Table of Contents

• Introduction• Reliability of Series Systems of “n” Identical and

Independent Components• Numerical Examples• The Case of Different Component Reliabilities• Reliability of Parallel Systems• Numerical Examples• Reliability of “K out of N” Redundant Systems with

“n” Identical Components• Numerical Example• Combinations of Configurations• Summary• For Further Study• About the Author• Other START Sheets Available

IntroductionReliability engineers often need to work with systems havingelements connected in parallel and series, and to calculatetheir reliability. To this end, when a system consists of acombination of series and parallel segments, engineers oftenapply very convoluted block reliability formulas and usesoftware calculation packages. As the underlying statisticaltheory behind the formulas is not always well understood,errors or misapplications may occur.

The objective of this START sheet is to help the reader bet-ter understand the statistical reasoning behind reliabilityblock formulas for series and parallel systems and provideexamples of the practical ways of using them. This knowl-edge will allow engineers to more correctly use the softwarepackages and interpret the results.

We start this START sheet by providing some notation anddefinitions that we will use in discussing non-repairable sys-tems integrated by series or parallel configurations:

1. All the “n” system component lives (X) areExponentially distributed:

2. Therefore, every ith component 1 < i < n Failure Rate(FR) is constant (λi(t) = λi).

3. All “n” system components are identical; hence, FR areequal (λi = λ; 1 < i < n).

4. All “n” components (and their failure times) are statisti-cally independent:

5. Denote system mission time “T”. Hence, any ith compo-nent (1 < i < n) reliability “Ri(T)”:

Summarizing, in this START sheet we consider the casewhere life is exponentially distributed (i.e., component FR istime independent). First, examples will be given using iden-tical components, and then examples will be consideredusing components with different FR. Independent compo-nents are those whose failure does not affect the performanceof any other system component. Reliability is the probabili-ty of a component (or system) of surviving its mission time“T.” This allows us to obtain both, component and systemFR, from their reliability specification.

We will first discuss series systems, then parallel and redun-dant systems, and finally a combination of all these configu-rations, for non-repairable systems and the case of exponen-tially distributed lives. Examples of analyses and uses ofreliability, FR, and survival functions, to illustrate the theory,are provided.

Reliability of Series Systems of “n” Identicaland Independent ComponentsA series system is a configuration such that, if any one of thesystem components fails, the entire system fails.Conceptually, a series system is one that is as weak as itsweakest link. A graphical description of a series system isshown in Figure 1.

START Selected Topics in AssuranceRelated Technologies

Volume 11, Number 5

Understanding Series and Parallel Systems Reliability

STA

RT

200

4-5,

S&

P S

YS

RE

L

A publication of the Reliability Analysis Center

( ) { } ( ) ( ) T-T- e TF dt

Tf ;e - 1 T XP TF λλ λ=δ==≤=

{ }T ...X and X and XP n21 >

{ } { } { }T X...P T XP T XP n21 >>>=

( ) ( ) ( )( )T

TRln- e T XP TR iT-

ii =λ⇒=>= λ

2

Figure 1. Representation of a Series System of “n”Components

Engineers are trained to work with system reliability [RS] con-cepts using “blocks” for each system element, each block hav-ing its own reliability for a given mission time T:

RS = R1 × R2 × … Rn (if the component reliabilities differ, or)

RS = [Ri ]n (if all i = 1, … , n components are identical)

However, behind the reliability block symbols lies a whole bodyof statistical knowledge. For, in a series system of “n” compo-nents, the following are two equivalent “events”:

“System Success” ≡ “Success of every individual component”

Therefore, the probability of the two equivalent events, thatdefine total system reliability for mission time T (denoted R(T)),must be the same:

The preceding assertion holds because Ri(T), the probability ofany component succeeding in mission time T, is its reliability.All system components are assumed identical with the same FR“λ” and independent. Hence, the product of all component reli-abilities Ri(T) yields the entire system reliability R(T). Thisallows us to calculate R(T) using system FR (λs = n×λ), or the“n×T” power of unit time component reliability [Ri (1)]nT, or the“nth” power of component reliability [Ri(T)]n, for any missiontime T. We will discuss, later in this START sheet, the casewhere different components have different reliabilities or FR.

