Selected Topics in Assurance Related Technologies START Volume 11, Number 5 Understanding Series and Parallel Systems Reliability Table of Contents 2. Therefore, every ith component 1 < i < n Failure Rate λ λ (FR) is constant ( i(t) = i). • Introduction 3. All “n” system components are identical; hence, FR are λ λ • Reliability of Series Systems of “n” Identical and equal ( i = ; 1 < i < n). Independent Components 4. All “n” components (and their failure times) are statisti- cally independent: • Numerical Examples • The Case of Different Component Reliabilities P{}X and X and ...X > T • Reliability of Parallel Systems 1 2 n • Numerical Examples = {}{}{}> > > • Reliability of “K out of N” Redundant Systems with P X1 T P X2 T ...P Xn T “n” Identical Components • Numerical Example 5. Denote system mission time “T”. Hence, any ith compo- • Combinations of Configurations nent (1 < i < n) reliability “Ri(T)”: • Summary ()() ()= (> )= -λT ⇒ λ = ln Ri T • For Further Study Ri T P Xi T e - • About the Author T • Other START Sheets Available Summarizing, in this START Sheet we consider the case where life is exponentially distributed (i.e., component FR is Introduction time independent). First, examples will be given using iden- Reliability engineers often need to work with systems having tical components, and then examples will be considered elements connected in parallel and series, and to calculate using components with different FR. Independent compo- their reliability. To this end, when a system consists of a nents are those whose failure does not affect the performance combination of series and parallel segments, engineers often of any other system component. Reliability is the probabili- apply very convoluted block reliability formulas and use ty of a component (or system) of surviving its mission time software calculation packages. As the underlying statistical “T.” This allows us to obtain both, component and system theory behind the formulas is not always well understood, FR, from their reliability specification. errors or misapplications may occur. We will first discuss series systems, then parallel and redun- The objective of this START Sheet is to help the reader bet- dant systems, and finally a combination of all these configu- ter understand the statistical reasoning behind reliability rations, for non-repairable systems and the case of exponen- block formulas for series and parallel systems and provide tially distributed lives. Examples of analyses and uses of examples of the practical ways of using them. This knowl- reliability, FR, and survival functions, to illustrate the theory, edge will allow engineers to more correctly use the software are provided. packages and interpret the results. Reliability of Series Systems of “n” Identical We start this START Sheet by providing some notation and and Independent Components definitions that we will use in discussing non-repairable sys- tems integrated by series or parallel configurations: A series system is a configuration such that, if any one of the system components fails, the entire system fails.
1. All the “n” system component lives (X) are Conceptually, a series system is one that is as weak as its 2004-5,START S&P SYSREL Exponentially distributed: weakest link. A graphical description of a series system is shown in Figure 1. λ δ λ F()T = P {X ≤ T } = 1 - e- T ; f ()T = F()T = λe- T dt A publication of the DoD Reliability Analysis Center λ 12 n • System FR s is then, the sum (“n” times) of all compo- nent failure rates (λ): λ λ λ λ λ λ Figure 1. Representation of a Series System of “n” R(T) = Exp{-( + + + …+ )×T} = Exp{-n× ×T}) = Exp{- sT} Components 3. Component FR (λ) can be obtained from system reliability R(T): λ Engineers are trained to work with system reliability [RS] con- • = [- ln (R(T))] / n×T (inverting the reliability results cepts using “blocks” for each system element, each block hav- given in 1) λ ing its own reliability for a given mission time T: • Component FR can also be obtained from component reliability Ri(T): λ n RS = R1 × R2 × … Rn (if the component reliabilities differ, or) = - ln [Ri(T)] / n×T = - ln [Ri(T)] /T λ n • Previous expression is used for allocating system FR s, RS = [Ri ] (if all i = 1, … , n components are identical) among the system components λ However, behind the reliability block symbols lies a whole body 4. Total system FR s can also be obtained from 3: λ n of statistical knowledge. For, in a series system of “n” compo- • s = [- ln (R(T))] / T = - ln [Ri(T)] / T λ λ nents, the following are two equivalent “events”: • s = n × remains time-independent in series configu- ration ≡ “System Success” “Success of every individual component” 5. Allocation of component reliability Ri(T) from systems requirements is obtained by solving for Ri(T) in the previ- Therefore, the probability of the two equivalent events, that ous R(T) equations. define total system reliability for mission time T (denoted R(T)), 6. System “unreliability” = U(T) = 1 - R(T) = 1 - reliability. must be the same: One can calculate the various reliability and FR values for the R(T) = P{}{System ↔ Succeeds = P Comp1 and Comp2...and Comp n ↔ Succeed }special case of unit mission time (T = 1) by letting “T” vanish from all the formulas (e.g., substituting T by 1). One can obtain -λT -λT -λT n reliability R(T) for any mission time T, from R(1), reliability for = P{}{}Comp1 ↔ Suc ...P Comp n ↔ Suc = R (T) ...R (T) = e ...e = (e ) 1 n unit mission time:
n n -λ nT nT nT λ λ = [R (T) ][ = Comp Reliability(T) ] = (e ) = [][R (1) = Comp Reliability(1) ] = ()> = - sT = - s T = [] T i i R(T) P X1 ..., Xn T e (e ) R(1)
The preceding assertion holds because Ri(T), the probability of Numerical Examples any component succeeding in mission time T, is its reliability. All system components are assumed identical with the same FR The concepts discussed are best explained and understood by “λ” and independent. Hence, the product of all component reli- working out simple numerical examples. Let a computer system abilities Ri(T) yields the entire system reliability R(T). This be composed of five identical terminals in series. Let the λ λ required system reliability, for unit mission time (T = 1) be R(1) allows us to calculate R(T) using system FR ( s = n× ), or the nT = 0.999. “n×T” power of unit time component reliability [Ri (1)] , or the “nth” power of component reliability [R (T)]n, for any mission i We will now calculate each component’s reliability, unreliabili- time T. We will discuss, later in this START Sheet, the case ty, and failure rate values. where different components have different reliabilities or FR. From the data and formulas just given, each terminal reliability From all of the preceding considerations, we can summarize the R (T) can be obtained by inverting the system reliability R(T) following results when all elements, which are identical, of a i equation for unit mission time (T = 1): system are connected in series:
λ λ λ 1. The reliability of the entire system can be obtained in one of = - s = -5 = - 5 = []5 = two ways: R(1) e (e ) (e ) Ri (1) 0.999 n • R(T) = [Ri(T)] ; i.e., the reliability (T) of any component “i” to the power “n” nT ⇒ = []1/ 5 = 1/ 5 = • R(T) = [Ri(1)] ; unit reliability of any component “i” to Ri (1) R(1) (0.999) 0.9998 the power “nT” 2. System reliability can also be obtained by using system FR λ λ Component unreliability is: Ui(1) = 1 - Ri(1) = 1 - 0.9998 = s: R(T) = exp{- σT}: λ λ λ λ λ λ λ 0.0002. • Since s = + + + …+ = n × (all component FR are identical)
