1、_ 6$(7HFKQLFDO6WDQGDUGV%RDUG5XOHVSURYLGHWKDW7KLVUHSRUWLVSXEOLVKHGE6$(WRDGYDQFHWKHVWDWHRIWHFKQLFDODQGHQJLneering sciences. The use of this report is entirely voluntary, and its applicability and suitability IRUDQSDUWLFXODUXVHLQFOXGLQJDQSDWHQWLQIULQJHPHQWDULVLQJWKHUHIURPLVWKHVROHUHVSRQVLELOLWRIWKHXVHU
2、 SAE reviews each technical report at least every five years at which time it may be revised, reaffirmed, stabilized, or cancelled. SAE invites your written comments and suggestions. Copyright 2013 SAE International All rights reserved. No part of this publication may be reproduced, stored in a retr
3、ieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of SAE. TO PLACE A DOCUMENT ORDER: Tel: 877-606-7323 (inside USA and Canada) Tel: +1 724-776-4970 (outside USA) Fax: 724-776-0790 Email: Custo
4、merServicesae.org SAE WEB ADDRESS: http:/www.sae.org SAE values your input. To provide feedback on this Technical Report, please visit http:/www.sae.org/technical/standards/JA6097_201305 SURFACE VEHICLE/ AEROSPACE RECOMMENDED PRACTICE JA6097 MAY2013 Issued 2013-05 Using a System Reliability Model to
5、 Optimize Maintenance Costs A Best Practices Guide RATIONALE Complex repairable systems that consist of many individual components can be designed to be quite reliable, at least ZKHQWKHUHVWLOOQHZ+RZHYHUDVVXFKDVVWHPFRQWLQXHVWREHXVHGDQGUHSDLUVDUHPDGHWRDGGUHVVIDLOXUHVWKDWoccur, the interval between fai
6、lures gets shorter and shorter. Eventually, that complex system becomes a collection of parts, each with a different amount of operating time and differing reliabilities. Maintenance intervals that may have aligned with one another when the system was new will eventually become misaligned, making ma
7、intenance more complicated, or at least sub-optimal and more expensive. It is not uncommon to have a repaired system returned to service only to be shut down shortly thereafter because of another failure. The classical Reliability Centered Maintenance (RCM) process does a good job of focusing attent
8、ion on the actual reliability and failure modes that is (or will be) exhibited by a particular system, and helps a system designer, implementer, or operator develop an effective maintenance strategy for that system. However, the RCM decision logic only looks at the LQGLYLGXDOFRPSRQHQWVDQGGRHVQWDGGUH
9、VVKRZWRUHFRQFLOHGLIIHUHQWPDLQWHQDQFHLQWHUYDOVIRUHDFKFRPSRQHQW across the entire system. Furthermore, RCM focuses more on preventive maintenance (lubrication and failure finding tasks) and GRHVQWUHDOODGGUHVVFRUUHFWLYHPDLQWHQDQFH So when a complex system is taken out service for maintenance, there is
10、no guidance as to what other optional maintenance should be performed to increase the time until the next failure and/or reduce long-term operating costs. 7KLV%HVW3UDFWLFHV*XLGHGHVFULEHVDSURYHQDSSURDFKIRUREMHFWLYHOGHWHUPLQLQJZKDWRWKHUPDLQWHQDQFHVKRXOGEHperformed when a system is being repaired to im
11、prove system reliability and reduce long-term operating costs. FOREWORD The classical RCM analysis process is designed to look at each individual component of a complex system (such as an aircraft, military tank, ship, radar system, factory, etc.), determine how each component can fail, identify the
12、 consequences of each failure mode, and decide which maintenance strategy best meets the needs of each component in order to maximize the reliability of the system. The RCM maintenance strategy options are: impose a scheduled discard event, running a component to failure, initiating failure-finding
13、tasks, implementing a preventive maintenance program, or a one-time change (redesigning a part if its failure would pose a safety or environmental risk and such a failure cannot be mitigated through other means, operational change, etc.). At the end of this process, the analyst will have a structure
14、d, thorough, maintenance plan that will account for all parts of the system being evaluated. SAE JA6097 Issued MAY2013 Page 2 of 19 However, classical RCM (as defined LQ WKH FXUUHQW 5&0 6WDQGDUGV GRHVQW WDNH LQWR DFFRXQW VLWXDWLRQV ZKHUHmaintenance on one piece of equipment may disturb or drive main
