In this chapter we shall consider a very important and growing area of application

Transcription

1 Reliabiliy Engineering In his chaper we shall consider a very imporan and growing area of applicaion of some of he probabiliy conceps discussed in he previous chapers, namely, Reliabiliy Engineering. Reliabiliy consideraions are playing an increasing role in almos all engineering disciplines. Generally he erm reliabiliy of a componen (or a sysem, ha is, a se of componens assembled o perform a cerain funcion) is undersood as is capabiliy o funcion wihou breakdown. The beer he componen performs is inended funcion, he more reliable i is. In he broader sense, reliabiliy is associaed wih dependabiliy, wih successful performance and wih he absence of breakdown or failures. However, from he engineering analysis poin of view, i is necessary o define reliabiliy quaniaively as a probabiliy. Technically, reliabiliy may be defined as he probabiliy ha a componen (or a sysem) will perform properly for a specified period of ime under a given se of operaing condiions. In oher words, i is he probabiliy ha he componen does no fail during he inerval (, ). We give below he formal mahemaical definiion of reliabiliy and discuss he relaed conceps. CONCEPTS OF RELIABILITY The definiions given below wih respec o a componen hold good for a sysem or a device also. If a componen is pu ino operaion a some specified ime, say, and if T is he ime unil i fails or ceases o funcion properly, T is called he life lengh or ime o failure of he componen. Obviously T ( ) is a coninuous random variable wih some probabiliy densiy funcion f(). Then he reliabiliy or reliabiliy funcion of he componen a ime, denoed by R(), is defined as R() P(T > ) or P(T ) F() () where F() is he cumulaive disribuion funcion of T, given by F() f() d

2 Probabiliy, Saisics and Random Processes Thus R() f() d f() d () Since F() and F( ) by he propery of cdf, R() and R( ) i.e., R(). Also since d d F() f(), dr() we ge f() (3) d Now he condiional probabiliy of failure in he inerval (, + ), given ha he componen has survived up o ime, viz., f() D P{ T + /T }, by he definiions of pdf and cdf F () f() /R() f() The condiional probabiliy of failure per uni ime is given by R () and is called he insananeous failure rae or hazard funcion of he componen, denoed by l(). Thus f() l() R () (4) Now, using (3) in (4), we have R () R () l() (5) Inegraing boh sides of (5) wih respec o beween and, we have R () d R () i.e., {log R()} i.e., log R() log R() i.e., log e [R()] - R() e Using (6) in (4), we have l() d l() d l() d l() d [Q R() ] l () d - l () d f() l() e (7) The expeced value of he ime he failure T, denoed by E(T) and variance of T, denoed by s T are wo imporan parameers frequenly used o characerise reliabiliy. E(T) is called mean ime o failure and denoed by MTTF. (6)

3 Reliabiliy Engineering 3 Now MTTF E(T) on inegraion by pars f() d R () d, by (3) [R()] + Rd () (8) é ù é ù - l () d ê ê ú ú Now [R()] êe ú ê ú (9) ê ê ú l () dú ë û ê ú ëe û and [R()] R() () Using (9) and () in (8), we ge MTTF Rd () () Var(T) s T E{T E(T )} or E(T ) {E()} f() d (MTTF) () Condiional reliabiliy is anoher concep useful o describe he reliabiliy of a componen or sysem following a wear-in period (burn-in period) or afer a warrany period. I is defined as R(/T ) P{T > T + /T > T } PT { > T + } RT ( + ) (3) PT { > T} RT ( ) T + - l () d T - l () d e e e T + - l () d T e T ét + ù -ê l() d- l() dú ê ú ë û (4) Some Special Failure Disribuions Some of he probabiliy disribuions describe he failure process and reliabiliy of a componen or a sysem more saisfacorily han ohers. They are he exponenial, Weibull, normal and lognormal disribuions. We shall now derive he

4 4 Probabiliy, Saisics and Random Processes reliabiliy characerisics relaing o hese failure disribuions using he formulas derived above. () The exponenial disribuion If he ime o failure T follows an exponenial disribuion wih parameer λ, hen is pdf is given by f() le l, (5) Then, from (), R() -l le d l f() le - -l -l ù -l û e é ë - e (6) From (4), l() l (7) R () e This means ha when he failure disribuion is an exponenial disribuion wih parameer l, he failure rae a any ime is a consan, equal o l. Conversely when l() a consan l, we ge from (7), - ld f() l. e le l, Due o his propery, he exponenial disribuion is ofen referred o as consan failure rae disribuion in reliabiliy conexs. We have already derived in he earlier chapers, ha MTTF E(T ) l (8) and Var(T ) s T l (9) Also - l( T + ) RT ( + ) e R(/T ) by (6) T RT ( ) -l e e l () This means ha he ime o failure of a componen is no dependen on how long he componen has been funcioning. In oher words he reliabiliy of he componen for he nex hours, say, is he same regardless of wheher he componen is brand new or has been operaing for several hours. This propery is known as he memoryless propery of he consan failure rae disribuion. () The Weibull disribuion The pdf of he Weibull disribuion was defined as f() ab b e ab, () An alernaive form of Weibull s pdf is b- b b f() æ ö é æ ö ù ç exp ê- ç ú, q >, b >, () q èqø ê èqø ë úû () is obained from () by puing a b q. b is called he shape parameer

5 Reliabiliy Engineering 5 and q is called he characerisic life or scale parameer of he Weibull s disribuion (). If T follows Weibull s disribuion (), hen b- b b R() æ ö é æ ö ù. ç exp ê- úd q è ç qø èqø êë úû b x e - æ ö dx, on puing ç q x è ø e x é b æ ö ù exp ê- ç ú ê èq ø ë úû (3) b- f() b æ ö l(). R () q ç èq ø (4) As derived in he chaper some special probabiliy disribuions, we have æ ö MTTF E(T ) qg ç + è b ø (5) and Var(T ) s T ì ü q ï æ ö é æ öù ï íg ç + - êg + ú ý è b ø ç è b ï ë ø î û ïþ (6) Now R ( + T ) R(/T ) RT ( ) é b æ + T ö ù exp ê- ç q ú ê ë è ø úû é b æt ö ù exp ê- ç ú è q ê ø ë úû é b b æ + T ö æt ù ö exp ê- ç + ú è q ø ç è q ê ø ë úû (7) (3) The normal disribuion If he ime o failure T follows a normal disribuion N(m, s) is pdf is given by é-( -m) ù f() exp ê, ú < < s p ë s û In his case, MTTF E(T ) m and Var (T ) st s. R() f() d is found ou by expressing he inegral in erms of sandard normal inegral and using he normal ables. l() is hen found ou by using (4).

