Grouping of mainteance activities Jrn Vatn NTNU 1

Grouping of mainteance activities Jørn Vatn, NTNU 1

Introduction to grouping n Grouping of maintenance activities is primarily done in order to take advantage of sharing so-called set-up costs, S, and to reduce the administrative effort when implementing a maintenance program n Usually initial maintenance intervals for components are established by analyzing each component separately n The easy way forward is then to group activities that have intervals in the same order of magnitude n Optimization methods exists, and we differentiate between n Static grouping where the groups are fixed thought the entire horizon n Dynamic grouping where the groups changes dynamically 2

Single component model n A single component model will be the basis for grouping models n Depending of failure mechanisms, possibilities to detect failure progression, type of maintenance etc. , we may in principle establish the relation between the effective failure rate, E(x), and the maintenance interval, x n The effective failure rate is the expected number of failures per unit time as a function of the maintenance interval n Typically the effective failure rate is an increasing function of the maintenance interval 3

BRP – Block replacement policy n One of the classical maintenance model is the block replacement policy, BRP n In the BRP a single component is maintained to a good as new condition at intervals of length x independent of whether failures has occurred or not in the interval n Time to failure when the component is not maintained is assumed to come from a statistical distribution, typically the Weibull distribution n The effective failure rate as a function of x may be found by dividing the renewal function, W(x), by the interval length x 4

Effective failure rate, Weibull situation n 5

Notation c Pi Planned maintenance cost, exclusive set-up cost. Typically the costs of replacing one unit periodically. c Ui Unplanned costs upon a failure. These costs include the corrective maintenance costs, safety costs, downtime costs, and costs due to material damage. Set-up costs, i. e. , the costs of preparing the preventive maintenance of a group of components maintained at the same time. We assume the same set-up costs for all activities. E, i(x) Effective failure rate for component i when maintained at intervals of length x. x is local time (age). S 6

Notation, cont Mi(x) = expected costs due to failures in a period [0, x) for a component maintained at time x, exclusive planned maintenance cost i(x, k) = [c. Pi + S/k + Mi(x)]/x = average costs per unit time if x is the length of the interval between planned maintenance, and the set-up costs are shared by totally k activities *i, k The minimum value of i(x, k), i. e. , minimization over x x*i, k The x-value that minimizes i(x, k) 7

Notation, cont t Planning horizon, or time between maintenance windows Running time (calendar time) t 0 Now, i. e. , time of planning ti Time of last maintenance of component i ki Number of activities sharing the setup cost (dynamic) li How often a maintenance occation is utilized T 8

Static grouping n In static grouping, the groups are fixed for the entire planning horizon n We have to stick to the grouping and the intervals even if new insight is gained, for example related to the failure rate n We differentiate between n Indirect grouping where the groups are formed by an indirect approach n Direct grouping where the groups are formed explicitly one way or another n Combination of indirect and direct 9

Indirect grouping n Assume there is an occasion for preventive maintenance every T time units n There altogether n maintenance activities to be carried out n c. Pi = individual PM cost n c. Ui = individual cost upon failure, CM + losses n S = setup cost n E, i(x) = effective failure rate n Each component is maintained every li. T time unit n Challenge: Obtain T and li, i = 1, . . , n that minimize total cost 10

Indirect grouping, cont n Average cost per time unit for given T, and l C(T, l) = S/T + i=1: n [c. Pi + Mi(li. T)]/(li. T) = S/T + i=1: n [c. Pi /(li. T)+ c. Ui E, i(li. T )] (*) n The problem is a mixed continious-integer programming problem which is a very difficult problem to solve n Proposed heuristic 1. 2. 3. 4. Choose an initial value of T that corresponds to the smallest individual maintenance interval i, i. e. , minimizing C( i) = (S+ c. Pi )/ i + c. Ui E, i( i) Choose li i /T Keep li fixed and minimize (*) wrt to T Go. To 2 and vary li slightly to check if a better solution may be obtained 11

Direct grouping n n Maintenance activities are partitioned into m groups Each group, say Gj, is a subset of {1, 2, . . , n} Gj Gk = Ø, and j Gj = {1, 2, . . , n} Activities in each group are maintained at the same interval, say Tj n Cost per time unit C(T) = j=1: m {S/Tj + i Gj [c. Pi /Tj + c. Ui E, i(Tj )] } (**) n Optimization problem 1. 2. Given the partitioning, Gj, j = 1, . . , m … it is straight forward to minimize (**) term by term 12