2 λ λ µ Component FR is obtained by solving for in the equation for We now calculate system FR ( s) and MTTF ( ) for the five- component reliability: engine system. These are obtained for mission time T = 10 hours and required system reliability R(10) = 0.9048: ln()R (T) - ln()0.9998 λ = - i = = 0.0002 () () T 1 λ = ln R(T) = - ln 0.9048 = 0.1001 s - T 10 10 Now, assume, that component reliability for mission time T = 1 1 is given: R (1) = 0.999. Now, we are asked to obtain total sys- = ⇒ = µ = = i 0.010005 MTTF λ 99.96 tem reliability, unreliability, and FR, for the (computer) system s and mission time T = 10 hours. First, for unit time: FR and MTTF values, equivalently, can be obtained using FR λ λ λ per component, yielding the same results: = - s = -5 = - 5 = []5 = 5 = R(1) e (e ) (e ) Ri (1) (0.999) 0.995 ln()R (T) - ln()0.9802 0.019999 λ = - i = = = 0.0019999 T 10 10 Hence, system FR is: ⇒ λ = ∑λ = λ = = ≈ () () s i 5 x 5 x 0.0019999 0.009999 0.01 λ = ln R(T) = - ln 0.995 = s - 0.005013 T 1 ∞ ∞ λ ⇒ = ∫ () = ∫ - sT = µ = 1 = MTTF R T dT e dT λ 99.96 If we require system reliability for mission time T = 10 hours, 0 0 s R(10), and the unit time reliability is R(1) = 0.995, we can use th λ λ λ either the 10 power or the FR s: Finally, assume that the required ship FR s = 5 × = 0.010005 is given. We now need component reliability, Unreliability and λ λ FR, by unit mission time (T = 1): R(10) = e-10 s = (e- s )10 = []R(1) 10 = (0.995)10 λ R(1) = Exp{- s} = Exp {-0.010005} = 0.99 λ λ λ 5 5 = e-10 s = (e-10x0.00501) = e-0.05 = 0.9512 = Exp{-5 × } = [Exp(- )] = [Ri(1)]
Component reliability: R (1) = [R(1)]1/5 = [0.99]0.2 If mission time T is arbitrary, then R(T) is called “Survival i Function” (of T). R(T) can then be used to find mission time “T” = 0.998 that accomplishes a pre-specified reliability. Assume that R(T) = 0.98 is required and we need to find out maximum time T: Component unreliability: Ui (1) = 1 - Ri (1) = 1 - 0.998 = 0.002
λ λ = - sT = -n T = ⇒ = lnR(T) = ln(0.98) = Component FR: λ = [- ln (R(1))]/n × 1 = [-ln(0.99)]/5 R(T) e e 0.98; T - λ - 4.03 s 0.005013 = 0.002
Hence, a Mission Time of T = 4.03 hours (or less) meets the The Case of Different Component Reliabilities requirement of reliability 0.98 (or more). Now, assume that different system components have different reliabilities and FR. Then: Let’s now assume that a new system, a ship, will be propelled by five identical engines. The system must meet a reliability λ λ Σ λ λ = = - 1T - nT = -T i i = - sT ⇒ λ = ∑n λ requirement R(T) = 0.9048 for a mission time T = 10. We need R(T) R1(T)...R n (T) e ...e e e s i=1 i to allocate reliability by engine (component reliability), for the required mission time T. We invert the formula for system reli- µ λ Σλ Then system Mean Time To Failure, MTTF, = = 1/ s = 1/ i ability R(10), expressed as a function of component reliability. Then, we solve for component reliability Ri(10): For example, assume that the five engines (components), in the above system (ship) have different reliabilities (maybe they λ λ λ = -10 s = -10x5 = -10 5 = [] 5 = R(10) e (e ) (e ) Ri (10) 0.9048 come from different manufacturers, or exhibit different ages). Let their reliabilities, for mission time (T = 10) be 0.99, 0.97, ⇒ ()= [] () 1/ 5 = ()0.2 = Ri 10 R 10 0.9048 0.9802
3 0.95, 0.93, and 0.9, respectively. Then, total system reliability = {}{}≤ ≤ 1 - P X1 T P X2 T R(T) for T = 10 and FR are:
-λ T -λ T ()= () ()= = = 1 - [][] 1 - e 1 1 - e 2 ; If R (T) ≠ R (T) ⇒ λ ≠ λ R T R1 T ...R n T 0.99 x 0.97 x 0.95 x 0.93 x 0.9 0.7636 1 2 1 2
()() {}() λ ⇒ λ = ln R T = - ln R 10 = 0.2697 = []{}> 2 = []- T 2 = s - 0.02697 1 - 1 - P X T 1 - 1 - e ; If R1(T) R 2 (T) T 10 10
