VARDHAMAN COLLEGE OF ENGINEERING

Transcription

1 VARDHAMAN COLLEGE OF ENGINEERING (Autoomous) Shamshabad, Hyderabad S. No UNITS BASIC PROBABILITY THEORY: Rules for combiig probability, Probability Distri butios, Radom variables, desity ad distributio fuctios. Mathematical expectatio. Biomial distributio, Poisso distributio, ormal distributio, expoetial distributio, Weibull distributio. 2 3 RELIABILITY: Defiitio of Reliability. Sigificace of the terms appearig i the defiitio. Compoet reliability, Hazard rate, derivatio of the reliability fuctio i terms of the Hazarad rate, Hazard models. FAILURES: Causes of failures, types of failures, Modes of failure, Bath tub curve, Effect of prevetive maiteace. Measures of reliability: mea time to failure ad mea time betwee failures. CLASSIFICATION OF ENGINEERING SYSTEMS: Series, parallel, series-parallel, parallelseries ad o-series-parallel cofiguratios. Expressios for the reliability of the basic cofiguratios. RELIABILITY LOGIC DIAGRAMS: Reliability evaluatio of No-series-parallel cofiguratios: miimal tie-set, miimal cut-set ad decompositio methods. Deductio of the miimal cut sets from the miimal path sets. 4 DISCRETE MARKOV CHAINS: Geeral modelig cocepts, stochastic trasitioal probability matrix, time depedet probability evaluatio ad limitig state probability evaluatio. Absorbig states. CONTINUOUS MARKOV PROCESSES: Modelig cocepts, State space diagrams, Stochastic Trasitioal Probability Matrix, Evaluatig limitig state Probabilities. Reliability evaluatio of repairable systems. 5 SERIES SYSTEMS AND PARALLEL SYSTEM: Series systems, parallel systems with two ad more tha two compoets, Network reductio techiques. Miimal cut set/failure mode approach UNIT-I

2 Basic Cocepts o Probability Probability meas whether a certai evet has a good chace of occurrig or ot. Its value lies betwee 0 ad. Rules for combiig probabilities ) Idepedet Evets: Two evets are said to be idepedet if the occurrece of oe evet does ot affect the probability of occurrece of the other evet. Example: Throwig a dice ad tossig coi are idepedet evets. 2) Mutually exclusive evets: Two evets are said to be mutually exclusive or disjoit if they caot happe at the same time. Example: (i) Whe throwig a sigle die, the evets, 2, 3, 4, 5 ad 6 spots are all mutually exclusive because two or more caot occur simultaeously (ii) Similarly success ad failure of a device are mutually exclusive evets sice they caot occur simultaeously. 3) Complimetary Evets: Two outcomes of a evet are said to be complemetary, if whe oe outcome does ot occur, the other must occur. If the outcomes A & B have probabilities P(A) ad P(B), the P(A) + P(B) = P(B) = P(A) Example: Whe tossig a coi, the outcomes head ad tail are complemetary sice P(head) + P(tail) = or P(head) = P(tail) P(tail) = P(head) Therefore we ca say that two evets that are complemetary evets are mutually exclusive also. But the coverse is ot ecessarily true i.e, two mutually exclusive evets are ot ecessarily complemetary. 4) Coditioal Evets; Coditioal evets are evets which occur coditioally o the occurrece of aother evet or evets. Cosider two evets A & B ad also cosider the probability of evet A occurrig uder the coditio that evet B has occurred. P(A/B) = P(A B) P(B) 5) Simultaeous occurrece of evets: Occurrece of both A & B - Mathematically it is represeted as A B, A AND B, AB. Case (i) Idepedet, the the probability of occurrece of each evet is ot iflueced by the probability of occurrece of the other.

3 P(A/B) = P(A) P(A B) = P(A) P(B) Ad P(B/A)- = P(B) Case (ii) Evets are depedet If two evets are ot idepedet, the the probability of occurrece of oe evet is iflueced by the probability of occurrece of the other Therefore, P(A B) = P(B/A).P(A) = (PA/B).P(B) Numerical Problem - A egieer selects two compoets A & B. The probability that compoet A is good is 0.9 & the probability that compoet B is good is 0.95.What is the probability of both compoets beig good. P(A good B good) = P(A good) (B good) = 0.9 x 0.95 = 0.80 Numerical Problem - 2 Oe card is draw from a stadard pack of 52 playig cards. Let A be the evet that it is a red card ad B be the evet that it is a face card. What is the probability that both A & B occur. P(A) = 26/52 Give that A has occurred The the sample space for B is 26 states, out of which 6 are those of a face card. Therefore, P(B/A = 6/26 P(A B) = 6/26 x 26/52 = 6/52 P(A B) = P(B/A) P(A) 6) Occurrece of atleast oe of two evets: The occurrece of atleast oe of two evets A ad B is the occurrece of A or B or BOTH. Mathematically it is the uio of the two evets ad is expressed as (AUB), (A or B) or (A x B) Case (i) Evets are idepedet but ot mutually exclusive. P(AUB) = P(A OR B OR BOTH A AND B) = P(NOT A AND NOT B) = P(A B) = P(A). P(B) = (-P(A)) (-P(B)) = P(A) + P(B) P(A).P(B) Usig Ve diagram P(AUB) = P(A) + P(B) P(A B) If P(A) = 0.9 ad P(B) = 0.95 P(AUB) = P(A) + P(B) P(A).P(B) = X 0.95 = 0.995

4 Case (ii) Evets are idepedet ad mutually exclusive I the case of evets A & B beig mutually exclusive, the the probability of their simultaeous occurrece P(A).P(B) must be zero by defiitio. P(AUB) = P(A) + P(B) Case (iii) Evets are ot idepedet If two evets are ot idepedet the P(AUB) = P(A) + P(B) P(A B) = P(A) + P(B) P(B/A).P(A) = P(A) + P(B) P(A/B).P(B) Numerical Problem - 3 A ciema hall gets electric power from a geerator ru by a diesel egie. O ay give day, the probability that the geerator is dow (evet A) is ad the probability that the diesel egie is doe (evet B) is What is the probability that the ciema house will have power o ay give day. Assume that the occurrece of evets A & B are idepedet of each other. Probability that the Ciema hall does ot have power give by the probability of the evet that either the diesel egie or geerator is dow. Q = P r (AUB) = P(A) + P(B) P(A) P(B) = x 0.04 = Therefore, the probability that the ciema house have power = R = = Numerical Problem - 4 I a sample of 60 mails, 0 of them cotais oly defective heads, five cotai oly defective tails ad ad five cotai both the defects. What is the probability that a mail that is selected radomly cotais either defective head or a defective tail? Let X deote the evet that a mail cotais a defective head ad Y deote the evet cotaiig a defective tail. Tha P(X) = = 0.25 P(Y) = = P(X Y) = 5/60 = The probability that a mail cotais either of the two defects is P(XUY) = P(X) + P(Y) P(X Y) = Therefore, probability that a mail cotais o defect is 40/60 = = -P(XUY) Radom Variables