From all of the preceding considerations, we can summarize thefollowing results when all elements, which are identical, of asystem are connected in series:

1. The reliability of the entire system can be obtained in one oftwo ways:• R(T) = [Ri(T)]n; i.e., the reliability (T) of any component

“i” to the power “n”• R(T) = [Ri(1)]nT; unit reliability of any component “i” to

the power “nT”2. System reliability can also be obtained by using system FR

λs: R(T) = exp{-λσT}:• Since λs = λ+λ+λ+ …+ λ = n × λ (all component FR λ

are identical)

• System FR λs is then, the sum (“n” times) of all compo-nent failure rates (λ):

R(T) = Exp{-(λ+λ+λ+ …+ λ)×T} = Exp{-n×λ×T}) = Exp{-λsT}

3. Component FR (λ) can be obtained from system reliabilityR(T):• λ = [- ln (R(T))] / n×T (inverting the reliability results

given in 1)• Component FR λ can also be obtained from component

reliability Ri(T):λ = - ln [Ri(T)]n / n×T = - ln [Ri(T)] /T

• Previous expression is used for allocating system FR λs,among the system components

4. Total system FR λs can also be obtained from 3:

• λs = [- ln (R(T))] / T = - ln [Ri(T)]n / T• λs = n × λ remains time-independent in series configu-

ration5. Allocation of component reliability Ri(T) from systems

requirements is obtained by solving for Ri(T) in the previ-ous R(T) equations.

6. System “unreliability” = U(T) = 1 - R(T) = 1 - reliability.

One can calculate the various reliability and FR values for thespecial case of unit mission time (T = 1) by letting “T” vanishfrom all the formulas (e.g., substituting T by 1). One can obtainreliability R(T) for any mission time T, from R(1), reliability forunit mission time:

Numerical ExamplesThe concepts discussed are best explained and understood byworking out simple numerical examples. Let a computer systembe composed of five identical terminals in series. Let therequired system reliability, for unit mission time (T = 1) be R(1)= 0.999.

We will now calculate each component’s reliability, unreliabili-ty, and failure rate values.

From the data and formulas just given, each terminal reliabilityRi(T) can be obtained by inverting the system reliability R(T)equation for unit mission time (T = 1):

Component unreliability is: Ui(1) = 1 - Ri(1) = 1 - 0.9998 =0.0002.

1 2 n

{ } { }Succeed n Comp dComp2...an and Comp1P Succeeds SystemP R(T) ↔=↔=

{ } { } n)

T-(e

T-...e

T-e (T)n...R (T)1R Suc n Comp ...P Suc Comp1P

λ=λλ==↔↔=

[ ] [ ] [ ] [ ] nTy(1)Reliabilit Comp

nT )1(iR

nT)

-e(

n y(T)Reliabilit Comp

n )T(iR ==λ=== ( ) [ ] T Ts-Ts-

n1 R(1) )e( e T X ..., XP R(T) ===>= λλ

[ ] 0.999 )1(R )e( )(e e R(1) 5i

5--5s- ===== λλλ

[ ] 0.9998 0.999)( R(1) (1)R 5/15/1i ===⇒

3

Component FR is obtained by solving for λ in the equation forcomponent reliability:

Now, assume, that component reliability for mission time T = 1is given: Ri(1) = 0.999. Now, we are asked to obtain total sys-tem reliability, unreliability, and FR, for the (computer) systemand mission time T = 10 hours. First, for unit time:

Hence, system FR is:

If we require system reliability for mission time T = 10 hours,R(10), and the unit time reliability is R(1) = 0.995, we can useeither the 10th power or the FR λs:

If mission time T is arbitrary, then R(T) is called “SurvivalFunction” (of T). R(T) can then be used to find mission time “T”that accomplishes a pre-specified reliability. Assume that R(T) =0.98 is required and we need to find out maximum time T:

Hence, a Mission Time of T = 4.03 hours (or less) meets therequirement of reliability 0.98 (or more).