15、tenance on one or more other pieces of equipment. It DOVRGRHVQWWHOOWKHDQDOVWZKHQWKH5&0DQDOVLVVSHFLILHVPDintenance intervals for various portions of the system that are all different from each other, how to reconcile those differences in such a way that will keep maintenance costs, down-time, or any
16、other system-wide maintenance optimization, to a minimum. Furthermore, it will not address how to best perform corrective maintenance i.e., how to decide what other optional tasks should be performed while the system is down for maintenance. Undoubtedly, there will be differences between different t
17、ypes of systems, users, operating locations, etc., and they all would have some kind of criteria to measure the cost-effectiveness of any maintenance decisions that are contemplated. The real issue is how to calculate the cost effectiveness in all those various situations. The costs themselves may b
18、e easy HQRXJKWRFRPSLOHEXWWKHHIIHFWLYHQHVVSDUWRIWKHHTXDWLRQLVXVXDOODELWPRUHGLIILFXOWWRGHWHUPLQH This guide is an attempt to document and explain one general method for determining the cost-effectiveness of various maintenance options that are possible when a complex, heavily integrated system is take
19、n off-line for maintenance. The techniques described herein have been developed for, and successfully applied to, aircraft and aircraft engines. They allow D FRPSDULVRQ RI D ZLGH YDULHW RI FRQVLGHUDWLRQV WR EH PDGH ZKHUHE WKH FRVW DQG WKH HIIHFWLYHQHVV DUHdetermined computationally (vice notionally)
20、, and gives system maintainers a more unambiguous recommendation for the maintenance that should be performed that will result in the lowest practical long-term maintenance cost. Since the techniques described herein have been successfully applied to aircraft and aircraft engines, the examples below
21、 will frequently refer to these applications. However, in the course of the following discussion, this guide will try to provide enough explanation of the issues involved so that the reader will be able to identify and apply the same techniques to their own situations and address any similar require
22、ments and constraints that pertain to them. TABLE OF CONTENTS 1. SCOPE 3 2. REFERENCES 3 2.1 Applicable Documents 3 2.2 Related Publications 3 2.3 Other Publications . 4 3. DEFINITIONS . 5 3.1 PROBABILISTIC . 5 3.2 STOCHASTIC . 5 3.3 SUNSHINE COSTS 5 3.4 WORKSCOPE 5 4. ACRONYMS 5 4.1 ATOW . 5 4.2 CB
23、M 5 4.3 CPFH/CPEFH . 5 4.4 ETOW . 5 4.5 LCF 5 4.6 LLP 5 4.7 OCM 6 4.8 PM . 6 4.9 RBD . 6 4.10 RCM 6 4.11 TSO/TSN . 6 5. BACKGROUND. 6 6. /,0,7$7,2162)&/$66,&$/5(/,$%,/,7RIWKHXVHIXOOLIHRIWKHEHOW The costs to defer the belt replacement would add-up as follows: x The cost of replacing the water pump $5
24、00 x Risk cost of in-service failure = $1-3k x F(t + dt | t) If we assume the conditional probability of failure is 5%, this would equate to $100 Risk Cost. SAE JA6097 Issued MAY2013 Page 15 of 19 On the next shop visit, the costs would be: x Separate visit to replace the belt $350 x Cost to replace
25、 the belt at next visit = $90 x Some sunshine risk cost = (Probability of finding an unexpected problem) x (cost to repair such a problem) If we assume the chance of finding an unexpected problem is 5%, and the average cost to correct such a problem in this area is $500, this would equate to a sunsh
26、ine risk cost of $25 The total expected cost of the option to defer replacement of the timing belt is approximately $600 (the cost of replacing the water pump plus the cost associated with the risk that the belt could fail sooner while it is still in service). If we assume the expected life until th
27、e subsequent shop visit to replace the belt is 35,000 miles, this equates to a long-term cost of $17.14 / 1000 miles. Even if it were based on the expected life after the belt was replaced, then it would be $1065 (the cost of this shop visit plus the next) divided by 35,000 + 90,000 miles or about $
28、8.52 / 1000 miles. The costs associated with replacing the timing belt now would be: x The cost of replacing the water pump $500 x Residual value of old belt $35 x Cost to replace the belt now = $90 x Additional labor to replace belt now $50 x Some sunshine risk cost = (Probability of finding an une