5 To study about a system s behavior for the applicatio of probability theory to reliability evaluatio, a series of experimets must be performed or a data collectio scheme should be deduced. To apply the probability theory to occurrece of these values or evets which are radom i ature, we eed to study these variables called as Radom Variables. Radom variable is a variable quatity which deotes the result or outcome of a give radom experimet. A radom variable is oe that ca have oly a discrete umber of states or coutable values. A radom variable ca be either discrete or cotiuous. A discrete radom variable is oe that ca have oly a discrete umber of states or coutable values. Ex:.Tossig a coi - Outcomes are heads or tails. 2. Rollig a dice - Outcomes are,2,3,4,5 or 6. A cotiuous radom variable is oe which takes a ifiite umber of values or if its rage forms a cotiuous set of real umbers. This does ot mea that the rage exteds from - to +. It oly meas that there are ifiite umber of possibilities of the value. Ex:. The life time of a light bulb. 2. If electric curret have values betwee 5A ad 0 A, the it idicates a cotiuous radom variable. Probability Desity Fuctio The probabilities associated with the radom variables ca be described by a formula called Probability desity fuctio or Probability mass fuctio. We use the otatio f(x) for the probability desity fuctio. Ex :. Cosider the throw of a dice Let the radom variable associated with the outcome be X. The value of X are, 2, 3, 4, 5 ad 6. f() = P(x=) =/6 f(2) = P(x=2) = /6 f() = f(2) = = f(6) = /6 f(x) = /6 = Costat desity fuctio. 2. Cosider the rollig of two dice. What is f(x) for the sum of dots facig up? X= Total Sum of dots P(x) = f(x) =? P(x=2) = f(2)

6 f(x) = x 3 6 f(x) = 3 x 3 6 for x= 2,3,4,5,6,7 for x= 8,9,0,,2 Probability Distributio Fuctio If x is a radom variable, the for ay real umber x, the probability that x will assume a value less tha or equal to x is called Probability distributio fuctios. It is idicated as F(x) f(x) = P(x) F(x) = P(X x) Ex: Cosider the rollig of a sigle dice. f(x) = /6 f() = f(2) =..f(6) = /6 F() = P (X ) = f() = /6 F(2) = P (X 2) = f() + f(2) = /6 + /6 = 2/6 F(3) = P (X 3) = f() + f(2) + f(3) = /6 + /6 + /6 = 3/6 F(4) = 4/6 F(5) = 5/6 F(6) = 6/6 = Suppose a radom variable X has the followig desity fuctio. X f(x) /32 5/32 0/32 0/32 5/32 /32 The, the Probability distributio fuctio is give by X F(x) /32 /32 + 5/32 =6/32 6/32 + 0/32 =6/32 6/32 + 0/32 =26/32 26/32 + 5/32 =3/32 3/32 + /32 =32/32 Relatio betwee Probability desity fuctio ad distributio fuctio: F(x) = f x (Discrete Radom variable)

7 F(x) = f x dx (Cotiuous Radom variable) A radom variable x ad the correspodig distributio fuctio F(x) are said to be cotiuous if the followig coditio is satisfied for all x. f(x) = d dx f(x) Mathematical Expectatio It is useful to describe the radom behavior of a system by oe or more parameters rather tha as a distributio. This is particularly useful i the case of system reliability evaluatio. This parametric descriptio ca be achieved usig umbers kow mathematically as momets of distributio. The most importat of these momets is the expected value, which is also referred to as average mea value. Mathematically it is the first momet of the distributio. Cosider a Probability model with outcome x, x 2, x 3 x ad the probability of each is P, P 2, P 3.P. the the expected value of the variable is E(x) = P x + P 2 X 2 + P 3 X 3 +.P X = x P Expected value E(x) of a discrete radom variable x havig outcomes x i each with a probability of occurrece P i is E(x) = i= x i P i where i P = I case of cotiuous radom variable, the equatio ca be modified from the summatio to itegratio. E(x) = x f x dx Expected value is the weighted mea of the possible value usig their Probability of occurrece as the weighig factor. Variace ad Stadard Deviatio The expected value is the most importat distributio parameters i reliability evaluatio. But to kow the amout of spread or dispersio of a distributio, the secod momet of distributio. i.e., variae V(x) should be deduced. The variae of a radom variable x is defied as the expectatio of the square of deviatio of x from E(x). m = E(x) = x f x dx Variace = σ 2 = (x m) 2 f x dx The quatity σ is called stadard Deviatio. The K th momet of a radom variable x about its expectatio is defied as M k = E[x E(x)] K. The secod momet of distributio is kow as variace V(x) (K=2) V(x) = E[x E(x)] 2 = E[x 2 2x E(x) + E 2 (x)] = E(x 2 ) E(2x E(x)) + E [ E 2 (x)] = E(x 2 ) 2 E(x) E(x) + E 2 (x) i=

8 = E(x 2 ) 2 E 2 (x) + E 2 (x) = E(x 2 ) - E 2 (x) = i= x 2 i P i - E 2 (x) Properties of the biomial distributio The biomial distributio ca be represeted by the geeral expressio: p + q For the expressio to be applicable, four specific coditios are required. These are: (a) There must be a fixed umber of trials, i.e. is kow (b) Each trial must result i either a success or a failure, i.e., oly two outcomes are possible ad p + q =. (c) All trials must have idetical probabilities of success ad therefore of failure, i.e., the values of p ad q remai costat, ad (d) All trials must be idepedet (this property follows from (c) sice the probabilities of success i trial i must be costat ad ot affected by the outcome of trials, 2,...., (i-)). I order to apply the biomial distributio ad to evaluate the outcomes ad their probability of occurrece of a give experimet or set of trials, the expressio p + q must be expaded ito the form of equatios ad p + q = p + p ( ) q + p 2 q 2 + 2! ( r + ) + p r q r +. + q r! If equatio is compared with, it is see that the coefficiet of the (r+)th term i the biomial expasio represets the umber of ways, i.e., combiatios, i which exactly r failures ad therefore (-r) successes ca occur i trials ad is equal to C r. Therefore each coefficiet i equatio ca be directly evaluated from the defiitio of C r as discussed ad the probability of exactly r successes or (-r) failures i trails ca be evaluated from p r =! r! r! pr q r = C r p r q r = C r p r ( p) r Substitutig of equatios gives Numerical example I (p + q) = r=0 c r p r q r =

