IV. A. 2. a. Reliability Block Diagrams and Models – Series Systems

(C) 0.010

A series model uses the following formula to calculate the system reliability (here the system is a two component system)

$$ R\left( t \right)=\sum\limits_{i=1}^{n}{{{R}_{i}}\left( t \right)}$$

Given only the mean time to fail we can only use the exponential distribution function for the estimate. The reliability of a component at t = 1 hour given the mean time to fail is 200 hours is

$$ {{R}_{1}}\left( 1 \right)={{e}^{-\frac{1}{200}}}=0.995$$

Therefore the system reliability at t = 1 hours is

$$ {{R}_{1}}\left( 1 \right)=0.995\times 0.995=0.990$$

The probability of failure for t = 1 hour is 1 – R_s(1) = 1 – 0.990 = 0.010

(B) 0.9670

The key here is the use of the reliability function for the exponential distribution along with the series system formula for system reliability. First calculate the reliability of one of the five components

$$ {{R}_{1}}\left( t \right)={{e}^{-\lambda t}}={{e}^{-0.00004\left( 168 \right)}}=0.9933$$

Then use the series system model which is the product of the reliability values in the series. Or, since all five components have the same reliability value, just raise the reliability value for one component to the 5^th power.

$$ {{R}_{sys}}\left( t \right)=\prod\limits_{i=1}^{n}{{{R}_{i}}=}{{0.9933}^{5}}=0.9670$$

Calculate the product of the individual part reliabilities.

A series system reliability is determined by calculating the product of the relaiblity of the elements of the system.

While it is possible to sum failure rates in some circumstances (all elements are described by an exponential distribution) summing is the wrong function for reliability values. Remember that a reliability value has to be between 0 and 1, and summing would quickly tally above 1. The same holds for summing of unreliabilities.

We do use the product of unreliabilities as part of the calculation concerning parallel systems, not series systems.

(B) 0.4198

The key formula is for the series system reliability calculation, which is simply the product of the system element reliabilities. Given only MTBF we can calculate the reliability of each component using the exponential distribution reliability function

$$ R\left( t \right)={{e}^{-\frac{t}{\theta }}}$$

where θ is the component’s MTBF. Once you calculate each reliability value, take the product of the four reliabilities for the solution.

Another method, which may be a little quicker involves determining the failure rates then calculating the system reliability with the sum of the failure rates. The failure rate for a component is the inverse of its MTBF value. Following this approach we sum the inverses of the four MTBF values as such

$$ \begin{array}{l}{{\lambda }_{sys}}=\sum\limits_{i=1}^{n}{\frac{1}{{{\theta }_{i}}}=\frac{1}{2,200}+}\frac{1}{3,400}+\frac{1}{1,700}+\frac{1}{1,200}\\{{\lambda }_{sys}}=0.00045455+0.00029412+0.00058824+0.00083333\\{{\lambda }_{sys}}=0.00217023\end{array}$$

The calculate the system reliability using the exponential distribution reliability function using the calculated system failure rate

$$ \begin{array}{l}{{R}_{sys}}\left( t \right)={{e}^{-\lambda t}}\\{{R}_{sys}}\left( 400 \right)={{e}^{-(0.00217023)400}}\\{{R}_{sys}}\left( 400 \right)=0.4198\end{array}$$

(A) The reliability is < 99% over an hour.

We do not know what else in the system nor over what time period the question of reliability applies. If we assume an hour duration, t = 1, I think this is a safe assumption since the failure rate is given as per hour. We can calculate the reliability over one hour with

$$ \begin{array}{l}{{R}_{sys}}\left( t \right)={{e}^{-\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)t}}\\{{R}_{sys}}\left( 1 \right)={{e}^{-\left( 0.0077+0.0077 \right)}}\\{{R}_{sys}}\left( 1 \right)={{e}^{-\left( 0.0077+0.0077 \right)}}=0.9847\end{array}$$

Any duration longer than an hour will further reduce this two component system further below 99%. Also, the calculation shows the reliability at an hour is 99.47% and not 99.23%.

Given the information available the only true statement is (A).