λ Since the system FR is s = 0.02697, then the system MTTF is µ λ Σλ X >T X >T = 1/ σ = 1/ i = 1/0.02697 = 37.077. 1 2
Reliability of Parallel Systems X1 and X2 >T A parallel system is a configuration such that, as long as not all Figure 3. Venn Diagram Representing the “Event” of Either of the system components fail, the entire system works. Device or Both Surviving Mission Time Conceptually, in a parallel configuration the total system relia- bility is higher than the reliability of any single system compo- This approach easily can be extended to an arbitrary number of nent. A graphical description of a parallel system of “n” com- “n” parallel components, identical or different. By expanding ponents is shown in Figure 2. the formula RS = 1 -(1 - R1)×(1 - R2)×…(1 - Rn) into products, the well-known reliability block formulas are obtained. For 1 example, for n = 3 blocks, when only one is needed:
RS = 1 -(1 - R1)×(1 - R2)×(1 - R3) = R1 + R2 +R3 - R1R2 - R1R3 2 - R2R3 + R1R2R3 or
2 3 RS = 1 -(1 - R)×(1 - R)×(1 - R) = 3R - 3R + R (if all compo- n nents are identical: Ri = R; i = 1, …, n
Figure 2. Representation of a Parallel System of “n” Using instead, the statistical formulation of the Survival Components Function R(T), we can obtain system MTTF (µ) for an arbitrary mission time T. For, say n = 2 arbitrary components: Reliability engineers are trained to work with parallel systems using block concepts: λ λ = [][]= - 1T - 2T R(T) 1 - 1 - R1(T) 1 - R 2 (T) 1 - 1 - e 1 - e Π RS = 1 - (1 - Ri) = 1-(1 - R1) × (1 - R2) ×… (1 - Rn); if the component reliabilities differ, or λ λ ()λ + λ = e- 1T + e- 2T - e- 1 2 T R = 1 - Π (1 - R ) = 1-[1 - R]n; if all “n” components are S i ∞ ∞ λ λ ()λ + λ identical: [Ri = R; i = 1, …, n] ⇒ = µ = ∫ = ∫ - 1T + - 2T - 1 2 T MTTF R(T) dT e e - e dT 0 0 However, behind the reliability block symbols lies a whole body of statistical knowledge. To illustrate, we analyze a simple par- = 1 + 1 1 λ λ - λ + λ allel system composed of n = 2 identical components. The sys- 1 2 1 2 tem can survive mission time T only if the first component, or λ the second component, or both components, survive mission Finally, one can calculate system FR s from the theoretical time T (Figure 3). In the language of statistical “events”: definition of FR. For n = 2:
()= { }= {> > > } R T P System Survives T P X1 T or X2 T or BOTH T δ - R(T) = {}{}{> + > > > } = λ = Density Function = dt P X1 T P X2 T - P X1 T and X2 T FR s Survival Function R(T) = + R1(T) R 2 (T) - R1(T) x R 2 (T) λ λ ()λ + λ λ - 1T + λ - 2T ()λ + λ - 1 2 T = R (T) []1 - R (T) + R (T) + (1 - 1) = 1 + R (T) [][]1 - R (T) - 1 - R (T) = 1e 2e - 1 2 e ≡ λ () 1 2 2 1 2 2 λ λ ()λ + λ s T e- 1T + e- 2T - e- 1 2 T = [][]= []{}> []{}> 1 - 1 - R1(T) 1 - R 2 (T) 1 - 1 - P X1 T 1 - P X2 T
4 Notice from this derivation that, even when every component Reliability of “K out of N” Redundant Systems FR(λ) is constant, the resulting parallel system Hazard Rate λ with “n” Identical Components s(T) is time-dependent. This result is very important! A “k” out of “n” redundant system is a parallel configuration where “k” of the system components, as a minimum, are Numerical Examples required to be fully operational at the completion time T of the Let a parallel system be composed of n = 2 identical compo- mission, for the system to “succeed” (for k = 1 it reduces to a λ nents, each with FR = 0.01 and mission time T = 10 hours, parallel system; for k = n, to a series one). We illustrate this only one of which is needed for system success. Then, total sys- using the example of a system operation depicted in Figure 5. tem reliability, by both calculations, is: The Probability “p” for any system unit or component “i”, 1 < i = {}> = -10λ = -0.1 = = Ri (10) P X 10 e e 0.9048; i 1,2 < n, to survive mission time T is: = [][][]= 2 ()= (> )= -λT = R(10) 1 - 1 - R1(10) 1 - R 2 (T) 1 - 1 - Ri (10) Ri T P Xi T e P
