CRE Sample Exam #1

(D) I, II, and III

Consider all the elements that lead to a financial gain or loss. The available answers include failure rates and the operating life of each system, which enables the calculation of the expected number of failures, and when this is multiplied by the cost of a failure, this provides the cost of failures.

Another financial element is the cost of maintaining the system, including preventative and corrective maintenance costs, which impact the overall financial performance of a system. Maintenance costs tend to focus on avoiding system failures or returning the system to service after a failure.

(A) early failure

The key wording is “most likely,” as all types of failure may occur. The best answer in this case corresponds to the type of failures that will most likely occur if the process of finding faults or errors is not done well (i.e., insufficient hardware debugging).

Since debugging or any product testing typically finds faults or errors that are quickly revealed as failure, the most likely issues left undiscovered will also occur soon after the hardware is placed into service, i.e. early failure.

It is possible to conduct insufficient hardware debugging that finds all types of failure except wear-out failure mechanisms. The question does not include information about the details of debugging conducted, just that it is insufficient. Don’t over think the question or potential answers.

Random or catastrophic failures may be detected in hardware testing; also, defects undiscovered may lead to these types of failures. The likelihood of random or catastrophic failures is also more difficult to discover via product testing than the failures that have a decreasing hazard rate (early failures).

(B) wear-out failure

The key words here are cycle fatigue, or even just fatigue. One of the characteristics of a wear-out failure mechanism is that it takes time for it to occur. The defect may build up, as with corrosion, or wear away, as with abrasion, or it may grow, as with crack propagation or tin whiskers. The chance of failure increases over time.

Cycle fatigue tends to alter the material properties involved, thus degrading the ability of the system to function. The accumulated damage with each cycle erodes the material’s ability to operate within design specifications.

The failures resulting from fatigue may occur at any time and thus may occur early or during mid-life, yet fatigue failure mechanisms tend to exhibit an increasing failure rate over time and are independent of when they occur so would be considered a wear-out failure.

Since fatigue failures tend to have an increasing failure rate, they would not be classified as having a constant failure rate (or constant failure).

(C) III only

Reliability assurance is not a common phrase in reliability engineering. So, consider the phrase quality assurance, which we may define as follows:

The planned and systematic actions that provide confidence a product or service will meet the given quality requirements.

If we substitute reliability for quality in the definition, response III has a very similar structure and meaning.

Response I is the definition of reliability alone and does not include the assurance element.

Response II may be related to the assurance concept, yet it does not closely match the quality assurance definition.

(A) Set numerical requirements for reliability and duration(s) in specific operating environments.

Each of the responses provides an element or approach to setting a reliability goal. The key words in this question are “best manner” and “set.” “Best manner” implies a rank ordering according to some general scope or capability. Here “set” is similar to “state” or “establish.”

(A) The correct response relies on three of the four elements of the definition of reliability. It is measureable and specific.

(B) Having management support is a definite plus, yet alone it does not provide a means to state a goal.

(C) Contract language may create a binding reliability goal, yet when “high reliability” is not specific.

(D) MTBF is a common measure of reliability yet alone it does not include the duration nor environment for the item to operate.

(C) I, III, and IV only

This question is from an old exam and body of knowledge. At one time assessing training needs and conducting training was an explicit element of the CRE Body of Knowledge.

In fact, many reliability professionals routinely conduct training assessments and perform training for their organizations. The assessment process is the gathering of information to determine either the need for training or the appropriate level needed to achieve the desired learning outcomes.

Costs analysis is not part of a needs assessment, yet it may be part of an assessment concerning the expected return on investment for the proposed training. Observations, interviews, and work sampling are examples of learning about the current situation and how the current work processes actually occur, and so are all candidates for a training needs assessment. 

(D) communication, cooperation, and technical knowledge.

Working with teams, communicating well, understanding business priorities, and possessing technical knowledge are elements of being an effective reliability engineer.

The key word in this question is “team.” Selections (A), (B), and (C) list common tools and tasks performed by a reliability engineer. The first three options do not address, specifically, the essential skill to employ when working with a team.

(C) $9,000

Lifecycle cost analysis hinges on the broad scope of the lifecycle. The lifecycle spans from concept through decommissioning and disposal. Thus cost of acquisition, transportation, installation, commissioning, training, operation, maintenance, repair, decommissioning, and disposal all contribute to the lifecycle cost of a piece of equipment.

What is not included is the cost of upgrades or the cost of adding new features to the equipment. In this question, all the costs listed except the $7,000 of upgrades count toward the lifecycle costs for the CNC machine.

(C) return on rate of expansion analysis 

An analysis of the rate of expansion plans for the production line may include marketing, demand, distribution, warehousing, capacity, and flexibility and may include one of the other three listed analysis approaches, yet it does not address the question of the new equipment being a good investment using only the given data.

Discounted Cash Flow (DCF) analysis is the most likely analysis to conduct in this case. DCF adjusts the future cash flow changes adjusted for the time value of money. Given the discount rate, this means that $100 today is worth 4% less a year from now.

Net Present Value (NPV) is the difference between the present value of cash inflows and the present value of cash outflows.

Internal Rate of Return (IRR) analysis is a metric useful for capital budgeting and is the discount rate. The analysis adjusts the IRR until the NPV of all cash flows equal zero. This allows a comparison of different investment options or projects, and the one with the highest IRR is then considered the best investment.

Return on Investment (ROI) analysis is a performance measure useful to understand the efficiency of an investment or to compare the efficiency of two or more investment options. ROI measures the amount of return relative to the investment’s cost. The ROI is the ratio of the benefit or return (calculated as the gain from the investment minus the cost of the investment) divided by the cost of the investment. ROI may or may not include the consideration of the time value of money by using NPV adjusted values.

(B) mission profile

The terms “operating” and “mission” in the question basically give this one away. The mission profile includes the activities, events, stresses, decisions, usage, capability, and environmental factors that fully describe what the system may experience over a use cycle as it accomplishes a specific goal. Think of a commercial airliner, with all the elements that make up a flight between cities, from fueling, boarding, take-off, to landing, disembarking, and cabin cleaning, as a mission profile associated with safely transporting passengers and cargo to the desired destination.

The operational profile is a subset of the mission profile, as it does not include the events associated with accomplishing the goal. Generally, operational profiles focus on the stresses and loads experienced during the mission or period of use.

The design and mission reliability relate to the probability of the system successfully completing the mission.

(B) 80%

System effectiveness is defined as follows:

A measure of the extent to which a system may be expected to achieve a set of specific mission requirements expressed as a function of availability, dependability, and capability.

McGraw-Hill Dictionary of Scientific & Technical Terms, 6E. S.v. “system effectiveness.” Retrieved April 19, 2016, from http://encyclopedia2.thefreedictionary.com/system+effectiveness.

