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Maintenance optimization at VAC An analysis of opportunistic/preventive models and policies Michael Patriksson, Chalmers

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Outline Maintenance activities at VAC Potential for opportunistic maintenance A deterministic maintenance model Illustration Extensions (spare parts; stochastics; work costs) Existing methods – policies Stochastic simulations Future research

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Maintenance activities at VAC Volvo Aero Corp. (VAC), Trollhättan Aircraft engines; 3 billion SEK (2004) Military and civil maintenance Fuel prices while operators compete with prices VAC improves competition by enabling an effective cost reduction Costs: # maintenance occasions; material (spare parts) and work costs Opportunistic/preventive maintenance necessary!

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More on fixed costs Maintenance of an engine can take up to 50 days Often reserve engines are used (are kept in store), especially in civil operations Each maintenance operation incurs a huge cost: administration, extensive testing, incl. probing and infrared measurements; staff intensive

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Potential Some spare parts cost more then 2 MSEK Total maintenance cost for an engine 15–30 MSEK Better maintenance can yield better contracts Research project since 2002 with the purpose to improve maintenance practice

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Logic Two types of components: safety-critical ones–life times determined at Solid Mechanics; others are on condition; replaced if broken OC parts responsible for 30% of maint. occasions Life measured in cycles – monitored extensively especially in military aircraft Engine composed of modules; individually shipped to VAC; simple maintenance done at some military sites Large civil airlines perform their own maintenance Current maintenance depend on cycles left and a measure of remaining value

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Stochastic parts, I Life times = maximal crack lengths Measurements over time can yield good approximations of expected remaining life times: Fatigue life group at Chalmers/FCC (Fraunhofer- Chalmers Center for industrial mathematics) crack length Weibull distributed

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Stochastic parts, II If we have an exponential distribution (light bulb) Typically, By gathering info from measurements the distribution of remaining life times can yield better and better predictions (i.e., conditional expected life times) Over time the inherently stochastic model becomes more and more deterministic Idea: replace distribution by conditional expectation, solve deterministic optimization model to determine maintenance actions today Compare with stochastic programming used in finance

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Deterministic model, I At time t part i is assumed to have the remaining life time. We assume that a new part has life time We have T time steps. At t=0 (that is, today) the module is assumed to be at the work shop Let be equal to one if maintenance is performed, zero otherwise. Further, let be equal to one if part i is replaced at time t, zero otherwise Part i costs and there is a fixed maintenance cost d

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Deterministic model, II

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Explanations Objective function sum of fixed costs (none incurred today) and replacement costs Constraints: replace part i at the very latest when it has reached its life time; if any replacements are made in the future, the module must be taken to the work shop and therefore incurs the fixed cost d

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Notes on the integer program The basic model (without additional spare parts – see below) is quite special: 1. The x variables can be assumed to be continuous – for fixed binary z values the problem has the integrality property! (coefficient matrix totally unimodular) 2. All the constraints define facets of the convex hull of the feasible set – a strong formulation, and therefore (probably) relatively easy to solve! The above is no longer true for more general problems with work costs and spare parts included

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The dependence of fixed costs (Note: c = our d) Four parts with various life times

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Notes No fixed cost: replace exactly according to the life times; no opportunistic maintenance is performed – a very large number of maintenance occasions, and a minimum number of spare parts replaced Very large fixed cost: replace more parts way before their life time is over; opportunistic maintenance rules! – a higher spare parts cost but a minimum number of maintenance occasions Both of these extremes are easy cases – the truth, however, lies somewhere in the middle; therefore, optimization is necessary!

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Current policy Value policy: introduce a cost per time measure: If is the remaining life of part i its value is the product of these two values Value policy: compare value with fixed cost d; replace if value is less Problem: cheap parts are always replaced; remedy: include a minimum time between replacements for cheap parts – often a requirement from clients

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Age replacement policy Common in practice to base an opportunistic policy on a simple measure of an age limit for each part Replace part i if and only if the age of the part is above the value where is a common parameter The value of the parameter is determined through a line search – calculate the total cost for each integer value and take the best solution as the policy

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A test example, I A low pressure turbine in the RM12 engine, consisting of ten parts Costs realistic for military maintenance Time steps of 50 flight hours We test the integer model against two policies and a policy based on no opportunistic maintenance at all; the latter maximizes the total number of maintenance occasions

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A test example, II We test for different values of the parameter and simulate over 200 scenarios We also introduce the possibility of utilizing a set of aged, cheaper, spare parts