6 6 Probabiliy, Saisics and Random Processes (4) The lognormal disribuion If X log T follows a normal disribuion N(m, s), hen T follows a lognormal disribuion whose pdf is given by é ì æ öü ù f() exp ê ï ï - log í ú s p s ç ý, ê ïî è m øïþ ú ë û (8) where s s is a shape parameer and m, he median ime o failure is he locaion parameer, given by log m m. I can be proved ha æ s ö MTTF E(T) m exp ç è ø (9) and Var(T ) st m exp(s ) [exp (s ) ] (3) In his case, F() P(T ) P{log T log } ì P log T m log mü í ý, since log T follows a N(m, s) î s s þ ì P ï log T - log m æ öüï í log s s ç ý ïî è m øïþ ì P ï æ öüï íz log ý, where Z follows he sandard s ç è ïî m øïþ normal disribuion N(, ). Then R() and l() are compued using () and (4). Example The densiy funcion of he ime o failure in years of he gizmos (for use on widges) manufacured by a cerain company is given by f(),. 3 ( + ) (a) Derive he reliabiliy funcion and deermine he reliabiliy for he firs year of operaion. (b) Compue he MTTF. (c) Wha is he design life for a reliabiliy.95? (d) Will a one-year burn-in period improve he reliabiliy in par (a)? If so, wha is he new reliabiliy? Soluion (a) f(), 3 ( + ) R() R() f() d ( + ) é - ù ê ú ë( + ) ( + ).864. û

7 Reliabiliy Engineering 7 (b) MTTF Rd () d ( + ) æ - ö ç è + ø years. (c) Design life is he ime o failure ( D ) ha corresponds o a specified reliabiliy. Now i is requeired o find D corresponding o R.95 ( + ).95 i.e., ( D + ).63 D.598 year or 95 days (d) R ( + ) R(/) R() ( + ) ( + ) Now R(/) > R(), if > ( + ) ( + ) ( + ) i.e., if > ( + ) + i.e., if > + i.e., >, which is rue, as One year burn-in period will improve he relaibiliy. Now R(/).843 >.864. ( + ) Example The ime o failure in operaing hours of a criical solid-sae æ ö power uni has he hazard rae funcion ll().3 ç, for. è5ø (a) Wha is he reliabiliy if he power uni mus operae coninuously for 5 hours? (b) Deermine he design ofe if a reliabiliy of.9 is desired. (c) Compue he MTTF. (d) Given ha he uni has operaed for 5 hours, wha is he probabiliy ha i will survive a second 5 hours of operaion? Soluion é ù (a) R() exp ê- l() dú êë úû é 5.5 æ ö ù R(5) exp ê-.3 ç d ú ê è 5 ø ë úû 5 é.3 3ù exp ê ú ë û.5

8 8 Probabiliy, Saisics and Random Processes é.3 ù exp ê ú ë û exp[.36].9689 (b) R( D ).9 i.e., i.e., i.e., D (c) MTTF é D.5 æ ö ù exp ê-.3 ç d ú ê è 5 ø ë ú.9 û D.3 - d Rd () 3 D ìï üï í ý ïî.3 ïþ æ.3 3ö - ç è 5 3 ø e - 3 a x 3 e -. x a d 3 e d, where a hours. dx, on puing x a 3/ 3a 3 ( 3) 3 3 (5 3) 3a.933, from he able of values of Gamma funcion. 3 a hours. PT ( ³ ) R() (d) P(T /T 5) PT ( ³ 5) R(5) é ù exp ê³ l() dú êë 5 úû éì. 3ü ì-. 3üù exp êí- ý - í 5 ýú ëî 5 þ î 5 þû exp[{.8944} {.36}].9438

9 Reliabiliy Engineering 9 æ ö Example 3 The reliabiliy of a urbine blade is given by R() ç - è ø,, where is he maximum life of he blade. (a) Show ha he blades are experiencing wear ou. (b) Compue MTTF as a funcion of he maximum life. (c) If he maximum life is operaing hours, wha is he design life for a reliabiliy of.9? Soluion æ ö (a) R() ç - è ø, R () Now l() R () æ ö - ìï æ öüï ç - í- - ý è ø ç ïî è øïþ - l () > and so l() is an increasing funcion of. ( - ) When he failure rae increases wih ime, i indicaes ha he blades are experiencing wear ou. æ (b) MTTF ö Rd () ç - d è ø 3 é æ ö ù ê- - ú ê 3 ç è ø ë ú û (c) When, R( D ).9 i.e., æ D ö ç -.9 è ø D.9487 D.63 hours. 3. Example 4 Given ha R() e -, (a) Compue he reliabiliy for a 5 hours mission. (b) Show ha he hazard rae is decreasing. (c) Given a -hour wear-in period, compue he reliabiliy for a 5-hour mission. (d) Wha is he design life for a reliabiliy of.95, given a -hour wear-in period?

10 Probabiliy, Saisics and Random Processes (a) R() R(5) (b) l() - R () R (). e -. 5 e -.95 e e -. (c) R(/T )., R ( + T ) RT ( ) which is a decreasing funcion of. R(6) R(5/) R() (d) R( D /).95 e e i.e., R ( D + ) R().95 i.e.,. ( D ) e ( D + ).59 D.89 hours. Example 5 A one-year guaranee is given based on he assumpion ha no more han % of he iems will be reurned. Assuming an exponenial disribuion, wha is he maximum failure rae ha can be oleraed? If T is he ime o failure of he iem,. e - Soluion hen P(T ).9 (Q no more han % will be reurned) i.e., R().9 i.e., -l le d.9 (Q T follows an exponenial disribuion) i.e., ( e l ).9 i.e., e l.9 l.54 l.54/year Example 6 A manufacurer deermines ha, on he average, a elevision se is used.8 hours per day. A one-year warrany is offered on he picure ube having a MTTF of hours. If he disribuion is exponenial, wha percenage of he ubes will fail during he warrany period?

11 Reliabiliy Engineering Soluion Since he disribuion of he ime o failure of he picure ube is exponenial, R() e l where l is he failure rae Given ha MTTF hours i.e., -l e d i.e., or l.5/hour l P(T year) P(T hours) [Q he T.V. is operaed for.8 hours/day] P{T > 657} R(657) e.8 i.e., 8% of he ubes will fail during he warrany period. Example 7 Nigh wachmen carry an indusrial flashligh 8 hours per nigh, 7 nighs per week. I is esimaed ha on he average he flashligh is urned on abou minues per 8-hour shif. The flashligh is assumed o have a consan failure rae of.8/hour while i is urned on and of.5/hour when i is urned off bu being carried. (a) Esimae he MTTF of he ligh in working hours. (b) Wha is he probabiliy of he ligh s failing during one 8-hour shif? (c) Wha is he probabiliy of is failing during one monh (3 days) of 8-hour shifs? Noe: In he earlier problems, we have no aken ino accoun he possibiliy ha an iem may fail while i is no operaing. Ofen such failure raes are small enough o be negleced. However, for iems ha are operaed only a small fracion of ime, failure during non-operaion may be quie significan and so should be aken ino accoun. In his siuaion, he composie failure rae λ c is compued as l c fl + ( f )l N, where f is he fracion of ime he uni is operaing and l and l N are he failure raes during operaing and non-operaing imes] Since he flashligh is used only a small fracion of ime ( minues ou of 8 hours), we compue he composie failure rae l c and use i for oher calculaions. l c fl + ( f )l N /hour 4 4 (a) MTTF 3 hours lc.85 8 (b) P(T 8) c e l c l - d e 8lc.69 (c) p probabiliy of failure in one day (nigh).69