This dictionary definition derives from the work of the Weapons Systems Effectiveness Industry Advisory Committee (WSEIAC) report in 1964.

The ARINC Research Corporation (1965 report) defines system effectiveness as a function of mission reliability, operational readiness, and design adequacy.

Figure 7.7, page 271, of Reliability Engineering Handbook by Dodson and Nolan, CRC Press 1999.

These aspects are similar to the common definition with three elements. All three terms in ARNINC’s definition are within the problem.

The other two values, availability and maintainability, are distractors from the common definition. Knowing only the common definition we could substitute operational readiness for availability, mission reliability for dependability, and design adequacy for capability, as the concepts are similar. Yet in this case the three terms from the ARNIC definition are within the problem:mission reliability, operational readiness, and design adequacy.

The tricky part is that there is a value for availability stated in the problem and it’s not the same as the value for operational readiness.

Thus the solution is 0.99 × 0.87 × 0.93 = 0.80.

(D) I, II, III, IV, and V

All five terms represent either ways to describe distributions (PDF, hazard rate, CDF, and conditional reliability) or the mean of a set of life data.

(B) redundant code programming

Replicating the same code would replicate any faults within the code. Instead, use redundancy with program diversity, where the elements in parallel are different sets of code, and employ a voting routine to determine the correct result if the alternative sets of code do not agree.

Structured programming constrains the programmers to use specific clear, well-defined design practices. Fault tolerance programming attempts to identify and mitigate, or if possible avoid, software failures. Modular programs breaks down complex software projects into smaller, easier to define portions of code, making debugging less complex.

Practical Reliability, 5^th ed., Section 10.5, pp. 268–270.  

(D) The reliability engineering function concerns failures over time.

There is quite a bit of overlap between the two functions. The element that is unique to reliability engineering is the consideration and working with failures over time. We work to determine what will fail and when.

(D) Reliability engineering is more concerned with product design and failure rate.

There is quite a bit of overlap between the two functions. Both quality and reliability engineers work to reduce variability and defect rates. Both work closely with design engineers to improve the quality or reliability of a product or system. The element that is unique to reliability engineering is the consideration and working with failures over time. We work to determine what will fail and when (in other words, the failure rate).

(C) I, II, and IV only

Product cost and the customer’s perception of the cost are primarily roles for the quality function, not the reliability function. Both departments use a range of statistical tools and one or the other department may provide oversight of tools germane to their department’s work. The other two functions generally fall to the reliability group.

(B) MTTF = 44; failure rate = 0.0227

The formula for MTTF is the total test time divided by the number of failures. In this case, add the time-to-failure times of the four units that failed and the total time for the one unit that did not fail. That is 12 + 22 + 30 + 37 + 75 = 176 hours of total test time. Then divide by the number of failures: 176 /4 = 44 hours MTTF. The inverse of MTTF is an estimate of the failure rate: 1/44 = 0.0227.

The question did not mention the units being replaced or quickly repaired after failures but only that the test time ends for each unit upon failure. Also, you are looking for the MTTF. So these aspects imply nonrepairable units. If you assumed that all units ran for 75 hours, there is an answer listed.

Also, if you divided by 5 instead of the number of failures, there are responses for that error too.

(A) exponential

The key here is the term “constant.” Although other distributions can model time between failures at a constant failure rate, this is the primary characteristic of the exponential distribution.

(D) 0.0000009791

Use the Poisson distribution as directed in the question and calculate the probability of exactly zero or one failure, then subtract the combined values from 1 to find the chance of two or more failures.

The expected number of failures over 7 hours given a failure rate of 2 per 10,000 hours is (2/10,000 ) 7 = 0.0014.

P(0,0.0014) = exp[−0.0014]×(0.0014)⁰ )/0! = 0.99860098.

P(1,0.0014) = exp[−0.0014]×(0.0014)¹/)1! = 0.00139902.

For two or more, we subtract the chance of zero and one from 1 (complement): 1 − 0.99860098 − 0.00139902 = 0.0000009791.

(B) 0.0.0513

Given a distribution PDF you can derive the reliability function with

$$ R(t)={{e}^{-\int_{-\infty }^{x}{h\left( \tau  \right)d\tau }}}$$

Then substitute the given hazard rate function and simplify to get

$$ R(t)={{e}^{-\int_{0}^{t}{0.66\left( \tau  \right)d\tau }}}={{e}^{-0.66\int_{0}^{t}{\left( \tau  \right)d\tau }}}$$

We can move the constant out of the integral and then solve the integral:

$$ R(t)={{e}^{-0.66\int_{0}^{t}{\left( \tau  \right)d\tau }}}={{e}^{-0.66\left( \frac{{{t}^{2}}}{2} \right)}}={{e}^{-0.33{{t}^{2}}}}$$

Finally, solving when t = 3 gives

$$ R(3)={{e}^{-0.33{{\left( 3 \right)}^{2}}}}=0.0513$$

The distractors are answers when the integral is done incorrectly a few different ways. 

(B) 2X

There are a few clues in the question that help narrow down the problem to the correct formula to examine. First, “randomly” implies the exponential distribution. The testing terminated when it reached “X failures,” suggesting a failure-terminated test design. It calls for the calculation of the “mean” “confidence limits.” Implying the use of the confidence interval for the mean, or MTBF, for a failure terminated test.

MTBF is the mean of the exponential distribution and the lower one-sided confidence interval is

$$ \frac{2T}{\chi _{\left( \alpha ,\text{ }2r \right)}^{2}}\le \theta $$

where 2r (or, in terms of this problem, 2X) is the number of degrees of freedom.

(B) 37%

The reliability function of the exponential distribution is

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

Setting t = θ we have

$$ R\left( \theta  \right)={{e}^{-\frac{t}{\theta }}}={{e}^{-1}}=0.3679$$ 

(C) 0.010

For a series model, use the following formula to calculate the system reliability (where 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 that 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

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

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

(C) 5,040

A factorial, commonly denoted with an “!” or in this problem would be 7!, is the product of an integer and all the integers below it to one. For example, 4! = 4 × 3 × 2 ×1 = 24.

Thus for this problem the answer is 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040.  

(A) coefficients

Knowing how to quickly construct Pascal’s triangle allows you to quickly determine the coefficients for a binomial expansion.

The binomial expansion is a mathematical theorem that specifies the expansion (multiplying out…) of any power (a + b)^m of a binomial (a + b) as a specific sum of products aⁱb^j. For example,

$$ {{\left( a+b \right)}^{2}}={{a}^{2}}{{b}^{0}}+2{{a}^{1}}{{b}^{1}}+{{a}^{0}}{{b}^{2}}$$

Recall that something raised to the 0 power is equal to 1. Also, the coefficients for the resulting three terms are 1, 2, and 1. There are m + 1 terms and m + 1 corresponding coefficients in the expanded form.