= 1 - ()1 - 0.9048 2 = 0.9909
λ λ ()λ + λ λ λ 1 R(T) = e- 1T + e- 2T - e- 1 2 T = 2e- T - e-2 T λ λ R(10) = 2e-10 - e-20 = 2e0.1 - e-0.2 = 0.9909; for T = 10;
Mean Time to Failure (in hours):
= µ = 1 + 1 1 = 2 1 = (Lives) MTTF λ λ - λ + λ - 150 1 2 1 2 0.01 0.02 i
The failure (hazard) rate for the two-component parallel system Times Operation Component is now a function of T: n λ λ ()λ + λ λ - 1T + λ - 2T ()λ + λ - 1 2 T Time 1e 2e - 1 2 e λ (T) = X Mission s -()λ + λ T i -λ T + -λ T - e 1 2 e 1 e 2 Figure 5. Units Either Fail/Survive Mission Time
0.02e-0.01T - 0.02e-0.02T ≡ All units are identical and “k” or more units, out of the “n” total, 2e-0.01T - e-0.02T are required to be operational at mission time T, for the entire system to fulfill the mission. Therefore, the Probability of λ This system hazard rate s(T) can be calculated as a function of Mission Success (i.e., system reliability) is equivalent to the any mission time T, as shown in Figure 4. probability 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 reliability Ri(T) of any independent unit or component “i”, for the required mission time “T”. Therefore, total system reliability R(T), for an arbitrary mission time T, is calculated by:
= ∑n ()= = = R(T) j=k P Succ. j; Tot. n; Unit Rel. p
= ∑n n j()n-j = ∑n ( ) j=k C j p 1 - p j=k B j; n, p − λ 1 - ∑k-1 Cnp j()1 - p n j Figure 4. Plot of the Hazard s(T) as a Function of Mission Sometimes the formula: j = 0 j is used instead. λ Time T. Hazard Rate s(T) increases as time T increases. This This holds true because: λ plot can be used to find the s(T) required to meet a Mission − − Time of T. Say T = 10, then λ (T) about 0.0018 = ∑k-1 n j()n j + ∑n n j ()n j s 1 j=0 C j p 1 - p j=k C j p 1 - p
5 The “summation” values are obtained using the Binomial = = -λ ⇒ λ = {} Distribution tables or the corresponding Excel algorithm (for- Ri (1) 0.9 e - ln Ri (1) mula). = - ln(0.9) = 0.105361 Following the same approach of the series system case, we 1 n 1 1 1 1 1 1 obtain the MTTF (µ). MTTF = µ = ∑ = x + + + λ j=k j 0.105361 2 3 4 5 = ∑n n -λTj -λT n− j ⇒ = µ R(T) j=k C j e (1 - e ) MTTF = 9.491 x 1.283 = 12.177
∞ ∞ Now, assume that a less expensive design is being considered, λ λ 1 n 1 = ∫ = ∑n n ∫ - Tj - T n-j = ∑ consisting of n = 8 identical components in parallel. The new R(T)dt j=k C j e (1 - e ) dt λ 0 0 j=k j design requires that at least k = 5 units are working for a suc- cessful completion of the mission. Assume that mission time is T = 1 and the new component FR λ = 0.223144. Compare the We can obtain all parameters for an arbitrary T, by recalculating two system reliabilities and MTTFs. probability p = e-λT of a component surviving this new mission time “T”. In the special case of mission time T = 1, the “T” van- First, we need to obtain the new component reliability Ri (T) = p ishes from all these formulas (e.g., substituted T by 1). for T = 1:
Applying the immediately preceding assumptions and formu- = > = -λ = -0.223144 = ≈ = las, we obtain the following results: Ri (1) P(X 1) e e 0.79999 0.8 p