12 Probabiliy, Saisics and Random Processes q p.937 n 3 Probabiliy of failure in 3 days Probabiliy of no failure in 3 days nc p q n, by binomial law (.937) Example 8 A relay circui has an MTBF of.8 year. Assuming random failures, (a) Calculae he probabiliy ha he circui will survive year wihou failure. (b) Wha is he probabiliy ha here will be more han failures in he firs year? (c) Wha is he expeced number of failures per year? Since he failures are random evens, he number of failures in an inerval of e lengh follows a Poisson process, given by P{N() n}, where l, failure rae. -l l ( ) n! n, n,,, Then he ime beween failures follows an exponenial disribuion wih mean l. Now MTBF Mean ime beween failures l.8 year l per year.5 per year.8 - e l l (a) P{N () } e.5.865! (b) P{N() > } e l ì ïl l l üï í + + ý ïî!!! ïþ ì (.5).865 ï üï í+.5+ ý ïî ïþ.35 (c) E{N()} l E{Number of failures per year} l.5 Example 9 For a sysem having a Weibull failure disribuion wih a shape parameer of.4 and a scale parameer of 55 days, find (a) R( days); (b) he B life; (c) MTTF; (d) he sandard deviaion, (e) he design life for a reliabiliy of.9. Soluion b- b b æ ö ìï æ ö üï The pdf of he Weibull disribuion is given by f(). ç exp í- ý, q è ç qø ï èqø î ï. Now b.4 and q 55 days þ

13 Reliabiliy Engineering 3 (a) R() (b) ìï æ ö R() exp í- ç ï èq ø î b üï ý ïþ ìï æ ö exp í- ç ï è55 ø î.4 üï ý ïþ.9 Noe: The life ime corresponding o a reliabiliy of.99 is called he B life. Similarly ha corresponding o R.999 is called he B. life. Le R be he B life of he sysem Then R( R ).99 i.e., ìï æ exp R ö í- ç è55 ï ø î.4 æ R ö ç è 55ø üï ý ïþ R 55.4 (.5).6 days æ ö (c) MTTF q. ç + è b ø 55 æ ö ç + è.4 ø 55 (.7) , from he Gamma ables 5.8 days (d) (S.D.) q ì ü ï æ ö é æ öù ï í ç + - ê + ú ý è b ø ç è b ï ê ø ë úû ï î þ 55 é ì ü ù ê æ ö ï æ öï + - í + ý ú ê ç è ç.4 ø ï è.4 ø î ïþ ú ë û é (.43) - ê (.7) ë 55 { } ù úû é.43 (.43) - ê (.7) ë 55 { } 55 [ (.957) ] S.D days ù úû

14 4 Probabiliy, Saisics and Random Processes (e) Le D be he required design life for R.9 ì.4 ï æ R( D ) exp D ö üï í- ç ý.9 ï è55 ø î ïþ æ D ö ç è 55ø D 55 (.536). days Example A device has a decreasing failure rae characerised by a woparameer Wibull disribuion wih q 8 years and b.5. The device is required o have a design life reliabiliy of.9. (a) Wha is he design life, if here is no wear-in period? (b) Wha is he design life, if here is a wear-in period of monh in he beginning? Answer (a) Le D be he required design life for R.9 R( D ) ìï æ exp D ö í- ç è8 ï ø î.5 üï ý ïþ.9.5 æ D ö ç.536 è8ø D 8 (.536).998 ; years. (b) If T is he wear-in period, hen ì b b ï æ + T ö æt ö üï R(/T ) exp í- ç + ý è q ø ç è q ï ø î ïþ Le W be he required design life wih wear-in. ì.5.5 æ ö æ ö ü W + ï ç Then exp ç ï í- ç + ý 8 ç 8.9 ï ç ç ï ï è ø è ø î ïþ.5.5 æ i.e., W + ö æ ö ç è 8 + ø ç è ø.5 i.e., æ W + ö ç è 8 ø.688 W i.e., W.8 years Thus a wear-in period of monh increases he design life by nearly.8 year or monhs.

15 Reliabiliy Engineering 5 Example Two componens have he same MTTF; he firs has a consan failure rae l and he second follows a Weibull disribuion wih b. (a) Find q in erms of l. (b) If for each componen he design life reliabiliy mus be.9, how much longer (in percenage) is he design life of he second (Weibull) componen? (a) MTTF is he same for boh he componens. For he consan failure rae disribuion (viz., exponenial disribuion), MTTF () l For he Weibull disribuion wih b (viz., Rayleigh disribuion), R() e MTTF Rd () - e q - q d q - x e dx, on puing q x p q From () and (), we ge q p l q l p (b) Le and be he required design lives for he wo componens. Then R( ) e - l.9, for he firs. l l () (3) (4) or q - R( ) e.9, for he second..536 q.3459 q.3459 l p, using (3).3666 l (5) Increase in design life.69/l

16 6 Probabiliy, Saisics and Random Processes ( - ) % increase in design life Example The wearou (ime o failure) of a machine par is normally disribued wih 9% of he failures occurring symmerically beween and 7 hours of use. (a) Find he MTTF and sandard deviaion of failure imes. (b) Wha is he reliabiliy if he par is o be used for hours and hen replaced? (c) Deermine he design life if no more han a % probabiliy of failure prior o he replacemen is o be oleraed. (d) Compue he reliabiliy for a -hour use if he par has been operaing for hours. Le he ime o failure T follow N(m, s), where s is he MTTF and s is he S.D. Soluion (a) 9% of he failures lie symmerically beween and 7 35 f() d 7 f() d.45, where m 35 hours s - 35 Puing z, we ge s f( z ) dz.45 From he normal ables, 35 s s.8 hours.645 (b) R() f() d R() f() d f( zdz ).5 + f( zdz ) dz (c)if D is he required design life, hen D - From he normal ables, f() d. or D D 85.6 hours ( D -35).8 f( zdz ). -

17 Reliabiliy Engineering 7 (d) R ( + T ) RT ( ) R() R() f() d f() d f( zdz ) f( zdz ) Example 3 A cuing ool wears ou wih a ime o failure ha is normally disribued. I is known ha abou 34.5% of he ools fail before 9 working days and abou 78.8 % fail before working days. (a) Compue he MTTF. (b) Deermine is design life for a reliabiliy of.99. (c) Deermine he probabiliy ha he cuing ool will las one more day given ha i has been in use for 5 days. (a) Le T follow a N(m, s) Soluion Given 9 f () d n-m s i.e., f( zdz ).345, on puing z - m s i.e., - m -9 s f( zdz ).55 m - 9.4, () s using he normal ables. Also i.e., i.e., f () d m s f( zdz ) m f( zdz ).88 - m s.8 ()

18 8 Probabiliy, Saisics and Random Processes using he normal ables. Solving equaions () and (), we ge m and s.5 i.e., MTTF days. (b) Le R be he required design life for R.99 f() d.99 or f( zdz ).99 R - R.5 f( zdz ).49 (c) P(T 6/T > 5) R R.3, using he normal ables..5 R 4. days f( zdz ) f( zdz ) PT ( ³ 6) PT ( > 5) f( zdz ).5 + f( zdz ) f() d f() d Example 4 A complex machine has a high number of failures. The ime o failure was found o be log normal wih s.5. Specificaions call for a reliabiliy of.95 a cycles. (a) Deermine he median ime o failure ha mus be achieved by engineering modificaions o mee he specificaions. Assume ha design changes do no affec he shape parameer s. (b) Wha are he corresponding MTTF and sandard deviaion? (c) If he desired median ime o failure is obained, wha is he reliabiliy during he nex cycles given ha i has operaed for cycles? (a) R() f() d, where f() is he pdf of he lognormal disribuion ælog -log M ö ç è s ø f(z) dz, where f(z) is he pdf of he sandard normal disribuion