Pascal’s triangle is a triangular array with 1’s at the ends of each row (and 1 at the apex or first row, too) and with the other values in a row being the sum of the two numbers in the row above.

The two arrows point to the first 3 on the fourth row, which is the sum of 1 and 2 found on the row above.

The coefficients for the expansion of (a + b)² above are the values in the third row, 1, 2, and 1. For (a + b)³ the coefficients are 1, 3, 3,and 1 as found in the fourth row. The binomial expansion for (a + b)³ becomes

$$ {{\left( a+b \right)}^{3}}={{a}^{3}}+3ab+3ab+{{b}^{3}}$$

(C) rejecting a null hypothesis when it is true 

The null hypothesis is rejected if the p value is less than the significance or α level. The α level is the probability of rejecting the null hypothesis given that it is true (Type I error).

The null hypothesis is a statement about a belief. We may doubt that the null hypothesis is true, which might be why we are “testing” it. The alternative hypothesis might, in fact, be what we believe to be true. The test procedure is constructed so that the risk of rejecting the null hypothesis, when it is in fact true, is small. This risk, α is often referred to as the significance level of the test. By having a test with a small value of α, we feel that we have actually “proved” something when we reject the null hypothesis. (NIST Engineering Statistical Handbook)

(C) 63.2

“β is the shape parameter and η is the scale parameter, or characteristic life —it is the life at which 63.2% of the population will have failed.” (Practical Reliability Engineering, 5^th ed., p. 38)

(C) 80

The failure rate is constant, and assumed constant over the entire time of operation. Thus the chance to fail in any given hour of operation is 4/40 = 0.10. Given the assumption of a constant failure rate, then over 800 hours we would expect 10% of the hours to have a failure; thus there would be 80 failures. This is not a question about the probability of surviving over 800 hours but instead about how many failures will occur over 800 hours. We treat each hour as a separate chance of failing, as if we have 800 units running for one hour each.

(C) 0.3370

This one may trip you up if you assume only four tires are at risk at a time, assuming the spare is installed after the first failure. The question is worded such that there is a 0.4 chance of failure over the duration of the trip for all 5 tires on the specially equipped vehicle. In short, the spare has the same chance of failure whether or not it is in use.

Since there are only four working tires on the four-wheeled vehicle, if there are two failures we don’t make it (i.e., we get stranded). We can use the binomial distribution PDF function to solve this by calculating the probability of exactly 0 failure and 1 failure, then sum those probabilities to get the chance of a successful trip (not stranded). The binomial PDF is

$$ P\left( x,n,p \right)=\left( \begin{array}{l}n\\x\end{array} \right){{p}^{x}}{{\left( 1-p \right)}^{n-x}}$$

where x is the exact number of failures of the number, n, of tires; here n = 5, and p is the probability of failure, p = 0.4.

First let’s calculate the probability of having none of the five tires fail (x = 0):

$$ P\left( 0,5,0.4 \right)=\left( \begin{array}{l}5\\0\end{array} \right){{0.4}^{0}}{{\left( 1-0.4 \right)}^{5-0}}=0.0778$$

Next when there is one of the five tires failing (x = 1) we have

$$ P\left( 1,5,0.4 \right)=\left( \begin{array}{l}5\\1\end{array} \right){{0.4}^{1}}{{\left( 1-0.4 \right)}^{5-1}}=0.2592$$

The sum of the probabilities of zero and one failed tire is the probability of successfully completing the trip, 0.0778 + 0.2592 = 0.3370.

(A) 0.354

The key word here is “constant.” Thus we should use the exponential distribution. The exponential distribution reliability function is

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

and setting θ = 130 and t = 135, we find the reliability at 135 hours as

$$ R\left( 135 \right)={{e}^{-\frac{135}{130}}}=0.354$$

(C) 4,348 hours

For the exponential distribution (assuming a constant failure rate) the MTBF is the inverse of the failure rate. Thus

$$ \theta =\frac{1}{\lambda }=\frac{1}{0.00023}=4,348\text{ hours}$$

(C) to simulate operations when random variations are an essential consideration

The Monte Carlo method is a computational algorithm that relies on repeated random sampling to obtain numerical results. The results of thousands (or more) random samples are then analyzed to get probabilities of different outcomes.

(C) 0.3510

You could use the k-out-of-n redundancy formula. It’s possible to use that formula if you convert the probability of failure, 0.23, to a reliability. The formula reduces to the binomial though and adds an extra step. So, just use the binomial directly.

You need at least 7 units to operate of the 9. You can calculate the probability that none fail (x = 0) and the probability that exactly 1 unit will fail. Then add those together, which is the probability that 7 or more will operate for the day of production. Thus,

$$ P\left( 0,9,0.23 \right)=\left( \begin{array}{*{35}{l}}9 \\ 0 \\ \end{array} \right){{0.23}^{0}}{{\left( 1-0.23 \right)}^{9-0}}=0.0952$$

and, for exactly one unit failing,

$$ P\left( 1,9,0.23 \right)=\left( \begin{array}{*{35}{l}}9 \\1 \\\end{array} \right){{0.23}^{1}}{{\left( 1-0.23 \right)}^{9-1}}=0.2558$$

So, adding the two results to find the probability of at least 7 units operating for a day gives 0.0952 + 0.2558 = 0.3510. 

(D) t test

The key phrase here is “sample means.” Thus the t-test is the best of the options. While the t-test is primarily for use with small samples (less than 30) drawn from normal populations, it is fairly robust to non-normal populations for the comparison of means.

The exponential distribution is not used for the comparison of means using hypothesis testing. The normal or z-test is used to compare the population (not sample) means. The chi-square distribution is used to compare population variances or to compare the observed and expected frequencies of test outcome.  

(A) The median is used to measure the central tendency.

Mean, median, and mode are measures of central tendency.

Variance is a measure of variability. Skewness is the third moment and kurtosis is the fourth moment. Also, the mean is the first moment and variance is the second moment.

(D) Weibull

This a classification or terminology problem. Binomial, hypergeometric,and Poisson are useful with count data and thus are considered discrete distributions. The Weibull distribution is useful with time, length, cycles, or other continuous datasets and thus is classified as continuous.

(D) 1 − β

β is the probability of accepting the lot or batch when we shouldn’t. It is the chance of a false acceptance given the sample data when the population would not be acceptable. Therefore, the meaning of 1 − β is the probability of rejecting (properly so) an unacceptable lot of material. The sample indicates that the population is bad.

1 − β is also known as the power of the test, which means the ability to sample to detect an unacceptable population. β is also called the Type II error or consumer’s risk. In contrast, α is the Type I error or producer’s risk.

(B) The loss function is assumed to be a step function relative to the specification limits.

There are two tenets emphasized by Taguchi’s approach. 1. Reducing product variation reduces the loss to society. 2. A proper development strategy includes intentionally reducing variation.