Let’s now assume that a new system, a ship, will be propelled byfive identical engines. The system must meet a reliabilityrequirement R(T) = 0.9048 for a mission time T = 10. We needto allocate reliability by engine (component reliability), for therequired mission time T. We invert the formula for system reli-ability R(10), expressed as a function of component reliability.Then, we solve for component reliability Ri(10):

We now calculate system FR (λs) and MTTF (µ) for the five-engine system. These are obtained for mission time T = 10hours and required system reliability R(10) = 0.9048:

FR and MTTF values, equivalently, can be obtained using FRper component, yielding the same results:

Finally, assume that the required ship FR λs = 5 × λ = 0.010005is given. We now need component reliability, Unreliability andFR, by unit mission time (T = 1):

R(1) = Exp{-λs} = Exp {-0.010005} = 0.99

= Exp{-5 × λ} = [Exp(-λ)]5 = [Ri(1)]5

Component reliability: Ri (1) = [R(1)]1/5 = [0.99]0.2

= 0.998

Component unreliability: Ui (1) = 1 - Ri (1) = 1 - 0.998 = 0.002

Component FR: λ = [- ln (R(1))]/n × 1 = [-ln(0.99)]/5 = 0.002

The Case of Different Component ReliabilitiesNow, assume that different system components have differentreliabilities and FR. Then:

Then system Mean Time To Failure, MTTF, = µ = 1/λs = 1/Σ λi

For example, assume that the five engines (components), in theabove system (ship) have different reliabilities (maybe theycome from different manufacturers, or exhibit different ages).Let their reliabilities, for mission time (T = 10) be 0.99, 0.97,

( ) ( )0.0002

1

0.9998ln-

T

)T(Rln- i ===λ

[ ] 0.995 0.999)( )1(R )e( )(e e R(1) 55i

5--5s- ====== λλλ

( ) ( )0.005013

1

0.995ln-

T

R(T)ln- s ===λ

[ ] 101010s-s-10 0.995)( R(1) )e( e R(10) ==== λλ

0.9512 e )(e e -0.051-10x0.0050s-10 ==== λ

4.03 0.005013

ln(0.98)-

lnR(T)- T 0.98; e e R(T)

s

Tn-Ts- ==λ

=⇒=== λλ

[ ] 0.9048 )10(R )e( )(e e R(10) 5 i

5 -10-10x5s-10 ===== λλλ

( ) ( )[ ] ( ) 0.9802 0.9048 10R 10R 2.05/1 i ===⇒

( ) ( )10

0.1001

10

0.9048ln-

T

R(T)ln- s ===λ

99.96 1

MTTF 0.010005 s

=λ

=µ=⇒=

( ) ( )0.0019999

10

0.019999

10

0.9802ln-

T

)T(Rln- i ====λ

0.01 0.009999 0.0019999 x 5 x 5 is ≈==λ=∑λ=λ⇒

( ) 99.96 1

dT e dT TR MTTF s0

Ts-

0=

λ=µ=∫=∫=⇒

∞ λ∞

∑ = λ=λ⇒λ=λΣ=λλ== n1i i s

Ts-e ii-T

e Tn-

...eT1-

e )T(nT)...R(1R R(T)

4

0.95, 0.93, and 0.9, respectively. Then, total system reliabilityR(T) for T = 10 and FR are:

Since the system FR is λs = 0.02697, then the system MTTF isµ = 1/λσ = 1/Σ λi = 1/0.02697 = 37.077.

Reliability of Parallel SystemsA parallel system is a configuration such that, as long as not allof the system components fail, the entire system works.Conceptually, in a parallel configuration the total system relia-bility is higher than the reliability of any single system compo-nent. A graphical description of a parallel system of “n” com-ponents is shown in Figure 2.

Figure 2. Representation of a Parallel System of “n”Components

Reliability engineers are trained to work with parallel systemsusing block concepts:

RS = 1 - Π (1 - Ri) = 1-(1 - R1) × (1 - R2) ×… (1 - Rn); ifthe component reliabilities differ, or

RS = 1 - Π (1 - Ri) = 1-[1 - R]n; if all “n” components areidentical: [Ri = R; i = 1, …, n]

However, behind the reliability block symbols lies a whole bodyof statistical knowledge. To illustrate, we analyze a simple par-allel system composed of n = 2 identical components. The sys-tem can survive mission time T only if the first component, orthe second component, or both components, survive missiontime T (Figure 3). In the language of statistical “events”:

Figure 3. Venn Diagram Representing the “Event” of EitherDevice or Both Surviving Mission Time

This approach easily can be extended to an arbitrary number of“n” parallel components, identical or different. By expandingthe formula RS = 1 -(1 - R1)×(1 - R2)×…(1 - Rn) into products,the well-known reliability block formulas are obtained. Forexample, for n = 3 blocks, when only one is needed:

RS = 1 -(1 - R1)×(1 - R2)×(1 - R3) = R1 + R2 + R3 - R1R2 - R1R3

- R2R3 + R1R2R3 or

RS = 1 -(1 - R)×(1 - R)×(1 - R) = 3R - 3R2 + R3 (if all compo-nents are identical: Ri = R; i = 1, …, n

Using instead, the statistical formulation of the SurvivalFunction R(T), we can obtain system MTTF (µ) for an arbitrarymission time T. For, say n = 2 arbitrary components:

Finally, one can calculate system FR λs from the theoretical def-inition of FR. For n = 2:

( ) ( ) ( ) 0.7636 0.9 x 0.93 x 0.95 x 0.97 x 0.99 T...RTR TR n1 ===

( )( ) ( ){ }0.02697

10

0.2697

10

10Rln-

T

TRln- s ===λ⇒

1

2

n

( ) { } { }T BOTHor T 2Xor T 1XP T Survives SystemP TR >>>==

{ } { } { }T 2X and T 1XP - T 2XP T 1XP >>>+>=

)T(2R x (T)1R - T)(2R (T)1R +=

[ ] [ ] [ ])T(2R - 1 - )T(2R - 1 (T)1R 1 1) - (1 (T)2R )T(2R - 1 (T)1R +=++=

[ ][ ] { }[ ] { }[ ]T 2XP - 1 T 1XP - 1 - 1 )T(2R - 1 )T(1R - 1 - 1 >>==

{ } { }T XPT XP - 1 21 ≤≤=

[ ][ ] 2 1 T)(2R (T)1R If ; T2-

e - 1 T1-

e - 1 - 1 λ≠λ⇒≠λλ=

{ }[ ] [ ] )T(2R (T)1R If ;2 T-e - 1 - 1 2 T X P - 1 - 1 =λ=>=

X1 > T X2 > T

X1 and X2 > T

[ ][ ]

== λλ T2-T1-

21 e - 1e - 1 - 1 )T(R - 1 )T(R - 1 - 1 R(T)

( )T 2 1-T2-T1- e - e e λ+λλλ +=

( ) dT e - e e dT R(T) MTTF 0

T 2 1-T2-T1-

0∫

+=∫=µ=⇒

∞ λ+λλλ∞

2121

1 -

1

1

λ+λλ+

λ=

R(T)

R(T)dt

-

Function Survival

FunctionDensity FR s

δ

==λ=

( ) ( )( ) ( )T

e - e e

e - e e sT 2 1-T2-T1-

T 2 1-21

T2-2

T1-1 λ≡

+

λ+λλ+λ=λ+λλλ

λ+λλλ

5

Notice from this derivation that, even when every componentFR(λ) is constant, the resulting parallel system Hazard Rateλs(T) is time-dependent. This result is very important!

Numerical ExamplesLet a parallel system be composed of n = 2 identical compo-nents, each with FR λ = 0.01 and mission time T = 10 hours,only one of which is needed for system success. Then, total sys-tem reliability, by both calculations, is:

Mean Time to Failure (in hours):

The failure (hazard) rate for the two-component parallel systemis now a function of T:

This system hazard rate λs(T) can be calculated as a function ofany mission time T, as shown in Figure 4.

Figure 4. Plot of the Hazard λs(T) as a Function of MissionTime T. Hazard Rate λs(T) increases as time T increases. Thisplot can be used to find the λs(T) required to meet a Mission

Time of T. Say T = 10, then λs(T) about 0.0018

Reliability of “K out of N” Redundant Systemswith “n” Identical ComponentsA “k” out of “n” redundant system is a parallel configurationwhere “k” of the system components, as a minimum, arerequired to be fully operational at the completion time T of themission, for the system to “succeed” (for k = 1 it reduces to aparallel system; for k = n, to a series one). We illustrate thisusing the example of a system operation depicted in Figure 5.