19 Now R ().95 f( z) dz.95 æö log s ç è M ø - æö From he normal ables, log.5 ç è M ø.645 M log æ ö ç è ø M Median exp ( cycles (b) MTTF M s e 787 e cycles s M s e (e s ) (787) e.565 (e.565 ) s 3355 cycles R ( + T ) (c) R(/T ) RT ( ) R() R() f() d f() d Reliabiliy Engineering 9 f( zdz ) æ ö logç.5 è 787 ø æö logç.5 è 787 ø f( zdz ) f( zdz ) f( zdz ) Example 5 Faigue wearou of a componen has a log normal disribuion wih M 5 hours and s.. (a) Compue he MTTF and SD. (b) Find he reliabiliy of he componen ofr 3 hours. (c) Find he design life of he componen for a reliabiliy of.95.

20 Probabiliy, Saisics and Random Processes Soluion (a) MTTF M s e 5 e. 5 hours s M s s e ( e - ) (b) R(3) 3 5 e.4 (e.4 ) s 3 hours f() d, where f() is he pdf of he lognormal disribuion f( zdz ) æ3 ö logç. è 5 ø f( zdz ) (c) If D is he required design life, R( D ).95 i.e., f( zdz ).95 æ log D ö ç. è 5 ø From he normal ables, D æ ö log. ç è5 ø.645 D hours Par A (Shor-answer quesions). Wha do you undersand by reliabiliy of a device?. Wrie down he mahemaical definiion of reliabiliy. 3. Prove ha R(), where R() is he reliabiliy funcion. 4. Wha is hazard funcion? How is i relaed o reliabiliy funcion? 5. Derive he relaion ha expresses R() in erms of l(). 6. Derive he relaion ha expresses he pdf of he ime o failure in erms of hazard funcion. 7. Wha is MTTF? How is i relaed o reliabiliy funcion? 8. Express he condiional reliabiliy R(/T ) in erms of he hazard funcion. 9. Prove ha he hazard funcion is a consan for he exponenial failure disribuion.. When he hazard funcion is a consan, prove ha he failure ime disribuion is an exponenial disribuion.. Wrie down he pdf of he consan failure rae disribuion.

21 Reliabiliy Engineering. When he hazard funcion is a consan l, wha are he MTTF and variance of he ime o failure? 3. Prove ha, for an exponenial disribuion, R(/T ) R(). 4. Wha is memoryless propery of he consan failure rae disribuion? 5. Wrie down he pdf of Weibull disribuion wih shape parameer m and scale parameer n. 6. Wrie down he formulas for R() and l() for a Weibull disribuion wih shape parameer b and characerisic life q. 7. When he ime o failure T follows a Weibull disribuion wih parameers b and q, wha are MTTF and Var(T)? 8. Derive he formula for R(/T ) when T follows a Weibull disribuion (b, q). 9. Wrie down he densiy funcion of a lognormal disribuion in erms of is shape parameer and median ime o failure.. Wrie down he values of MTTF and s T for a lognormal disribuion (s, m ). Par B. The densiy funcion of he ime o failure of an appliance is 3 f() ; (> ) is in years. ( + 4) (a) Find he reliabiliy funcion R(). (b) Find he failure rae l(). (c) Find he MTTF.. A household appliance is adverised as having more han a -year life. If is pdf is given by f().( +.5) 3 ; deermine is reliabiliy for he nex years, if i has survived a -year warrany period. Wha is is MTTF before he warrany period? Wha is is MTTF afer he warrany period assuming ha i has sill survived? 3. A logic circui is known o have a decreasing failure rae of he form l() year, where is in years (a) If he design life is one year, wha is he reliabiliy? (b) If he componen undergoes wearin for one monh before being pu ino operaion, wha will he reliabiliy be for a one-year design life? 4. A componen has he following hazard rae, where is in years: l().4 4, (a) Find R(). (b) Deermine he probabiliy of he componen failing wihin he firs monh of is operaion. (c) Wha is he design life if a reliabiliy of.95 is desired? 5. The pdf of he ime o failure of a sysem is given by f()., days. Find.

22 Probabiliy, Saisics and Random Processes (a) R(); (b) he hazard rae funcion; (c) he MTTF; (d) he sandard deviaion. 6. The failure-disribuion is defined by 3 f(), hours 9 (a) Wha is he probabiliy of failure wihin a -hour warrany period? (b) Compue he MTTF. (c) Find he design life for a reliabiliy of.99. (d) Compue he average failure rae over he firs 5 hours. [Hin: Average failure rae of a componen over he inerval (, ) is defined as ìr ( ) ü AFR(, ) l() d - log í ý - îr( ) þ 7. The pdf of he ime o failure of a new fuel injecion sysem which experiences high failure rae is given by f().5/( + ).5,, where is measured in years. The reliabiliy over is inended life of years is unaccepable. Will a wear-in period of 6 monhs significanly improve upon his reliabiliy? If so, by how much? 8. A home compuer manufacurer deermines ha his machine has a consan failure rae of l.4 per year in normal use. For how long should he warrany be se, if no more han 5% of he compuers are o be reurned o he manufacurer for repair? [Hin: Find o such ha R(T ).95] 9. A more general exponenial reliabiliy model is defined by R() a b ; a > ; b >, where a and b are parameers. Find he hazard rae funcion and show how his model is equivalen o R() e l. 3. Show ha, for an exponenial disribuion (l), he residual mean life is, regardless of he lengh of he ime he sysem has been operaing. l [Hin: MTTF (/T ) RT ( ) d] T 3. The one-monh reliabiliy on an indicaor lamp is.95 wih he failure rae specified as consan. Wha is he probabiliy ha more han wo spare bulbs will be needed during he firs year of operaion (Ignore replacemen ime). 3. A urbine blade has demonsraed a Weibull failure paern wih a decreasing failure rae characerised by a shape parameer of.6 and a scale parameer of 8 hours. (a) (b) Compue he reliabiliy for a -hour mission. If here is a -hour burn-in of he blades, wha is he reliabiliy for a -hour mission? 33. Two componens have he same MTTF; he firs has a consan failure rae l.4 and he second follows a Weibull disribuion wih shape parameer equal o.

