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1 Stracener_EMIS 7305/5305_Spr08_01.31.08 Reliability Models & Applications (continued) Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis Systems Engineering Program Department of Engineering Management, Information and Systems

Reliability Models & Applications (continued)

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Systems Engineering Program. Department of Engineering Management, Information and Systems. EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis. Reliability Models & Applications (continued). Dr. Jerrell T. Stracener, SAE Fellow. Leadership in Engineering. - PowerPoint PPT Presentation

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Page 1: Reliability Models & Applications (continued)

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Reliability Models & Applications (continued)

Dr. Jerrell T. Stracener, SAE Fellow

Leadership in Engineering

EMIS 7305/5305Systems Reliability, Supportability and Availability Analysis

Systems Engineering ProgramDepartment of Engineering Management, Information and Systems

Page 2: Reliability Models & Applications (continued)

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The Normal or Gaussian Model:

• DefinitionA random variable T is said to have the Normal (Gaussian) Distribution with parameters and , where > 0, if the density function of T is:

, for - < t <

• DefinitionIf T ~ N(,) and if , then Z ~ N(0,1)

the Standard Normal Distribution and Cumulative Probability is tabulated for various values of z.

22

t2

1

e2

1 )t(f

-T

Z

Page 3: Reliability Models & Applications (continued)

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Properties of the Normal Model:

• Probability Distribution Function

Where (Z) is the Cumulative Probability DistributionFunction of the Standard Normal Distribution.

• Reliability Function

provided that P(T < 0) 0

t

)t(F

t

-1 )t(R

Page 4: Reliability Models & Applications (continued)

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Properties of the Normal Model:

• MTBF (Mean Time Between Failure)

• Standard Deviation of Time to Failure =

• Failure Rate

MTBF

)t(R

)t(f )t(h

Page 5: Reliability Models & Applications (continued)

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Properties of the Normal Model - Failure Densities:

Page 6: Reliability Models & Applications (continued)

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The Normal Model - Example

Example=1,000=100

Page 7: Reliability Models & Applications (continued)

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0

z

z

Normal Distribution:

Standard Normal Distribution ~ X ~ N (, )

Page 8: Reliability Models & Applications (continued)

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Properties of the Normal Model - Standard Normal Distribution:

-t

Z

Table of Probabilities p

Page 9: Reliability Models & Applications (continued)

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Standard Normal Distribution0 0.01 0.02 0.03 0.04 0.09 0.06 0.07 0.08 0.09

-5.4 3.3396E-08 3.1585E-08 2.9868E-08 2.8243E-08 2.6703E-08 2.0146E-08 2.3864E-08 2.2556E-08 2.1318E-08 2.0146E-08-5.3 5.8022E-08 5.4928E-08 5.1994E-08 4.9211E-08 4.6574E-08 3.5308E-08 4.1702E-08 3.9455E-08 3.7326E-08 3.5308E-08-5.2 9.9834E-08 9.4602E-08 8.9635E-08 8.4921E-08 8.0447E-08 6.1285E-08 7.2173E-08 6.835E-08 6.4724E-08 6.1285E-08-5.1 1.7012E-07 1.6136E-07 1.5304E-07 1.4513E-07 1.3762E-07 1.0535E-07 1.237E-07 1.1726E-07 1.1115E-07 1.0535E-07

-5 2.871E-07 2.7258E-07 2.5877E-07 2.4564E-07 2.3315E-07 1.7934E-07 2.0998E-07 1.9924E-07 1.8904E-07 1.7934E-07-4.9 4.7987E-07 4.5604E-07 4.3335E-07 4.1175E-07 3.912E-07 3.0237E-07 3.53E-07 3.3528E-07 3.1841E-07 3.0237E-07-4.8 7.9435E-07 7.5564E-07 7.1874E-07 6.8358E-07 6.5007E-07 5.0489E-07 5.8774E-07 5.5877E-07 5.3117E-07 5.0489E-07-4.7 1.3023E-06 1.24E-06 1.1806E-06 1.1239E-06 1.0699E-06 8.3497E-07 9.6916E-07 9.2228E-07 8.7758E-07 8.3497E-07-4.6 2.1146E-06 2.0155E-06 1.9207E-06 1.8303E-06 1.7439E-06 1.3676E-06 1.5828E-06 1.5077E-06 1.436E-06 1.3676E-06-4.5 3.4008E-06 3.2444E-06 3.0949E-06 2.952E-06 2.8154E-06 2.2185E-06 2.5602E-06 2.4411E-06 2.3272E-06 2.2185E-06-4.4 5.417E-06 5.1728E-06 4.9392E-06 4.7156E-06 4.5018E-06 3.5644E-06 4.1016E-06 3.9145E-06 3.7355E-06 3.5644E-06-4.3 8.546E-06 8.1687E-06 7.8072E-06 7.461E-06 7.1295E-06 5.6721E-06 6.5082E-06 6.2172E-06 5.9387E-06 5.6721E-06-4.2 1.3354E-05 1.2777E-05 1.2223E-05 1.1692E-05 1.1183E-05 8.94E-06 1.0228E-05 9.7804E-06 9.3512E-06 8.94E-06-4.1 2.0669E-05 1.9794E-05 1.8954E-05 1.8148E-05 1.7375E-05 1.3956E-05 1.5922E-05 1.5239E-05 1.4584E-05 1.3956E-05