9 A coi is tossed 5 times. Evaluate the probability of each possible outcome ad draw the probability mass (desity) fuctio ad the probability distributio fuctio. Solutio I this example =5, p=q=/2. Usig the biomial expasio the outcomes, the probability of exactly r heads or (-r) tails ad the cumulative probability are determied as show i below table. Number of Idividual probability Cumulative heads Tails Expressio Value probability 0 5 5C 0 (/2) o (/2) 5 /32 /32 4 5C (/2) (/2) 4 5/32 6/ C 2 (/2) 2 (/2) 3 0/32 6/ C 3 (/2) 3 (/2) 2 0/32 26/32 4 5C 4 (/2) 4 (/2) 5/32 3/ C 5 (/2) 5 (/2) 0 /32 32/32 = I the above table the values of idividual probability have bee summated ad a value of uity obtaied. The results are plotted as a probability mass (desity) fuctio ad probability distributio fuctios i figures. The probability desity fuctio is symmetrical. This oly occurs whe p=q=/2 sice i this case the success ad failure evets ca be iterchaged without ay alteratio i the umerical value of ay of the idividual outcomes. This will ot be the case whe p ad q are uequal. Example

10 Numerical example II Cosider the case i which the probability of success i a sigle trial is ¼ ad four trials are to be made. Evaluate the idividual ad cumulative probabilities of success i this case ad draw the two respective probability fuctios. Solutio =4, p=/4, q=3/4 Number of Idividual probability Cumulative successes Failures probability 0 4 (3/4) 4 = 8/256 8/ (/4) (3/4) 3 = 08/256 89/ (/4) 2 (3/4) 2 = 54/ / (/4) 3 (3/4) = 2/ / (/4) 4 = / /256 = Numerical example III A die is throw i 6 times. Evaluate the probability of gettig 2 spots o the upper face 0,, 2,..., 6 times ad draw the probability mass (desity) fuctio ad the probability distributio fuctio. Solutio: O each throw, the probability of gettig 2 spots o the upper face is /6 ad the probability of ot gettig 2 spots is 5/6. If these two evets are defied as success ad failure respectively, the, although there are six possible outcomes o each throw, the problem has bee costraied to have two outcomes ad the biomial distributio becomes applicable.

11 Cosequetly =6, p=/6 ad q=5/6. The probability results are show i below table. Number of successes Idividual probability Cumulative probability 0 (5/6) 6 = 5625/ / (/6) (5/6) 5 = 8750/ / (/6) 2 (5/6) 4 = 9375/ / (/6) 3 (5/6) 3 = 2500/ / (/6) 4 (5/6) 2 = 375/ / (/6) 5 (5/6) = 30/ / (/6) 6 = / /46656 = Expected value ad Stadard Deviatio for Biomial Distributio: The two most importat parameters of a distributio are the expected or mea value ad the stadard deviatios. The biomial distributio is a discrete radom variable ad therefore the expected value ad stadard deviatio ca be evaluated usig equatio. E(x) = i= x i P i where i P i = = x=0 x ( cx p x q x ) [P(r,) = cr p r ( p) r )] = x=0 x! x! x! px q x As the cotributio to this summatio made by x=0 is zero, the E(x) =! x= p x q x (x )! x! E(x) = = x ( )! x= p p x q x x(x )! x! ( )! x= p p x q x (x )! x! = p ( )! x= p x q x (x )! x!

12 Let - = m ad x- = y E(x) = p E(x) = p m ( )! y=0 p y q y m y! y! m! y=0 p y q m y y! m y! Sice, m m! y=0 p y q m y = y! m y! E(x) = p Expected value or Mea value ad Stadard Deviatio for Expoetial Distributio: The expected value of a cotiuous radom variable havig a rage (0, ) is give by E(x) = t. f t dt 0 E(x) = t. λ e λt dt 0 This ca be itegrate by parts Let u = t ad v = e λt du = dt E(t) = dv= λ e λt dt udv = [uv] 0 - vdu 0 = [ t e λt ] 0 - e λt dt 0 = t e λt 0 λ e λt 0 = 0 + λ E(t) = λ σ 2 = u = t 2 du = 2tdt t 2. λ e λt dt E 2 (t) 0 v = e λt dv = λ e λt dt σ 2 = udv E 2 (t) Itegratig by parts = [uv] 0 - vdu -E 2 (t) 0 = [ t 2 e λt ] 0-2t e λt dt 0 E 2 (t) σ 2 = λ. λ λ 2 = 2 λ 2 λ 2 = λ 2 σ 2 = λ 2

13 σ = λ Expected value ad Stadard deviatio of a Expoetial Distributio are equal. Mea Time To Failure (MTTF) The expected value of a failure desity fuctio is ofte desigated as the mea time to failure MTTF. I case of expoetial distributio this is equal to the reciprocal of the failure rate λ. p o E(t) = t f t dt f t = d dt R(t) MTTF p = t dr(t) o p o = R t dt p o = t R(t) o p + R t dt MTTF = p R t dt o Reliability Aalysis of series etworks usig expoetial distributio Let R (t), R 2 (t).... R (t) be the reliabilities of compoets coected i series. R s (t) = R (t) R 2 (t).. R (t) = R i t i= Let R (t) = e λ t & R 2 t = e λ 2t R (t) = e λ t R s (t) = e λ λ 2 λ 3..+ λ t = e i= λ it Total failure rate λ s t = i= λ i t Therefore, Hazard rate fuctio for the system is determied by summig the hazard rate fuctio of the idepedet compoets. p o p o MTTF = R s t dt = e i= λ it dt MTTF = λ = i= λ i = i=mttf i If all the compoets coected i series have the same failure rates λ = λ 2 = λ λ the λ = λ MTTF = λ