23 (a) (b) Reliabiliy Engineering 3 Find he characerisic life of he second componen. If for each componen he design life reliabiliy mus be.8, how much longer (in percenage) is he design life of he second componen han he firs? 34. A pressure gauge has a Weibull failure disribuion wih a shape parameer of. and a characerisic life of, hours. Find (a) R(5); (b) he Bl and B. lives; (c) he MTTF and (d) he probabiliy of failure in he firs year of coninuous operaion. 35. A componen has a Weibull failure disribuion wih b.86 and q 45 days. By how many days will he design life for a.9 reliabiliy specificaion be exended as a resul of a 3-day burn-in period? 36. For a hree-parameer Weibull disribuion wih b.54, q 85 and 5 hours, find (a) R(5 hours); (b) MTTF; (c) SD and (d) he design life for a reliabiliy of.98. [Hin: The pdf of a 3-parameer Weibull disribuion is f() b- b b æ - ö é æ - ù ö. exp q ç ê- ú è q ø ç è q ; ê ø. All formulas for his disribuion ë úû can be had from he corresponding formulas for he -parameer Weibull disribuion by replacing by ( ). éæ öù MTTF + q êç + ú è b êë øú û 37. A failure PDF for an appliance is assumed o be a normal disribuion wih a mean of 5 years and an S.D. of.8 year. Find he design life of he appliance for (a) a reliabiliy of 9%, (b) a reliabiliy of 99%. 38. A lahe cuing ool has a life ime ha is normally disribued wih an S.D. of. (cuing) hours. If a reliabiliy of.99 is desired over hours of use, find he corresponding MTTF. If he reliabiliy has o lie in (.8,.9), find he range wihin which he ool has o be used. 39. A roor used in A.C. moor has a ime o failure ha is lognormal wih an MTTF of 36 operaing hours and a shape parameer s equal o. (a) Deermine he probabiliy ha he roor will survive he firs hours. (b) If i has no failed ill he firs hours, wha is he probabiliy ha i will survive anoher hours? 4. The life ime of a mechanical valve is known o be lognormal wih median 36 hours and s.4. (a) Find he design life for a reliabiliy of.98. (b) (c) Compue he MTTF and S.D. of he ime o failure. The componen is used in pumping device ha will require 5 weeks of coninuous use. Wha is he probabiliy of a failure occurring because of he valve? Reliabiliy of Sysems As was poined ou, a sysem is generally undersood as a se of componens

24 4 Probabiliy, Saisics and Random Processes assembled o perform a cerain funcion. In he previous secion we have considered a number of imporan failure models (disribuions) for he individual com ponens. To evaluae he reliabiliy of a complex sysem, we may apply a paricu lar-failure law o he enire sysem. Bu i will be proper if we deermine an appropriae reliabiliy model for each componen and hen compue he reliabiliy of he sysem by applying he relevan rules of probabiliy according o he configuraion of he componens wihin he sysem. In his secion, we shall discuss he sysem reliabiliy in respec of a few simple bu relaively imporan cases. Serial Configuraion Series or nonredundan configuraion is one in which he componens of he sysem are conneced in series (or serially) as shown in he following reliabiliy block diagram (Fig..). Each block represens a componen. Fig.. In series configuraion, all componens mus funcion for he sysem o funcion. In oher words he failure of any componen causes sysem failure. Le R (), R () and R s () be he reliabiliies of he componens C and C and he sysem (assuming ha here are only componens in series), Then R P(C ) probabiliy ha C l funcions and R P(C ) probabiliy ha C funcions Now R s probabiliy ha boh C and C funcion P(C C ) P(C ), P(C ), assuming ha C and C funcion independenly. R R This resul may be exended. If C, C,..., C n be a se of n independen componens in series wih reliabiliies R (), R (),..., R n (), hen R s () R () R ()... R n () min {R (), R (),..., R n ()} [Q < R i () < ] i.e., he sysem reliabiliy will no be greaer han he smalles of he componen reliabiliies. Deducions. If each componen has a consan failure rae l i, hen R s () e -l. e -l L e -l n ( ) e - l + l + L + l n s e -l This means ha he sysem also has a consan failure rae l s. If he componens follow he Weibull failure law he parameers b i and q i, hen n å i é b b bn æ ö ù é æ ö ù é æ ö ù R s () exp ê- ç exp exp q ú ê- ç ú ê- ú q L ê è ø ú ê è ç ø ú ê èq n ø ë û ë û ë ú û é exp ê- ë n å i æ ö ù ç bi ú èq i ø û l. i

25 Reliabiliy Engineering 5 This means ha he sysem does no follow Weibull failure law, even hough every componen follows a Weibull failure disribuion. Parallel Configuraion Parallel or redundan configuraion is one in which he componens of he sysem are conneced in parallel as shwn in he following reliabiliy block diagram (Fig..). In parallel configuraion, all componens mus fail for he sysem o fail. This means ha if one or more componens funcion, he sysem coninues o funcion. Taking n and denoing he sysem reliabiliy by Rp(p p for parallel configuraion), we have Rp P(C or C or boh funcion) P(C C ) P(C ) + P(C ) P(C C ) P(C ) + P(C ) P(C ) P(C ), since C and C are independen R + R R R ( R ) ( R ) Exending o n componens, we have R p ( R ) ( R )... ( R n ) Max {R, R,..., R n } Noe: For a wo componen sysem in parallel having consan failure rae, i R i () e - l ; i,. l -l Hence R p () ( + e )( -e ) and MTTF -l -l e e Rp () d e -l ( ) e - l + l + -l d + e d l l e + - l l l + l. - ( + ) d

26 6 Probabiliy, Saisics and Random Processes Parallel Series Configuraion A sysem, in which m subsysems are conneced in series where each subsysem has n componens conneced in parallel as shown in Fig..3, is said o be in parallel series configuraion or low-level redundancy. Fig..3 If R is he reliabiliy of he individual componen, he reliabiliy of each of he subsysems ( R) n (In he diagram n ) Since m subsysems are conneced in series [In he Fig..3, m 3], he sysem reliabiliy for he low-level redundancy is given by R Low { ( R) n } m Series-Parallel Configuraion A sysem, in which m subsysems are conneced in parallel where each subsysem has n componens conneced in series as in Fig..4, is said o be in series parallel configuraion or high-level redundancy. Fig..4 If R is he reliabiliy of each componen, he reliabiliy of each of he subsysems R n (In Fig..4, n ) Since m subsysems are conneced in parallel (In he diagram m 3), he sysem reliabiliy for he high-level redundancy is given by R High ( Rn) m Noe: When m n, R Low R High, since R Low R High { ( R) } { ( ( R ) } (R R ) (R R 4 ) R 4R 3 + R 4 R ( R). Two Componen-sysem Reliabiliy of Markov Analysis Markov analysis can be applied o compue sysem reliabiliy. Fopr simpliciy, we consider a sysem conaining wo componens. To use Markov analysis, he

27 Reliabiliy Engineering 7 sysem is considered o be in one of he four saes a any ime as deailed in Table.: Table. Sae Componen Componen operaing operaing failed operaing 3 operaing failed 4 failed failed Since he probabiliy ha he sysem undergoes a ransiion from one sae o anoher depends only on he presen sae bu no on any of is previous saes, we can use Markov analysis o compue P i (), he probabiliy ha he sysem is in sae i a ime and hence he sysem reliabiliy. If we assume ha he componens have consan failure raes l and l, he sae ransiion diagram of he sysem will be as shown in Fig..5: The nodes in he diagram represen he saes of he sysem and he branches represen he ransiion raes from one node o anoher which will be he same as insananeous failure raes. Fig..5 Now P ( + ) P{he sysem is in sae a ime and neiher Componen nor Componen fails during (, + )} P () P{Componen does no fail in inerval} P{Componen does no fail in inerval} P () { l D} { l }, [since P{Componen i fails in inerval} l i and P{Componen i does no fail in inerval} l i ] P () { (l + l ) }, omiing ( ) erm. P ( +D) -P( ) (l + l ) P () D Taking limis on boh sides as, we ge dp( ) (l D + l ) P () () Noe: Equaion () and similar equaions corresponding o he oher nodes may be obained mechanically as follows:

28 8 Probabiliy, Saisics and Random Processes L.H.S. of he equaion is P i (). To obain he R.H.S. of he equaion, we noe down he arrows enering and/or leaving he node i. If i is an enering arrow, we muliply he ransiion rae corresponding o ha arrow wih he probabiliy corresponding o he node from which i emanaes. If i is a leaving arrow, we muliply he negaive of he ransiion rae corresponding o ha arrow wih he probabiliy corresponding o he node from which i emanaes (viz., he curren node). Then he R.H.S. is obained by adding hese producs. For he oher saes, he corresponding equaions are dp( ) l.p () l. P () () d dp3( ) l P () l p 3 () (3) d dp4( ) l P () + l P 3 () (4) d Solving equaion (), we ge ( ) P () c e - l + l (5) Iniially, P () P(he sysem is in sae a ) P (boh he componens funcion a ) Using his iniial condiion in (5), we ge c ( ) P () e - l + l (6) Using (6) in (), we have ( ) P () + l P () l e - l + l (7) I.F. equaion (7) e l Soluion of equaion (7) is l e. P ( ) c e - l (8) Using he iniial condiion P () in (8), we ge c Soluion of equaion () becomes P () e - l ( ) e - l + l (9) Similarly, he soluion of equaion (3) is P 3 () e - l ( ) e - l + l () To ge P 4 (), we need no solve equaion (4), bu we may use he relaion P () + P () + P 3 () + P 4 () () as he sysem has o be in any one of he four muually exclusive and exhausive saes. Now he sysem reliabiliy depends on he configuraion. For series configuraion, R s () P(boh componens are funcioning) P(he sysem is in sae )

29 Reliabiliy Engineering 9 ( ) P () e - l + l, which agrees wih he resul derived already. For parallel configuraion, R p () P(a leas one of he componens funcion) P(he sysem is in sae, or 3) P () + P () + P 3 () e - l + e - l ( ) e - l + l which agrees wih he resul derived already. Reliabiliy of a Sand by Sysem by Markov Analysis In parallel configuraion wih wo componens, we have assumed so far ha boh he componens are operaing from he sar. Such a sysem is called an acive parallel (redundan) sysem. Now we shall consider he parallel configuraion wih wo componens in which only one componen will be operaing from he sar and he oher, called he sandby or backup componen, will be kep in reserve and brough ino operaion only when he firs fails. Such a sysem is called a passive parallel sysem or sandby redundan sysem. Le us now compue he reliabiliy of such a sandby sysem wih he simplifying assumpion ha he sandby uni does no fail while i is kep in sandby mode (in reserve) Le l and l be he consan failure raes of he acive uni and he sandby uni (when operaive) respecively. This sysem will be in any of hree saes described below: Sae Main uni is funcioning and he oher in sandby mode Sae Main uni has failed and he sandby uni is funcioning Sae 3 Boh unis have failed. The sae ransiion diagram for he sandby sysem is given in Fig..6: Fig..6 From Fig..6, we ge he following Markov equaion easily as explained earlier. P () l P () () P () l P () l P () () P 3 () l P () or P () + P () + P 3 () (3) Solving equaion (), we ge P () ce l (4) Using he iniial condiion P (), we ge C Soluion of equaion () in P () Using (5) in (), i becomes e - l (5)

30 3 Probabiliy, Saisics and Random Processes P () + l P () l e - l (6) I.F. of equaion (6) Soluion () is e -l e l l ( l-l) P () e + c l - l Using he iniial condiion P () in (7), l we ge c l - l Using his value in (7), The soluion of () is go as l -l -l P () ( e - e ) l - l Using (5) and (8) in (3), we can ge he value of P 3 (). Now he reliabiliy of he sandby sysem is given by R() P(main uni or sandby uni is funcioning) P(sysem is in sae or ) P () + P () -l l -l -l e + ( e -e ) l - l l -l -l le - le l - l ( ) Noe:. When he failure raes of he main and sandby unis are equal, viz., when l l l, é -l -( l+ h) R() lim {( l+ he ) -le } h ê ëh ù ú û (7) (8) e -l é æ -h - e öù lim ê + l ç ú h êë è øúû h e é ì ï h üïù lim + íh. + Lýú êë h ïî! ïþúû -l l ê h ( + l) e l. When l l l, (i) MTTF (for acive redundan sysem) -l - l ( e - e ) d 3 l

31 Reliabiliy Engineering 3 (ii) MT TF (for sandby redundan sysem) -l ( + le ) d l. Example An elecronic circui consiss of 5 silicon ransisors, 3 silicon diodes, composiion resisors and ceramic capaciors conneced in series configuraion. The hourly failure rae of each componen is given below: Silicon ransisor : l 4 5 Silicon diode : l d 3 5 Composiion resisor : l r 4 Ceramic capacior : l c 4 Calculae he reliabiliy of he circui for hours, when he componens follow exponenial disribuion. Soluion Since he componens are conneced in series, he sysem (circui) reliabiliy is given by R s () R (). R (). R 3 (). R 4 () -l -l -l3 -l4 e. e. e. e -(5l + 3ld + lr + lc) e ( ) R s () e -4 -( ) e e Example There are 6 componens in a non-redundan sysem. The average reliabiliy of each componen is.99. In order o achieve a leas his sysem reliabiliy by using a redundan sysem wih 4 idenical new componens, wha should be he leas reliabiliy of each new componen? For he non-redundan sysem, Soluion R s R 6 (.99) 6.85 Le he new componens have a reliabiliy of R each. Then for he redundan sysem wih 4 componens, R p.85 i.e., ( R ) 4.85 i.e., ( R ) R (.5) or.6 R.38 i.e., he reliabiliy of each of he new componens should be a leas.38. Example 3 Thermocouples of a paricular design have a failure rae of.8 per hour. How many hermocouples mus be placed in parallel if he sysem is o run for hours wih a sysem failure probabiliy of no more han.5? Assume ha all failures are independen.

32 3 Probabiliy, Saisics and Random Processes Soluion If T is he ime o failure of he sysem, i is required ha P(T ).5 i.e., R p ().5 () Le he number of hermocouples o be conneced in parallel be n. Then R p () ( R) n () where R is he reliabiliy of each couple. The failure rae of each couple.8 (consan) R e.8 (3) Using (3) in (), we have R p () ( e.8 ) n R p () ( e.8 ) n (4) Using (4) in (), we have ( e.8 ) n.5 i.e., (.5567) n.5 (5) By rials, we find ha (5) is no saisfied when n,,, 3, 4 and 5. When n 6, (.5567) n.788 <.5 Hence 6 hermocouples mus be used in he parallel configuraion. Example 4 Two parallel, idenical and independen componens have consan failure rae. If i is desired ha R s ().95, find he componen and sysem MTTF. In addiion o he independen componens if a common mode componen wih a consan failure rae of. is conneced o he sysem, find he new sysem MTTF. Soluion Le R and l be he reliabiliy and failure rae of each of he wo independen componens. Then R() e l Since he wo componens are conneced in parallel, R s () { R()} { e l } i.e., e l e l R s () e l e l R s ().95 i.e., x x +.95, on puing x e l x ± or i.e., e l.36 or l.8 or.53 Rejecing he value.9, we ge l.53 Componen MTTF l and sysem MTTF Rs () d