The Taguchi design of experiments approach uses orthogonal arrays to optimize performance or minimize cost with equivalent performance. The concept of the loss function is the squared deviation of the objective characteristic from its target value; it is a continuous parabolic function and does not include the tolerances or specification limits.

All of the other statements are true about the Taguchi DOE approach.

(B) The confidence interval normally encompasses the statistical tolerance limits of the population parameter.

Confidence intervals apply to the limits of the statistical parameter being estimated such as the mean or variance. Tolerance intervals provide a set of limits for the future individual values, not the mean or variance. Tolerance intervals do not apply to estimates of population parameters.

(A) deterministic process

Algorithms, models, procedures, processes, etc. having only one outcome for a given set of inputs are said to be deterministic because their outcome is predetermined.

In probability theory, a stochastic process, or often a random process, is a collection of random variables representing the evolution of some system of random values over time. This is the probabilistic counterpart to a deterministic process (or deterministic system).

A predictive link is not a common phrase and may refer to regression analysis or another stochastic relationship.

(C) 1.533

The key element here is that the calculation is for a sample, not a population. Thus, be sure your calculator is providing the sample standard deviation result, based on the formula

$$ s=\sqrt{\frac{\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\bar{X} \right)}^{2}}}}{n-1}}$$

and not the formula for the population standard deviation,

$$ s=\sqrt{\frac{\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\bar{X} \right)}^{2}}}}{n}}$$

The difference of dividing by n or n − 1 does make a difference for small samples. At about n = 30 there is little difference in the results.

(D) probabilistic

In statistical inference, we are dealing with variation and using a sample to estimate parameters for a population. We base our ability to estimate confidence intervals and hypothesis testing on the tenets of probability.

(A) 0.00

This is sort of a trick question. For a continuous distribution the chance of any one specific (exact) value existing approaches zero. We use small intervals or less than or greater than statements to say something meaningful about a continuous distribution.

For example, we may be interested in the percentage of males taller than 2 meters based on a sample of 100 people. The chance that someone is exactly, precisely, 2 meters tall is very, very small given there is an infinite set of values between 1.99 meters and 2.01 meters tall on a continuous scale. Of course, we are not able to measure to the absolute precision this implies, so within our measurement capability, which is a small range (say 1 mm wide), we do have a finite probability of someone being 2 meters tall, within 1 mm.

(D) Conduct a demonstration at the customer’s facility.

Reviews and predictions are common approaches to evaluate a design during the development process. They do not include the end customer’s use environment or fully reflect the (often unknown) customer expectations. The key phrase here is “most accurate,” which means that the various methods are all potential methods to verify the meeting of a requirement. In this case, the verification that includes the fewest assumptions or limitations on understanding or applying the customer’s requirements is the best or most accurate.

(C) 0.918

Although this problem is easy to solve, it takes time. Hold off solving it till the end after you have answered all the quick questions 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 let’s 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||R₄ to get

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

That sets up another simple parallel structure to reduce:

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

This leaves two elements in series:

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

(C) 0.993

The formula for a series system and a little algebra is all that is needed here. The formula to determine system reliability given a series system is

$$ {{R}_{sys}}={{R}_{1}}\times {{R}_{2}}\times {{R}_{3}}\times {{R}_{4}}$$

Since we’re given R_sys and three of the four other reliability values, we can solve for the missing value, say R₄:

$$ {{R}_{4}}=\frac{{{R}_{1}}\times {{R}_{2}}\times {{R}_{3}}}{{{R}_{sys}}}$$

Plug in the values for the three component reliabilities and divide by the system reliability to find the minimum reliability value of the last component:

$$ {{R}_{4}}=\frac{0.992\times 0.994\times 0.991}{0.97}=0.993$$

(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 fifth power:

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

C) reliability prediction

This could be defining the program plan, if limited to just these activities,or it could be the process to create a reliability prediction, which certainly describes a common approach for predictions.

The process described is not a demonstration as it did not describe any working units or test conditions. We also can rule out a design review, as the process does not include providing feedback on a specific design.

Calculate the product of the individual part reliabilities.

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

Although it is possible to sum failure rates in some circumstances (if 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.

(D) Determine whether product requirements can be met with subassemblies assumed at their worst combination of tolerances. 

The process to analyze a design by worst-case analysis is defined by setting the individual component values at not the maximum values rather at the values that create the least desirable situation. Worst-case tolerance analysis is a conservative method to evaluate the impact of tolerances on product performance.

Worst here does not imply failures or rejects; instead, looking at failures or rejects is part of a failure analysis process or potential improvement project.

(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 describes multiplying the unreliabilities, then subtracting the result from 1 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 (and assuming that t is the same for each value) for this problem we are able to quickly calculate the result:

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

(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, you 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 and 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:

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

Then, 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. Assuming an hour duration (t = 1) 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 that the reliability at an hour is 99.47% and not 99.23%.

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

(B) I and IV only

Musa and others have developed a range of software reliability models. Two are the basic execution time model and the logarithmic Poisson execution time model. There are variations of these models using other distributions, yet these two tend to be commonly used along with being easy to use.

Markov analysis is a method to analyze system reliability and availability when strong dependencies exist between elements of the system. These are rarely used specifically for software. Simulation modeling is a technique of emulating a system; although it is possible to implement with software, it generally wouldn’t be useful nor suitable for reliability modeling.

Musa, J. D, A. Iannino, and K. Okumoto. 1987. Software Reliability: Measurement, Prediction, Application. New York: McGraw-Hill.       

(C) the shape parameter

There are two or three statistics determined during a Weibull analysis: the shape parameter estimated by the shape statistic, β, the scale parameter estimated by the scale statistic, η, and the (less commonly used or useful) location parameter, representing a failure-free period, represented with γ.

MTBF is the mean of the distribution or dataset and is not the same as the scale parameter unless the shape parameter is equal to 1. Expected life may be MTBF or mean, yet not part of a Weibull analysis.

Product quality is important and may be the subject or reason for doing Weibull analysis, yet it is not directly a part of a Weibull analysis.

(A) 1 + 0 = 1

Recall your Boolean truth table for A or B (A + B):

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 1

Boolean values include only zero’s or one’s. Thus response (B) is common base-10 integer addition. The minus sign is a way to convert the “not” logic element, making 1 – 0 = 1 + 1, which is equal to 1, and 1 – 1 = 1 + 0, which is also equal to 1.

(D) 0.966

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

The best way 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 enables us to determine the system reliability given that E is working. Then, add the system reliability given that E is not working. Here’s the formula:

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

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

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

Start with reducing the two parallel elements:

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

Then reduce the other parallel element:

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

Then 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 find the system reliability given that E is not working starting by 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) 0.99997

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:

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

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:

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

(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) Is it continuous process starting in the concept and planning stage.