The Probability “p” for any system unit or component “i”, 1 < i< n, to survive mission time T is:

Figure 5. Units Either Fail/Survive Mission Time

All units are identical and “k” or more units, out of the “n” total,are required to be operational at mission time T, for the entiresystem to fulfill the mission. Therefore, the Probability ofMission Success (i.e., system reliability) is equivalent to theprobability of obtaining “k” or more successes out of the possi-ble “n” trials, with success probability p.

This probability is described by the Binomial (n, p) Distribution.In our case, the probability of success “p” is just the reliabilityRi(T) of any independent unit or component “i”, for the requiredmission time “T”. Therefore, total system reliability R(T), foran arbitrary mission time T, is calculated by:

Sometimes the formula: is used instead.This holds true because:

{ } 1,2 i 0.9048; e e 10 XP 10)(R -0.1-10i ====>= λ

[ ][ ] [ ] 2 i21 )10(R - 1 - 1 )T(R - 1 )10(R - 1 - 1 R(10) ==

( ) 0.9909 0.9048 - 1 - 1 2 ==

( ) T2-T-T 2 1-T2-T1- e - 2e e - e e R(T) λλλ+λλλ =+=

10; Tfor 0.9909; e - 2e e - 2e R(10) -0.20.1-20-10 ==== λλ

150 0.02

1 -

0.01

2

1 -

1

1 MTTF

2121==

λ+λλ+

λ=µ=

( ) ( )( )

e e

e - e e T)(

T 2 1-e - T2-T1-

T 2 1-21

T2-2

T1-1

s λ+λλλ

λ+λλλ

+

λ+λλ+λ=λ

0.02T-0.01T-

-0.02T-0.01T

e - 2e

0.02e - 0.02e ≡

Haz

ard(

T)

0.003

0.002

0.001

0.000

0 10 20

Mission T

Com

pone

nt O

pera

tion

Tim

es

(Liv

es)

Time

1

i

n

Xi Mission

( )∑ ==== =n

kj p Rel. Unit n; Tot. j; Succ.P R(T)

( ) ( )∑=∑= ==n

kjj-nn

kjjn

j p n, j;B p - 1p C

( ) ( )∑+∑= =−−

=n

kjjnjn

jjnj1-k

0jnj p - 1pC p - 1pC 1

( ) ( ) P e T XP TR T-ii ==>= λ

( ) jn1-k0j

jnj p - 1pC - 1 −

=∑

The “summation” values are obtained using the BinomialDistribution tables or the corresponding Excel algorithm (for-mula).

Following the same approach of the series system case, weobtain the MTTF (µ).

We can obtain all parameters for an arbitrary T, by recalculatingprobability p = e-λT of a component surviving this new missiontime “T”. In the special case of mission time T = 1, the “T” van-ishes from all these formulas (e.g., substituted T by 1).

Applying the immediately preceding assumptions and formu-las, we obtain the following results:

• The reliability R(T) of the entire system, for specified T, isobtained by:- Providing the total number of system components (n)

and required ones (k)- Providing the reliability (for mission time T) of one

component: Ri(T) = p- Alternatively, providing the Failure Rate (FR) λ of one

unit or component• System MTTF can be obtained from R(T) using the preced-

ing inputs and:

-

• The “Unreliability” = U(T) = 1 - Reliability = 1 - R(T)

Numerical ExampleLet there be n = 5 identical components (computers) in a system(shuttle). Define system “success” if k = 2 or more components(computers) are running during re-entry. Let every component(computer) have a reliability Ri(1) = 0.9. Let mission “re-entry”time be T = 1. If each component has a reliability Ri(T) = p =0.9, then total system (shuttle) reliability R(T), the componentFR (λ) and the MTTF (µ) are obtained as:

= 9.491 x 1.283 = 12.177

Now, assume that a less expensive design is being considered,consisting of n = 8 identical components in parallel. The newdesign requires that at least k = 5 units are working for a suc-cessful completion of the mission. Assume that mission time isT = 1 and the new component FR λ = 0.223144. Compare thetwo system reliabilities and MTTFs.