33 Reliabiliy Engineering 33 Noe: -l - l ( e - e ) d 3 l Common Mode Failure Componen We have assumed ha he componens conneced in series and parallel configuraions are independen. This assumpion of independence of failures among he componens wihin a sysem may be easily violaed. For example, he componens in he sysem may share he same power source, environmenal dus, humidiy, vibraion, ec., resuling in wha is known as common mode failure. This common mode failure is represened by including an exra (common mode) componen in series wih hose componens ha share he failure mode. Le l be he failure rae of common mode componen conneced in series wih he already exising redundan sysem. Then he sysem reliabiliy becomes R s() R s () R (), where R () is he reliabiliy of he common mode componen. e l e l ) e l R s () (e.53 e.56 ) e. (Q l.53 and l.).94 Now he new sysem MTTF is given by (MTTF) -( l+ l ) -( l+ l ) [ e - e ] d l + l l + l Example 5 I is known ha he reliabiliy funcion for a criical solid sae power uni for use in a communicaion saellie is R() +, and measured in years. (a) How many unis mus be placed in parallel in order o achieve a reliabiliy of.98 for 5 year operaion? (b) If here is an addiional common mode wih consan failure rae of. as a resul of environmenal facors, how many unis should be placed in parallel? Soluion (a) Le n unis be placed in parallel. n æ ö Then R s () ç - è + ø

34 34 Probabiliy, Saisics and Random Processes As æ ö ç - è 3ø R s (5).98, we have n.98 i.e.,. or 3 n 5 n 4 3 n (b) When he addiional common mode uni is also used. As ìï æö í - ç ï è3ø î n R s () é æ ö ê - ç è + ê ø ë R s (5).98, we have üï ýe.98 ïþ n ù ú ú û e. i.e., 3 n i.e., æ.98ö 3 n ç - è.99ø i.e., 3 n 99 (nearly) n 5. EXAMPLE 6 If wo idenical componens having a guaraneed life of monhs and a consan failure rae of.5 per year are conneced in parallel, wha is he sysem reliabiliy for, hours of coninuous operaion? If R c () is he reliabiliy of each componen, hen R c () e.5 ( in years) Each componen has a guaraneed life (wear-in period) of monhs or 6 year. Rc ( + ) e R c (/ ) Rc ( ) e -.5( + ) e.5 If R s () is he sysem reliabiliy, hen R s () R c (/ ) {R c (/ )} e.l5 e.3 R s ( hours) R s (.l4 year) e.5.4 e.3 l Example 7 For a redundan sysem wih n independen idenical componens wih consan failure rae l, show ha he MTTF is equal o n i- (-) å nci l i i

35 Reliabiliy Engineering 35 If A./hour, wha is he minimum value of n so ha he MTTF is a leas hours? Soluion If R s () is he sysem reliabiliy, R s () ( e l ) n, since he componen reliabiliy e l MTTF n n å i i (-). nc e - l i å (-) -. nci e -l i Rs () d n (-) å nci l i i When n, MTTF i i i n i- -il å (-) i i i- é ù ê - ë ú û 5 When n 3, MTTF nc e d i- (-) å Ci i i 3 i- (-) å 3Ci i i æ ö ç è 3ø i- (-) When n 4, MTTF å 4Ci i i 5 8 Hence he required minimum value of n 4. Example 8 A je engine consiss of 5 modules (conneced in series) each of which was found o have a Weibull failure disribuion wih a shape parameer of.5. Their characerisic lives are (in operaing cycles) 36, 7, 585, 478 and 93. Find he reliabiliy funcion of he engine and he MTTF. If R() is he reliabiliy funcion of he engine (sysem) hen R() R (). R ()... R 5 ().5.5 -( q) -( q) -( q 5) e e Le ( - - L - ) 5 - e q + q + + q

36 36 Probabiliy, Saisics and Random Processes.5 e æ ö -ç è 84.7 ø e This shows ha he engine failure ime also follows a Weibull s disribuion wih b.5 and q æ ö MTTF of he engine q ç + è b ø æ ö 84.7 ç + è.5ø cycles. Example 9 A signal processor has a reliabiliy of.9. Because of he lower reliabiliy a redundan signal processor is. o be added. However, a signal splier mus be added before he processors and a comparaor mus be added afer he signal processors. Each of he new componens has a reliabiliy of.95. Does adding a redundan signal processor increase he sysem reliabiliy? Fig..7 R R(firs subsysem) R(S ) R(P ) R(C ) R R(second subsysem) R(S ) R(P ) R(C ) (.95) R(sysem) ( R ) ( R ) The addiion of a redundan signal processor increases he reliabiliy. Example Six idenical componens wih consan failure raes are conneced in (a) high level redundancy wih 3 componens in each subsysem (b) low level redundancy wih componens in each subsysem. Deermine he componen MTTF in each case, necessary o provide a sysem reliabiliy of.9 afer hours of operaion.

37 Reliabiliy Engineering 37 Le l be he consan failure rae of each componen. Then R e l, for each componen. For high level redundancy, R s () [ {R()} 3 ] ( e 3l ) R s () ( e 3l ).9 i.e., ( e 3l ). i.e., e 3l.363 e 3l l.383 MTTF of each componen hours. l.383 For low level redundancy R s () [ { R()} ] 3 [ { e l } ] 3 R s () [ { e l } ] 3.9 i.e., ( e l ) ( e l ).345 e l.8577 e l.843 l.55 MTTF of each componen hours. l.55 Example A sysem consiss of wo subsysems in parallel. The reliabiliy of æ ö -ç èq each sub sysem is given by (Weibull failure) R() e ø. Assuming ha common mode failure may be negleced, deermine he sysem MTTF. Soluion The sysem reliabiliy is given by R s () é ê - e ê êë æ ö ù -ç è q ø ú ú úû é ê - e + e ë - q - q ù úû - e q - e q - Sysem MTTF Rs () d - q - e d- e d q -x -y q e qdx e dy,

38 38 Probabiliy, Saisics and Random Processes (on puing q x in he firs inegral and q y in he second inegral.) p q p q -.5 q. Example A sysem is designed o operae for days. The sysem consiss of hree componens in series. Their failure disribuions are () Weibull wih shape parameer. and scale parameer 84 days; () longnormal wih shape parameer(s).7 and median 435 days; (3) consan failure rae of.. (a) Compue he sysem reliabiliy. (b) If wo unis each of componens and are available, deermine he high level redundancy reliabiliy. Assume ha he componens and can be configured as a sub-assembly. (c) If wo unis each of componens and are available, deermine he low level redundancy reliabiliy. Soluion (a) For he firs (Weibull) componen, ì b ï æ ö üï ì. ï æ ö üï R () exp í- ç ý exp í- ç ý ï èq ø è î ï 84 ø þ ïî ïþ ì. ï æ ö üï R () exp í- ç ý.95 ï è84 ø î ïþ For he second (longnormal) componen, R () f() d, where f() is he lognormal pdf R () f() d f( zdz ), wher f( z) is he sandard normal densiy æ ö logç.7 è 435ø f( zdz ) For he hird (consan failure rae) componen, R 3 () e l e. R 3 () e..99 Since he hree componens are conneced in series, R s () R () R () R 3 ()

