(1)The analytical approach presented in the prior chapter of this on-line reference, RBDs and Analytical System Reliability, involved the determination of a mathematical expression that describes the reliability of the system, expressed in terms of the reliabilities of its components. So far, we have estimated only static system reliability (at a fixed time). For example, in the case of a system with three components in series, the system's reliability equation was given by:
(1)
The values of R1, R2 and R3 were given for a common time and the reliability of the system was estimated for that time. However, since the component failure characteristics can be described by distributions, the system reliability is actually time-dependent. In this case, Eqn. (1) can be rewritten as:
The reliability of the system for any mission time can now be estimated. Assuming a Weibull life distribution for each component, Eqn. (1) can now be expressed in terms of each component's reliability function, or:
In the same manner, any life distribution can be substituted into the system reliability equation. Suppose that the times-to-failure of the first component are described with a Weibull distribution, the times-to-failure of the second component with an exponential distribution and the times-to-failure of the third component with a normal distribution. Then, Eqn. (1) can be written as:
It can be seen that the biggest challenge is in obtaining the system's reliability function in terms of component reliabilities, which has already been discussed in depth. Once this has been achieved, calculating the reliability of the system for any mission duration is just a matter of substituting the corresponding component reliability functions into the system reliability equation.
The primary advantage of the analytical solution is that it produces a mathematical expression that describes the reliability of the system. Once the system's reliability function has been determined, other calculations can then be performed to obtain metrics of interest for the system. Such calculations include:
Determination of the system's pdf.
Determination of warranty periods.
Determination of the system's failure rate.
Determination of the system's MTTF.
In addition, optimization and reliability allocation techniques can be used to aid engineers in their design improvement efforts. Another advantage in using analytical techniques is the ability to perform static calculations and analyze systems with a mixture of static and time-dependent components. Finally, the reliability importance of components over time can be calculated with this methodology.
The biggest disadvantage of the analytical method is that formulations can become very complicated. The more complicated a system is, the larger and more difficult it will be to analytically formulate an expression for the system's reliability. For particularly detailed systems, this process can be quite time-consuming, even with the use of computers. Furthermore, when the maintainability of the system or some of its components must be taken into consideration, analytical solutions become intractable. In these situations, the use of simulation methods may be more advantageous than attempting to develop a solution analytically. Simulation methods are presented in later chapters of this on-line reference.
The complexity involved in an analytical solution can be best illustrated by looking at the simple "complex" system with 15 components, as shown in Figure 5.1.
Figure 5.1: An RBD of a complex system.
The system reliability for this system (computed using BlockSim) is shown next. The first solution is provided using BlockSim's symbolic solution. In symbolic mode, BlockSim breaks the equation into segments, identified by tokens, that need to be substituted into the final system equation for a complete solution. This creates algebraic solutions that are more compact than if the substitutions were made.
Substituting the terms yields:
BlockSim's automatic algebraic simplification would yield the following format for the above solution:
(2)
In this equation, each Ri represents the reliability function
of a block. For example, if RA has a Weibull distribution, then
each 
and so forth. Substitution of each component's
reliability function in Eqn. (2) will result in an analytical expression
for the system reliability as a function of time, or Rs(t),
which is the same as (1 - cdfSystem).
Once the system reliability equation (or the cumulative density function, cdf) has been determined, other functions and metrics of interest can be derived. Consider the following simple system:
Figure 5.2: Simple two component system.
Furthermore, assume that component 1 follows an exponential distribution with a mean of 10,000 (μ = 10,000, λ = 1/10,000) and component 2 follows a Weibull distribution with β = 6 and η = 10,000. The reliability equation of this system is:
(3)
The system cdf is:
Once the equation for the reliability of the system has been obtained, the system's pdf can be determined. The pdf is the derivative of the reliability function with respect to time or:
(4)
For the system shown in Figure 5.2, this is:
(5)
Figure 5.3 shows a plot of Eqn. (5).
Figure 5.3: pdf plot of the two component system shown in Figure 5.2 and given in Eqn. (5).
Conditional reliability is the probability of a system successfully completing another mission following the successful completion of a previous mission. The time of the previous mission and the time for the mission to be undertaken must be taken into account for conditional reliability calculations. The system's conditional reliability function is given by:
(6)
Eqn. (6) gives the reliability for a new mission of duration t having already accumulated T hours of operation up to the start of this new mission and the system is checked out to assure that it will start the next mission successfully.
For the system in Figure 5.2, the reliability for mission of t = 1,000 hours, having an age of T = 500 hours, is:
Now in this formulation, it was assumed that the accumulated age was equivalent for both units. That is, both started life at zero and aged to 500. It is possible to consider an individual component that has already accumulated some age (used component) in the same formulation. To illustrate this, assume that component 2 started life with an age of 100. Then the reliability equation of the system, as given in Eqn. (3), would need to be modified to include a conditional term for 2, or:
(7)
In BlockSim, the start age input box may be used to specify this.
Once the distribution of the system has been determined, the failure rate can also be obtained by dividing the pdf by the reliability function:
(8)
For the system in Figure 5.2:
(9)
Figure 5.4 shows a plot of Eqn. (9).
Figure 5.4: Failure rate function plot of the system in Figure 5.2.