First, we need to obtain the new component reliability Ri (T) = pfor T = 1:

Proceeding as before, we obtain the new total system reliabilityfor unit mission time:

= 4.481 x 1.283 = 5.7497

The cheaper (second) design is, therefore, less reliable (and hasa lower MTTF) than the first design.

Combinations of ConfigurationsSome systems are made up of combinations of several series andparallel configurations. The way to obtain system reliability insuch cases is to break the total system configuration down intohomogeneous subsystems. Then, consider each of these subsys-tems separately as a unit, and calculate their reliabilities. Finally,put these simple units back (via series or parallel recombination)into a single system and obtain its reliability.

For example, assume that we have a system composed of thecombination, in series, of the examples developed in the previ-ous two sections. The first subsystem, therefore, consists of twoidentical components in parallel. The second subsystem consistsof a “2 out of 5” (parallel) redundant configuration, composed ofalso five identical components (Figure 6). Assume also thatMission Time is T = 10 hours.

6

µ=⇒∑= −λλ= MTTF )e - (1 eC R(T) jnT-Tj-n

kjnj

∑λ

=∫∑=∫==

λ∞ λ=

∞ n

kj

j-nT-

0

Tj-nkj

nj

0 j

1

1 dt )e - 1( e C dtR(T)

∑λ

==

n

kj j

1

1 MTTF

∑ ==== =5

2j 0.9) Rel. Unit 5; Tot. j; Succ.(P R(1)

j-5 j52j

nj 0.9) - 1(0.9C ∑= =

j-5 j1-20j

nj 0.9) - 1(0.9C - 1 ∑= =

s-e 0.99954 0.00046 - 1 λ===

{ })1(Rln- e 0.9 1)(R i-

i =λ⇒== λ

0.105361 ln(0.9)- ==

+++=∑

λ=µ=

= 5

1

4

1

3

1

2

1 x

0.105361

1

j

1

1 MTTF

n

kj

p 0.8 0.79999 e e 1) P(X (1)R -0.223144-i =≈===>= λ

j-81-50j

jnj

j-8j8

5k

nj 0.8) - 1(0.8C - 1 0.8) - 1( 0.8C R(1) ∑=∑= =

=

s-e 0.94372 0.05628 - 1 λ===

+++=∑

λ=µ=

= 5

1

4

1

3

1

2

1 x

0.223144

1

j

1

1 MTTF

n

kj

7

Figure 6. A Combined Configuration of Two ParallelSubsystems in Series

Using the same values as before, for subsystem, A (two identicalcomponents in parallel, with FR λ = 0.01 and mission time T =10 hours), we can calculate reliability as:

Similarly, subsystem B (“2 out of 5” redundant) has five identi-cal components, of which at least two are required for the sub-system mission success. R3(1) = R4 (1) = R5 (1) = R6 (1) = R7(1)= 0.9, for T = 1. We first recalculate the component reliabilityfor the new mission time T = 10 and then calculate subsystem Breliability as follows:

Recombining both subsystems, we get a series system, consist-ing of subsystems A and B. Therefore, the combined systemreliability, for mission time T = 10, is:

This result immediately shows which subsystem is driving downthe total system reliability and sheds light about possible meas-ures that can be taken to correct this situation.

SummaryThe reliability analysis for the case of non-repairable systems,for configurations in series, in parallel, “k out of n” redundantand their combinations, has been reviewed for the case of expo-nentially-distributed lives. When component lives follow otherdistributions, we substitute the density function in the corre-sponding reliability formulas R(T) and redevelop the algebra.Of particular interest is the case when component lives have anunderlying Weibull distribution:

Here, we substitute these values into equations 1 through 5 ofthe first section and 1 through 6 of the second section and rede-velop the algebra. Due to its complexity, this case will be thetopic of a separate START sheet. Finally, for those readers inter-ested in pursuing these studies at a more advanced level, we pro-vide a useful bibliography For Further Study.

For Further Study1. Kececioglu, D., Reliability and Life Testing Handbook,

Prentice Hall, 1993.2. Hoyland, A. and M. Rausand, System Reliability Theory:

Models and Statistical Methods, Wiley, NY, 1994.3. Nelson, W., Applied Life Data Analysis, Wiley, NY,

1982.4. Mann, N., R. Schafer, and N. Singpurwalla, Methods for

Statistical Analysis of Reliability and Life Data, JohnWiley, NY, 1974.