14 Parallel cofiguratio Let R (t) R 2 (t)... R (t) be the reliabilities of the compoets coected i parallel. R s (t) = - [(-R (t)) (-R 2 (t))..... (-R (t))] = - i= [ R i t ] e λ it Rs(t) = - i= Where λ i = failure rate of its compoet Two compoets i parallel R s (t) = - e λ (t) e λ 2(t) = e λ t + e λ 2t e λ +λ 2 t P o P o MTTF = R s t = e λ t dt + e λ 2t dt e λ + λ 2 t If λ = λ 2 = λ The R s (t) = 2e λt e 2λt MTTF = 2 λ 2λ = λ + λ 2 λ +λ 2 P o p o Numerical example A aircraft egie cosists of three modules havig costat failure rates of λ = λ 2 = 0.05 ad λ 3 = failures per operatig hour. What is the reliability fuctio for the egie ad what is the MTTF? Solutio: R(t) = e = e 0.095t MTTF = t = 5.28 operatig ours Numerical example 2 Cosider a four compoet system of which the compoets are idepedet ad idetically distributed with Costat Failure Rate (CFR). If R s (00) = 0.95, fid the idividual compoet MTTF? Solutio: R s (00) = e 00λ s = e 00(4xλ) = 0.95 λ = l (0.95) 400 =

15 MTTF = = Numerical example 3 A Simple electroic circuit cosists of 6 trasistors each havig a failure rate of 0-6 f/hr, 4 diodes each havig a failure rate of 0.5x0-6 f/hr, 3 capacitor each havig a failure rate of 0.2x 0-6 f/hr, 0 resistors each havig a failure rate of 5x0-6 f/hr. Assumig coectors ad wirig are 00% reliable (these ca be icluded if cosidered sigificat ), evaluate the equivalet failure rate of the system ad the probability of the system survivig 000hr if all compoets must operate for system success. Solutio λ s = Equivalet failure rate of the system = 6 X ( X 0-6 ) + 4 X (0.5 X 0 6 ) + 3 X (0.2 X 0-6 ) + 0 X(5 X 0-6 ) + 2 X (2 X 0-6 ) = 6.26 x 0-5 f/hr R s (t) = e -λs t R s (000) = exp (-6.26 X 0-5 x 000) = Sice Q s (t) = -R s (t) Q s (000) = = Poisso distributio Poisso distributio is a discrete probability distributio that expresses the probability of a give umber of evets occurrig i a fixed iterval of time ad/or space if these evets occur with a kow average rate ad idepedetly of the time. The Poisso distributio ca also be used for the umber of evets i other specified itervals such as distace, area or volume. Poisso distributio is a approximatio to biomial distributio. It is used for large values of ad small p P x, λ = e λ λ x x! λ = shape parameter which idicates the average umber of evets i the give time iterval. = Mea value Numerical example - A rare disease has a icidece of i 000 perso-years. Assumig that members of the populatio are affected idepedetly, fid the probability of k cases i a populatio of 0,000 for k=0,, 2 Solutio: Expected mea λ = 0.00 X 0,000 = 0 P x = 0 = e = !

16 P x = = e 0 0 = P x = 2 = e = ! 2! Numerical example - 2 I a large system the average umber of cable faults per year per 00 km of cable is 0.5. Cosider a specified piece of cable 0km log ad evaluate the probabilities of 0,, 2 etc, faults occurrig i (a) a 20 year period, ad (b) a 40 year period Solutio: Assumig the average failure rate data to be valid for the 0km cable ad for the two periods beig cosidered, the expected failure rate λ is, λ = 0.5 x 0 00 = 0.05 f/yr Ad (a) For a 20 year period, E(x) = 0.05 X 20 =.0 P x =.0x e.0 x! Ad for x = 0,, 2, (b) For a 40 year period, E(x) = 0.05 X 40 = 2.0 P x = 2.0x e 2.0 x! for x = 0,, 2, UNIT-II Reliability : The reliability of a device is cosidered high if it had repeatedly performed its fuctio with success ad low if it had teded to fail i repeated trials. The reliability of a system is defied as the probability of performig the iteded fuctio over a give period of time uder specified operatig coditios. The above defiitio ca be broke do o ito four parts: i) Probability ii) Iteded fuctio

17 iii) iv) Give period of time Specified operatig coditios Probability: Because, the reliability is a Probability, the reliability of system R j is govered by the equatio 0 R s. The equality sig hold good i case of equipmet called oe shot equipmet. Iteded fuctio: It is also defied to as the successful operatio. Example-: As a example, let us cosider the buildig up of voltage by a dc shut geerator. For some reasos, let us assume that the voltage is ot build up. We say that the dc shut geerator has failed to do its job. The failure i this cotest does t imply ay physical failure, but oly the operatioal failure. Example-2: Lightig arrester : The lightig arrestor should burst i the evet of occurrece of a lightig stroke. O the occurrece of a lightig stroke, if the lightig arrestor bursts, there is the physical failure or damage but operatioally it is successful. O the other had, if it does t burst there is o physical failure, but yet there is a operatioal failure ad we say that the lightig arrestor has failed (operatioally). Give period of time: Ay compoet has some useful life period, withi which time the compoet should operate successfully. For example, a power trasformer has a useful life of at least 20 to 25 years. If, it fails withi this time period, the the istrumet is said to be ureliable ad if it fails after its useful life period, the we say it is reliable. Specified operatig or evirometal coditios: Ay equipmet is supposed to perform its duty satisfactorily uder cotai specified operatig coditio such as temperature, humidity, pressure ad altitude. Though a equipmet is able to perform its duty satisfactorily i a cold coutry yet it may fail whe used uder hot climatic coditios. Compoet Reliability It is usual for a large system to be divided ito compoets for the purpose of reliability evaluatio. A compoet is that part of a system which is treated as a sigle etity for the purpose of reliability evaluatio. There is o clear distictio betwee compoet ad system. The same uit ca be cosidered as compoet (or) system depedig o circumstaces. For istace, a geeratig uit is cosidered as a compoet while dealig with the reliability of etire power system. Same ca be treated as a complex system cosistig of several compoets like the boiler, Turbie ad Geerator, etc. Compoets ca be classified ito two groups No-repairable compoets ad Repairable compoets. No-repairable compoets are compoets that caot be repaired or the repair is uecoomical. Repairable compoets are compoets which ca be repaired upo failure ad thus their life histories cosist of alteratig operatig ad repair periods. I the reliability evaluatio of power systems, it is the repairable type that is of greater iterest. For several reasos, a compoet put ito service fails after sometime, called the TIME TO FAILURE (T), this ca be recogized as a radom variable ad the reliability of a compoet at ay time ca be expressed as R(t) = P(T>t) = F(t) = - P(T t)

