\mathbf{A}\) and the vector \(\mathbf{B}\) for a deterministic vector \(\mathbf{C}=\{c_1,c_2,\ldots,c_n\}\), as a result we arrive at the following linear programming problem: it is required to compute the maximum of the linear form
\[ L=\sum_{i=1}^{n} c_i t_{pi}(p_i)=c_1 t_{p1}(p_1)+c_2 t_{p2}(p_2)+\ldots+c_n t_{pn}(p_n)=\max \]
under the following constraints:
\[ \begin{gathered} \text{a)}\quad a_{11}c_1 t_{p1}(p_1)+a_{12}c_2 t_{p2}(p_2)+\ldots+a_{1n}c_n t_{pn}(p_n)\leqslant b_1,\\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\\ a_{j1}c_1 t_{p1}(p_1)+a_{j2}c_2 t_{p2}(p_2)+\ldots+a_{jn}c_n t_{pn}(p_n)\leqslant b_j,\\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\ \cdot\\ a_{\kappa 1}c_1 t_{p1}(p_1)+a_{\kappa 2}c_2 t_{p2}(p_2)+\ldots+a_{\kappa n}c_n t_{pn}(p_n)\leqslant b_\kappa,\\ \text{b)}\quad t_{pi}\geqslant 0,\qquad i=\overline{1,n}. \end{gathered} \]
Computational schemes for determining the optimal vector \(T_p^*\) are presented, for example, in the monographs [2, 3].
Example. Suppose that, for a hypothetical system, the following are given:
\[
P=e^{-\frac{t}{100}},\qquad
T_{\mathrm{pr}}\leqslant 100;\qquad n=4;\qquad t_y=1\ \text{hour};\qquad C\equiv 1;\qquad \kappa=1.
\]
It is required to determine the optimal working vector \(T_p^*\).
Proceeding from the specified time resource for preventive maintenance, let us take, for example, the following set \(\{T_{\mathrm{pr}}^{j}\}\):
\[ T_{\mathrm{pr}}^{1}=\{10,20,30,40\},\qquad T_{\mathrm{pr}}^{2}=\{20,20,20,40\},\qquad T_{\mathrm{pr}}^{3}=\{20,40,30,10\}. \]
For the worst operating mode of the system,
\[ t_p=\frac{200}{4}=50;\qquad t_{\mathrm{pr}}=\frac{100}{4}=25. \]
Proceeding from the adopted set of vectors \(\{T_{\mathrm{pr}}^{1},T_{\mathrm{pr}}^{2},T_{\mathrm{pr}}^{3}\}\), as well as \(t_p\) and \(t_{\mathrm{pr}}\), we determine from (5) and (10) the condition matrix \(A\) and the loss vector \(B\):
\[ A\simeq \left\| \begin{array}{cccc} 0{,}03 & 0{,}06 & 0{,}07 & 0{,}15\\ 0{,}02 & 0{,}07 & 0{,}15 & 0{,}17\\ 0{,}02 & 0{,}03 & 0{,}08 & 0{,}20 \end{array} \right\|, \qquad B\simeq (15,20,16)^T. \]
Having determined the condition matrix \(A\) and the loss vector \(B\), we arrive at the following linear programming problem: it is required to determine the maximum of the linear form
\[ L=t_{p1}+t_{p2}+t_{p3}+t_{p4}=\max \]
under constraints a) and b)
\[ \begin{gathered} \text{a)}\quad 0{,}03t_{p1}+0{,}06t_{p2}+0{,}07t_{p3}+0{,}15t_{p4}\leqslant 15,\\ 0{,}02t_{p1}+0{,}07t_{p2}+0{,}15t_{p3}+0{,}17t_{p4}\leqslant 20,\\ 0{,}02t_{p1}+0{,}03t_{p2}+0{,}08t_{p3}+0{,}20t_{p4}\leqslant 16;\\ \text{b)}\quad t_{pi}\geqslant 0,\qquad i=1,2,3,4. \end{gathered} \]
Using the first computational algorithm [2] of the method of successive improvement of the vector, we determine the optimal working vector
\[ T_p^*=(150,0,0,65), \]
and in this case \(L\geqslant 200\). The preventive-maintenance time in the first approximation may be divided in the ratio
\[ \frac{t_{p1}}{t_{p2}}\simeq \frac{3}{1}; \]
as a result we obtain \(t_{\mathrm{pr}1}=65;\ t_{\mathrm{pr}2}=20\).
In the present work the problem of selecting the optimal working vector has been solved for one total preventive-maintenance time. This problem ...
can be generalized if it is solved for the set of maintenance times \(\{T_{\mathrm{pm}k}\}\), \(k=1,N\).
In this case the optimal solution will consist in choosing such a \(T_{\mathrm{pm}k}\) from the set \(\{T_{\mathrm{pm}k}\}\) and the corresponding optimal working vector \(T_{pk}^{*}\), which give the maximum of the linear form:
\[ (T_{pk}^{*}, T_{\mathrm{pm}k})=\max_{T_{\mathrm{pm}k}\in\{T_{\mathrm{pm}k}\}}\{L_k\}. \]
Appendix. In the case of an exponential reliability law, dependence (8) can be obtained exactly under the following assumptions:
a) the system consists of \(n\) elements of the same type;
b) \(p_1=p_2=\ldots=p_n=p=e^{-\frac{t}{T}}\), i.e., all elements have the same reliability;
c) during maintenance the elements are restored to \(p=1\);
d) the maintenance process obeys an exponential law,
\[ H=e^{-\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}}. \]
Under these assumptions the reliability of the system is:
\[ p_c(t)=p^n=e^{-\frac{nt}{T}}. \]
Suppose that at time \(t_1\), \(n_{\mathrm{pm}}\) elements are subjected to maintenance. In this case
\[ p_c^{*}(t+t_1)=p_c(t)e^{-\frac{n-n_{\mathrm{pm}}}{T}t_1} =p_c(t)p_c(t_1)e^{\frac{n_{\mathrm{pm}}}{T}t_1}, \tag{11} \]
where \(p_c^{*}(t+t_1)\) is the reliability of the system after maintenance;
\(p_c(t_1)\) is the reliability of the system at time \(t_1\).
On the other hand:
\[ p_c(t_1)=e^{-\frac{n}{T}t_1} =e^{-\frac{n_{\mathrm{pm}}}{T}t_1}e^{-\frac{n}{T}t}e^{\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}}. \]
Proceeding from the obvious equality:
\[ e^{-\frac{n_{\mathrm{pm}}}{T}t_{\mathrm{pm}}}e^{\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}} =1 \quad\text{or}\quad e^{\frac{n_{\mathrm{pm}}}{T}t} =e^{\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}}, \]
and using relation (11), we obtain:
\[ p_c^{*}(t+t_1)=p_c(t)p_c(t_1)e^{\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}}. \]
Expanding \(e^{\frac{n_{\mathrm{pm}}}{T_{\mathrm{pm}}}t_{\mathrm{pm}}}\) in a series and restricting ourselves to the first two terms of the expansion, we finally have:
\[ p_c^{*}(t+t_1)\simeq \left(1+n_{\mathrm{pm}}\frac{t_{\mathrm{pm}}}{T_{\mathrm{pm}}}\right)p_c(t_1)p_c(t). \]
References
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