(D) 0.966

There three (maybe more) ways to solve a keystone RBD. Keystone configurations do not resolve by simplifying parallel and series constructions alone.

The best way I know to quickly solve this type of RBD for system reliability is by first identifying the keystone element. That means if you say a block will work (or not work) the remaining elements are easy to reduce and resolve.

In this case component E is the keystone element.

Using Bayes’ Theorem permits us to determine the system reliability given E is working, then add the system reliability given E is not working. Here’s the formula:

$$ {{R}_{sys}}=P({{R}_{sys-E}}|E)\times P(E)+P({{R}_{sys-\bar{E}}}|\bar{E})\times P(\bar{E})$$

where

$$ \bar{E}=1-{E}$$

First let’s sort out the reliability of the system when component E is working. This creates a structure

Start with reducing the two parallel elements

$$ \begin{array}{l}{{R}_{A||C}}=(1-(1-{{R}_{A}})(1-{{R}_{C}}))\\{{R}_{A||C}}=(1-(1-0.85)(1-0.88))\\{{R}_{A||C}}=0.982\end{array}$$

and, the other parallel element

$$ \begin{array}{l}{{R}_{B||D}}=(1-(1-{{R}_{B}})(1-{{R}_{D}}))\\{{R}_{B||D}}=(1-(1-0.96)(1-0.81))\\{{R}_{B||D}}=0.9924\end{array}$$

The solve for the situation when component E is working

$$ \begin{array}{l}{{R}_{sys-E}}={{R}_{A||C}}\times {{R}_{B||D}}\times {{R}_{E}}\\{{R}_{B||D}}=0.982\times 0.9924\times 0.7\\{{R}_{B||D}}=0.6822\end{array}$$

Then the system reliability given E is not working starting with redrawing the RBD structure

Start with reducing the pairs of series structures

$$ \begin{array}{l}{{R}_{AB}}={{R}_{A}}\times {{R}_{B}}=0.85\times 0.96=0.816\\{{R}_{CD}}={{R}_{C}}\times {{R}_{D}}=0.88\times 0.81=0.7128\end{array}$$

The reduce the remaining parallel structure

$$ \begin{array}{l}{{R}_{AB||CD}}=1-\left( 1-{{R}_{AB}} \right)\left( 1-{{R}_{CD}} \right)\\{{R}_{AB||CD}}=1-\left( 1-0.816 \right)\left( 1-0.7128 \right)\\{{R}_{AB||CD}}=0.2841\end{array}$$

The system reliability is then

$$ \begin{array}{l}{{R}_{sys}}=P({{R}_{sys-E}}|E)\times P(E)+P({{R}_{sys-\bar{E}}}|\bar{E})\times P(\bar{E})\\{{R}_{sys}}={{R}_{sys-E}}+{{R}_{sys-\bar{E}}}\\{{R}_{sys}}=0.6822+0.2841=0.9663\end{array}$$

(C) 0.918

If you see this type of problem, while it’s easy to solve, it takes time. Hold till the end after you have answered all the quick one you know.

The approach to solve for the system reliability value is to reduce the parallel elements to form series elements. The equation to reduce two elements in parallel to determine the reliability of the parallel elements is

$$ {{R}_{a||b}}=1-\left( 1-{{R}_{a}} \right)\left( 1-{{R}_{b}} \right)$$

Looking at the diagram lets start by reducing the block R3 and R4

$$ \begin{array}{l}{{R}_{3||4}}=1-\left( 1-{{R}_{3}} \right)\left( 1-{{R}_{4}} \right)\\{{R}_{3||4}}=1-\left( 1-0.75 \right)\left( 1-0.7 \right)\\{{R}_{3||4}}=0.925\end{array}$$

Next, let’s reduce the parallel structure involving R5 and R6

$$ \begin{array}{l}{{R}_{5||6}}=1-\left( 1-{{R}_{5}} \right)\left( 1-{{R}_{6}} \right)\\{{R}_{5||6}}=1-\left( 1-0.6 \right)\left( 1-0.5 \right)\\{{R}_{5||6}}=0.8\end{array}$$

Then reduce the series string of R2 and R3||4

$$ \begin{array}{l}{{R}_{2-3||4}}={{R}_{2}}\times {{R}_{3||4}}\\{{R}_{_{2-3||4}}}=0.9\times 0.925\\{{R}_{_{2-3||4}}}=0.8325\end{array}$$

That sets up another simple parallel structure to reduce

$$ \begin{array}{l}{{R}_{(2-3||4)||(5||6)}}=1-\left( 1-{{R}_{2-3||4}} \right)\left( 1-{{R}_{5||6}} \right)\\{{R}_{(2-3||4)||(5||6)}}=1-0.8325\times 0.8\\{{R}_{(2-3||4)||(5||6)}}=0.9665\end{array}$$

Which leaves two elements in series

$$ \begin{array}{l}{{R}_{sys}}={{R}_{1}}\times {{R}_{(2-3||4)||(5||6)}}\\{{R}_{sys}}=0.95\times 0.9665\\{{R}_{sys}}=0.918\end{array}$$

(A) Take the reliability for A and combine it with the reliability for C and then combine that result with the reliability for E.