A reliability prediction is an estimate of the future reliability performance of the system in question. During each stage of the product lifecycle, there are estimates of reliability performance including the following:

engineering-based guesses,

simulations,

parts count,

parts stress count,

vendor data,

life-testing data,

physics of failure modeling, and

field data analysis.

(A) part stress analysis

Of the options provided, only the earliest possible methods, not requiring working prototypes, are the parts count and part stress analysis methods. Part stress analysis attempts to account for the stress experienced by each part, providing little improvement over the parts count method.

(D) I, II, III, and IV

Check the definitions (see glossary) for each type of censoring.

(A) Failure mode and effects analysis

Failure mode and effects analysis is a risk identification and prioritization tool and may be a useful tool as part of root cause analysis.

8D is Eight Disciplines (with 7D and 9D being variations), a structured problem-solving process developed and used widely in the automotive industry and beyond.

Management oversight and risk tree analysis is “a comprehensive analytical procedure that provides a disciplined method for determining the causes and contributing factors of major accidents.” [Mort User’s Manual, U.S. Department of Energy, SSDC-4, Revision 3, February 1992, http://www.osti.gov/scitech/servlets/purl/5254810/]

Human performance evaluation system is “a resource for U.S. Nuclear Regulatory Commission inspectors to use when reviewing licensee problem identification and resolution programs with regard to human performance.” [The Human Performance Evaluation Process: A Resource for Reviewing the Identification and Resolution of Human Performance Problems (NUREG/CR-6751), http://pbadupws.nrc.gov/docs/ML0209/ML020930054.pdf]

(C) V, II, IV, III, I

The phases are from the 8D or eight disciplines model, which is a structured problem-solving process. The disciplines are as follows:

D0: Plan: Plan for solving the problem and determine the prerequisites.

D1: Use a Team: Establish a team of people with product and process knowledge.

D2: Describe the Problem: Specify the problem by identifying in quantifiable terms: the who, what, where, when, why, how, and how many (5W2H) for the problem.

D3: Develop Interim Containment Plan: Define and implement containment actions to isolate the problem from any customer.

D4: Determine, Identify, and Verify Root Causes and Escape Points: Identify all applicable causes that could explain why the problem has occurred. Also identify why the problem was not noticed at the time it occurred. All causes shall be verified or proved. One can use five whys or Ishikawa diagrams to map causes against the effect or problem identified.

D5: Verify Permanent Corrections (PCs) for Problem: Using pre-production programs, quantitatively confirm that the selected correction will resolve the problem. (Verify that the correction will actually solve the problem.)

D6: Define and Implement Corrective Actions: Define and Implement the best corrective actions.

D7: Prevent System Problems: Modify the management systems, operation systems, practices, and procedures to prevent recurrence of this and all similar problems.

D8: Congratulate Your Team: Recognize the collective efforts of the team. The team needs to be formally thanked by the organization.

(C) IV, II, I, III, V

In part, this is common sense, as is a systematic failure reporting and corrective action system (FRACAS). The basic steps in a closed-loop FRACAS may include the following:

Identify a problem.

Control or contain the problem (also include safety steps).

Report the failure.

Log any associated data.

Conduct failure analysis.

Determine and implement short- and long-term corrective and preventative actions.

Determine the effectiveness of corrective actions and monitor preventative actions.

(C) field support data

Although all the options could provide reliability information, this is a ranking questions , so you need to select the “best.” The most accurate data that reflect the product use and reliability are those obtained directly from customers. Ideally, a product will have onboard tracking and reporting capability. In most cases, customers report issues or defects, especially if they have a warranty or service contract.

The other options are removed from the actual customer using the product, and although they may be done well they will not contain the user expectations and often cover an unknown full range of use variability.

(A) deterioration of manufacturing processes and procedures

The most likely cause is the huge error associated with reliability predictions. We can rule out the ‘initial estimation’ because, although it may be the resulting reliability prediction,that is rarely the case for comparison to field performance. Lack of training or accumulation of process variation may increase the difference, yet in this case, given this set of options, the most likely cause is option (A).

(C) long-term movement

Keep in mind that the purpose of corrective action is to systematically and conclusively solve problems and keep them from happening in the future. A long-term solution is the goal.

(B) Kepner-Tregoe

“Is”/”Is Not” is a method to clearly define a problem by describing the problem boundaries using what issues and elements are part of the problem and which are not.

(D) operational data

Data gathered about component failures, especially when they includedetailed failure analysis, are the data that truly reflect the use and conditions experienced by the product and components. Using actual (operational) data has the least assumptions about system use.

(D) I, II, and III

The question is asking you to judge each option on whether or not it is part of a FRACAS. They all are, but the list is not all encompassing. The basic steps in a closed-loop FRACAS may include the following:

Identify a problem.

Control or contain the problem (also include safety steps).

Report the failure.

Log any associated data.

Conduct failure analysis.

Determine and implement short- and long-term corrective and preventative actions.

Determine the effectiveness of corrective actions and monitor preventative actions. 

(B) engineering specifications

Note the key word “not.” The engineering specification may establish a reliability goal and criteria for reliability testing, yet it is not a source of reliability data.

(A) I only

Being familiar with MIL-STD-785 is essential here. The two other options can be logically ruled out. FRACAS does require competent staff, yet it does not require a computer (although in some projects a database application is certainly helpful). Also, the word “always” limits the correctness of the last option.

(A) 1

A failure is “gold”: It is the product telling you that there is an opportunity to prevent future problems. The key word in this question is “should,” as in practice for a complex system, given the volume of failures considered, not every issue will receive attention or corrective actions. The aim is to do so, nonetheless.

(B) customer satisfaction

Customer satisfaction reflects the customers’ acceptance of a product. Does the product meet their often-unstated needs and expectations. Tracking customer complaints is a reasonable method to assess satisfaction. Other methods include surveys, comment forms, polls, and focus groups. None are perfect, yet complaints provide a measure that reflects implemented product changes and/or customer demographic or expectation changes.

(C) how to provide all key people with useful information

Keep in mind the primary purpose of the collected data. It is to assist members of your organization in making decisions. Thus a key aspect of the physical storage system is accessibility by those needing access to the data.

This is an older question and implies that PC or desktop machines are less available across a network. Although a mainframe (or network server) is often more available, that is no longer always true. Another option is to use “cloud” storage services to provide data access, making the server or desktop question moot.

Cost and security are considerations, just not the primary aspects to consider.

(B) identifying, investigating, and analyzing failures

Of the options, (B) is best. (A) if done to assign ‘blame’ should not be part of a FRACAS. (D) is a good thing to do, yet not necessary as part of a FRACAS. The primary role of a FRACAS is to effectively implement correction and preventative actions that reduce failures rates. (B) has essential elements for an effective FRACAS.

