In other words, in order to achieve a high system reliability, the component reliability must be high also, especially for systems with many components arranged reliability-wise in series. \\ To calculate system availability for a certain period of time, divide an asset’s total amount of uptime by the sum of total uptime and total downtime. 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. However, the component with the highest reliability in a parallel configuration has the biggest effect on the system's reliability, since the most reliable component is the one that will most likely fail last. Doing so yields ${{I}_{7}}\,\! & +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\ The reliability of the system is simply the probability of the union of these paths. The first row of the table shows the given reliability for each component and the corresponding system reliability for these values.$, \begin{align} It is calculated by dividing the total operating time of the asset by the number of failures over a given period of time. The unreliability of the system is then given by: In the case where the failure of a component affects the failure rates of other components, then the conditional probabilities in equation above must be considered. for $k=4\,\! Example: Effect of the Number of Components in a Series System. & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}-{{R}_{2}}\cdot {{R}_{9}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{5}}\cdot {{D}_{1}}-{{R}_{9}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\ 1.2.1 Reliability Reliability is the probability of an item to perform a required function under stated conditions for a specified period of time. The following figure shows the equation returned by BlockSim. In this method, all possible operational combinations are considered in order to obtain the system's reliability. The same methodology and principles can also be used for other applications.$ into the equation . Measurement 3. English-Chinese dictionary of mining (英汉矿业大词典). R_S = & P({{X}_{1}}\cap {{X}_{2}}\cap ...\cap {{X}_{n}}) \\ For example, consider an airplane that has four engines. This type of a configuration is also referred to as a complex system. \end{align}\,\! & +{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\ & +{{R}_{5}}\cdot ({{R}_{7}}\cdot {{I}_{7}})+{{R}_{8}}\cdot ({{R}_{7}}\cdot {{I}_{7}})) \ Since at least two hard drives must be functioning at all times, only one failure is allowed. \\ X5= & \overline{AB}C-\text{Units 1 and 2 fail}\text{.} \end{align}\,\! [/math] at 100 hours? For equipment that is expected to be oper… Another Example for Failure Mode Analysis. \end{align}\,\! 1893 0 obj <>/Filter/FlateDecode/ID[]/Index[1887 15]/Info 1886 0 R/Length 52/Prev 264556/Root 1888 0 R/Size 1902/Type/XRef/W[1 2 1]>>stream [/math], \begin{align} Thecombined system is operational only if both Part X and Part Y are available.From this it follows that the combined availability is a product ofthe availability of the two parts. ρ x x ′. The reliability of a series system with three elements with R 1 = 0.9, R 2 = 0.8, and R 3 = 0.5 is R = 0.9 × 0.8 × 0.5 = 0.36, which is less than the reliability of the worst component (R 3 = 0.5). \end{align}\,\! 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. \end{align}\,\! In the figure below, blocks 1, 2 and 3 are in a load sharing container in BlockSim and have their own failure characteristics. @�מ���[����\f�q/˾�l�ދǭ��?0�-�^��V�5����T��&*[��э����ă]�k��֏�����!�BR�瘖D�iY���oH)]U��SJˤOɘθ��4�4�i)9 ��F���=m����捈/�0r׏R��@��%s%�~̂-�hf�Sּ��W���gO�F1�ƭ�M���� ��o� By using multi blocks within BlockSim, a single block can represent multiple identical blocks in series or in parallel configuration. Do the following for the RBD shown below: The event space method is an application of the mutually exclusive events axiom. \end{align}\,\!, ${{D}_{1}}=+{{R}_{7}}\cdot {{I}_{7}} \ \,\! Complex systems are discussed in the next section.$ and {{R}_{3}}\,\ = 80%\,\! X6= & \overline{A}B\overline{C}-\text{Units 1 and 3 fail}\text{.} {{R}_{s}}= & 1-[(1-0.982065)\cdot (1-0.973000)] \\ At least two of them must function in order for the computer to work properly. = & P(ABD)+P(ACD)-P(ABCD) \end{align}\,\!, \begin{align} While multi blocks allow the analyst to represent multiple items with a single block in an RBD, BlockSim's mirrored blocks can be used to represent a single item with more than one block placed in multiple locations within the diagram., \begin{align} The corresponding reliability for the system is [math]{{R}_{s}} = 97.6%\,\!. {{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\ In this mode, portions of the system are segmented. [/math], \frac{1}{{{r}_{eq}}}=\frac{1}{\infty }+\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\,\! The rate of change of the system's reliability with respect to each of the components is also plotted. {{R}_{1,2}}= & 