• The reliability R(T) of the entire system, for specified T, is Proceeding as before, we obtain the new total system reliability obtained by: for unit mission time: - Providing the total number of system components (n) and required ones (k) 8 = ∑ n j 8-j = ∑5-1 n j 8-j - Providing the reliability (for mission time T) of one R(1) C j 0.8 (1 - 0.8) 1 - j=0 C j 0.8 (1 - 0.8) k=5 component: Ri(T) = p - Alternatively, providing the Failure Rate (FR) λ of one λ = = = - s unit or component 1 - 0.05628 0.94372 e • System MTTF can be obtained from R(T) using the preced- 1 n 1 1 1 1 1 1 ing inputs and: MTTF = µ = ∑ = x + + + n λ = j 0.223144 2 3 4 5 = 1 ∑ 1 j k - MTTF λ j=k j = 4.481 x 1.283 = 5.7497 • The “Unreliability” = U(T) = 1 - Reliability = 1 - R(T) The cheaper (second) design is, therefore, less reliable (and has Numerical Example a lower MTTF) than the first design. Let there be n = 5 identical components (computers) in a system (shuttle). Define system “success” if k = 2 or more components Combinations of Configurations (computers) are running during re-entry. Let every component Some systems are made up of combinations of several series and (computer) have a reliability Ri(1) = 0.9. Let mission “re-entry” parallel configurations. The way to obtain system reliability in time be T = 1. If each component has a reliability Ri(T) = p = such cases is to break the total system configuration down into 0.9, then total system (shuttle) reliability R(T), the component homogeneous subsystems. Then, consider each of these subsys- FR (λ) and the MTTF (µ) are obtained as: tems separately as a unit, and calculate their reliabilities. Finally, put these simple units back (via series or parallel recombination) = ∑5 = = = into a single system and obtain its reliability. R(1) j=2 P(Succ. j; Tot. 5; Unit Rel. 0.9) For example, assume that we have a system composed of the = ∑5 n j 5-j j=2 C j 0.9 (1 - 0.9) combination, in series, of the examples developed in the previ- ous two sections. The first subsystem, therefore, consists of two identical components in parallel. The second subsystem consists = 1 - ∑2-1 Cn 0.9 j(1 - 0.9) 5-j j=0 j of a “2 out of 5” (parallel) redundant configuration, composed of also five identical components (Figure 6). Assume also that λ = 1 - 0.00046 = 0.99954 = e- s Mission Time is T = 10 hours.
6 This result immediately shows which subsystem is driving down R3 the total system reliability and sheds light about possible meas- R 1 R4 ures that can be taken to correct this situation.
R5 Summary The reliability analysis for the case of non-repairable systems, R R6 2 for configurations in series, in parallel, “k out of n” redundant R and their combinations, has been reviewed for the case of expo- Subsystem 7 A nentially-distributed lives. When component lives follow other Subsystem distributions, we substitute the density function in the corre- B sponding reliability formulas R(T) and redevelop the algebra. Figure 6. A Combined Configuration of Two Parallel Of particular interest is the case when component lives have an Subsystems in Series underlying Weibull distribution:
Using the same values as before, for subsystem, A (two identical β λ T components in parallel, with FR = 0.01 and mission time T = - α 10 hours), we can calculate reliability as: F()T = P { X ≤ T } = 1 - e
R (10) = 1 - [][]1 - R (10) 1 - R (10) β A 1 2 T δ β - ()= ()= β-1 α = [] 2 = 2 = f T F T β T e 1 - 1 - Ri (10) 1 - (1 - 0.9048) 0.9909 dt α
Similarly, subsystem B (“2 out of 5” redundant) has five identi- Here, we substitute these values into equations 1 through 5 of cal components, of which at least two are required for the sub- the first section and 1 through 6 of the second section and rede- system mission success. R3(1) = R4 (1) = R5 (1) = R6 (1) = R7(1) velop the algebra. Due to its complexity, this case will be the = 0.9, for T = 1. We first recalculate the component reliability topic of a separate START Sheet. Finally, for those readers for the new mission time T = 10 and then calculate subsystem B interested in pursuing these studies at a more advanced level, we reliability as follows: provide a useful bibliography For Further Study.
