corrective maintenance with spare parts

8/7/2019 Corrective Maintenance with Spare Parts

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Corrective Maintenance and

Spare Parts Decisions

• No preventive maintenance or

preventive replacement will be

conducted.

• Repair and maintenance will be

conducted whenever a failure occurs.

• This strategy may be the cheapest one

for non-critical components, devices,

and systems.

CorrectiveMaintenanceCorrective

Maintenance

• What is the expected number of

failures in a certain time interval?

• What is the optimal level of spare

parts inventory?

• How often and how much should one

order such spare parts?

• How to make these decisions for

different lifetime distributions?

Relevant DecisionsRelevant Decisions

• There are several units of such a

device that are used in the

organization.

• Their failures are independent.

• Whenever there is a failure, a spare

part is used to replace the device.

• We don’t like to see a stockout.

General AssumptionsGeneral Assumptions

Data NeededData Needed

• How expensive is each item?

• What is the average lifetime?

• What is the lifetime distribution?

• How long does it take to order and get

one?

• How high is the ordering cost?

• How high is the storage cost for spares?

The Exponential DistributionThe Exponential Distribution



Exponential DistributionExponential Distribution

• f(t)

• h(t)

• The memoryless properly

• x = 0, 1, 2, …

• E(X) = ρ, Var(X) = ρ• If a device has exponential lifetime

distribution, and each failed unit is

replaced by a new one right away with

negligible time, then the number of

replacements in any interval of length t

follows the Poisson distribution with ρ =

λt.

Poisson DistributionPoisson Distribution

( )!

xe

f xx

ρ ρ −

=

When N Units Are UsedWhen N Units Are Used

• The number of failure replacements

needed in interval t follows the

Poisson distribution with ρ = Nλt.

• When N is large (greater than 30), the

Poisson distribution here can be

approximated by the normal

distribution.

Example 3.1Example 3.1

• You are looking after 20 units of a

certain device. It is used 24 hours a

day and has a constant failure rate of

0.004/hr. You are considering

ordering spare parts once a month (30

days). What is the quantity to order to

ensure the probability of running out

of spare parts in a month to be less

than 1%? (see Excel file too).

Use the Normal ApproximationUse the Normal Approximation



Practical IssuesPractical Issues

• Assume the starting inventory is zero.

Order 82 units for the first month.What could happen during this

month? What do we do

correspondingly?

• What should we do in the following

months?

• Lead time considerations?

Annual Cost of this Practice?Annual Cost of this Practice?

• Ordering cost: $1000

• Storage cost: $0.05/unit/hour

• Shortage cost: $0

Optimal Ordering PolicyOptimal Ordering Policy

• How many times to order? Or what is

the ordering interval?

• What are the issues? How to find such

a policy?

• What are the assumptions? Are they

valid?

• Remember: Exponential Distribution

The Weibull DistributionThe Weibull Distribution

Weilbull DistributionWeilbull Distribution

• PDF f(x):

• Mean value: E(X) = αΓ(1+1/ β)

• Variance:

Var(X) = α2{Γ(1+2/ β)–[Γ(1+1/ β)]2}

β

β 1 , x 0f(x)

0 ,

x

-x

Otherwise

e α

β

β

α

⎛ ⎞−⎜ ⎟⎝ ⎠

⎧⎪ >= ⎨⎪⎩



Weilbull DistributionWeilbull Distribution

• Failure rate function h(x):

• h(x) = β xβ-1/αβ

• β< 1:

• β = 1:

• β > 1:

If Only One Device Is UsedIf Only One Device Is Used

• Suppose β = 2, α=1000, 24 hr/day

Weibull Probability Density Function

0.000000

0.000100

0.000200

0.000300

0.000400

0.000500

0.000600

0.000700

0.000800

0.000900

0.001000

0 5 00 1 00 0 1 5 00 2 00 0 2 5 00 3 00 0 3 5 00

Time (x)

pdf function f(x)

• Run to failure• How many

failures a year?

• How many to

order each year?

• What is the probability of shortage?

One Device AnswersOne Device Answers

• Analytical solutions would be very complicated.

• Use computer simulation

• Generate a random variable value x1 following the pdf

function, x2, etc, until the sum is longer than a year’s

operation time, now you have N1. Repeat until you reach

Nn. From these Ni values, you find the probability

distribution function of N, the # of failures in a year.

• From this distribution function, you can find the

probability that the number of failures is greater than a

specified value. You can then find the annual order

quantity to ensure a certain shortage probability.

• See next page for illustration

One Device SituationOne Device Situation

• Relevant decision questions:

• What is the cost of the current

practice?

• What is the optimal order quantity?

• What is the optimal order interval?

• What if more than one device is used?

We can use the same approach.

Consider n DevicesConsider n Devices

• Suppose n is relatively large: > 30

• Let Yi denote the number of failures

that may be experienced by device i in

a certain time interval

• Y = Y1 + Y2 + … + Yn, may be

approximated by the normal

distribution



ExampleExample

• Consider 30 devices. The lifetime of

each one follows the Weibull

distribution with α= 3000, β=3. We

use the run to failure strategy. Consider

ordering once a year. Ordering cost =

$500. Inventory cost = $0.012/unit/hr.

Shortage probability < 2%. How many

to order? What is the total cost?

DiscussionsDiscussions

• Assumptions: Weibull distribution?

• Other issues to be considered?

corrective maintenance with spare parts

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