0.982065\text{ or }98.2065% %%EOF \end{align}\,\! As the number of components connected in series increases, the system's reliability decreases., {{R}_{Computer1}}={{R}_{Computer2}}\,\! \end{align}\,\! \\, \begin{align} The failure times and all maintenance events are the same for each duplicate block as for the original block. & -{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot {{I}_{7}}) \\ 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. Put another way, [math]{{r}_{1}}\,\! However, for a complex system, determination of the system reliability becomes more involved. Reliability at a given time: The failure rate can be expressed as λ = NF / No t = No - Ns / (No t)(2) where NF = No - Ns = number of failing components at time t Ns= number of live surviving components at time t No= initial number of live surviving components at time zero 25362 is correct, how ever to add some more details for the system mentioned by you i.e. Selecting Unit 1, the probability of success of the system is: That is, if Unit 1 is operating, the probability of the success of the system is the probability of Units 2 and 3 succeeding. This page was last edited on 5 January 2016, at 18:52. When BlockSim constructs the equation internally, it does so in what we will call a symbolic mode. The following figure illustrates the effect of the number of components arranged reliability-wise in series on the system's reliability for different component reliability values. It involves choosing a "key" component and then calculating the reliability of the system twice: once as if the key component failed ( [math]R=0\,\!, ${{R}_{s}}=P(C)+{{R}_{1}}{{R}_{2}}P(\overline{C})={{R}_{3}}+{{R}_{1}}{{R}_{2}}(1-{{R}_{3}})\,\!$, and thus not affect the outcome. [/math], \begin{align} To illustrate this configuration type, consider a telecommunications system that consists of a transmitter and receiver with six relay stations to connect them. {{I}_{7}}= & -{{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}} \\ Clearly, the reliability of a system can be improved by adding redundancy. During this correct operation, no repair is required or performed, and the system adequately follows the defined performance specifications. It is widely used in the aerospace industry and generally used in mission critical systems., ${{R}_{s}}(k,n,R)=\underset{r=k}{\overset{n}{\mathop \sum }}\,\left( \begin{matrix} Consequently, the analysis method used for computing the reliability of a system will also depend on the reliability-wise configuration of the components/subsystems. In order to construct a reliability block diagram, the reliability-wise configuration of the components must be determined.$ and ${{X}_{8}}\,\! Consider three components arranged reliability-wise in series, where [math] { {R}_ {1}}=70%\,\! In life data analysis and accelerated life testing data analysis, as well as other testing activities, one of the primary objectives is to obtain a life distribution that describes the times-to-failure of a component, subassembly, assembly or system.$, $-{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\!$, \begin{align}, {{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}}\,\! It should be pointed out that the complete equation can get very large. Create a block diagram for this system. X8= & \overline{ABC}-\text{all units fail}\text{.} \end{align}\,\! Reliability Testing can be categorized into three segments, 1. \end{align}\,\! System availability is calculated by dividing uptime by the total sum of uptime and downtime.Availability = Uptime ÷ (Uptime + downtime)For example, let’s say you’re trying to calculate the availability of a critical production asset. \end{align}\,\!, ${{R}_{2}}=80%\,\!$, \begin{align} The following graphic demonstrates the RBD for the system. %PDF-1.4 %���� In addition, the weakest link in the chain is the one that will break first. & +\left( \begin{matrix}, ${{Q}_{s}}=\underset{i=1}{\overset{n}{\mathop \prod }}\,P({{X}_{i}})\,\! {{R}_{s}}={{R}_{3}}+{{R}_{1}}{{R}_{2}}-{{R}_{1}}{{R}_{2}}{{R}_{3}} \frac{1}{{{r}_{eq}}}= & \frac{1}{{{r}_{1}}}+\frac{1}{{{r}_{2}}}+\frac{1}{{{r}_{3}}} \\ h�bbdb��A�Q�@�s#�; ���7��W�/� ���$, $+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! What would the reliability of the system be if there were more than one component (with the same individual reliability) in series?$ must succeed in order for the system to succeed. The following operational combinations are possible for system success: The probability of success for the system (reliability) can now be expressed as: This equation for the reliability of the system can be reduced to: If all three hard drives had the same reliability, $R\,\! It is most often expressed as a percentage, using the following calculation: Availability = 100 x (Available Time (hours) / Total Time (hours)) For equipment and/or systems that are expected to be able to be operated 24 hours per day, 7 days per week, Total Time is usually defined as being 24 hours/day, 7 days/week (in other words 8,760 hours per year). {{R}_{s}}=95.86%$ and ${ {R}_ {3}}=90%\,\! r \\ {{R}_{s}}=99.95% Subsystem 1 has a reliability of 99.5%, subsystem 2 has a reliability of 98.7% and subsystem 3 has a reliability of 97.3% for a mission of 100 hours.