Spare Parts Provisioning

Spare Parts Provisioning

Andrew K S Jardine CBM Laboratory Department of Mechanical & Industrial Engineering University of Toronto Canada [email protected] August 2006

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Economic Order Quantities

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The stores controller wants to determine which order quantity will minimize the total cost.

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This total cost can be plotted and used to solve the problem. A much more rapid solution, however, is to construct a mathematical model of the decision situation. The following parameters can be defined: •D •Q •Co •Ch

total annual demand order quantity ordering cost / order stockholding cost per item / year

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• Solution: 2 DCo ∴Q = Ch *

Q =order quantity Co =ordering costs Ch =holding cost/item/year D =annual demand www.ipamc.org Andrew Jardine, CBM Lab

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Example Let D = 1,000 items Co = $ 5.00 Ch = $ 0.25

2 x1000 x 5 ∴Q = 0 .25 = 200 items *

Thus, each time the stock level reaches zero, the stores controller should order 200 items to minimize the total cost per year of ordering and holding stock www.ipamc.org Andrew Jardine, CBM Lab

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Extensions • • • • •

Non-zero lead time Quantity discounts Back orders allowed Uncertainty in demand Useful reference to inventory control: Nahmias, S., (1997), Production and Operations Analysis, Chicago, Irwin/McGraw-Hill www.ipamc.org Andrew Jardine, CBM Lab

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Spare Parts Provisioning: Preventive Replacement Spares

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If preventive maintenance is being conducted on a regular basis according to either the constant interval or age-based replacement models then a spare part is required for each preventive replacement, but in addition, spare parts are required for any failure replacements. The goal of this section is to present a model that can be used to forecast the expected number of spares required over a specified period of time, such as a year, for a given preventive replacement policy. www.ipamc.org Andrew Jardine, CBM Lab

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Construction of Model tp is the preventive replacement time (either interval or age). f(t) is the probability density function of the item’s failure times. T is the planning horizon, typically one year. N (T, tp) is the expected number of spare parts required over the planning horizon, T, when preventive replacement occurs at time tp. www.ipamc.org Andrew Jardine, CBM Lab

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The Constant Interval Model N (T, tp) = Number of preventive replacements in interval (0,T) + number of failure replacements in interval (0,T) = T / tp + H (tp ) (T / tp ) where H (tp) is the expected number of failure replacements during an interval of length tp www.ipamc.org Andrew Jardine, CBM Lab

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The Age-Based Preventive Replacement Model N (T, tp) = Number of preventive replacements in interval (0,T) + number of failure replacements in interval (0,T). In this case the approach to take is to calculate the expected time to replacement (either preventive or failure) and divide this time into the planning horizon, T. This gives: N (T , t p ) =

T t p × R(t p ) + M (t p ) × [1 − R(t p )] www.ipamc.org Andrew Jardine, CBM Lab

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An Application: Cylinder Head Replacement – Constant Interval Policy A cylinder head for an engine costs $1,946 and the policy employed is to replace the 8 cylinder heads in an engine as a group at age 9,000 hours, plus failure replacement as necessary during the 9,000-hour cycle. In the plant there were 86 similar engines in service. Thus, over a 12 month period there is total component utilization of 8 × 86 × 8,760 = 6,026,880 hours worth of work. www.ipamc.org Andrew Jardine, CBM Lab

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Estimating the failure distribution of a cylinder head, and taking the cost consequence of a failure replacement as ten times that of a preventive replacement, it was estimated that with the constant interval replacement policy , the expected number of spare cylinder heads required per year to service the entire fleet was 799 (576 due to preventive replacement and 273 due to failure replacement). www.ipamc.org Andrew Jardine, CBM Lab

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The constant interval and age-based models will be illustrated using the OREST software

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OREST Software Educational version from CRC Press, publisher of Maintenance, Replacement & Reliability: Theory and Applications, by A K S Jardine & A H Tsang, 2006 www.crcpress.com/e_products/downloads.asp?cat_no=DK9669 Or

www.banak-inc.com www.ipamc.org Andrew Jardine, CBM Lab

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OREST Educational Version Limitations

Cost ratio fixed at: Cf = $1000, Cp = $ 100.

At most 6 observations can be analyzed (mixture of preventive replacements and suspensions). www.ipamc.org Andrew Jardine, CBM Lab

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1 Records from two heavy-duty dumper trucks show that fan belt failures occurred at the odometer readings (kilometers, from new) listed in the following table: Truck 1 51,220 68060

Truck 2 45,380 103,510

At present, the odometer readings are 115,680 km for truck 1 and 132,720 km for truck 2. (a) Prepare reliability data in a form suitable for analysis by OREST. (b) Determine the following Weibull parameters: Shape Parameter β Scale Parameter η Mean Life (c) What type of failure pattern is indicated (EARLY LIFE, RANDOM, WEAROUT?) www.ipamc.org Andrew Jardine, CBM Lab