5. O’Connor, P., Practical Reliability Engineering, Wiley,NY, 2003.

6. Romeu, J.L. Reliability Estimations for Exponential Life,RAC START, Volume 10, Number 7. <http://rac.alionscience.com/pdf/R_EXP.pdf>.

About the AuthorDr. Jorge Luis Romeu has over thirty years of statistical andoperations research experience in consulting, research, andteaching. He was a consultant for the petrochemical, construc-tion, and agricultural industries. Dr. Romeu has also worked instatistical and simulation modeling and in data analysis of soft-ware and hardware reliability, software engineering, and ecolog-ical problems.

Subsystem

ASubsystem

B

R1

R7

R6

R5

R4

R3

R2

[ ][ ])10(R - 1 )10(R - 1 - 1 10)(R 21A =

[ ] 0.9909 0.9048) - 1( - 1 )10(R - 1 - 1 2 2 i ===

{ } 0.9 e 1 XP 1)(R -i ==>= λ

{ } 0.105361 ln(0.9)- )1(Rln- i ===λ⇒

{ } p e 10 XP (10)R T-i ==>= λ

p 0.3487 e e -1.0536110-0.105361x ====

0.3487) p 5; Tot. j; Succ.(P (10)R 52jB ===∑= =

j-5 1-20j

jnj 0.3487) - 1(0.3487C - 1 ∑= =

s-10e 0.5691 0.4309 - 1 λ===

0.5639 0.5691 x 0.9909 10)(R x (10)R R(10) BA ===

( ) { }β

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Dr. Romeu has taught undergraduate and graduate statistics,operations research, and computer science in several Americanand foreign universities. He teaches short, intensive profession-al training courses. He is currently an Adjunct Professor ofStatistics and Operations Research for Syracuse University anda Practicing Faculty of that school’s Institute for ManufacturingEnterprises.

For his work in education and research and for his publicationsand presentations, Dr. Romeu has been elected CharteredStatistician Fellow of the Royal Statistical Society, Full Memberof the Operations Research Society of America, and Fellow ofthe Institute of Statisticians.

Romeu has received several international grants and awards,including a Fulbright Senior Lectureship and a SpeakerSpecialist Grant from the Department of State, in Mexico. Hehas extensive experience in international assignments in Spainand Latin America and is fluent in Spanish, English, and French.

Romeu is a senior technical advisor for reliability and advancedinformation technology research with Alion Science andTechnology previously IIT Research Institute (IITRI). Sincerejoining Alion in 1998, Romeu has provided consulting forseveral statistical and operations research projects. He has writ-ten a State of the Art Report on Statistical Analysis of Materials

Data, designed and taught a three-day intensive statistics coursefor practicing engineers, and written a series of articles on sta-tistics and data analysis for the AMPTIAC Newsletter and RACJournal.

Other START Sheets AvailableMany Selected Topics in Assurance Related Technologies(START) sheets have been published on subjects of interest inreliability, maintainability, quality, and supportability. STARTsheets are available on-line in their entirety at <http://rac.alionscience.com/rac/jsp/start/startsheet.jsp>.

For further information on RAC START Sheets contact the:

Reliability Analysis Center 201 Mill StreetRome, NY 13440-6916Toll Free: (888) RAC-USER Fax: (315) 337-9932

or visit our web site at:

<http://rac.alionscience.com>

About the Reliability Analysis CenterThe Reliability Analysis Center is a world-wide focal point for efforts to improve the reliability, maintainability, supportabilityand quality of manufactured components and systems. To this end, RAC collects, analyzes, archives in computerized databas-es, and publishes data concerning the quality and reliability of equipments and systems, as well as the microcircuit, discretesemiconductor, electronics, and electromechanical and mechanical components that comprise them. RAC also evaluates andpublishes information on engineering techniques and methods. Information is distributed through data compilations, applica-tion guides, data products and programs on computer media, public and private training courses, and consulting services. Alion,and its predecessor company IIT Research Institute, have operated the RAC continuously since its creation in 1968.