(C) a mouse

Mice, trackpads, or styluses are pointing devices and used as input, like a keyboard, to a computer. [This question was derived from a previously published CRE exam. ASQ used to publish the exams after use. It is not clear why it was part of the exam or which part of the body of knowledge it evaluates.]

(B) limited sizes

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

(C) part 3

This takes some careful reading and knowledge to understand that, when the strength curve is higher (stronger) than the stress curve for most scales of stress/strength application, then the part is more reliable. Parts 2 and 4 have the mean stress higher than the mean strength, suggesting a lower reliability than with part 1, which still isn’t a great situation. Part 3 has a higher mean strength than stress, which will result in a higher reliability than that of the other parts.

(A) Design control techniques are applicable generally throughout industry.

Even when not documented or not a formal process, the product generation process generally starts with an idea, then proceeds into design and development, assembly or manufacturing, then distribution. The product lifecycle and specifically the control of the design process generally uses a phase gate approach with reviews at the end of each phase assessing readiness to proceed to the next phase.

Your organization may have different names for the process or phases of the process, yet the design control techniques apply across industries and types of products.

(C) I, II, and III only

Criticality is a relative measure of failure mode consequence and its frequency. Thus all the options apply except efficiency. If “efficiencies” here apply to how well the failure manifests or to the manufacturing process, neither concept applies to the concept of FMECA or specifically the criticality element.

(C) III and IV only

The Arrhenius model describes the chemical rate of reaction, and it has been used to empirically fit data on occasion independent of understanding the underlying chemical reaction. The primary use of the Arrhenius model is to relate time to failure to temperature and it is useful for “significant” thermal stresses, which is valid if the temperature is not beyond a phase transition for the material involved.

(D) A different FTA is needed for each defined top event.

The structure of an FTA starts with a unique top event, whereas an FMEA/FMECA starts by listing potential failure modes within one study.

using a 100% rating

The derating concept entails using components at less than 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 that 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) 0.1025

This is a normal-based stress-strength calculation, assuming both distributions are normal and independent. Use the following formula to determine the z value representing the probability of failure:

$$ z=\frac{{{\mu }_{x}}-{{\mu }_{y}}}{\sqrt{\sigma _{x}^{2}+\sigma _{y}^{2}}}$$

where the x subscripts are for the strength parameters and y is for the stress parameters. Note that the equation uses variance, not the standard deviation. Inserting the values and calculating we find

$$ z=\frac{2200-1750}{\sqrt{190_{{}}^{2}+300_{{}}^{2}}}=0.1025$$

Then use the z-table to determine the probability of failure. In this case you should find that the probability value in the body of the table is between z values of 1.26 and 1.27, with probabilities of 0.1038 and 0.1020, respectively. Use interpolation to determine the probability for a z value of 1.27.

Interpolation uses the idea of equivalent ratios:

$$ \frac{{{z}_{between}}-{{z}_{lower}}}{{{z}_{higher}}-{{z}_{lower}}}=\frac{{{P}_{between}}-{{P}_{lower}}}{{{P}_{higher}}-{{P}_{lower}}}$$

In this case, insert all the known values and solve for the unknown probability corresponding to a z value of 1.27:

$$ \begin{array}{l}\frac{1.267-1.26}{1.27-1.26}=\frac{{{P}_{between}}-0.1038}{0.1020-0.1038}\\{{P}_{between}}=\left( \frac{1.267-1.26}{1.27-1.26} \right)\left( 0.1020-0.1038 \right)+0.1038\\{{P}_{between}}=0.1025\end{array}$$

(B) to determine functional capability under specified environmental conditions

Testing is done for many reasons, and from the question there is little information on the nature or purpose of the testing. Yet one thing a vendor has to provide is component and part specifications. These are often based on component testing done by the vendor prior to listing performance and environmental conditions on a datasheet.

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

The key word here is “minimize”: Knowing stress and strength values or that they are fixed or that the values meet a safety margin does 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.

A) knowledge of the usage environment

There are four elements to a reliability specification: the function, the environment, which includes usage, the probability of successful operation, and a duration. MTBF is only the inverse of a failure rate and alone provides little useful information.

(A) Define the system requirements.

According to The Basics of FMEA by Robin E. McDermott et. al. there are ten steps to FMEA:

1. Review the product.
2. Brainstorm potential failure modes.
3. List potential effects of each failure mode.
4. Assign a severity rating for each effect.
5. Assign an occurrence rating for each failure mode.
6. Assign a detection rating for each failure mode and/or effect.
7. Calculate the risk priority number for each effect.
8. Prioritize the failure modes for action.
9. Take action to eliminate or reduce the top priorities.
10. Calculate the resulting RPN as improvements occur.

Other references place the scoring after listing causes and detection elements related to each failure mode. All start with understanding the system or component or process under consideration.

(D) an improvement in reliability

Testing does not improve reliability; it only reflects the current state,which is useful for the other three options. One could argue that screening out bad units improves field reliability, yet the actual reliability does not really change; we just find and remove failures before they reach customer (or try to as it is a very ineffective process and expensive in most cases).

(A) 1,000 lb

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

(D) I, II, and III

This question 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 10, 15, and 20 volts, we should select the 20-volt-rated component.

(B) The system can withstand high component parameter drift.

Component parameter values may change over time, beyond the process related variability. A design the functions correctly even when using the extreme worst set of component parameters indicates a robust design with ample margin for the nominal set of components. In general, the components that have parameter drift may do so over wider ranges than a design unable to demonstrate functionality using this method.

(C) to lower failure rates

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

(C) the person most capable of making design decisions

Every organization will establish a policy (written or not) on the makeup of design review teams. The leader or chair of the review committee may change to suit the nature of the review, and in each case itshould be someone able to make decisions as needed to address issues raised and discussed during the review. This may be any of the named positions in the responses, yet it best serves the purpose of the review if the person is able to make decisions and allocate resources to implement action items.

(A) A machine should perform all the operations.

Machines are very good at repetitive, precise, and fast operations. Humans are not. If the operations are to be done by a person, a good practice is to rotate the people though the position regularly within a shift to prevent boredom and errors, as well as minimize the risk of repetitive-motion injuries.

(B) 3

Clear communication among all parties early in the product and process development process means less of a need to adjust an initial design. Getting early involvement of the various technical disciplines allows the design team to understanding the range of needs and constraints and allows the team to create a solution that meets the needs of manufacturing, purchasing, reliability, quality, service, sales, marketing, etc.

(B) 15 degrees

This is tough if you do not know the standard or have experience working with human factor considerations. You can quickly narrow down the options by assuming a requirement that the person not have to strain his or her eyes or turn his or her head. Without having the standard available, it is best to make an educated guess.

(C) II and III only

In quantification of a hazard, one does not consider detection as is done in FMEA. How often and the consequence of a failure to humans or environment are the primary factors in hazard analysis.