$, ${{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}(-{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot ({{R}_{7}}\cdot (-{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{6}}+{{R}_{3}}\cdot {{R}_{5}}\cdot {{R}_{4}}+\ \,{{R}_{3}}\cdot {{R}_{6}}\cdot {{R}_{4}}+{{R}_{5}}\cdot {{R}_{6}}\cdot {{R}_{4}}-{{R}_{3}}\cdot {{R}_{5}}-{{R}_{3}}\cdot {{R}_{6}}-{{R}_{3}}\cdot {{R}_{4}}-{{R}_{5}}\cdot {{R}_{6}}-{{R}_{5}}\cdot {{R}_{4}}-{{R}_{6}}\cdot {{R}_{4}}+{{R}_{3}}+{{R}_{5}}+{{R}_{6}}+{{R}_{4}}))\,\! This is illustrated in the following example. {{R}_{1,2}}= & 0.9950\cdot 0.9870 \\ There are other multiple redundancy types and multiple industry terms.$, \begin{align} The MTTR is the time (on average) needed to fix this system failure. While many smaller systems can be accurately represented by either a simple series or parallel configuration, there may be larger systems that involve both series and parallel configurations in the overall system., {{R}_{System}}=+{{R}_{1}}\cdot {{R}_{11}}\cdot {{I}_{11}} \ \,\! \end{align}\,\! This means that the engines are reliability-wise in a k-out-of- n configuration, where k = 2 and n = 4. and $n=6\,\! The RBD is analyzed and the system reliability equation is returned.$, \begin{align} Reliability is the probability that a system performs correctly during a specific time duration., \begin{align} and $D\,\! Assume starting and ending blocks that cannot fail, as shown next. So, if there are three units tested for 500 hours and one fails at 400 hours (not replaced), the … If Unit 3 fails, then the system is reduced to: The reliability of the system is given by: Example: Using the Decomposition Method to Determine the System Reliability Equation. In the case of the parallel configuration, the number of components has the opposite effect of the one observed for the series configuration. \end{matrix} \right){{0.85}^{5}}{{(1-0.85)}^{1}} \\$ units succeeds, then the system succeeds. Units in load sharing redundancy exhibit different failure characteristics when one or more fail. {{R}_{s}}= & \underset{r=4}{\overset{6}{\mathop \sum }}\,\left( \begin{matrix} One block within the container must be operating, otherwise the container will fail, leading to a system failure (since the container block is part of a series configuration in this example). [/math], \begin{align} = & 0.1762+0.3993+0.3771 \\ \end{align}\,\! The RBD is shown next, where blocks 5A, 7A and 1A are duplicates (or mirrored blocks) of 5, 7 and 1 respectively. The equivalent resistance must always be less than [math]1.2\Omega \,\!. The symbolic solution for the system in the prior case, with the Use IBS option selected and setting equal reliability block properties, is: When using IBS, the resulting equation is invalidated if any of the block properties (e.g., failure distributions) have changed since the equation was simplified based on those properties. For example, a block that was originally set not to fail can be re-set to a failure distribution and thus it would need to be used in subsequent analyses. MIL-HDBK-338, Electronic Reliability Design Handbook, 15 Oct 84 2. n \\ Determine the reliability equation of the same system using BlockSim. Redundancy is a very important aspect of system design and reliability in that adding redundancy is one of several methods of improving system reliability. Ea�Ǽ��|/�Y��5 These structural properties, however, refer to the system's state of success or failure based on the states of its components. All three must fail for the container to fail. {{I}_{11}}= & -{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}}+{{R}_{2}}\cdot {{R}_{9}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{D}_{1}} \\ & +{{R}_{A}}\cdot {{R}_{B}}\cdot {{R}_{C}}\cdot {{R}_{D}}\cdot {{R}_{F}} \\ \end{align}\,\! One approach, described in detail later in this chapter, is to use the event space method. [/math] in series, as shown next: In the diagram shown below, electricity can flow in both directions. \end{align}\,\! The reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. [/math] (for a given time). Total System Reliability is a calculation which allows you to combine the reliabilities of several components to give a new value for syystem reliability. & +{{R}_{2}}\cdot {{R}_{5}}\cdot {{R}_{10}}\cdot {{R}_{8}}\cdot {{D}_{1}} \\ [/math], \begin{align} \end{matrix} \right){{0.85}^{r}}{{(1-0.85)}^{6-r}} \\ What is the overall reliability of the system for a 100-hour mission?, [math]+{{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}})))\,\!