\end{align}\,\! P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\ X2= & \overline{A}BC-\text{only Unit 1 fails}\text{.} • Parallel System This is a system that will fail only if they all fail. For this purpose, the container can be defined with its own probability of successfully activating standby units when needed. [/math], \begin{align} 1 Citations; 171 Downloads; Abstract., {{R}_{s}}={{R}_{B}}\cdot {{R}_{F}}(-{{R}_{A}}\cdot {{R}_{C}}\cdot {{R}_{E}}-{{R}_{A}}\cdot {{R}_{D}}\cdot {{R}_{E}}+{{R}_{A}}\cdot {{R}_{C}}+{{R}_{A}}\cdot {{R}_{E}}+{{R}_{D}}\cdot {{R}_{E}})\,\! Furthermore, suppose that the design of the aircraft is such that at least two engines are required to function for the aircraft to remain airborne. Example: Effect of a Component's Reliability in a Parallel System. \end{align}\,\! Redundancy is a very important aspect of system design and reliability in that adding redundancy is one of several methods of improving system reliability. \end{align}\,\! In mirrored blocks, the duplicate block behaves in the exact same way that the original block does. = & \left( \begin{matrix} Standby redundancy configurations consist of items that are inactive and available to be called into service when/if an active item fails (i.e., the items are on standby). Consider a system that consists of a single component. As a result, the reliability of a series system is always less than the reliability of the least reliable component. = & P(s|A)P(A)+P(s|\overline{A})P(\overline{A}) 162 Downloads; Part of the Engineering Applications of Systems Reliability and Risk Analysis book series (EASR, volume 1) Abstract. To better illustrate this consider the following block diagram: In this diagram [math]Bm\,\! In this case then, and to obtain a system solution, one begins with [math]{{R}_{System}}\,\!. 0000039484 00000 n Since at least two hard drives must be functioning at all times, only one failure is allowed. However, in the case of independent components, equation above becomes: Or, in terms of individual component reliability: In other words, for a pure series system, the system reliability is equal to the product of the reliabilities of its constituent components. constant failure rates) arranged in series.The goal of these standards is to determine the system failure rate, which is computed by summation of the component failure rates. Suppose a system is composed of two sub-systems say, A and B are connected in series as shown in the figure below. The following figure demonstrates the use of multi blocks in BlockSim. Bazovsky, Igor, Reliability Theory and Practice 3. First, the reliability of the series segment consisting of Units 1 and 2 is calculated: The reliability of the overall system is then calculated by treating Units 1 and 2 as one unit with a reliability of 98.2065% connected in parallel with Unit 3. In this case, the reliability of the system with such a configuration can be evaluated using the binomial distribution, or: Example: Calculating the Reliability for a k-out-of-n System. 3 \\ P({{X}_{1}}\cup {{X}_{2}})= & P({{X}_{1}})+P({{X}_{2}})-P({{X}_{1}}\cap {{X}_{2}}) \\ Configuration with inherited subdiagrams. [/math], \begin{align} \end{align}\,\! 6 \\, ${{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})\,\! Unless explicitly stated, the components will be assumed to be statistically independent. Thus the probability of failure of the system is: Since events [math]{{X}_{6}}\,\! X7= & A\overline{BC}-\text{Units 2 and 3 fail}\text{.}$ is a mirrored block of $B\,\!$. The rate of change of the system's reliability with respect to each of the components is also plotted. Each item represented by a multi block is a separate entity with identical reliability characteristics to the others. [/math], \begin{align} Reliability block diagrams are created in order to illustrate the way that components are arranged reliability-wise in a system. \end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ \,\! What is the reliability of the system if [math]{{R}_{1}} = 99.5%\,\! Effect of Component Reliability in a Series System, Effect of Number of Components in a Series System, Effect of Component Reliability in a Parallel Configuration, Effect of Number of Components in a Parallel System, Difference Between Physical and Reliability-Wise Arrangement, RBDs for Failure Modes and Other Applications, Identical Block Simplification Option ("Use IBS"), [math]\begin{align} Complex Systems and Redundancy 6. Consider three components arranged reliability-wise in series, where R 1 = 70%, R 2 = 80% and R 3 = 90% (for a given time). Once all the parameters are received, it then will become the Reliability Engineer’s responsibility to model the system using the appropriate reliability math models. (for a given time). This can be removed, yielding: Several algebraic solutions in BlockSim were used in the prior examples. [/math], \begin{align} Put another way, [math]{{r}_{1}}\,\! Note the slight difference in the slopes of the three lines. 