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(d) The Preventive Replacement Cost is $100 and the Failure Replacement Cost is $1,000. Determine the optimal preventive replacement age, the cost under this policy, and the savings under this policy when compared with a policy of replacement-only-on failure. (e) Preventive replacement can only be carried out at odometer readings which are multiples of 5,000 km. Select an appropriate preventive replacement age. What is the cost ($/km) for this policy? How does this compare with the cost for the optimal policy? (f) If the company has a fleet of 30 similar dump trucks, each of which averages 50,000 kilometers per year, estimate the number of replacement fan belts that will be needed per year, under an appropriate replacement policy. (g) If 30 dump trucks average 50,000 kilometers per year, estimate the number of in-service fan belt failures that will occur, given that the policy is to replace fan belts on a preventive basis at 20,000 kilometers. www.ipamc.org Andrew Jardine, CBM Lab

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2. The cloth filter on a sugar centrifuge is currently replaced on a preventive basis if a suitable opportunity occurs and the cloth has been in use for at least 20 hours. The cloth is also replaced on failure. The centrifuge cloth failure data provided in the following Table are available for 10-hour time intervals of cloth life.

Age in Hours Failure Replacement Preventive Replacement 0 – 9.99 14 0 10 – 19.99 5 0 20 – 29.99 2 4 30 – 39.99 1 8

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(a) Use OREST to analyze the failures and estimate the following parameters: Shape Parameter β Scale Parameter η Mean Life (b) Is the current policy correct? What policy do you recommend? (c) The company has three centrifuges each of which runs an average of 400 hours per month. Estimate the number of replacement cloths required per month under the existing and under the recommended replacement policies.

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Spare Parts Provisioning: Insurance Spares

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Real world research Managing Risk: A CBM Optimization Tool

Securing Canada's Energy Future

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Research Team

Research Students

Research Staff

Diederik Lugtigheid (Repairable Systems) Darko Louit (Spare Parts Optimization) Jean-Paul Haddad (Research Topic TBA) Andrey Pak (Maintenance & Repair Contracts)

Dr. Dragan Banjevic, Project Director Wei Hua (Walter) Ni, Programmer/Analyst Dr. Daming Lin, Research Associate Dr. Ali Zuaskiani, Post doctoral Fellow Neil Montgomery, Research Associate Susan Gropp, Research Assistant

Principal Investigator

Collaborating Researchers

Prof. Andrew K.S. Jardine

Dr. Xiaoyue Jiang, Assistant Professor Louisiana State University

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Spares Optimization Software

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Spares Management Software

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Spares Management Software (SMS) Optimization criteria

Non-Repairable Spares

Interval Stock Reliability Instant. Stock Reliability

Optimal Spares

Repairable Spares

Cost Minimization

Requirement

Availability

Stock Supportability

Remaining Life www.ipamc.org

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Non-Repairable Spares

failures

stock

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With non-repairable components, when a component fails or has been preventively removed, it is immediately replaced by one from the stock (the replacement time is assumed to be negligible), and the replaced component is not repaired (i.e. it is discarded, see Figure 2.34). It is assumed that the demand for spares follows a Poisson process, which, for emergency parts demand, has found wide application. Several references describe models based on this principle www.ipamc.org

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# items in service

Repairable Spares Failures time

stock failed units

repaired units Repair shop

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Criteria for Decision Making 1.Instant reliability 2.Interval reliability 3.Cost minimization 4.(Process) Availability www.ipamc.org Andrew Jardine, CBM Lab

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Definitions 1. Instantaneous reliability. This is the probability that a spare is available at any given moment in time. In some literature this is known as availability of stock, fill rate or point availability in the long run. 2. Interval reliability. This is the probability of not running out of stock at any moment over a specified period of time, such as one year. 3. Cost minimization. This takes into account costs associated with purchasing and stocking spares, and the cost of running out of a spare part. 4. Availability. This is the percentage of non-downtime (“uptime”) of a system/unit where the downtime is due to shortage of spare parts. www.ipamc.org Andrew Jardine, CBM Lab

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Scenario • Plant has 62 electric motors on their conveyor systems (Mining company) • MTBReplacements of motors is 3000 days (8 years) • Planning horizon is 1825 days (5 years) • Cost of spare motor is 15,000 $ • Value of unused spare is 10,000 $ • Cost of emergency spare is 75,000 $ • MTTRepair a motor is 80 days • Cost of plant downtime for a single motor is 1000 $ per day • Holding cost of a spare is 4.11 $ per day (10% of value of part/annum)

QUESTION: HOW MANY SPARE PARTS TO STOCK? www.ipamc.org Andrew Jardine, CBM Lab

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Results: Repairable Parts Randomly failing motors • Interval reliability: 95% reliability requires 7 spares • Instant reliability: 95% reliability requires 4 spares • Cost minimization: requires 6 spares. Associated plant availability is 100.00% • Availability of 95%: requires 0 spares. Associated electric motor availability is 97.4% [Note: If availability of 99% was required (rather than the specified 95%) then spares required would be 2] www.ipamc.org Andrew Jardine, CBM Lab

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Reference Case Population

100 transformers

Failure Rate Repair Time

0.005 failures/transformer/yr 1 yr

Replacement Time

0.001yr

Interval

1 yr www.ipamc.org Andrew Jardine, CBM Lab

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Repairable Instantaneous Reliability Vary Spares 1 0.99 0.98

Reliability

0.97 0.96 0.95 0.94 0.93 0.92 0.91 0.9 1

2

3

4

Spares Andrew Jardine, CBM Lab

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•

Fume fan shaft – steel mill Spares provisioning optimization project • • •

Part: fume fan shaft used in a Blast Furnace Decision: should there be 0 or 1 spares? Complication: • Part has long lifespan (25-40 years). • Long lead time (22 weeks). • If part fails, results are catastrophic (loss of almost $6 million per week). • Inventories are trying to be minimized.