The failure mode is a symptom presented when a failure occurs. The failure mechanism is the underlying event or series of events that manifest as a failure mode.

(D) how manufacturing failures affect the product operation

The key word here is “process.” FMEA or FMECA are tools to analyze designs or processes from system to component level.

(A) 0.0026

This is a normal-based stress-strength calculation, assuming both distributions are normal and independent. Use the following formula to determine the z value representing the probability of failure:

$$ z=\frac{{{\mu }_{x}}-{{\mu }_{y}}}{\sqrt{\sigma _{x}^{2}+\sigma _{y}^{2}}}$$

where the x subscripts are for the strength parameters and y is for the stress parameters. Note that the equation uses variance, not the standard deviation. Inserting the values and calculating we find that z is

$$ z=\frac{7000-4500}{\sqrt{800_{{}}^{2}+400_{{}}^{2}}}=2.795 $$

Then use the z-table to determine the probability of failure. In this case, you should find that the probability value in the body of the table is between z values of 2.79 and 2.80, with probabilities of 0.0026 and 0.0026, respectively. There is no need to use interpolation to determine the value for a z value of 2.795, as it is the same for the bounding values, thus the resulting probability of failure is 0.0026.

(D) I, II, III, and IV

Human factors include any interaction of a human with the equipment and thus include installation, operation, and maintenance.

(A) the details of the system

FMEAs and FMECAs are versatile tools to determine risks and prioritize improvement efforts. A system study focuses on the system. A common practice is to use the results of the system study to identify subsystems, components, or processes for a detailed FMEA or FMECA study.

(B) Implement a design change.

Inspections and improving maintenance practices may improve availability performance to some degree by reducing the repair time, yet a design can reduce the number of failures and the repair time, thus constituting a more effective method to improve inherent availability.

(D) a representation of the failure mechanism

Understanding and describing the failure mechanism behavior over time with a distribution enable an accurate model to be constructed for decision making, test planning, and analysis. When detailed failure mechanism data are not available you may resort to empirical modeling, with the assumption that each failure in the analysis is from the same failure mechanism.

Ease of use and convenience are important considerations and should not overrule the use of the appropriate model to describe the failure mechanism time-to-failure behavior.

(B) Identify the problems causing the majority of downtime.

An initial step in any problem-solving task is to understand the available information and define the boundaries of the study. Starting with a Pareto of the problem may provide a starting point to narrow the focus of the study to the most common or serious problems. Then move to the step of understanding the causes of the problems.

(A) Availability means being in an operable and committed state at the beginning of a mission.

A definition of availability (from O’Connor and Kleyner, 2012) is

“the probability that an item will be available when required, or as the proportion of total time that the item is available for use.”

(B) modeling the acquisition of computer software systems

As with 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.

(C) monitor sensor data

Monitoring sensor data is a key feature of a prognostic health management approach to accurately forecast time to failure for specific failure mechanisms. All of the other options are elements of the reliability-centered maintenance approach.

(B) CCM >> CPM

The investment in preventative maintenance enables the avoidance of unplanned failures. If the cost of prevention is as much or more than the cost of a failure, then it is not a good investment. Ideally, we could cost effectively prevent failures from occurring.

(C) hybrid automobiles

Memory, satellites, and DVD players are either not cost effective to maintain or repair or they are unavailable to do so. The hybrid automobile is a complex, expensive system with many elements that require regular preventative maintenance, along with being suitable for performing cost effective corrective maintenance.

(A) initial design and development

The idea is to set up and create the right systems for your program that will enable effective and efficient execution of the maintenance program.

(B) I, II, and III only

The mean time to failure is a poor measure of reliability, not maintainability. All the others are useful for measuring a maintainability program.

(A) Maintainability is the probability of a system being restored to functional operation within a given period of time.

A complete definition is:

The measure of the ability of an item to be retained or restored to a specified condition when maintenance is performed by personnel having specified skill levels, using prescribed procedures and resources, at each prescribed level of maintenance and repair. 

(B) downtime for routine or scheduled tasks to retain an item in operable condition

All the times listed except the downtime for routine or schedule tasks are part of corrective maintenance time. Gathering parts and equipment along with travel time for preventative maintenance do require resources, yet they do not count against the downtime of the equipment, unlike when conducting corrective maintenance.

(C) System function should be preserved.

The focus of reliability-centered maintenance is on complex systems,and keeping the system in operating shape is a key tenet. Focusing on doing the maintenance task, which actually supports the increase of operating time versus maintenance time, relies on understanding the specific failure mechanisms and how they manifest over time.

(C) a rocket motor

By definition, MTBF is a metric for a repairable system and MTTF is a metric for a nonrepairable systems. MTTF measures the average time to the first (only) failure, whereas MTBF is a measure ofthe average time from one failure of a system to the next failure.

(D) Follow a preventive maintenance program with scheduled diagnostic checks.

Avoiding the occurrence or identifying early signals of impending failure allows the team to prepare and execute repairs with minimal unscheduled downtime.

(B) II only

The idea of preventive maintenance is to avoid failures by properly maintaining and repairing equipment.

Predictive maintenance uses monitoring or time-to-failure estimates to replace items before failure. Corrective maintenance is the approach that identifies and repairs existing failures.

(C) prognostic health management

Prognostic health management focuses on monitoring either directly or indirectly indicators of specific failure mechanisms that will eventually lead to failure.

(B) design readiness reviews 

Although readiness reviews are important as part of the design process, they do not specifically focus on system safety.

(B) A “negative” result means that, given the state of the technology, no peanuts can be detected.

Testing includes the capability of the measurement system including measurement error. A test that does not find evidence is only valid to the limits of the measurement capability.

(D) to accelerate material property test results

The testing of materials may provide information concerning product safety, yet it is not a human factors element and thus not necessary for an assessment of human factors for safety planning.

(A) performing a contingency analysis

The key word here is “contingency,” which implies that the failure has occurred and the analysis is for a backup or work-around process.

(D) I, III, and IV only

The product specifications are generally listed in a product requirements document. The product requirements document then may include or reference the mission profile, which describes the where and how of product operation.

(C) information to enable management to allocate blame

Although field data could help in assigning blame, relying on such data would be a misuse of the information and generally considered of little value to solve the problems that led to the field failures.

(A) XYZ’s president or CEO

The legal responsibility of an organization’s output is the person with the legal authority and responsibility to oversee, manage, and approve any output.

(A) Preliminary hazard analysis is best accomplished with actual customer complaint data of the product.

The key word is “preliminary,” which implies early or first analysis. Using customer complaint information implies that the product is produced and shipped to the customer before the analysis occurs. In general,the purpose of a hazard analysis is to avoid providing a product to customers that pose a potential hazard.

