Four Methods for Maintenance Scheduling

We had a problem to be solved: the thermal generator maintenance scheduling problem [Yam82]. We wanted to look at stochastic methods and this paper will present three methods and discuss the pros and cons of each. We will also present evidence that strongly suggests that for this problem, tabu search was the most effective and efficient technique. The problem is concerned with scheduling essential maintenance over a fixed length repeated planning horizon for a number of thermal generator units while minimising the maintenance costs and providing enough capacity to meet the anticipated demand. Traditional optimisation based techniques such as integer programming [DM75], dynamic programming [ZQ75, YSY83] and branch-and-bound [EDM76] have been proposed to solve this problem. For small problems these methods give an exact optimal solution. However, as the size of the problem increases, the size of the solution space increases exponentially and hence also the running time of these algorithms. To overcome this difficulty, modern techniques such as simulated annealing [Č85, KGV83], stochastic evolution [SR91], genetic algorithms [Gol89] and tabu search [Ree93] have been proposed as an alternative where the problem size precludes traditional techniques. The method explored in this paper is tabu search and a comparison is made with simulated annealing (the application of simulated annealing to this problem is given in [SN91]), genetic algorithms and a hybrid algorithm composed with elements of tabu search and simulated annealing. 1 Problem Description Consider I generating units producing output over a planning horizon of J periods. Each unit 1 i I must be maintained for Mi contiguous periods during the horizon. However the starting period denoted by xi for each unit i is unconstrained even in the case that xi = J and Mi > 1 for some i. Since we are considering a rolling plan, the maintenance period would wrap around to the start of our planning horizon. The operating capacity of each unit is denoted by Ci. Under no circumstances is it possible for a unit to exceed this limit. In order to avoid random factors in the problem, such as unit random outages, a reserve capacity variable proportional to the demand is incorporated into the problem description. This problem is classified as a deterministic cost-minimisation problem and can be solved using an optimisation-based technique. Therefore in period j where 1 j J , the anticipated demand for the system as a whole will be denoted by Dj and the reserve capacity required by Rj . Fuel costs can also be estimated for each period as a constant, fj per unit output. Finally, let pij represent the generator output of uniti at period-j, ci(j) be the maintenance cost of unit-i if committed at period-j and let yij be a state variable, equal to one if unit-i is being maintained in period-j and otherwise zero. The objective of the problem is to minimise the sum of the overall fuel cost and the overall cost of maintenance: Minimise J Xi=j (fj : I Xi=1 pij) + I Xi=1 ci(xi) Once the maintenance of unit-i starts, the unit must be in the maintenance state for Mi contiguous periods. yij = 8<: 0 if j = 1; 2; : : : ; xi 1 1 if j = xi; : : : ; xi +Mi 1 0 if j = xi +Mi; : : : ; J If xi + Mi > J then the maintenance wraps around to the next repitition of the planning horizon. This formulation captures the notion of continual maintenance. It would be possible to arrange matters so that overlap never happens; the difference is minor. The generator output must not exceed the upper limit; the output of the generator is set to zero during maintenance. 0 pij Ci(1 yij) The total output must equal the demand in each period, In Proceedings of the International Conference on Artificial Neural Networks and Genetic Algorithms, pp264–269, Springer, ISBN: 3-211-83087-1 I Xi=1 pij = Dj where j = 1; 2; : : : ; J and the total capacity must not be less than the required reserve. I Xi=1(1 yij)Ci (Dj +Rj) Our formulation of the problem is based closely on that of [SN91]. To simplify the operation of the algorithm, all solutions in the solution space are considered valid. A solution could be infeasible if the demand and reserve constraints cannot be met. In this case the solution is penalised by the addition of a penalty function: J Xj=1 uj + J Xj=1 vj Where and are tunable parameters and uj and vj are derived from the shortfall in output: ( I Xi=1 pij) + uj = Dj and the shortfall in capacity. ( I Xi=1 Ci(1 yij)) + vj = Dj +Rj uj and vj are not permitted to be negative, thus a feasible solution incurs no penalty function. Thus any initial solution can be chosen and the optimisation algorithm will be directed towards feasible solutions through the choice of sufficiently high and . 2 Problem Solution and Discussion Simulated annealing [KGV83], genetic algorithms [Hol75], tabu search [Ree93] and a hybrid algorithm composed from elements of simulated annealing and tabu search were implemented to solve this problem. The problem as described in the previous sections can be reduced to that of finding the optimal values of xi. 2.1 Implementation of the methods 2.1.1 Simulated Annealing Each iteration consists of generating n states in the neighbourhood of the current state. A solution is accepted with probability exp =Tk where Tk is the current “temperature” and if accepted, it replaces the current state from which the new states are derived during the current iteration and so the current solution can change several times during one iteration. Tk is initialised to some value T0 and varies according to a cooling schedule Tk+1 = pTk; 0 p 1. Once n states have been generated, if the number of solutions which were accepted during the last iteration is less than some value then the algorithm terminates. 2.1.2 Genetic Algorithms The encoding used is a binary representation of each unit’s starting period concatenated together. This gives a suitable representation for use with a genetic algorithm. After generating a starting population consisting of individuals generated at random, for a predetermined number of iterations the genetic algorithm performs roulettewheel selection to choose a pair of individuals. Singlepoint crossover is then applied, followed by the application of a mutation operator. Selection favours the better individuals by ensuring that each individual’s probability of being selected is proportional to the fitness of that individual. The probability of crossover taking place is predetermined and if crossover does occur, single-point crossover is applied from a random position in the individual. Finally, each bit in a new individual is flipped, again with a predetermined probability.