18 Reliability Fuctio: f(x) = Probability desity fuctio F)x) = Probability distributio All compoets have a differet failure rate, hece these time-to-failure obey a probability distributio, thus probability value is a fuctio of time that is specified or cosidered. f(t)= desity fuctio which idicate the rate of failures per hour. R(t)= Reliability fuctio A additioal fuctio which is oe of the most extesively used fuctio i reliability evaluatio is the hazard rate h(t). I terms of failure the hazard rate is a measure of the rate at which failures occur or the istataeous failures/hour. o of failures f(t) = o of compoet x operatig hours o of failures h(t) = o of compoet at the begiig of iteral x operatig hours Thus the hazard rate is depedet o the umber of failures i a give time period ad the umber of compoets exposed to failures. PROBLEMS:. The field test data i respect of 72 compoets is as give below. Calculate failure desity rate ad hazard rate Time iteral hrs Failure i iteral Solutio : f(t) h(t) 59/72 X /72X000 24/72X00 24/3x000 29/72X00 29/89x000 30/72X00 30/60X00 7/72X00 7/30X00 3/72X00 3/3X00 2. The compoet failure data for te compoets subjected to a life test are give below. Fid the failure desity rate ad hazard rate. Failure Operatig time hrs Solutio : Time iteral f(t) 0x8 0x23 0x25 0x20 = x2 0x4 0x2 0x7 0x45 0x20

19 h(t) 0x8 = x2 8x4 7x2 6x7 5x23 4x25 3x20 2x45 x20 HAZARD FUNCTION : h(t) The probability fuctio of the radom variable T ca be determied by f(t) =lim t 0 P( t<t t+ t) t t = icremet time Suppose at t=0, N(0) compoets are put to work. At ay time t suppose the umber of compoets is N(t), the f(t) = N t N(t+ t) N 0. t If the probability i the above case is calculated as coditioal probability coditio beig that the compoet should be workig at t, the the fuctio is Hazard fuctio h(t). h(t) = P( t<t t+ t) lim t 0 t at (T > t) h(t) = N t N(t+ t) N t. t Geeral Reliability fuctios All compoets have a differet failure rate, hece these times to failure obey a probability distributio. This Probability value is a fuctio of time that is specified or cosidered. Let f(t) = failure desity fuctio which idicates the rate of failures per hour R(t) = Reliability fuctio A additioal fuctio which is oe of the most extesively used i reliability evaluatio is the hazard rate h(t). I terms of failure, the hazard rate is a measure of the rate at which failures occur. It idicates the istataeous failures / hour. f(t) = Number of failures Number of compoets X operatig hours h(t)= Number of failures Number of compoets at the begiig of iterval X operatig hours The hazard rate is depedet o the umber of failures i a give time period ad the umber of compoets exposed to failure.

20 Failure desity fuctio f(t) is the rate of failures per hour. Hazard rate h(t) is the istataeous failures per hour. Derivatio of Reliability fuctio R(t) i terms of hazard rate h(t) A o - repairable compoet is of use oly till the failure occurs ad if the compoet fails, we have to replace it with a ew compoet Such a compoet is described by its life time T, a radom variable. Sice R is a fuctio of t (operatig time), the reliability ca be defied as R (t) = P (T>t) () = P ( T t) But P( T t) = F(t) = failure distributio fuctio R t = F t (2) Failure desity fuctio f(t) = d F(t) dt = d R t dt (3) Cosider a case i which fixed umber N 0 of idetical compoet are tested. Let N s (t) = Number of compoets survivig at time t N f (t ) = Number of compoets failed at time t N s (t) + N f (t) = N 0 At ay time t the reliability fuctio R(t) = N s(t) N o (4) = N o N f (t) = - N f (t) (5) N o N o Similarly the probability of failure or cumulative failure distributio F(t) = N f (t) N o (6) From equatios 5 ad 6 we get equatio 2 R(t) = F(t) d R t dt = d F(t) dt = - N o d N f (t) dt (7) f(t) = d R t dt

21 = N o d N f (t) dt (8) The failure desity fuctio ad hazard rate are idetical oly at t= 0. The geeral expressio for hazard rate at time t is h(t) = N s (t) d N f (t) dt (9) = N o. N o N s (t) d N f (t) dt Let us cosider = = N o N s (t) R t t = From equatio 8 t = R t N o d N f (t) dt f t = f t R t f t (0) R(t ) d R t dt f t = d R(t) dt () = d dt [l R(t)] = R t = R t d dt R t d dt d F(t) dt d R t dt [ F(t)] = - f(t) R t = -h(t) l R t = (t) t l R(t) = t dt 0 R(t) = e t t dt (2) For a costat hazard rate, h(t) = λ= umber of areas R(t) = e t 0 λ dt = e λt R(t) = e λt (3)

22 Relatio betwee R(t), Q(t), F(t), f(t) ad h(t) R(t) = Reliability fuctio Q(t) = Ureliability fuctio h(t) = hazard rate fuctio F(t) = failure Distributio fuctio f(t) = failure desity fuctio which idicates the rate of failures per hour R(t) Q(t) = F(t) f(t) h(t) R(t) - Q(t) f t dt e t 0 t dt Q(t) - R(t) f t dt e t 0 t dt f(t) d R t dt d Q t dt (t)e t 0 t dt h(t) d dt l R t d Q(t) dt Q(t) f(t) f t dt Measures of Reliability: Cosider a sigle repairable compoet for which the failure rate ad repair rate are costat. The state trasitio diagram for this compoet is show below. Sigle compoet system (a) State space diagram (b) Mea time/ state diagram Let, λ = failure rate of the compoet µ = repair rate of the compoet m = mea operatio time of the compoet r = mea repair time of the compoet

23 The two system states ad their associated trasitios ca be show chroologically o a time graph. The mea values of up ad dow times ca be used to give the average performace of this two state system. This is show i figure b. I figure b, the period T is the system cycle time ad is equal to the sum of the mea time to failure (MTTF) ad mea time to repair (MTTF). This cycle time is defied as the mea time betwee failures (MTBF). Some times, MTBF is used i place of MTTF. It is evidet however that there is a sigificat coceptual differece betwee MTTF ad MTBF. The umerical differece betwee them will deped o the value of MTTR. I practice the repair time is usually very small compared with the operatig time ad therefore the umerical values of MTTF ad MTBF are usually very similar. The followig relatioships ca therefore be defied m = MTTF = / λ r = MTTR = / µ T = MTBF = m+r = / f Where f = cycle frequecy, i.e., the frequecy of ecouterig a system state. The failure rate λ is the reciprocal of the mea time to failure, MTTF, with the times to failure couted from the momet the compoet begis to operate to the momet it fails. Similarly, the repair rate µ is the reciprocal of the mea time to repair, MTTR, with these times couted from the momet the compoet fails to the momet it is retured to a operable coditio. λ = umber of failures of a compoet i the give period of time total period of time the compoet was operatig µ = umber of repairs of a compoet i the give period of time total period of time the compoet was repaired Bath Tub Curve The plot of hazard rate versus time is referred to as Bath tub curve. Most of the compoets have a high failure desity rate at the begiig ad the failure rate decreases with time. The failures i the begiig are maily due to defects i desig ad due to the improper maufacturig techiques, etc. These will be detected ad corrected so that failure rate or hazard rate decreases with time. This is idicated by the portio 0 t of the curve. Betwee t ad t 2 the hazard rate is more or less costat ad beyod t 2 the hazard rate icreases with time due to ormal wear ad tear.