0000056098 00000 n This is a 2-out-of-3 configuration. \end{align}\,\! 1. in a series system - all devices in the system must work for the system to work 2. in a parallel system - the system works if at least one device in the system works 0000069925 00000 n X5= & \overline{AB}C-\text{Units 1 and 2 fail}\text{.} = & 0.999515755 The reason that BlockSim includes all items regardless of whether they can fail or not is because BlockSim only recomputes the equation when the system structure has changed., $+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! Let the reliabilities of the two components be RA and RB. \end{matrix} \right){{0.85}^{r}}{{(1-0.85)}^{6-r}} \\ 3. P(s| \overline{A})= &{{R}_{B}}{{R}_{D}}{{R}_{E}}{{R}_{F}} \\ \end{matrix} \right){{R}^{2}}(1-R)+\left( \begin{matrix} {{R}_{system}}= & (1\cdot 1(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \\ & -{{R}_{2}}\cdot {{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}})-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\ 0000006600 00000 n What would the reliability of the system be if the system were composed of two, four or six such components in parallel? & +{{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}\cdot {{R}_{2/3}}\cdot {{R}_{E}}\cdot {{R}_{F}} \\ The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. • Series System This is a system in which all the components are in series and they all have to work for the system to work. In this mode, portions of the system are segmented. Consequently, the analysis method used for computing the reliability of a system will also depend on the reliability-wise configuration of the components/subsystems. Consider the four-engine aircraft discussed previously.$ units must succeed for the system to succeed. The failure characteristics of each block in a load sharing container are defined using both a life distribution and a life-stress relationship that describe how the life distribution changes as the load changes. Combined (series and parallel) configuration. HD #3 fails while HDs #1 and #2 continue to operate. \end{align}\,\! In this example it can be seen that even though the three components were physically arranged in parallel, their reliability-wise arrangement is in series. The relays are situated so that the signal originating from one station can be picked up by the next two stations down the line. {{R}_{s}}= & P(s\cap A)+P(s\cap \overline{A}) \\ The next figure includes a standby container with three items in standby configuration where one component is active while the other two components are idle. \end{align}\,\! As the number of components connected in series increases, the system's reliability decreases. M. L. Shooman, Probabilistic reliability: an engineering approach, McGraw-Hill, … Reliability optimization and costs are covered in detail in Component Reliability Importance. Put another way, if unit 1 succeeds or unit 2 succeeds or any of the n\,\! To address this issue, Hu and Du [9, 10] proposed a physicsbased reliability method for component adopted in new series systems. & +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\ 6 \\ To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. As an example, consider the complex system shown next. The paper is devoted to the combining the results on reliability of the two-state series and consecutive “m out of n: F” systems into the formulae for the reliability function of the series-consecutive “m out of k: F” systems with dependent of time reliability functions of system components Guze (2007a), Guze (2007b), Guze (2007c). \end{align}\,\!, $-{{R}_{9}}\cdot {{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! Another Example for Failure Mode Analysis. Successful system operation requires at least one output (O1, O2 or O3) to be working. Note that the system configuration becomes a simple parallel configuration for k = 1 and the system is a six-unit series configuration [math]{{((0.85)}^{6}}= 0.377)\,\! There are other multiple redundancy types and multiple industry terms.$, {{r}_{3}}\,\! Consider a system consisting of n components in series. {{X}_{1}}=1,2\text{ and }{{X}_{2}}=3 startxref N. Reliability analysis of a series system 365 The Laplace transform of the system reliability is R(s) = P0(s). & +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot {{D}_{1}}+{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}}) \ In many cases, it is not easy to recognize which components are in series and which are in parallel in a complex system. X4= & AB\overline{C}-\text{only Unit 3 fails}\text{.