-4 3.1686E-05 3.0374E-05 2.9113E-05 2.7902E-05 2.6739E-05 2.158E-05 2.4549E-05 2.3519E-05 2.253E-05 2.158E-05-3.9 4.8116E-05 4.6167E-05 4.4293E-05 4.2491E-05 4.0758E-05 3.3052E-05 3.7491E-05 3.5952E-05 3.4473E-05 3.3052E-05-3.8 7.2372E-05 6.9507E-05 6.6749E-05 6.4094E-05 6.1539E-05 5.0142E-05 5.6715E-05 5.4438E-05 5.2248E-05 5.0142E-05-3.7 0.00010783 0.00010366 9.9641E-05 9.5768E-05 9.2038E-05 7.5349E-05 8.4983E-05 8.165E-05 7.844E-05 7.5349E-05-3.6 0.00015915 0.00015313 0.00014734 0.00014175 0.00013635 0.00011216 0.00012614 0.00012131 0.00011665 0.00011216-3.5 0.00023267 0.0002241 0.00021582 0.00020782 0.0002001 0.00016538 0.00018547 0.00017853 0.00017184 0.00016538-3.4 0.00033698 0.00032487 0.00031316 0.00030184 0.00029091 0.00024156 0.00027013 0.00026028 0.00025075 0.00024156-3.3 0.00048348 0.00046654 0.00045014 0.00043429 0.00041895 0.00034952 0.00038977 0.00037589 0.00036248 0.00034952-3.2 0.0006872 0.00066374 0.00064102 0.00061901 0.00059771 0.000501 0.00055712 0.0005378 0.0005191 0.000501-3.1 0.00096767 0.0009355 0.00090432 0.0008741 0.00084481 0.00071143 0.00078891 0.00076226 0.00073644 0.00071143

-3 0.00134997 0.00130631 0.00126394 0.00122284 0.00118296 0.00100085 0.00110675 0.00107036 0.00103507 0.00100085-2.9 0.00186588 0.00180721 0.00175022 0.00169488 0.00164113 0.00139496 0.00153826 0.00148907 0.00144131 0.00139496-2.8 0.00255519 0.00247714 0.00240124 0.00232746 0.00225574 0.00192628 0.00211827 0.00205242 0.00198844 0.00192628-2.7 0.00346702 0.00336421 0.00326415 0.00316677 0.00307201 0.00263546 0.00289012 0.00280287 0.002718 0.00263546-2.6 0.00466122 0.00452715 0.00439653 0.00426928 0.00414534 0.00357265 0.00390708 0.00379261 0.00368115 0.00357265-2.5 0.00620968 0.00603657 0.00586776 0.00570315 0.00554265 0.00479883 0.00523363 0.00508495 0.00494005 0.00479883-2.4 0.00819753 0.00797626 0.00776025 0.00754941 0.00734363 0.00638717 0.00694686 0.00675566 0.00656913 0.00638717-2.3 0.01072408 0.01044405 0.01017041 0.00990305 0.00964185 0.00842418 0.00913745 0.00889403 0.00865631 0.00842418-2.2 0.0139034 0.01355253 0.01320934 0.01287368 0.01254542 0.01101063 0.01191059 0.01160376 0.01130381 0.01101063-2.1 0.01786436 0.01742912 0.01700296 0.01658575 0.01617733 0.01426207 0.01538628 0.01500337 0.01462868 0.01426207