24 The period 0- t is referred to as debuggig period or bur i period ad the failures are refied to as ifat mortality. The period t t 2 is called as the useful life period. I this period the failures are chace failures or radom failures. The period beyod t 2 is called the wear-out period ad the failures are maily due to agig effect. These failures are called wear-out failures. Because of the shape of the curve, it is called as Bath- Tub- Curve. The Bath Tub Curve ca be divided ito three regios amely i) Decreasig hazard rate regio ii) Costat hazard rate regio iii) Icreasig hazard rate regio

25 UNIT-III The reliability evaluatio of egieerig systems ca be obtaied by drawig RLD Reliability Logic Diagram or Reliability Block Diagram RBD or RLG Reliability Logic Graph. RLD or RBD : I a RBD each of the egieerig system compoets is idicated by a block. The RLD shows the compoets i order that the iteded fuctio ca be accomplished. It may or may ot have ay resemblace to the actual arragemet of the compoets. Ex:- Let us cosider a hypothetical geeratig statio cosistig of three geerators A, B, ad C rated at 5Mw, 4Mw ad 3Mw respectively to feed a load coected at the statio bus- bars. Case : System load is 0Mw The geeratig statio is successful if a power to 0Mw ca be supplied. This requires that (A) ad, (B) ad (C) must be i a operatig coditio. Each of the geerators is represeted by a block ad they are all coected i series. The AND is reflected as a series coectio both the compoets cocered. The RLD is as show i figure Case 2: System load is 2MW A power of 2mw ca be applied by geerator (A)or (B) or (C) This OR is reflected as a parallel coectio amog the compoets i the RLD is as show i the figure Case 3: Required power 6 MW. This power requiremet ca be met by the power fuctioig of geerators., (A) AND (B) -. OR. {(B). AND. (C)}, OR. {(C) AND. (A)} From a examiatio of the RLD s we ca observe the followig:. The RLD may or may ot have resemblace to the physical arragemet of the compoets. 2. Oe ad the same eergy system will have differet RLD based o differet criteria for the success of the system.

26 For example,i the series RLD A, B, C may represet a DC shut geerator where A- Armature widig B- Field widig ad C- Commutator The same diagram may represet the RLD of a ball where A= Baal B= Riffle ad C=ib of the riffle Reliability Logic Diagram (RLD ) or Reliability Block Diagram (RBD): The RBD shows the logical itercoectios amog the various compoets, rather tha the actual coectios, it may or may ot have ay resemblace to the physical arragemet of the compoets. I the RBD each compoet is represeted by a block, whereas i the RLG each compoet is represeted by a brach (edge) coected betwee two odes, i additio to two odes, viz., the source ode(s/i) ad sik ode (t/out). Oce the RLD is draw, the actual system loses its sigificace. Oe ad the same system may have differet RLD s based o the differet success criteria ad coversely, oe ad the same RLD may correspod to differet system. Classificatio of Egieerig Systems: All other parameters remaiig the same, the reliability of a system depeds o how the system is to be modeled. As for as the reliability evaluatio is cocered, Egieerig systems are classified based o their Reliability Logic Diagram as. i) Series system (cofiguratio). ii) Parallel system (cofiguratio). iii) Series-parallel system (cofiguratio). iv) Parallel- series system (cofiguratio). v) K out of m: Good systems v) No-series-parallel system (cofiguratio) or Complex System. Series Cofiguratio: If a system is successful, i.e., if it is able to perform its iteded fuctio, if ad oly if each ad every oe of the compoets is operative, the the system is to be modeled as a series cofiguratio as show i figure. The system reliability Fig: Series Cofiguratio R S (t) = j= r j (t) () Where r j t is the reliability of the compoet at the j th stage (or) subsystem. The reliability of a system, which is modeled as a series cofiguratio, is less tha the least amog the compoet reliabilities, sice the reliability of each compoet is less tha uity ad the system reliability is the product the compoet reliabilities. A series system is referred to as -out of : Good System or equivaletly -out of : Failed system.

27 Parallel Cofiguratio: If the success of a system is esured by the operatig coditio of ay oe of the compoets, the the system is modeled as a parallel cofiguratio as show i fig. Parallel Cofiguratio Sice the system is successful eve if oe of the compoets is operative, the system fails if ad oly if each ad every oe of the compoets fails, i.e., the system ureliability is Q S (t) = j= q j (t)..(2) Where q j (t) =-r j (t) ad the system reliability, R S t = - Q S (t) The ureliability of a parallel system is less tha the least amog the compoet ureliabilities or the reliability is higher tha the highest amog the compoet reliabilities. A parallel cofiguratio is referred to as -out of : Good system (or) equivaletly, -out of : Failed system. Series- Parallel cofiguratio: If the iteded fuctio ca be achieved by differet alteratives ad if each of the alteratives requires the fuctioig of all the compoets coected i series i that alterative, the system is to be modeled as a series parallel cofiguratio as show i fig. Series- Parallel Cofiguratio Let R k (t) be the reliability of the K th alterative (of the series- parallel cofiguratio) obtaied by usig equatio the the system ca be reduced the oe show i fig.

28 Equivalet Parallel System The reliability of the system show i fig ca be obtaied by makig us of equatio () by replacig q j (t) by Q j (t) =(-R j (t). Parallel Series Cofiguratio: If the fuctio of ay subsystem ca be achieved by oe or more compoets coected i parallel ad if all of the subsystems must be operative to esure the success of the system, the it is to be modeled as a parallel-series cofiguratio as show i fig. Parallel-Series System The system is successful if each subsystem has at least oe operative compoet. The reliability is evaluated by replacig each subsystem by a sigle uit of equivalet reliability as show i fig. Equivalet series system Where R S (t) = j= R j (t)..(3) Where the various Rj s obtaied by makig use of equatio(2). r out of : Good systems: If the compoet of a system are arraged such that the system works if ay r compoets out of compoets preset i the system are workig. Let the reliabilities of all the compoets be equal to R. R r =P(r) = Cr R r ( R) r R r+ = P(r+) = Cr+ R r+ ( R) r..

