DYNAMIC PROGRAMMING FOR OPTIMAL ALLOCATION OF MAINTENANCE RESOURCES ON POWER DISTRIBUTION NETWORKS

Transcription

1 Eduardo Tadeu Bacalhau - Unicamp DYNAMIC PROGRAMMING FOR OPTIMAL ALLOCATION OF MAINTENANCE RESOURCES ON POWER DISTRIBUTION NETWORKS Eduardo Tadeu Bacalhau STATE UNIVERSITY OF CAMPINAS IFORS - Melbourne, 2011

2 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

3 Labore Laboratory of Power Networks Optimisation Christiano Lyra Filho - State University of Campinas - christi@densis.fee.unicamp.br; Celso Cavellucci - State University of Campinas - celsocv@densis.fee.unicamp.br; Fábio Luiz Usberti - State University of Campinas - fusberti@densis.fee.unicamp.br.

4 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

5 Radial Network Figure: Reference Radial Network

6 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

7 Failure Rates where: λ t s = λ s + e E s λ t e λ t e = λ (t 1) e δ kenxen t n N ke xen t = 1, n N ke λ t e is the failure rate for equipment e in the period t; λ (t 1) e is the failure rate for equipment e in the last year or the initial failure rate for equipment e(t = 1); N ke is a set of all preventive maintenance actions; δ ken is the failure rate multiplier for equipment k e for action level n; xen t is a boolean decision variable denoting whether the equipment e received (xen t = 1) or not (x en t = 0) maintenance level n in the period t.

8 Reliability Constraints Thus, the SAIFI (The System Average Interruption Frequency Index) of system can be calculated: where: S is the set of all sections; SAIFI t = 1 λ t N sn s, T s S λ t s is failure rate of section s in the period t; N s is the number of customers into section s; N T is the number of all customers into the Network.

9 Optimisation Problem min x t en HP t=1 (p kenxen) t + λ t ec ke α t n N ke e E s.t : SAIFI t SAIFI perm t = 1,...HP, where: E is a set that contains all the equipment which can receive preventive maintenance; SAIFI perm is the maximum permitted for SAIFI; p ken is the cost for action preventive level n for equipment k e; c ke is the cost for action corrective level for equipment k e; α t is a parameter which is related to each period.

10 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

11 State Space Search Example 30 equipments and HP=3 S 0 = Chosen_action(S,S ) Depth Search with Simulated Annealing Constructive Heuristic Depth Search Simulated Annealing

12 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

13 Knapsack Problem on Dynamic Programming F n (b) = max s.t. n c j x j j=1 n a j x j b j=1 x j {0, 1}, j = 1,..., n Where: n c j x j : total value of the selected elements to the Knapsack; j=1 n a j x j : total volume of the selected elements to the Knapsack. j=1

14 The aim is get n-th value as from basic cases Get F n (a 0 ) Where F k (a) = max {F k 1 (a), F k 1 (a a k ) + c k } With F 0 (a) = 0 a To determine the optimal solution: Create an indicator p k that is equal 0 if F n (b) = F n 1 (b), and 1 otherwise. Analyzes all indicators from p n up to p 1. If the indicator p k = 0 then x k = 0, else x k = 1.

15 Problem Adapted Get F n (M 0 ) Where F k (M) = min { F k 1 (M) + Cp k + ( λ cm ) k Cc k, F k 1 (M v k ) + ( λ sm )} k Cc k With F 0 (M) = 0 M Where: Cp k is the maintenance preventive cost for equipment k; Cc k is the maintenance corrective cost for equipment k; λ cm k is the failure rate for equipment k which was received preventive maintenance; is the failure rate for equipment k which was not received preventive maintenance; v k is the volume of the equipment k which was selected to the knapsack M. λ sm k

16 Important Steps of the Algorithm Developed 1 Calculating the boundary given by the reliability constraints; 2 Calculating the volume of equipments: v k = δ (λ sm k 3 Calculating knapsack size: λ cm k ) M = δ (SAIFI perm SAIFI min )

17 Pseudocode Pseudocode Knapsack Problem (n,m) 1: Calculate SAIFI min SAIFI max 2: Calculate v k k = 1..n 3: knapsack(n,m) p 4: Calculate F 0 (M) = 0 M 5: For k = 1..n do 6: For m = 1..M do 7: F k (M) = min {F k 1 (M) + Cp k + (λ cm k Cc k ), F k 1 (M v k ) + (λ sm k Cc k )} 8: End For 9: End For 10: OptimalSolution(p,M) S Return: S

18 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

19 Studied Cases Instances Six intances were created for the problem: 1 Instance with 30 equipments; 2 Instance with 50 equipments; 3 Instance with 100 equipments; 4 Instance with 150 equipments; 5 Instance with 300 equipments; 6 Instance with 400 equipments; All instances were executed for only one period.

20 SAIFI permitted For each instance five values of constraints were chosen; Calculating via Equation: SAIFI α = SAIFI min + (SAIFI max SAIFI min ) α where α is 0.2, 0.4, 0.6, 0.8 and 1.0.

21 Equipments Values Tipo CMC CMP Mtx MP CSM Mtx SM TF Initial Cable infrastructure infrastructure Post Post Regulator Recloser Primary Pruning Secondary Pruning Transformer

22 Results Instance with 30 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) (x 1000) (s) (%)

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24 Results Instance with 50 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) ( x 1000) (s) (%)

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26 Results Instance with 100 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) (x 1000) (s) (%)

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28 Results Instance with 150 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) (x 1000) (s) (%)

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30 Results Instance with 300 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) (x 1000) (s) (%)

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32 Results Instance with 400 equipments: DP DSA Cost Time Cost Time Profit SAIFI (x 1000) (s) (x 1000) (s) (%)

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34 Summary 1 Introduction 2 Motivation 3 Optimisation Model 4 Heuristic Method 5 Dynamic Programming 6 Experiments 7 Conclusions

35 Conclusions In all instances the dynamic programming algorithm achieved the best results; In small instances, the state space search algorithm achieved good results when the constraints were looser, but the results deteriorates when the number of equipment grows; Dynamic programming has maintained a standard result on the all values of time achieved, increasing as the number of equipments grows. The same not happened with the state space search algorithm, increasing a lot of the computational time.

36 Future Works As Future Works: Increase the period of optimisation; The development of equations dividing the problem in 2 HP subproblems.

37 Acknowledgements CNPq CAPES