-2 0.02275006 0.02221552 0.02169162 0.0211782 0.02067509 0.01830884 0.0196992 0.01922611 0.0187627 0.01830884-1.9 0.02871649 0.02806654 0.02742888 0.02680335 0.02618978 0.0232954 0.02499783 0.02441912 0.02385169 0.0232954-1.8 0.03593027 0.03514784 0.03437945 0.03362491 0.03288406 0.02937891 0.0314427 0.03074184 0.03005397 0.02937891-1.7 0.04456543 0.0436329 0.04271618 0.0418151 0.04092947 0.0367269 0.03920386 0.03836352 0.03753793 0.0367269-1.6 0.05479929 0.05369892 0.05261613 0.05155074 0.05050257 0.04551395 0.04845721 0.04745966 0.04647863 0.04551395-1.5 0.06680723 0.06552174 0.06425551 0.06300838 0.06178019 0.0559174 0.05937995 0.05820756 0.05705344 0.0559174-1.4 0.08075671 0.07926989 0.07780389 0.07635856 0.07493374 0.06811215 0.07214508 0.07078091 0.06943666 0.06811215-1.3 0.09680055 0.09509798 0.09341757 0.0917592 0.09012273 0.08226449 0.08691502 0.08534351 0.08379338 0.08226449-1.2 0.11506973 0.11313951 0.1112325 0.10934862 0.10748776 0.09852539 0.10383475 0.10204238 0.10027263 0.09852539-1.1 0.1356661 0.13349956 0.13135693 0.12923816 0.1271432 0.11702326 0.12302446 0.12100054 0.11900017 0.11702326

-1 0.15865526 0.15624765 0.15386424 0.15150502 0.14916997 0.13785661 0.14457233 0.14230969 0.14007112 0.13785661-0.9 0.18406009 0.18141123 0.17878635 0.17618552 0.17360876 0.16108706 0.1685276 0.16602324 0.16354306 0.16108706-0.8 0.21185533 0.20897003 0.20610799 0.20326933 0.20045414 0.18673291 0.19489447 0.19215016 0.18942961 0.18673291-0.7 0.24196358 0.23885199 0.23576242 0.23269502 0.22964992 0.21476382 0.22362722 0.22064988 0.21769537 0.21476382-0.6 0.27425306 0.27093085 0.26762883 0.26434723 0.26108623 0.24509702 0.25462685 0.25142882 0.24825216 0.24509702-0.5 0.30853753 0.30502572 0.30153177 0.29805594 0.29459849 0.27759528 0.28773968 0.28433881 0.28095726 0.27759528-0.4 0.3445783 0.34090301 0.33724276 0.33359785 0.32996858 0.31206695 0.32275813 0.31917752 0.3156137 0.31206695-0.3 0.38208864 0.37828054 0.37448423 0.37070005 0.36692833 0.34826832 0.35942363 0.3556913 0.35197276 0.34826832-0.2 0.42074031 0.41683387 0.41293561 0.40904593 0.40516518 0.38590818 0.39743194 0.39358019 0.38973881 0.38590818-0.1 0.4601721 0.45620464 0.45224153 0.44828318 0.44432997 0.42465458 0.43644053 0.43250507 0.42857629 0.42465458

0 0.5 0.49601062 0.49202165 0.48803347 0.4840465 0.46414354 0.47607775 0.47209676 0.46811856 0.46414354

Cumulative ProbabilityDistribution Function F(x)

Page 10: Reliability Models & Applications (continued)

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The Lognormal Model:

• DefinitionA random variable T is said to have the Lognormal Distribution with parameters and , where - < < and > 0, if the density function of T is:

, for t > 0

, for t 0

• RemarkThe Lognormal Model is often used as the failure distribution for mechanical items and for the distribution of repair times.

22

tln2

1

e2t

1 )t(f

0

Page 11: Reliability Models & Applications (continued)

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Properties of the Lognormal Model:

• Failure Distribution

where (z) is the cumulative distribution function

• Reliability Function

• If T ~ LN(,), then Y = lnT ~ N(,)

tln

)t(F

tln

-1 )t(R

Page 12: Reliability Models & Applications (continued)

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Properties of the Lognormal Model:

• MTBF (Mean Time Between Failures)

•Variance of Time to Failure

• Failure Rate

2

2

1

eMTBF

tln

1

)t(f )t(h

1ee)T(Var222

Page 13: Reliability Models & Applications (continued)

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The Lognormal Model:

Failure rate functions for various values of and

Page 14: Reliability Models & Applications (continued)

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The Lognormal Model:

Page 15: Reliability Models & Applications (continued)

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A theoretical justification based on a certain materialfailure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. Suppose the parameters are = 5 and = 0.1

(a) Compute E(X) and Var(X)

(b) Compute P(X > 120)

(c) Compute P(110 X 130)

(d) What is the value of median ductile strength?

(e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120?