BlockSim uses numerical methods to estimate the failure rate. It should
be pointed out that as 
, numerical evaluation of Eqn. (8)
is constrained by machine numerical precision. That is, there are limits
as to how large t can get before floating point
problems arise. For example, at t = 5,000,000 both numerator and
denominator will tend to zero (e.g.
). As this numbers become very small they will start looking
like a zero to the computer, or cause a floating point error, resulting
in a 
or 
operation. In these cases, BlockSim
will return an N/A for
the result. Obviously this does not create any practical constraints.
The mean life (or mean time to failure, MTTF) can be obtained by integrating the system reliability function from zero to infinity:
(10)
The mean time is a performance index and does not provide any information about the behavior of the failure distribution of the system.
For the system in Figure 5.2:
Sometimes it is desirable to know the time value associated with a certain reliability. Warranty periods are often calculated by determining what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail. Similarly, engineering specifications may call for a certain BX life, which also represents a time period during which a certain proportion of the population will fail. For example, the B10 life is the time in which 10% of the population will fail.
This is obtained by setting RS(t) to the desired value and solving for t.
For the system in Figure 5.2:
(11)
To compute the time by which reliability would be equal to 90%, Eqn. (11) is recast as follows and solved for t.
In this case, t = 1053.59. Equivalently, the B10 life for this system is also 1053.59.
Outside of some trivial cases, a closed form solution for t cannot be obtained. (Note: A trivial example is a case where all items are exponentially distributed, i.e. have a constant failure rate, and are in a series configuration.) Thus, it is necessary to solve for t using numerical methods. BlockSim uses numerical methods.
Consider a system consisting of three exponential units in series with the following failure rates (in failures per hour): λ1 = 0.0002, λ2 = 0.0005 and λ3 = 0.0001.
Obtain the reliability equation for the system.
What is the reliability of the system after 150 hours of operation?
Obtain the system's pdf.
Obtain the system's failure rate equation.
What is the MTTF for the system?
What should the warranty period be for a 90% reliability?
The analytical expression for the reliability of the system is given by:
(12)
At 150 hours of operation, the reliability of the system is:
In order to obtain the system's pdf, the derivative of the reliability equation given in Eqn. (12) is taken with respect to time, or:
(13)
The system's failure rate can now be obtained simply by dividing the system's pdf given in Eqn. (13) by the system's reliability function given in Eqn. (12), or:
Combining Eqn. (10) and Eqn. (12), the system's MTTF can be obtained:
Solving Eqn. (12) with respect to time will yield the corresponding warranty period for a 90% reliability. In this case, the system reliability equation is simple and a closed form solution exists. The warranty time can now be found by solving:
Thus, the warranty period should be 132 hours.
Consider the system shown in Figure 5.5.
Figure 5.5: Complex bridge system in Example 2.
Components A through E are Weibull distributed with β = 1.2 and η = 1230 hours. The starting and ending blocks cannot fail. Determine the following:
The reliability equation for the system and its corresponding plot.
The system's pdf and its corresponding plot.
The system's failure rate equation and the corresponding plot.
The MTTF.
The warranty time for a 90% reliability.
The reliability for a 200 hour mission, if it is known that the system has already successfully operated for 200 hours.
The first step is to obtain the reliability function for the system. The methods described in the previous chapter can be employed. For example, the event space or path tracing methods can be used. Using BlockSim, the following reliability equation is obtained:
(14)
Note that since the starting and ending blocks cannot fail, RStart = 1 and REnd = 1, Eqn. (14) can be reduced to:
(15)
Where RA is the reliability equation for Component A, or:
(16)
RB is the reliability equation for Component B, etc.
Since the components in this example are identical, the system reliability equation can be further reduced to:
(17)
Or, in terms of the failure distribution:
(18)
The corresponding plot is given in Figure 5.6.
Figure 5.6: Reliability plot for the system in Figure 5.5.
In order to obtain the system's pdf, the derivative of the reliability equation given in Eqn. (18) is taken with respect to time, resulting in:
(19)
The pdf can now be plotted for different time values, t, as shown in Figure 5.7.
Figure 5.7: pdf plot for the system in Figure 5.5.
The system's failure rate can now be obtained by dividing the system's pdf given in Eqn. (19) by the system's reliability function given in Eqn. (18), or:
The corresponding plot is given in Figure 5.8.
Figure 5.8: Failure rate for the system in Figure 5.5.
The MTTF of the system is obtained by integrating the system's reliability function given by Eqn. (18) from time zero to infinity, as given by Eqn. (10). Using BlockSim's Analytical QCP, an MTTF of 1007.8 hours is calculated, as shown in Figure 5.9.
Figure 5.9: MTTF of the system in figure 5.5.
The warranty time can be obtained by solving Eqn. (18) with respect to time for a system reliability Rs = 0.9. Using the Analytical QCP and selecting the Warranty Time option, a time of 372.72 hours is obtained, as shown in Figure 5.10.
Figure 5.10: Time at which R = 0.9 or 90% for the system in Figure 5.5.
Figure 5.11: Conditional reliability calculation for the system in Figure 5.5.
Lastly, the conditional reliability can be obtained using Eqn. (6) and Eqn. (18), or:
This can be calculated using BlockSim's Analytical QCP, as shown in Figure 5.11.
See Also:
Time-Dependent System Reliability (Analytical)
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