} \end{align}\,\!, and thus not affect the outcome. [/math], ${{R}_{Computer1}}={{R}_{Computer2}}\,\! Each hard drive is of the same size and speed, but they are made by different manufacturers and have different reliabilities. 0000060521 00000 n$ ) and once as if the key component succeeded $(R=1)\,\!$. 114 53 [/math], \begin{align} \end{matrix} \right){{R}^{r}}{{(1-R)}^{n-r}} \ \,\! For this configuration, the system reliability, Rs, is given by: where R1, R2,..., Rn … Therefore: The k-out-of- n configuration is a special case of parallel redundancy. 0000131504 00000 n The weakest link dictates the strength of the chain in the same way that the weakest component/subsystem dictates the reliability of a series system. 0000002881 00000 n Mirrored blocks can be used to simulate bidirectional paths within a diagram. & +{{R}_{2}}\cdot {{R}_{10}}+{{R}_{9}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\ & +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\, \begin{align} The network shown next is a good example of such a complex system. \end{align}\,\! Since the reliabilities of the subsystems are specified for 100 hours, the reliability of the system for a 100-hour mission is: When we examined a system of components in series, we found that the least reliable component has the biggest effect on the reliability of the system., \begin{align} \end{align}\,\! The bidirectionality of this system can be modeled using mirrored blocks. = & \left( \begin{matrix} For example, a motorcycle cannot go if any of the following parts cannot serve: engine, tank with fuel, chain, frame, front or rear wheel, etc., and, of course, the driver., ${{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+\ \,{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! {{R}_{s}}= & 1-0.000484245 \\ 0000055283 00000 n As long as there is at least one path for the "water" to flow from the start to the end of the system, the system is successful. Issue 108, February 2010. & -{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}} \\ With this technique, it is possible to generate and analyze extremely complex diagrams representing the behavior of many subsystems in a manageable way. {{R}_{System}}= & +{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\$, \begin{align} \end{align}\,\! The container serves a dual purpose. • Reliability of a product is defined as the probability that the product will not fail throughout a prescribed operating period. The system steady-state availability is given by Av = lim sP0(s)., B\,\! \end{align}\,\!, \begin{align} What is the reliability of the series system shown below? The effective reliability and availability of the system depends on the specifications of individual components, network configurations, and redundancy models., \begin{align} \end{align}\,\! For this configuration, the system reliability, R s, is given by: where R 1, R 2, ..., R n are the values of reliability for the n components. & +{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}} \\ 0000065810 00000 n The plot illustrates the same concept graphically for components with 90% and 95% reliability. 0000036160 00000 n 0000003248 00000 n contains the token ${{I}_{11}}\,\!$. [/math] in series, as shown next: In the diagram shown below, electricity can flow in both directions. [/math], \begin{align} Create a block diagram for this system. P({{X}_{7}})= & P(A\overline{B}\overline{C})=({{R}_{1}})(1-{{R}_{2}})(1-{{R}_{3}}) \\ {{R}_{s}}={{R}_{2}}{{R}_{3}}P(A)={{R}_{1}}{{R}_{2}}{{R}_{3}} Draw the reliability block diagram for this circuit., then the equation for the reliability of the system could be further reduced to: The example can be repeated using BlockSim. In a simple parallel system, as shown in the figure on the right, at least one of the units must succeed for the system to succeed. For example, consider an airplane that has four engines. Figure 9. & +{{R}_{5}}\cdot {{D}_{1}}+{{R}_{8}}\cdot {{D}_{1}} \ Clearly, the reliability of a system can be improved by adding redundancy. [/math] (for a given time). [/math], -{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! \\ & -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}}-{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\ & +\left( \begin{matrix} 0000006051 00000 n {{R}_{s}}=95.86% Using [math]{{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix} \end{align}\,\! \end{align}\,\! IT systems contain multiple components connected as a complex architectural., \begin{align}, \begin{align} We assume the strengths of these k components are subjected to a common stress which is independent of the strengths of the k components. BlockSim uses a 64K memory buffer for displaying equations., ${{R}_{system}}=({{R}_{S}}\cdot {{R}_{E}}(-{{R}_{1}}\cdot {{R}_{2}}\cdot {{R}_{3}}+{{R}_{1}}\cdot {{R}_{2}}+{{R}_{3}})) \ \,\! This is illustrated in the following example.