39 Reliabiliy Engineering 39 (a) Fig..8 (b) Fig..9 R HL ( R R ) {.95.98}.997 R LL { ( R ) } { ( ( R ) } Example 3 Find he variance of he ime o failure for wo idenical unis, each wih a failure rae l, placed in sandby parallel configuraion. Compare he resul wih he variance of he same wo unis placed in acive parallel configuraion. Ignore swiching failures and failures in he sandby mode. Soluion For sandby parallel configuraion: R() ( + l) e l f() R () [le l l( + l ) e l ] l e l MTTF E(T), where T is he ime o failure of he sysem é ù f () d êor R() d ú êë úû é l æ -l e ö æ -l l e ö æ - e öù ê ç l ç ç ú êë è - ø è l ø è -l øúû l E(T ) f() d l l 3 e -l d é 3 æ - e l ö æ - e l ö æ - e l ö æ - e l öù ê ç l - ç + ç - ç ú êë è - ø è l ø è -l ø è l øúû

40 4 Probabiliy, Saisics and Random Processes 6 l Variance of T E(T ) E (T ) 6 4 () l l l For he acive parallel configuraion: R() e l e l f() R () le l le l MTTF E(T) é ù R () dêor f() dú êë úû E(T ) æ -l -l e e ö ç + -l l è ø f() d - 3 l l l é ù -l l - l ê e d- e dú êë úû é ì æ -l e ö æ -l e ö æ -l e öü l êï ï í ý ê ç l ç ç ï è - ø è l ø è -l ø ë î ï þ ìï í ïî æ - l e ö æ - l e ö æ - l e öü ù ï ú ç 3 l - ç + ç ý è - ø è 4l ø è-8l øï ú þ û é ù l ê ël 4l ú û 7 l Var (T) E(T ) E (T) 7 9 l 4l 4l Comparing () and (), we see ha he variance is greaer for sandby parallel sysem Example 4 A compuerised airline reservaion sysem has a main compuer online and a secondary sandby compuer. The online compuer fails a a consan rae of. failure per hour and he sandby uni fails when online a he consan rae of.5 failure per hour. There are no failures while he uni is in he sandby mode. 5 ()

41 Reliabiliy Engineering 4 (a) Deermine he sysem reliabiliy over a 7 hours period. (b) The airline desires o have a sysem MTTF of hours. Deermine he minimum MTTF of he main compuer o achieve his goal, assuming ha he sandby compuer MTTF does no change. (a) l./hour; l.5/hour R s () -l -l { le - le } l - l.4 {.5 e.. e.5 } R s (7) 4 {5e.7 e.36 } (b) MTTF The requiremen is Rd () l -l ( l - l ) l - l e e d æl l ö - l - l ç è l l ø l + l + ll l l + l l l.5 (Q MTTF and hence l do no change for sandby uni) 8 i.e., MTTF of he main compuer should be increased o 8 hours. Example 5 A fuel pump wih an MTTF of 3 hours is o operae coninuously on a 5 hour mission. (a) Wha is he mission reliabiliy? (b) Two such pumps are pu in sandby parallel configuraion. If here are no failures of he back up pump while in sandby mode, wha are he sysem MTTF and he mission reliabiliy? (c) If he sandby failure rae is 75% ha of he main pump (when operaional), wha are he sysem MTTF and he mission reliabiliy? Soluion (a) MTTF 3 l l 3 R() e l R(5) e (5/3).8465

42 4 Probabiliy, Saisics and Random Processes (b) l l l 3 R s () ( + l)e l R s (5) æ 5 ö ç + è 3ø e (5/3).9876 MTTF l 6 hours (c) l 3 ; l l -l R s () ( le - le ) l - l, R s (5) (4e /8 3e /6 ).995 MTTF + 7 hours. l l æ - - ö e 4 e 3 ç - è3 4 ø Par A (Shor answer quesions). Wha do you mean by non-redundan and redundan configuraions?. Derive he reliabiliy of a sysem consising of wo componens in series in erms of reliabiliies of he componens. 3. Derive he reliabiliy of a sysem consising of wo componens in parallel in erms of he reliabiliies of he componens. 4. Prove ha he reliabiliy of a non-redundan sysem canno exceed he leas componen reliabiliy. 5. Prove ha he reliabiliy of a redundan sysem is no less han he maximum componen reliabiliy. 6. Find he MTTF of (a) a non-redundan sysem (b) a redundan sysem consising of wo consan failure rae componens. 7. Explain low level redundancy wih a diagram. 8. Obain he reliabiliy of a low level redundancy wih m subsysems each of which consiss of n componens. 9. Explain high level redundancy wih a diagram.. Obain he reliabiliy of a high level redundancy wih m subsysems each of which consiss of n componens.

43 Reliabiliy Engineering 43. Wha is he difference beween acive and sandby redundan sysems?. Wrie down he formulas for he reliabiliy of a sandby redundan sysem, when he main and sandby componens have (i) unequal consan failure raes and (ii) equal consan failure raes. 3. Wha are he MTTF s for acive and sandby redundan sysems when he main and he sandby componens have unequal consan failure raes. 4. Wha are he MTTF s for acive and sandby redundan sysems when he main and sandby componens have equal consan failure raes. 5. Find he reliabiliy of an aircraf elecronic sysem consising of sensor, guidance, compuer and fire conrol subsysems wih respecive reliabiliies.8,.9,.95 and.65 are conneced in series. 6. In a lead here are 6 glow bulbs conneced in series and he average reliabiliy of each bulb is.99. Compue he reliabiliy of he lead. 7. An equipmen consiss of 3 componens A, B and C in parallel and he respecive reliabiliies are R A.9, R B.95 and R C.96. Calculae he equipmen reliabiliy. 8. If he reliabiliy of each of 3 componens conneced in parallel is.9, calculae he reliabiliy of he sysem. 9. If R A.8, R B.9 and R C.85 and if A, B, C are conneced in series o form a subsysem, wha is he reliabiliy of he high level redundan sysem wih wo such subsysems.. If wo A s, wo B s and wo C s in Quesion (9) are conneced in parallel and he hree subses are conneced o form a low level redundan sysem, find he reliabiliy of he sysem. Par B. The ime o failure (in years) of a cerain brand of compuer has he pdf f() /( + ) ; > (a) If hree of hese compuers are placed in parallel aboard he proposed space saion, wha is he sysem reliabiliy for he firs 6 monhs of operaion? (b) Wha is he sysem design life in days if a reliabiliy of.999 required?. Ten Weibull componens, each having a shape parameer of.8 mus operae in series. Deermine a common characerisic life in order ha hey have a design life of year wih a reliabiliy of If hree componens, each wih consan failure rae are placed in parallel, deermine he sysem reliabiliy for. year and he MTTF. Their failure raes are 5 per year, per year and 5 per year. 4. Specificaions for a power uni consising of hree independen and serially conneced componens require a design life of 5 years wih a.95 reliabiliy. (a) If he consan failure raes l, l, l 3 are such l l and l 3 3l, wha should be he MTTF of each componen of he sysem?