29 R = P() = C R ( R) 0 Total reliability r out of = P(r) + P(r+)+ P() (if r= the it is a series system ad if r= the it is a parallel system) No-series-parallel system (cofiguratio) or Complex System: If the RBD does ot fall uder ay oe of the four basic cofiguratio, viz. series, parallel, series- Parallel ad parallel-series, the the system is to be modeled as a complex cofiguratio. The bridge type cofiguratio show i fig. is a example of a complex cofiguratio. The presece of the elemet C reders the system to be complex. Complex Cofiguratio Problems:. Calculate the reliability of the system show usig etwork reductio techique? R s = R s R 7 = -{(-0.8)(-0.9)} = =0.98 R 8 = R 7 R 4 =0.98x0.8 =0.784 R 9 = -{(-R 5 )(-R 8 ) = - (0.2x 0.26) = = R 0 =R R 9 =0.9x = 0.862

30 R - {(-0.862)(-0.9)} =-(0.3888x0.) = = A system cosists of 0 idetical compoets, all of which must work for system success. What is the system reliability it each compoet has a reliability of R s = = A two compoet series system cotais idetical compoets each havig a reliability of Evaluate the ureliability of the system. R s (0.99) 2 Q s = -(0.99) 2 = A system desig required 200 idetical compoets i series. It the overall reliabilitys must ot be less tha 0.99, what is the miimum reliabilitys of each compoet. R 200 = 0.99 R= 0.99 /200 = A system cosists of from compoets i parallel havig reliabilities of 0.99, 0.95, 0.98 & what is the reliability ad ureliability of the system. Q p = (-0.99)(-0.95)(-098)(-0.97) = 3x0-7 R p = A system is to be desiged with a overall reliability of 0.99 usig compoets havig idividual reliabilities of 0.7. what is the miimum umber of compoets that must be coected i parallel. Q p = (Q i ) (-0.999) =(-0.7) 0.0 = (0.3) N=5.75 =6 7. A series system has 0 idetical compoets. It the overall system reliability must be at least 0.99, what is the miimum reliability reqired of each compoet (R) 0 = 0.99 R = (0.99) /0 8. A parallel system has 0 idetical compoet. If the overall system reliability must be at least 0.99, how power ca these compoets be (-0.99) = (-R) 0 (-R) = (-0.99) /0 = A parallel system has idetical compoets havig a reliability of 0.5. what is the miimum umber of compoet if the system reliability must be at least Derive a geeral expressio for the reliability of the model show below ad hece evaluate the system reliability if all compoets have a reliability of 0.9. R = R s R 9 =R R 2 R 3 R 4

31 R 0 = R 5 R 6 R 7 R 8 R = - Q R =-Q 9 Q 0 = -{(-R 9 ) (-R 0 )} = R 9 + R 0 R 9 R 0 = R R 2 R 3 R 4 + R 5 R 6 R 7 R 8 - R R 2 R 3 R 4 R 5 R 6 R 7 R 8 R = R R 2 R 3 R 4 R 5 R 6 R 7 R 8 = 0.9 R = = R =R s = Fid the reliability of the system below if at lest 2 should for system such ad if the reliability of each compoet is 0.8? R 4 is evaluated by applyig the biomial distributio to compoet,2, ad 3. r=2, = 3 =P 2 = 3 C = 3 X P 3 = 3 C3 = (0.8) 3 P r = P 2 + P 3 = (0.8) (0.2) Reliability Evaluatio of No-series-parallel system or Complex System If the Reliability block diagram does ot fall uder ay of the four basic cofiguratio, viz. series, parallel, series-parallel ad parallel-series, the the system is to be modeled as a complex cofiguratio. The bridge type cofiguratio show i fig. is a example of a complex cofiguratio. The presece of the elemet C reders the system to be complex. Complex Cofiguratio Reliability Evaluatio of Complex Systems

32 Though the reliability of a complex system ca be determied by various methods such as the exhaustive search for the successful states, direct caoical expasio, probability map method, probability calculus method etc., oe of the followig methods is commoly used: Miimal path set (or) Miimal tie set method Miimal cut set method Decompositio method Miimal Pathset Method: A path is a sequece of the braches of the reliability block diagram from the iput ode to the output ode such that the succeedig ode of ay brach is the same as the preceedig ode of the ext brach. A miimal path is oe which satisfies the above property. Accordigly, a miimal pathset may be defied as the set of the elemets the proper fuctioig of which esures the success of the system ad o proper subset of which ca esure the success of the system. With referece to the RLG, the set of the elemets which esures the coectivity betwee the odes S ad t ad o subset of which [other tha the set itself] esures this coectivity is a miimal pathset. I the case of small systems the miimal pathsets ca be obtaied by ispectio. Simple algorithms are available for the determiatio of the miimal pathsets of large systems, ivolvig more umber of braches. Thus, the miimal path sets (tie sets) of a bridge etwork are T = {AB} T 2 = {DE} T 3 = {ACE} T 4 = {DCB} A structure fuctio X is such that X = {T U T 2 U T 3 U T 4 U} The reliability of the cofiguratio is probability of X, i.e., R s (t) = Pr(X) The symbolic reliability expressio of the system is R s (t) = ab + de + ace + bcd abde abce abcd acde bcde + 2 (abcde), Where the lower case letters represet the reliabilities of the correspodig elemets represeted by upper case letters, i.e., a = R A (t) etc. The reliability expressio thus obtaied is the u-complemeted symbolic reliability expressio. The umerical value of the reliability is obtaied by substitutig the values of the compoet reliabilities.

