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).