RELIABILITY ENGINEERING UNIT ASST4403
LECTURE 32 SPARE PARTS CALCULATIONS
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• • Determ Determine ine optima optimall stock stock holdin holdings gs for for preven preventiv tive e • Determ Determine ine optima optimall stock stock holdin holdings gs for for insura insurance nce (emergency) spares
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Main considerations of spares provisioning 1. Failure rate – determines quantity and perhaps location of 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-line repair time.
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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 be stocked 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 parts requ re over a me per o • T
the planning horizon e.g., 1 year
• E[N(T, tp)] expected number of spare parts required over 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)
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1
1 1 , for 1 and 0 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)=0
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1
1 1 1 1 1 0 0
•
1 0
, , it is equivalent to finding the value of standard normal distribution in (-5,-4), that is 1 4 5 0 0 0 0
1 1 0
1 0 0
.
Numeric example (cont’d) • 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
Numeric example • Failure = N(5,1), i.e., =5 weeks and =1 week. For tp=1
Calculating directly or from the normal ordinate table 4 0.0001
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1 0.0001 5 0.00003 0.00005
∞ Similarly for tp=4, we have
=0.551
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
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 11
Series of events (point processes) • Discrete events that occur randomly in a continuum such • The arrival values x1, x2, …, xn are the values of the independent variables e.g. time from x=0 at which each event occurs • The inter-arrival values X1, X2, …, Xn are the intervals between successive events 1, 2, …,n from x=0. • Considering time as the independent variable, we use the concept of counting process. • We denote by N(t) the number of events occurring in important concept is the time until the kth event S(k).
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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 rth failure
• Disregard the time it takes for the replacement 14
Model development
• B relatin N T m to S k m the robabilit of not runnin out of stock of size k, for a fleet of m components in a time period T, is • For a probability target and failure distribution, we can determine a stock size k or how long a stock of size k will last, i.e., maximum T for a given stock size k. • N(T,m) or S(k,m),which is complicated. Approximation is often 15 suggested
S ecial case when T is lar e • When the time period T is large compared to /m, we can use norma str ut on to approx mate • S(k,m) has an approximate normal distribution with mean k/m and variance 2k/m for large k
• If we demand an availability of stock p, then we can find
• where Zp is the value corresponding to cumulative standard 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 re the 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 , the num er o a ures ,m o ows exact y o sson process i a a for any m, i.e. P ( N (T , m) i )
e
• Where a= expected number of failures in [0,T]. For m=1, a=T/ and for m components, a=mT/ (a is assumed not to e very arge • We calculate k such that k
P ( N (T , m) k ) P ( S (k 1, m) T )
a
i
i! e
a
p
i 0
• The obtained value of k will be the minimum stock level that 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 KT, then follow the steps in above bullet
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