33 Thus,, if the reliabilities of the compoets A, B, C, D ad E are, respectively 0.70, 0.75, 0.80, 0.85 ad 0.90, the overall system reliability is Miimal Cutset Method: A Cutset is a set of braches which whe cut will ot allow ay path from iput ode to output ode. A miimal cutest is a set of braches which satisfies this property, but o subset of this has the same property. A miimal cutest is defied as the set of the elemets, the failure of which i.e, the ioperative coditio of which causes the failure of the system, ad o proper subset of which ca cause the failure of the system. With referece to the RLG, a miimal cutest destroys the coectivity betwee the source ad the sik odes. Thus, referrig to the miimal cutsets are C = {A D} C 2 = {B E} C 3 = {A CE} C 4 = {BC D} The structure fuctio X = {C U C 2 U C 3 U C 4 } The ureliability of the system is Q s (t) = P(X) ad the reliability, R s (t) = Q s (t). The symbolic ureliability of the system ca be obtaied as Q s (t) = a d + b e + a c e + b c d a c d e a b c d a b c e b c d e + 2(a b c d e ), where a = (-a) etc. Usig the same values of the compoet reliabilities give i the miimal pathset approach, the ureliabilities of the compoets A through E are 0.30, 0.25, 0.20, 0.5 ad 0., respectively. The overall system ureliability is ad the overall system reliability is , as i the previous case. While the miimal pathsets ca be obtaied i a straight-forward maer, there is o simple algorithm by which the miimal cut sets ca be obtaied. The miimal cutsets are deduced from the miimal path sets by makig use of DeMorga s laws. While the reliability ca be obtaied from a kowledge of miimal pathsets, the ureliability is obtaied from a kowledge of the miimal cutsets ad hece the reliability is obtaied. While it caot be geeralized that the umber of miimal cutsets is less tha the umber of miimal pathsets, for most of the complex cofiguratios the umber of miimal cutsets is less tha the umber of miimal pathsets. For example, the umber of miimal pathsets for the cofiguratio, whereas the umber of miimal cutsets is oly 6. To evaluate the reliability by the miimal tie set method the umber of terms to be evaluated is ie., 27. O the other had the umber of terms to be cosidered is oly i.e., 63, if the miimal cutest method were to be used. Decompositio Method

34 The reliability of a complex system ca be evaluated by cosiderig the combiatios that reders the system a simple series-parallel or a parallel-series cofiguratio. I this method, oe of the elemets preferably the oe that reders the system complex is chose as the corer stoe or key elemet. If K is the keystoe elemet, the the reliability of the system R s = (Reliability of modified system/give K is good) (Reliability of K) + (Reliability of modified system / give K is failed) (ureliability of K) A B s E t C D For the RLD show, the presece of elemet E reders the system complex ad E is a bidirectioal elemet. Thus to evaluate the reliability of this system, we cosider two states which are mutually exclusive with respect to E amely E is operative ad E is failed. Let us choose E as the corer stoe elemet, so that the reliability of system = R e + R 2 (- e) where R = Reliability of modified system with E good R 2 = reliability of modified system with E failed. R s = R e + R 2 (-e) where e is the reliability of compoet E. Whe E is workig - Calculatio of R Whe E is operative, the two odes & 2 are always coected together or they ca be merged with each other to form a combied ode (, 2). The RLD ca be redraw as below A (, 2) B s C D t A B C D Reliability of system R = (a+c-ac) (b+d-bd) Whe E is ot - workig - Calculatio of R 2 E is failed state implies a ope circuit betwee odes & 2 ad the RLD is modified as below.

35 A B C D R 2 = -[(-ab) (-cd)] = -[-ab-cd+abcd] =ab+cd-abcd Overall Reliability = R s = R e + R 2 (-e) R s = [(a+c-ac)(b+d-bd)]e + [(ab+cd-abcd)(-e)] = (ab+ad-abd+cd+cb-bcd-abc-acd+abcd)e + (ab+cd-abcd-abe-cde+abcde) = abe+ade-abde+cde+cbe-bcde-abce-acde+abcde+ab+cd-abcd-abe-cde+abcde R s = ab + cd + ade +bce abde bcde abce acde abcd + 2abcde Uit IV Trasitio rate cocepts MARKOV PROCESS Cosider the case of a sigle repairable compoet for which the failure rate ad repair rate are costat, i.e. they are characterized by the expoetial distributio. The state trasitio diagram for this compoet is show i below figure

36 Let P 0 (t) = probability that the compoet is operable at time t P (t) = probability that the compoet is failed at time t λ = failure rate μ = repair rate The failure desity fuctio for a compoet with a costat hazard rate of was give i equatio as f t = λe λt The desity fuctios for the operatig ad failed states of the system show i above figure are therefore f o t = λe λt ad f t = μe μt, respectively The parameters λ ad μ are referred to as state trasitio rates sice they represet the rate at which the system trasits from oe state of the system to aother. The failure rate λ was foud to be the reciprocal of the mea time to failure, MTTF, with the times to failure couted from the momet the compoet begis to operate to the momet it fails. Similarly, the repair rate μ is the reciprocal of the mea time to repair, MTTR, with these times couted from the momet the compoet fails to the momet it is retured to a operable coditio. The correct iterpretatio of state residece time is a importat poit, as the failure ad repair rates are sometimes icorrectly evaluated by coutig the umber of failures or repairs i a give period of time, ad dividig by the elapsed time. The correct time value to use i the deomiator is the portio of time i which the compoet was i the state beig cosidered. This is less tha the actual elapsed period of time uless o trasitios occurred from the state. Cosequetly λ = umber of failures of a compoet i the give period of time total period of time the compoet was operatig μ = umber of repairs of a compoet i the give period of

37 time total period of time the compoet was beig repaired This cocept of a trasitio rate leads to the defiitio Trasitio rate = umber of times a trasitio occurs from a give state/time spet i that state Evaluatig time depedet probabilities The relevat state space diagram for the simple sigle compoet is show i above figure. The trasitios i these diagrams were represeted by the value of trasitioal probability. I the case of cotiuous Markov processes they are usually represeted by a trasitio rate by the trasitios λ ad μ from the operatig ad failed states respectively. Cosider ow a icremetal iterval of time dt which is made sufficietly small so that the probability of two or more evets occurrig durig this icremet of time is egligible. This cocept was first used i coectio with the Poisso distributio. The probability of beig i the operatig state after this time iterval dt, that is the probability of beig i state 0 at time (t + dt) is [Probability of beig operative at time t AND of ot failig i time dt] + [probability of beig failed at time t AND of beig repaired i time dt]. Usig a similar approach to that used to develop the Poisso distributio. P o t + dt = P o t λdt + P t μ dt Similarly, from equatio as dt 0 P t + dt = P t μdt + P o t λ dt P o t + dt P o t dt = λp o t + μp t P o t + dt P o t dt dt 0 = dp o t dt = P o (t) thus, Similarly, from equatio P o t = λp o t + μp t P t = λp o t + μp t Equatios may be expressed i matrix form as P o t P t = P o t P t λ μ λ μ The coefficiet matrix i equatio is ot a stochastic trasitioal probability matrix because the rows of this coefficiet matrix summate to zero whereas those of the stochastic trasitioal probability matrix summate to uity.