Direct grouping, cont n Proposed heuristic 1. 2. 3. 4. 5. Find individual maintenance interval i, i. e. , minimizing C( i) = (S+ c. Pi )/ i + c. Ui E, i( i) Sort the intervals in increasing order, i. e. , (1) < (2) < … Look for clusters in the intervals, and let these forms groups G 1, G 2, … Given this partitioning, Gj, j = 1, . . , m, minimize (**) wrt T, i. e. C(T) = j=1: m {S/Tj + i Gj [c. Pi /Tj + c. Ui E, i(Tj )] } Go. To 3 and vary the groups slightly to check if a better solution may be obtained 13

Dynamic grouping n In dynamic grouping the groups are not fixed n The idea is to establish the groups “on the fly” n This will enable n To update the strategy when new information becomes available, e. g. , new failure rate estimates n Reschedule the plan if opportunities arise, e. g. , upon a failure there will be an opportunity for advancing the next planned preventive maintenance n The repair of a component failure may be postponed to the next preventive maintenance n Take into account that the usage of a component is not fixed 14

Proposed heuristic Step 0 Step 1 Setp 2 Step 3 - Initialization Tentative plan Establish the candidate groups Optimize execution time for each candidate group, and choose the candidate group with the lowest cost Step 4 - Proceed with the next group, and Go. To Step 1 15

Step 0 - Initialization n Let i(x, k) be the expected cost per unit time when component i is maintained together with k-1 components at intervals of length x: n The value of x that minimises i(x, k), say xi, k*, could be found by a fix point iteration scheme (Weibull modell): n Where ’() is the derivative of the correction term: 16

Step 0 - cont n We do not know ki, hence, xi, k* and i, k*= i i(xi, k*, k) are calculated for an initial guess of the average ki n xi* = “average” xi, k* n i* = “average” i, k* n The values of xi* and i* are calculated only once 17

Step 1 - Tentative plan n Find the individual component due dates, ti* according to the formula ti*= xi* + ti 18

1 - Tentative plan, cont. 19

2 – Candidate groups 20

2 – Candidate groups 21

2 – Candidate groups n As we add more activities to the 1 st candidate group we save set-up cost n However, there are penalties of shifting each individual point of execution n At some point these penalties exceeds the savings in set-up cost n At this point we stop adding more activities to the 1 st candidate group 22

2 – Candidate groups, cont n Note n We also have to stop searching for further candidate groups if one activity tentatively is repeated twice within the range of the interval for the candidate group 23

3 - Optimal execution time n For a given candidate group, Kk we find the next execution time, t*, by minimizing the following cost elements n Set-up cost n Component specific preventive maintenance cost n Deterioration cost from now to t n Average maintenance and deterioration cost from t to T 24

3 - Optimal candidate group size n Components not included in a candidate group are assumed to be maintained at their average optimal interval, yielding a cost from t 0 to T: n Total cost used to compare candidate groups is thus 25

4 – Next group n To proceed, we now set the clock: t* t 0 26

4 – Next group n Then we start forming candidate group 2 and so forth 27

Comments n Wildeman (1996) proposed to see at least one group ahead when fixing the first group (maintenance package) n We only consider the first group, and assume that subsequent activities are executed at average optimal point of times n This simplifies the procedure significantly n In some cases this may lead to a non optimal first group n Wildeman did not consider the situation of repeated activities n In our heuristic this is rather easy n Wildeman used c. Pi + S instead of c. Pi + S/k to establish the tentative plan n This will in some situation give different ordering of due activities, and hence give weaker results in our opinion 28

Comments, continued n Initially, the ki’s were set more or less arbitrary n After “running” the procedure, we may record the average values of the ki’s n These values may be used to iterate for a better solution n Also note that the procedure may be improved: n When two subsequent groups have been established, we may reduce cost by moving one activity from one of the group to the other group 29

Opportunities n The process of forming groups will in principle establish a plan for maintenance through the entire planning horizon T n The plan is followed if nothing special happens n If new evidence become available, or an opportunity for reducing maintenance cost, e. g. , a failure that may share set-up cost we may advance the execution of the next group of preventive maintenance activities n Recalculation of the grouping 30