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RELIABILITY ENGINEERING UNIT

ASST4403

LECTURE 32 SPARE PARTS CALCULATIONS

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Determine optimal stock holdings for preventive

Determine optimal stock holdings for insurance

(emergency) spares

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Main considerations of spares provisioning

1. Failure rate determines quantity and perhaps locationof s ares.

2. Acceptable probability of stockout fixes spares level.

3. Turnaround of second-line repair affects lead time. .

item 2.

. affects number of different spares to be held.

6. Lead time on ordering effectively part of second-linerepair time. 3

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Given a reventive re lacement olic either interval or

age) applied on a regular basis, what is the expected

number of spare parts required over a certain time period?

How many critical items (with/without PR) should bestocked in order to assure the followin ?

Availability of stock Interval availability Minimised cost Equipment availability

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Preventive replacement spares

A model to predict the expected number of spare partsrequ re over a me per o

T the planning horizon e.g., 1 year

E[N(T, t p)] expected number of spare parts requiredover T when reventive re lacement occurs at time t

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The interval/age replacement situation

For interval replacement:

E[ N(T, t )] =number of preventive replacements in(0,T) + expected number of failures in (0,T)

1 1 , for 1110

and 0 0 For age replacement, the expected cycle length is

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Numeric exam le Failures follow N(5,1) (weeks) and T=12 months=52 weeks

The optimal interval is 4 weeks

Solution:

We need to find H(4). Note H(0)=011 1 1

1 1 1 1 0 1 0 0

, ,it is equivalent to finding the value of standard normal

distribution in (-5,-4), that is 10 4 5 0 0 01

1 1 0 0 0

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Numeric example (contd). Similarly

2 1 21 3 4 0.00135 0 0.00135

2 1 2 1 0 1 2 0 1 1 1 4 5 1 0 3 4 1 0 0 1 0 0.00135 0.00135

H(4) =0.16

Now we need to find, where is the ordinate, =CDF

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Numeric example Failure = N(5,1), i.e., =5 weeks and =1 week. For t p=1

Calculating directly or from the normal ordinate table

1 4 0.0001

1 0.0001 5 0.00003 0.00005Similarly for t p=4, we have =0.551

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Numeric exam le Failures follow N(5,1) (weeks) and T=12 months=52 weeks

The optimal interval is 4 weeks

Solution:

With interval replacement, we have earlier

,

=0.84

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nsurance emergency spares

the failure replacement, not preventive replacement

running out stock of a spare part against the cost of

The stocked spare parts will be optimal with regards

Availability of stock

Minimised cost Equipment availability

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Model in uts assum tions

T length of a time period under consideration mean time to failure of component

N(T,m) total number of failures in interval (0,T)

S(k,m) time until the r th failure

Disregard the time it takes for the replacement

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S ecial case when T is lar e

When the time period T is large compared to /m, wecan use norma str ut on to approx mate

S(k,m) has an approximate normal distribution withmean k/m and variance 2k/m for large k

If we demand an availability of stock p, then we can find

where Z p is the value corresponding to cumulativestandard normal distribution value being 1 p.

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Failures follow N(5,1) (weeks) and T=12 months=52

wee s. e assume t e eet s ze m= an requ rethe probability of having spares in stock in T=52

,

For p=0.95, 1 p=0.05, which gives us Zp= 1.65

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Special case by a Poisson

s r u on approx ma on e a ure s r u on s exponen a w a e ng e

expected number of failures in (0,T) and assume a not a,

from which we can solve the k for a specified probability

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Applying Poisson distribution

If the underlying failure distributions are exponential , thenum er o a ures ,m o ows exact y o sson processfor any m, i.e. ai

ea

imT N P

)),(( Where a= expected number of failures in [0,T]. For m=1,

a=T/ and for m components, a=mT/ (a is assumed not toe very arge

We calculate k such thatk

i

ai

peia

T mk S Pk mT N P0 !

)),1(()),((

The obtained value of k will be the minimum stock levelthat ensures a reliability of p (probability of not having a

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Spare part requirement

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Exam le Equipment has K=20 parts, operating 24 h per day

=0.1 /1000h

S ares to be calculated for T=3 month interval

P=95% availability of having spare when required

Answer: = . ,

P=95% on the far-right vertical, thus finding S = wanted

number of spares on scale (7), i.e., 8 Or 1) locate K on the far-left vertical scale, go to scale (2)

project to cross point on (3) to extend to (5) to find K T,then follow the steps in above bullet21