29、xpected problem) x (cost to repair such a problem) If we assume the chance of finding an unexpected problem is 5%, and the average cost to correct such a problem in this area is $500, this would equate to a sunshine risk cost of $25 Therefore, the total expected cost of the option to replace the tim
30、ing belt now would be about $700. With an expected life of 90,000 miles, this yields a long-term operating cost of $7.78 / 1000 miles which is the lowest cost option in the long run. The decision is ultimately based on a comparison between the expected cost of a minimal repair divided by the expecte
31、d life for that workscope vs. the cost of replacing the timing belt divided by the expected life after replacing the belt. This would be expressed as follows: t)Replacemen(BeltEt)Replacemen(BeltEvs.Repair)(MinimalERepair) (MinimalELifeCostLifeCost(Eq. 4) The term that yields the lower value (on the
32、left or the right side of the expression above) would be the optimum choice between these 2 options. 11. COST OPTIMIZATION EXAMPLE #2 REPAIR OF AN AIRCRAFT ENGINE 1RZLWVQRWKDUGWRXQGHUVWDQGWKDWWKHSURFHVVLVPXFKPRUHLQYROYHGIRUFRPSOHVVWHPVOLNHDQDLUFUDIWRUDLUFUDIWengine, but the basic elements are the sa
33、me, it just needs to be more methodical and structured since there are so many more options to consider. It also benefits from a graphical representation of all the different combinations of possible workscopes and the corresponding long-term costs. 5HFDOOWKH0DLQWDLQHUVLOHPPDZKLFKVKRZVVKRSYLVLWFRVWY
34、VZRUNVFRSH7KHSUREOHPLVWKDWWKLVFKDUWGRHVQWLGHQWLIDQRSWLPXPZRUNVFRSHVRLWFDQWDQVZHUWKHPDLQWDLQHUVTXHVWLRQ What other repairs should be done at this time? SAE JA6097 Issued MAY2013 Page 16 of 19 FIGURE 3 The problem is that shop visit cost is not the right basis for comparison the comparison should be l
35、ooking at the long-term return-on-investment. For aircraft operations, the standard basis is Cost-Per-Flying-Hour (CPFH). For ground vehicles (cars, trucks, etc.), it could be cost per mile or some other measure of goodness. FIGURE 4 When the CPFH of the various workscope options are plotted against
36、 Average Time On-:LQJ $72: WKH EHVWworkscope option (in terms of long-term cost per flying hour) can be clearly identified. In order to do this, it is necessary to first determine the costs for each possible workscope. Recall that the costs include (at a minimum) the material & labor to effect repai
37、rs to address the primary reason for removal plus any additional optional work, whether it is performed to improve reliability or not. The costs may also include depreciation over the expected life of the component, the residual value of parts removed while they still have some useful life remaining
38、, costs associated with risk, and Sunshine effect costs. :KHQDOORIWKHZRUNVFRSHRSWLRQVDUHVKRZQZLWK&3)+SORWWHGDJDLQVW(72:WKHUHVXOWVDSSHDUVDVDVFDWWHUSORWZKHUHWKHORZHVWSRLQWUHSUHVHQWVWKHRSWLPXPFKRLFH1RWHWKDWWKHSUHYLRXVFKDUWUeally represents the lower bounds of the scatter plot below. Fix only what broke
39、 Overhaul everything whether it needs it or not Too Little Maintenance Poor Reliability Too Much Maintenance High Cost Shop Visit Cost High Low Maintenance Workscope Light Heavy SAE JA6097 Issued MAY2013 Page 17 of 19 FIGURE 5 The following chart is an example of a mid-thrust turbofan engine that ha
40、d previously been inducted for repair. All of the data points shown represent the expected life (Expected Time On-Wing or ETOW) and the long-term cost (CPFH) for each possible workscope option for this engine, given the accumulated operating time and condition of all the modules and components insta
41、lled in that engine. Comparing the CPFH and ETOW for the repairs actually performed (shown in the JUHHQFLUFOH DQGWKHRSWLPXPZRUNVFRSH VKRZQLQWKHUHGFLUFOH WKHRSWLPXPZRXOGKDYHUHVXOWHGLQDORZHUshop visit cost (about $1.1M in this case) and a 33% reduction in CPFH, while giving up only 7% in terms of ETOW
42、. FIGURE 6 Estimated Time on Wing (ETOW) Estimated Cost per HourOptimum Workscope X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X $1,050 $1,250 $1,450 $1,650 $1,850 Cost Per Hour 2000 2100 2200 2300 2400 Estimated Time On Wing (ETOW), Hours $850 1900 Optimal Workscope: ETOW = 2044 Cost/Hr = $888.53 Actual Workscope: ETOW = 2197 Cost/Hr = $1327.55 SAE JA6097 Issued MAY2013 Page 18 of 19 12. SUMMARY Traditional RCM d
