Spare Parts Provisioning

www.ipamc.org

Andrew K S JardineCBM Laboratory

Department of Mechanical & Industrial EngineeringUniversity of Toronto

Canadajardine@mie.utoronto.ca

August 2006

www.ipamc.orgAndrew Jardine, CBM Lab

“Spare Parts Provisioning”

Economic Order Quantities

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

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 total annual demand•Q order quantity•Co ordering cost / order•Ch stockholding cost per item / year

• Solution:

ChDCoQ 2* =∴

Q =order quantityCo =ordering costsCh =holding cost/item/yearD =annual demand

ExampleLet D = 1,000 itemsCo = $ 5.00Ch = $ 0.25

20025.0

510002*

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

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

Spare Parts Provisioning: Preventive Replacement Spares

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.

Construction of Modeltp 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.

The Constant Interval ModelN (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

The Age-Based Preventive Replacement ModelN (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:

)](1[)()(),(

ppppp tRtMtRt

TtTN−×+×

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.

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).

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

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

www.banak-inc.com

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).

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 Truck 251,220 45,38068060 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 βScale Parameter ηMean Life(c) What type of failure pattern is indicated (EARLY LIFE, RANDOM, WEAROUT?)

(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.

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 Replacement0 – 9.99 14 0

10 – 19.99 5 020 – 29.99 2 430 – 39.99 1 8

(a) Use OREST to analyze the failures and estimate the following parameters:

Shape Parameter βScale Parameter ηMean 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.

Spare Parts Provisioning: Insurance Spares

Real world research

Securing Canada's Energy Future

Managing Risk: A CBM Optimization Tool

Research Team

Collaborating ResearchersDr. Xiaoyue Jiang, Assistant Professor

Louisiana State University

Principal InvestigatorProf. Andrew K.S. Jardine

Research StaffDr. Dragan Banjevic, Project DirectorWei Hua (Walter) Ni, Programmer/AnalystDr. Daming Lin, Research AssociateDr. Ali Zuaskiani, Post doctoral FellowNeil Montgomery, Research AssociateSusan Gropp, Research Assistant

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

Spares Optimization Software

Spares Management Software

Interval StockReliability

Spares Management Software (SMS)

Optimal

Spares

Requirement

Optimization criteria

Instant. StockReliability

Availability

CostMinimization

SupportabilityStock

Remaining

NonNon--Repairable Repairable SparesSpares

Repairable Repairable SparesSpares

failures

Non-Repairable Spares

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

Repairable Spares# items in service

Repairshop

repaired units

Failures

failed units

Criteria for Decision Making1.Instant reliability

2.Interval reliability

3.Cost minimization

4.(Process) Availability

Definitions1. 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.

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?

Results: Repairable PartsRandomly 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]

Reference Case

1 yrInterval

0.001yrReplacement Time

1 yrRepair Time

0.005 failures/transformer/yr

Failure Rate

100 transformersPopulation

Repairable Instantaneous ReliabilityVary Spares

1 2 3 4 5

Spares

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

How many spares – Fume fan shaft?MTBF Vs Reliability with 22 week LT

0 5 10 15 20 25 30 35 40 45

Mean time between failures

0 spares1 spare2 spares

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.

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

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)

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)

SMS Problems1. 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 (roundedto 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

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

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

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

SMS referencesLouit, D, Banjevic, D and Jardine, A.K.S., (2005), Optimization of spare parts inventories composed of repairable or non-repairable 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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