$, \begin{align} A = .001, B = .002, mission time (t) = 50 hours . = & 95.26% 0000127981 00000 n, ${{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}}\,\! These are reliability-wise in series and a failure of any of these subsystems will cause a system failure. & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\ Consider a system of 6 pumps of which at least 4 must function properly for system success. Hello there, A)Have this Reliability Block Diagrams(RBD) of a simple series system with the following data. {{P}_{f}}=1-{{R}_{1}}{{R}_{2}}-{{R}_{3}}+{{R}_{1}}{{R}_{2}}{{R}_{3}}$ are identical, this diagram is equivalent to a diagram with $A\,\! Three components each with a reliability of 0.9 are placed in series. trailer$, \begin{align} Recommended for you Example: Effect of a Component's Reliability in a Series System. and ${{X}_{8}}\,\! In the second row, the reliability of Component 1 is increased by a value of 10% while keeping the reliabilities of the other two components constant.$, $+{{R}_{5}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! Example 1. The RBD is analyzed and the system reliability equation is returned. Selecting Unit 3 as the key component, the system reliability is: That is, since Unit 3 represents half of the parallel section of the system, then as long as it is operating, the entire system operates.$, $\frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{3}=\frac{1}{3}\,\! \end{matrix} \right){{0.85}^{6}}{{(1-0.85)}^{0}} \\ The RBD is shown next, where blocks 5A, 7A and 1A are duplicates (or mirrored blocks) of 5, 7 and 1 respectively. Reliability describes the ability of a system or component to function under stated conditions for a specified period of time. In this paper, we estimate the reliability of series system with k components. Parallel Configuration Systems 5. In a chain, all the rings are in series and if any of the rings break, the system fails. It can be seen that Component 1 has the steepest slope, which indicates that an increase in the reliability of Component 1 will result in a higher increase in the reliability of the system. These terms use tokens to represent portions of the equation. They will make you ♥ Physics. & -{{R}_{2}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\$ at 100 hours? The last step is then to substitute ${{R}_{System}}\,\! Authors; Authors and affiliations; Palle Thoft-Christensen; Yoshisada Murotsu; Chapter. A series system is a system that fails if any of its components fails. This is a very important property of the parallel configuration, specifically in the design and improvement of systems. Once again, this is the opposite of what was encountered with a pure series system. Mathematically, the reliability of this series system, R ss, is given by (1.1) where R 1, R 2, , R n are the reliabilities of Unit 1, Unit 2, ..., Unit n, respectively.$, \begin{align} Calculating Series System Reliability and Reliability for Each Individual Component. Three subsystems are reliability-wise in series and make up a system. and $C\,\! Lectures by Walter Lewin. Redundancy models can account for failures of internal system components and therefore change the effective system reliability and availability perfor… When computing the system equation with the Use IBS option selected, BlockSim looks for identical blocks (blocks with the same failure characteristics) and attempts to simplify the equation, if possible. All these elements are thus arranged in series.$ yielding: Now the equation above contains the token ${{D}_{1}}\,\!$. {{R}_{s}}= & P(s|A)P(A)+P(s|\overline{A})P(\overline{A}) \\ [/math], [math]\begin{align} n \\ where R s is the overall reliability of the system, and r n is the reliability of the n th component. What is the overall reliability of the system for a 100-hour mission? 2.3 Combination System . We have already discussed reliability and availability basics in a previous article. Thus, this arrangement would require two consecutive relays to fail for the system to fail. What would the reliability of the system be if there were more than one component (with the same individual reliability) in series? One can easily take this principle and apply it to failure modes for a component/subsystem or system. 3 \\ 0000063244 00000 n This page uses frames, but your browser doesn't support them. Apr 13, 2006 #1. The mutually exclusive system events are: System events [math]{{X}_{6}}\,\!