(f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

Example - Ductile Strength

Page 16: Reliability Models & Applications (continued)

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Example - Solution

59.223

01005.0*83.22247

)1()(

is X of variance theand

,16.149

)(

is X ofmean The

22

2

2

2

2

)1.0(5

2

eeXVar

e

eXE

(a)

Page 17: Reliability Models & Applications (continued)

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9834.0

0166.01

)13.2(1

)1.0

579.4(1

)1.0

5120ln(1

)120(1)120(

ZP

ZP

ZP

XPXP(b)

Example - Solution

Page 18: Reliability Models & Applications (continued)

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0921.0

0013.00934.0

)00.3()32.1(

)32.100.3(

)1.0

5130ln

1.0

5110ln()130110(

ZP

ZPXP(c)

41.1485.0 ex(d)

Example - Solution

Page 19: Reliability Models & Applications (continued)

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83.9

)9834.0(10

)(

and

4)B(10,0.983~Y Then,

120least at ofstrength with testsofnumber Y

(b)part from 9834.0)120(

np

YE

XPp

(e)

Example - Solution

Page 20: Reliability Models & Applications (continued)

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0.05,-1.645)P(Z Since

,05.0)XP(Xwhich

for,say X, of value thefind toneed We

0.05

05.0

x

(f)

Example - Solution

Page 21: Reliability Models & Applications (continued)

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9015.125

Finally,

8355.4

),645.1)(1.0(5ln

,645.1ln

05.0)ln

P(Z

and

8355.40.05

0.05

0.05

0.05

ex

x

x

x

Example - Solution

Page 22: Reliability Models & Applications (continued)

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The Binomial Model:

Definition

- If X is the number of successes in n trials, where:

(1) The trials are identical and independent,

(2) Each trial results in one of two possible outcomes

success or failure,

(3) The probability of success on a single trial is p, and is constant from trial to trial, then X has the binomial Distribution with Probability Mass Function given by:

Page 23: Reliability Models & Applications (continued)

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The Binomial Model

Probability Mass Function

, for x = 0, 1, 2, ... , n

, otherwise

where

x)(XPp)n,|b(x)x(b

0

!xn!x

!n

x

n

xnx ppx

n

1

=

Page 24: Reliability Models & Applications (continued)

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Binomial Distribution

Rule:

)p,n;x(bq

p

1x

xn )p,n;1x(b

Page 25: Reliability Models & Applications (continued)

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Binomial Distribution

• Mean or Expected Value

= np

• Standard Deviation

= (npq)1/2 ,

where q=1-p

Page 26: Reliability Models & Applications (continued)

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The Binomial Model - Example Application 1:

Four Engine Aircraft

Engine Unreliability Q(t) = p = 0.1

Mission success: At least two engines survive

Find RS(t)

Page 27: Reliability Models & Applications (continued)

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The Binomial Model - Example Application 1- Solution

X = number of engines failing in time t

RS(t) = P(x 2) = b(0) + b(1) + b(2)

= 0.6561 + 0.2916 + 0.00486 = 0.9963

Page 28: Reliability Models & Applications (continued)

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Number of Failures Model:

• Definition- If T ~ E() and if X is the number of failures occurringin an interval of time, t, then X ~ P(t/ ), the Poisson Distribution with Probability Mass Function given by:

for x = 0, 1, ...

Where is the failure rate

• The expected number of failures in time t is

x!

eλtx)P(Xp(x)

λtx

θ

θ

tλtμ

Page 29: Reliability Models & Applications (continued)

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The Poisson Model:

X ~ P(2)

x p(x) F(x)0 0.135335 0.1353351 0.270671 0.4060062 0.270671 0.6766763 0.180447 0.8571234 0.090224 0.9473475 0.036089 0.9834366 0.012030 0.9954667 0.003437 0.9989038 0.000859 0.999763

Page 30: Reliability Models & Applications (continued)

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The Poisson Model:

p(x)

Number of Failures ~ x

Page 31: Reliability Models & Applications (continued)

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The Poisson Model - example continued

Number of Failures ~ x

1 2 3 4 5 6 7 8

1.00

0

x

0y

ypxF0.2λtμ

Page 32: Reliability Models & Applications (continued)

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The Poisson Model - Example Application:

An item has a failure rate of = 0.002 failures per hourif the item is being put into service for a period of 1000hours. What is the probability that 4 spares in stock willbe sufficient?

Page 33: Reliability Models & Applications (continued)

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Expected number of failures (spares required) = t = 2

P(enough spares) = P(X 4)

= p(0) + p(1) + p(2) + p(3) + p(4) = 0.945

or about a 5% chance of not having enough spares!

The Poisson Model - Example Application - Solution