Reliability

The reliability module will compute the reliability of simple systems. If it is used repeatedly, complex systems can be developed. This module can easily be used to determine the appropriate number of backup (standby) pieces of components. The module has five submodels. The general framework for the first three is identical and for the last two is identical.

Data

The general framework for reliability is given by the number of simple systems that are in series and the maximum number of components in any simple system. By a simple system, we mean that it is a set of parallel components without any series. In the following example, we are setting up for a system with four simple systems in series. The largest number of components in any of these four simple systems is six. There is only one type data that needs to be entered.

Component reliability. The required information is the reliability of each component. It is used for computing the reliability of the simple parallel series represented by the column.

NOTE: This is a module where using the option to not display zeros will make the data display more readable.

Solution

A sample solution screen that also contains the data is given next. Notice that the row in which the probability is entered does not matter, as exemplified by the fact that systems 1 and 2 each have the same reliability of 99%.

Simple system reliability. Below each parallel system (column) its reliability is presented. The reliability, r, of n components in parallel is given by:

r = 1 - (1 - r1)(1 - r2)...(1 - rn)

where rj is the reliability of the jth individual component. In the example, the first parallel set has a reliability of its one component, which is .99; the same is true for the second set; the third has an overall reliability of .999856, as listed at the bottom of the third column; and the fourth has a reliability of .999872.

System reliability. The overall system reliability is given at the bottom. The overall reliability is the product of the individual parallel series reliabilities. The example has a system reliability of .979834.

Example 2 - Determining the number of backups

It is possible to compute the number of backups required in order to ensure a specified system reliability for a parallel system. For example, suppose that the reliability of an individual component is 50% and the desired system reliability is 99%. Then, by creating a table with no backups, 1 backup, 2 backups, etc., the appropriate number of backups can be found. (This is an enumeration method). For the reliabilities specified as 50% and 99% we see from below that the appropriate number of components is seven (six backups).

Identical parallel or identical serial components

Alternatively, a different submodel could be used. The third and fourth submodels in the reliability module may be used for computing the reliability of parallel systems with identical components or systems in series with identical components.

Below, we show a screen for identical parallel components. There are only two items to be entered.

Number in parallel. In the extra data panel above the data table there is a scroolbar/textbox combination into which we place the number of components. In this example we are indicating that there are 10 identical components (one original and nine backups).

Component reliability. The data table requires one piece of information. This is the reliability of the components. In the example we have placed a .5 indicating that each of the 10 parallel components has a reliability of 50%.

Solution

The solution indicates that the overall reliability is .9961 which agrees with the more detailed display in the previous screen.