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Low pressure turbine, RM12 1HTT tätsegmentOC 2LT turbinskivaLLB 3LufttätringLLB 4LLT tätsegmentOC 5LLT munstyckhusOC 6LedskenesegmentOC 7LT turbinnavLLB 8LT turbinskovelLLB 9HT turbintäthusOC 10LT turbinhusOC

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Results for the deterministic problem

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Notes No opportunistic maintenance 13 occasions – very costly. The other methods reduce this number significantly Current method replaces parts 6, 7, 8, and 9 too often in order to reduce the number of maintenance occasions Age replacement policy reduces the number of occasions to 3, but replaces parts 1, 9, and 10 too often The optimization method replaces exactly the same number of parts as the method of doing no opportunistic maintenance at all, but is much cheaper! This is (of course) the minimum number of replacements, and the cost is minimized by grouping them in the best possible way!

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Simulations

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Data Simulation I: OC parts have = 6 Simulation II: OC parts have = 4 Simulation III: OC parts have = 2 Simulation IV: parts 1, 4: = 2, parts 5,6: = 4, and parts 5,6: = 6 Note: with = 1 there is no point in performing any opportunistic maintenance; our choices test its relative benefit

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Total maintenance costs (Note: costs divided by the corresponding optimal deterministic cost)

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Notes Costs increase with lower values The age replacement policy is on one occasion even more expensive than doing no opportunistic maintenance! The optimization model always gives the best performance (which is not guaranteed); it is 35% cheaper than the current method and 10% cheaper than the age replacement policy

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Total number of maintenance occasions

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Notes Clear that the number of maintenance occasions increases with an increased stochasticity – this signals that planning maintenance then becomes more difficult All of the above signal the importance of obtaining reliable information about stochastic events!

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Additional spare parts Spare part 1: life time 2/3 of new; ½ price of new Spare part 2: life time 1/3 of new; price 1/8 of new We study case IV of the simulations – mixed parts 4 possible bars per part - # parts replaced in the model without extra spare parts; # new parts in the new model; # extra spare parts of the two types in the new model

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Results for the spare parts problem

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Notes The optimization model uses 2/3-life parts 6, 8, 9, and 10 more, and 1/3-life parts 2, 3, 8, and 9 less, lowering the cost by 6% We utilize old spare parts to some degree in all cases where they are available

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Main conclusions from experiments The optimization model always gives the best solutions It is especially good when the stochasticity is weak, but quite a lot better even when stochasticity is present to a large degree The age replacement policy copes much less well with stochasticity The benefit of using opportunistic maintenance increases with the fixed costs. In reality the fixed cost is in fact much higher than in the example, especially in the civil case – availability is then much more valuable

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Pros and cons – existing method + Reduces # maintenence occasions compared with no opportunistic maintenance + Simple – Total cost very high for high fixed costs – Not automatic, quality depending on manaul work – Difficult to extend to, e.g., extra spare parts or work costs

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Pros and cons – age replacement policy + Reduces # maintenance occasions compared with no opportunistic maintenance and to the existing method + Easy to implement – Heuristic; can yield very costly schedules – Diffucult to predict when it is bad – Difficult to extend to, e.g., extra spare parts

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Pros and cons, optimization model + Always the best schedule, not just compared with the other methods but with any other method + Reduces # maintenance occasions compared to all the other methods + Flexible; easy to extend to more complex problems – Demands more advanced mathematical methods and computer facilities +/– Is as strong as the availability of good estimates of the failure distribution

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Maintenance scheme presentation

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Future research Current model works with a single module The whole engine is more complex – modules can be separated in different ways work costs complex to evaluate Work with the polyhedral structure of the model to learn about the properties and potential of different algorithms Joint work with a new research associate

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Future vision

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Project assignment, I Solve the deterministic optimization problem for the case of only new spare parts (requires file modifications) as well as for the case of existing old spare parts (existing file) Analyze the solutions: how does the total cost/number of maintenance occasions vary with the problem parameters? How sensitive is the optimal solution? Are there alternative optimal solutions for the basic case? Analyze also the differences between the solutions to the two problems

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Project assignment, II Consider the following natural modification: The client wishes that at termination of the contract the module has a given property, such as a minimum remaining life time. Provide at least two natural such requirements, besides the life time one, and analyze how the optimal solution differs to the basic case Is the case with extra spare parts such that the integrality property is fulfilled?

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