1. BRIDGING THE GAP BETWEEN THEORY AND PRACTICE IN MAINTENANCE D.N.P. (Pra) MURTHY RESEARCH PROFESSOR THE UNIVERSITY OF QUEENSLAND

2. PART-3: BUSINESS FOCUS

3. OUTLINE Framework & modelling Case 1: Dragline Maintenance Outsourcing Case 2: Hydraulic pumps

4. FRAMEWORK & MODELLING

5. KEY ELEMENTS

6. MODELLING The elements that are relevant depends on the decision problem Need to model the relevant elements separately Link the models to build the model for solving the decision problem Data plays a critical part

7. DRAGLINE CASE STUDY [CONTINUATION FROM PART 2]

8. DECISION PROBLEM Commercial considerations dictate an increase in output Idea: Increase bucket size (100 tons to 140?) Greater load on components Implications for reliability and maintenance

9. LOAD DEGRADATION MAINTENANCE AVAILABILITY FAILURE DUTY CYCLE YIELD

10. MODELLING Modelling system in terms of its major components [Decomposition] Modelling degradation of each component Modelling effect of bucket load on component and system performance Involves reliability science, engineering and mathematics

11. SYSTEM PERFORMANCE Availability: Depends on up and down times Down times: To rectify minor failures and preventive maintenance to avoid major failures Up time: Productive time Cycle: Time between major maintenance

12. SYSTEM PERFORMANCE Bucket load affects both these variables Need to take into account preventive maintenance schedules for different components [Different time scales] Multiple objectives: Study different alternatives

13. OBJECTIVES Maximise total output per year Maximise revenue per year Minimise total cost per year Maximise yield [dirt moved per unit time] Need to take into account various constraints

14. SYSTEM FAILURE MODELLING System comprised of 25 components All components need to be working for the system to be working. System fails whenever a component fails. System failure distribution is given by a competing risk model involving the failure distribution of the 25 components

15. MODELLING THE SYSTEM Failure distribution for the system is given by Failure distributions of the individual components was discussed in Part 2. Minimal repairs for subsequent failure modelling

16. AVAILABILITY Cycle Time: Depends on load v the ratio of load to the base load Up time: T v Expected downtime (for minor and major preventive maintenance) – obtained from field data From this we can obtain availability

17. AVAILABILITY

18. Reliability  1 0.95 T1T1 T0T0 P.M. Interval (T) Bucket load V 1 Bucket load V 0 RISK CONSTRAINT

19. AVAILABILITY vs v v Availability

20. MAJOR PM INTERVAL vs v v Major PM Interval

21. YIELD vs BUCKET LOAD

22. SENSITIVITY STUDY (  )

23. CONCLUSIONS Study revealed increase in output yield with increase in bucket size Maximum yield corresponds to v  1.3 (dragline load = 182 tonnes or payload of 116 tonnes) as opposed to current payload of 74 tonnes Shutdown interval will need to be reduced from 43680 usage hours to 25000 usage hours (or 4.1 calendar years)

24. REFERENCE For more details, see Townson, P. Murthy, D.N.P. and Gurgenci, H. (2002), Optimisation of Dragline Load, in Case Studies in Reliability and Maintenance, WR Blischke and DNP Murthy [Editors], Wiley, New York.

25. MAINTENANCE OUT-SOURCING

26. CONCEPT Outsourcing of maintenance involves some or all of the maintenance actions (preventive and/or corrective) being carried out by an external service agent under a service contract. The contract specifies the terms of maintenance and the cost issues and can involve penalty and incentive terms.

27. KEY ELEMENTS

28. OVERALL FRAMEWORK

29. MAINTENANCE ACTIVITIES D-1: What (components) need to be outsourced for maintenance? D-2: When should the maintenance be carried out? D-3: How should the maintenance be carried out?

30. ALTERNATE CONTRACT SCENARIOS

31. DECISION PROBLEMS From a business perspective Well defined objective (or goal) Models to evaluate alternate options and for deciding on the optimal option Most businesses do not do this and outsource decisions are based on qualitative evaluation

32. EXCAVATORS CASE STUDY [Outsourcing Hydraulic Pumps]

33. EXCAVATORS Excavators are used in mining to load coal or ore on to dump trucks for transporting Hydraulic pumps operate the excavators Four pumps per machine Mine operator had four machines on site

34. MAINTENANCE OUTSOURCING The company selected on Scenario 1 where the owner decided on D-1 and D-2 Outsourcing the maintenance of hydraulic pumps PM action if a pump did not fail for 12,000 hours [based on manufacturer recommendation] CM action on failure

35. MAINTENANCE Both CM and PM maintenance results in the reconditioned pump being back to as- good-as new Some items were junked based on their condition whilst others were subjected either CM or PM action Customer used both new and reconditioned pumps

36. DATA ASPECTS Customer had failure data for items that failed and censored data (resulting from PM actions or discarding) No information on number of times a unit was subjected to maintenance action Some other information was also collected.

37. DATA ASPECTS There was no terms in the contract for the Service Agent to provide the owner with the state of items sent for PM action or the failure mode of items sent for CM action.

38. DECISION PROBLEM The cost of a CM action >> the cost of a PM action The owner was interested in seeing if the age for PM actions can be increased to 15,000 hours so as to reduce the maintenance costs paid to the Service Agent

39. DATA COLLECTION 6 year window yielded 103 data 46 failure data and 57 censored data. For each failure data, additional information relating to (i) the associated excavator (one of four different excavators), (ii) the pump position (one of four different positions) and, (iii) the engine (one of two) was also collected.

40. DATA COLLECTION For the 45 pumps that failed the following additional information was obtained. –15 are known to be new pumps –2 are suspected to be new pumps –8 are known to be reconditioned pumps –2 are suspected to be reconditioned pumps –19 are unknown

41. MODEL FORMULATION Based on WPP plot [Discussed in Part 2] The model selected was a mixture model Two cases: shape parameters (i) same and (ii) not same

42. WPP PLOTS – DATA AND MODEL [SHAPE PARAMETERS SAME]

43. WPP PLOTS – DATA AND MODEL [SHAPE PARAMETERS DIFFERENT]

44. MODEL PARAMETERS Model parameters obtained by least squares fit Select the one with the same shape parameters

45. MODEL ANALYSIS Two sub-populations MTTF given by ; Around 7.5 – 8.5% of items have early failures Reasons for early failures: –Particular machine and location? [some data available to test this] –Operating environment? [no data available]

46. OPTIMAL DECISION Optimum age for PM – can be derived using the well known PM policy Objective function: Asymptotic maintenance cost per unit time

48. IMPLICATIONS With current reliability the optimum age for PM is 15,000 hours with By proper understanding and identification of the root cause one can eliminate early failures In this case the reliability increases and the PM interval can be increased

49. REFERENCES Murthy, D.N.P., Xie, M. and Jiang, R. (2003), Weibull Models, Wiley, New York. –[Deals with many Weibull based models and the use of WPP plots for model selection.] Murthy, D.N.P. and Jack, N. (2008), Outsourcing of Maintenance, in Complex System Maintenance Handbook, K.A.H. Kobbacy and D.N.P. Murthy (eds), Springer Verlag, London,

50. Thank you Any Questions?