SMS was used to quantify the risk involved in not having a spare Decision support Andrew Jardine, CBM Lab

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How many spares – Fume fan MTBF Vsshaft? Reliability with 22 week LT 100.5 100

R eliability

99.5 0 spares

99

1 spare 98.5

2 spares

98 97.5 97 0

5

10

15

20

25

30

35

Mean time between failures Andrew Jardine, CBM Lab

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Problem: Dry Gas Seal- used in a compressor

• Repairable component

• 10 in use in petrochemical plant (2 in each compressor) • MTBReplacements: 60 months • MTTRepair: 10 months • Cost new: $200,000 • Planning horizon: 5 years Use SMS to determine how many spares to stock. www.ipamc.org Andrew Jardine, CBM Lab

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Spares Management Software (SMS) Educational version from CRC Press, publisher of Maintenance, Replacement & Reliability: Theory and Applications, by A K S Jardine & A H Tsang, 2006 www.crcpress.com/e_products/downloads.asp?cat_no=DK9669

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SMS Educational Version Limitations

Non-Repairable Spares

50 parts in use (fixed) Day time unit (fixed) Interval reliability (reliability required of 95% fixed, cannot select spares in stock) Cost calculation (cannot select spares in stock, $15,000 regular cost of spare part fixed, $ 75,000 emergency cost of spare part fixed, 0.0288617289% cost of capital per day fixed, $1,4913.6518075166 future value of unused spare fixed)

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SMS Educational Version Limitations (cont.)

Repairable Spares

50 parts in use (fixed) Day time unit (fixed) Interval reliability (reliability required of 95% fixed, cannot select spares in stock) Cost calculation (cannot select spares in stock,$ 8.22 holding cost of one spare part per day fixed) Instant reliability (reliability required of 95% fixed, cannot select spares in stock) Availability calculation (availability required of 98% fixed, cannot select spares in stock)

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SMS Problems 1. A manufacturing plant uses a total of 50 optical sensors to identify different part geometries. The sensors cannot be repaired easily so they must be replaced when they fail. On average, a sensor lasts 2 years, assuming the parts fail completely at random. (Assume 356 days a year) How many sensors are expected to fail over 4 months (rounded to nearest integer value)? How many spares will the company need to keep in stock if they require at least a 95% reliability over 6 months? How many spares will the company need to keep in stock if they require a value as close to 95% reliability as possible over 6 months? Answers: 8,19,18 www.ipamc.org Andrew Jardine, CBM Lab

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2. A factory uses 50 presses to manufacture shoes. The presses are repairable, and fail on average every 5 years. A press takes about a week to repair. The downtime cost is $15,000 an hour, and the holding cost is $3,000 per year (assume 52 weeks are in a year, the factory operates 24 hours a day, 7 days a week). How many spares are required to achieve a reliability of 95% over 30 weeks? If two presses are in stock, what is the probability that a shortage in spare parts will occur over 25 weeks? How many spare presses are required for a 95% instant reliability? How many spare presses kept in stock would result in the minimum cost? What is the minimum cost? If the company is only interested in at least a 98% availability, how many spares should be kept in stock? Answers: 3, 7.71%,1 ,4 ,$33.56 per day,0 Andrew Jardine, CBM Lab

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3. A clothing company uses 50 presses to put labels and graphics on t-shirts. A component of the presses was poorly designed, and it causes the presses to wear out and need to be replaced about every three years (the employee who chose these presses was promptly fired). The company requires 95% reliability over a year. How many presses should they keep in stock? Answer: 24

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4. A company uses the same size ball bearings on type machines A and type machines B. There are 5 type A machines and 2 type B machines in use at all times. Type A machines utilize 10 ball bearings each, while type B machines use 25 ball bearings each. If the company uses quarter year planning horizons and requires as close to a 95% reliability as possible, how many ball bearings should be kept in stock? Ball bearings need to be replaced on average every 6 months and 9 months for type A and B machines, respectively. Assume 365 days in a year. Answer: 56

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SMS references Louit, D, Banjevic, D and Jardine, A.K.S., (2005), Optimization of spare parts

inventories composed of repairable or nonrepairable parts. Proceedings, ICOMS,

Australia, 2005. Wong, J.Y.F., Chung, D.W.C., Ngai, B.M.T., Banjevic, D. and Jardine, A.K.S. (1997)

Evaluation of Spares Requirements Using Statistical and Probability Analysis Techniques, Transactions of Mechanical Engineering, IEAust. Vol.22(3 & 4), 77-84

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