The simplest approach is to simplify the parallel elements, A and C, into a single block, which is then in series with E

(B) active and standby 

Active redundancy means all the elements are active. Standby redundancy includes some components that are waiting to become active when necessary to replace the functionality of a primary (active) set of components.

Inactive redundancy is not a phrase used to describe a redundancy configuration. Parallel and series are reliability block diagram models used to describe the arrangement (reliability-wise) of elements of a design.

(D) 0.994

The formula to determine reliability of a parallel structure is

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

This formula is multiplying the unreliabilities then the last 1 minus converts back to reliability. This works for active redundancy when only one element in parallel is necessary for the system to operate. This is a 1-out-of-n system.

Inserting the reliability values (assuming t is the the same for each value) for this problem we are able to quickly calculate the result

$$ {{R}_{sys}}=1-\left( \left( 1-0.92 \right)\left( 1-0.83 \right)\left( 1-0.54 \right) \right)=0.994$$

(C) 0.99997

One way to solve this is to solve each component for reliability at 900 hours. For component A given a failure rate of 0.0005, we solve for R(900)

Calculate this system’s reliability at 900 hours.

$$ \begin{array}{l}{{R}_{A}}\left( t \right)={{e}^{-\lambda t}}\\{{R}_{A}}\left( 900 \right)={{e}^{-\left( 0.0005 \right)\times 900}}=0.956\end{array}$$

And for component C with 12,000 hours MTBF we have

$$ \begin{array}{l}{{R}_{A}}\left( t \right)={{e}^{-\frac{t}{\theta }}}\\{{R}_{A}}\left( 900 \right)={{e}^{-\frac{900}{12,000}}}=0.928\end{array}$$

The reliability of the system is now

$$ \begin{array}{l}{{R}_{sys}}=1-\left( 1-{{R}_{A}} \right)\left( 1-{{R}_{B}} \right)\left( 1-{{R}_{C}} \right)\left( 1-{{R}_{D}} \right)\\{{R}_{sys}}=1-\left( 1-0.956 \right)\left( 1-0.91 \right)\left( 1-0.928 \right)\left( 1-0.88 \right)\\{{R}_{sys}}=0.99997\end{array}$$

(C) 0.99999

An easy approach is to determine the reliability of one component with t = 1.

$$ \begin{array}{*{35}{l}}{{R}_{A}}\left( t \right)={{e}^{-\frac{t}{\theta }}} \\{{R}_{A}}\left( 1 \right)={{e}^{-\frac{1}{300}}}=0.928 \\\end{array}$$

Then determine the system reliability with the two components in parallel.

$$ \begin{array}{l}{{R}_{sys}}=1-\left( 1-{{R}_{A}} \right)\left( 1-{{R}_{A}} \right)\\{{R}_{sys}}=1-\left( 1-0.99667 \right)\left( 1-0.99667 \right)\\{{R}_{sys}}=0.99999\end{array}$$

(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).

(B) modeling the acquisition of computer software systems

As for any reliability model, the selection of the data, models, risks, and how the team will make trade-off decisions is key to creating a useful reliability model.

A) knowledge of the usage environment

There are four elements to a reliability specification, the function, the environment which includes useage, the probability of successful operation, and a duration. I personal do not like MTBF as it is only the inverse of a failure rate and alone provides little useful information.

using a 100% rating

The derating concept is to use components at less then the vendor’s listed ratings. Thus if a capacitor is rated for 10 volts, the design should use this capacitor on a circuit that has less than 10 volts occurring across the capacitor. A 50% voltage derating would imply we should use a 10 volt rated capacitor for a situation that would experience at most 5 volts. 5 / 10 = 0.5 or 50%.

There are many stresses and depending on the specific technology used within a component, stresses such as voltage, current, temperature, power, etc. may apply. It is common to have two or more stresses derated.

(B) that the load and strength values are statistically described

The key word here is “minimize” knowing stress strength values or that they are fixed or the values meet a safety margin do not allow the designer to minimize the risk of failure. Knowing the statistical distributions of both stress and strength, as well as how the distributions change over time are critical to know.

(A) 1,000 lb

The key word here is “derated” which implies below the rated value. The only option is less then the rated breaking strength of 2,000 is the (A) response.

(D) I, II, and III

This one requires careful reading as it uses the word “highest” meaning the most derated component for a specific application. For example, if selecting a capacitor for a 5 volt application, and the options that worked in the circuit include components rated at 10V, 15V and 20V, we should select the 20V rated component.

(C) to lower failure rates

Using components that have the strength to withstand applied stress with margin to handle variation operate without failure longer.

(B) limited sizes

The key here is the difference of the words “standard” which means common or known, and “standardization”, implies calibration, normalization or consistency. In context of this question the best response is (B) limited sizes. The benefit of such a program is few items to source, stock and monitor. This program improves reliability by reducing assembly errors and by having more time (few parts to evaluate) for reliability evaluations.

Limiting part numbers or vendors may also enhance reliability performance in a similar fashion and builds on the ability of the team to use components with a known reliability performance track record.