= {}> = -λ = Ri (1) P X 1 e 0.9 For Further Study
⇒ λ = {}= = 1. Kececioglu, D., Reliability and Life Testing Handbook, - ln Ri (1) - ln(0.9) 0.105361 Prentice Hall, 1993. 2. Hoyland, A. and M. Rausand, System Reliability Theory: = {}> = -λT = Ri (10) P X 10 e p Models and Statistical Methods, Wiley, NY, 1994. 3. Nelson, W., Applied Life Data Analysis, Wiley, NY, = e-0.105361x10 = e-1.05361 = 0.3487 = p 1982. 4. Mann, N., R. Schafer and N. Singpurwalla, Methods for = ∑5 = = = Statistical Analysis of Reliability and Life Data, John R B(10) j=2 P(Succ. j; Tot. 5; p 0.3487) Wiley, NY, 1974. 5. O'Connor, P., Practical Reliability Engineering , Wiley, = ∑2-1 n j 5-j 1 - j=0 C j 0.3487 (1 - 0.3487) NY, 2003. 6. Romeu, J.L. Reliability Estimations for Exponential Life, λ = = = -10 s RAC START, Volume 10, Number 7.
7 Dr. Romeu has taught undergraduate and graduate statistics, Data, designed and taught a three-day intensive statistics course operations research, and computer science in several American for practicing engineers, and written a series of articles on sta- and foreign universities. He teaches short, intensive profession- tistics and data analysis for the AMPTIAC Newsletter and RAC al training courses. He is currently an Adjunct Professor of Journal. Statistics and Operations Research for Syracuse University and a Practicing Faculty of that school’s Institute for Manufacturing Other START Sheets Available Enterprises. Many Selected Topics in Assurance Related Technologies (START) sheets have been published on subjects of interest in For his work in education and research and for his publications reliability, maintainability, quality, and supportability. START and presentations, Dr. Romeu has been elected Chartered sheets are available on-line in their entirety at
For further information on RAC START Sheets contact the: Romeu has received several international grants and awards, including a Fulbright Senior Lectureship and a Speaker Reliability Analysis Center Specialist Grant from the Department of State, in Mexico. He 201 Mill Street has extensive experience in international assignments in Spain Rome, NY 13440-6916 and Latin America and is fluent in Spanish, English, and French. Toll Free: (888) RAC-USER Fax: (315) 337-9932 Romeu is a senior technical advisor for reliability and advanced information technology research with Alion Science and or visit our web site at: Technology previously IIT Research Institute (IITRI). Since rejoining Alion in 1998, Romeu has provided consulting for
About the Reliability Analysis Center The Reliability Analysis Center is a Department of Defense Information Analysis Center (IAC). RAC serves as a government and industry focal point for efforts to improve the reliability, maintainability, supportability and quality of manufactured com- ponents and systems. To this end, RAC collects, analyzes, archives in computerized databases, and publishes data concerning the quality and reliability of equipments and systems, as well as the microcircuit, discrete semiconductor, and electromechani- cal and mechanical components that comprise them. RAC also evaluates and publishes information on engineering techniques and methods. Information is distributed through data compilations, application guides, data products and programs on comput- er media, public and private training courses, and consulting services. Located in Rome, NY, the Reliability Analysis Center is sponsored by the Defense Technical Information Center (DTIC). Alion, and its predecessor company IIT Research Institute, have operated the RAC continuously since its creation in 1968. Technical management of the RAC is provided by the U.S. Air Force's Research Laboratory Information Directorate (formerly Rome Laboratory).
8