(C) qualification or acceptance

There are too many terms describing quality and reliability testing. Some refer to the purpose of the testing, e.g., qualification or acceptance. Other terms, e.g., accelerated or degradation, refer to the testing approach or mechanics involved with the testing. Degradation testing monitors a performance parameter (most often) of the product as it undergoes some form of stress or use. The parameter deteriorates toward a failure condition, thus allowing us to ascertain the time-to-failure results for either a qualification or acceptance purpose.

(A) It can reduce premature failures in use.

Thermal cycling is a common form of stress that a product experiences; if there are early-life failures susceptible to thermal cycling stress, they will fall out in the testing. This enables the team to conduct detailed failure analysis and improve the design or assembly process as needed to minimize the specific failures.

Testing at normal operating conditions for 6 months may be the entire expected lifetime of a product, yet more often (and we’re not given any indication of the expected lifetime) a product is expected to last longer than 6 months. The product will likely experience thermal cycling in normal use and, if it experiences one such cycle per day, it is possible to test more cycles per day in thermal chambers, providing an acceleration factor.

Without more details, there is little to use to form a clear picture of the benefits of the testing. (C) implies that this test will ensure customer quality, which is unlikely to be accomplished by one test using a single stress factor. (D) has the word “all,” which is a key,as no single test is able to reveal all failure mechanisms.

(B) is possibly an answer, yet it is addressing constant failure rates and accelerated cycling generally addresses design or assembly-type mistakes or longer term wear-out or fatigue failure mechanisms, which are generally not considered contributors to a constant failure rate.

(B) sudden-death testing

The description of the test approach fits the definition of sudden-death testing. See the paper by Motyka for a comparison of sudden death testing versus traditional censored life testing. [http://matwbn.icm.edu.pl/ksiazki/cc/cc36/cc36111.pdf ]

(A) 100 units to be tested with one failure reported

A quick way to solve this problem is with the Beta table based on the work of Locks. The Beta table entry is the number of samples, n, and the number of successes, r. So, for (A) with n = 200 and r = 199 (since there is one failure) and under the column with Y = 0.90 for the 90% lower confidence bound, we find a value of 0.981 or 98.1%.

Given that we’re looking for the minimum testing results (or fewest samples) let’s check (B). Here n = 100 and r = 99 with 90% confidence so we find 96.2%, which just barely meets our requirement and uses 100 fewer samples.

For (C) we find 95.6% and for (D) we find 91.5%, both of which are below the target.

(D) to eliminate infant mortality failures

Screening in not an efficient method to achieve product reliability, yet in certain circumstances it is necessary. If the risk of an unavoidable (thru design, quality, or control approaches) and unacceptable defect rate, the use of stress screening may work to identify the units with defects before shipping to customers. In general, these defects are not always caused by component latent defects or assembly errors and would result in early-life failures when placed into service.

(C) I, III and IV only

The test, analyze, and fix program relies on finding, understanding, and designing out or mitigating as many failures as possible. No failures are irrelevant.

If the unit under test was not set up properly, your customer will likely have setup-related failures.

If the test technician drops a unit, it is likely that your customer will drop the one it has.

If an engineer doesn’t understand an error message, users of the product will also fail to understand too.

If the device stops working briefly at 65°C in a thermal chamber, the underlying root cause may reflect a design flaw and the device will just take longer to fail at lower temperatures.

(A) because test to provide adequate reliability performance information under normal operating conditions would take too long

Accelerated life testing is done to cheat time. It shortens the time it takes to learn about the relationship between stress and time to failure. In some cases, a failure mechanism occurs with the same fundamental pattern at a higher applied stress. If we have or can determine a relationship between stress and time to failure we can effectively shorten the time to failure in a meaningful manner.

(D) I, II, III, and IV

All are valid reasons for reliability testing. In general, we conduct testing to learn something about the design of a product or system. We want to reveal problems or check progress toward a goal or verify that design changes are effective. Every reliability test should include a clear statement about the information the test results will provide. Ideally, it will also include who needs to make a decision based on the results.

(D) the mechanisms of the failures

The time to failure and test design information do not provide details concerning the failure mechanisms. Understanding the failure mechanism involved for each failure allows the comparison of the failures to use-condition potential failures.

If the testing results in two more failure mechanisms, this makes the analysis more complex, yet both mechanisms may occur at use conditions, just at different rates, or a failure mechanism may only occur at a specific or high stress level. For example, over small displacements the metal in a solder joint is elastic to some degree and thus is resilient to failure (although it may accumulate damage,leading to eventual failure.) In contrast, if the displacement is too large the solder will separate as a result of the high shear stress. If the normal use conditions do not include the application of sufficiently high displacement to shear solder joints, these specific failures are not likely to occur under use conditions. 

(A) product function vs. time

This is from the definition of reliability: the function in an environment with a probability of success over a duration. You can often think of reliability as quality over time.

(B) Quality Function Deployment

Quality Function Deployment (QFD) is a structured approach used to define customer needs and requirements as specific engineering specifications. The process also includes cascading the requirements to the assembly process, thus linking customer needs to design specifications to assembly and component requirements. 

(B) function

A failure modes and effects analysis does include a listing and discussion of the function of the item or system. It is not scored or ranked. The severity of a failure mode, the probability of a cause, and the ability to detect a cause before the failure occurs are all scored and comprise elements used to calculate the risk priority number (RPN).

(B) active and standby

Active redundancy means that 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.

(C) Inform her employer of her business connection to the book.

Here’s an excerpt from the ASQ Code of Ethics:

Article 5 – Act as faithful agents or trustees and avoid conflict of interest and the appearance of conflicts of interest.

It is the conflict or appearance of a conflict of interest that is the issue here. If the only reason to recommend her book is the profit that she would receive, this would be a conflict of interest. The employer may or not be comfortable with that particular arrangement, so informing the employer of the situation is appropriate. In this case, by not informing the other party it at a minimum appears that she is withholding information pertinent to the selection of a book for the training program.

One way to avoid the issue is to not make the recommendation. The book may well be the right one for the program and deserves consideration. By informing her employer, all parties have the same information concerning the business connection to the book.

(B) critical

Hazard severity categories, originally defined in MIL-STD-1629A Procedures for Performing a Failure Mode, Effects and Criticality Analysis (canceled in 1998) and currently defined in MIl-STD 882D, Standard Practice for System Safety, February 10, 2008, p. 18, are defined as follows:

Catastrophic — Could result in death, permanent total disability, loss exceeding $1,000,000, or irreversible severe environmental damage that violates law or regulation.

Critical — Could result in permanent partial disability, injuries or occupational illness that may result in hospitalization of at least three personnel, loss exceeding $200,000 but less than $1,000,000, or reversible environmental damage causing a violation of law or regulation.

Marginal — Could result in injury or occupational illness, resulting in one or more lost work days, loss exceeding $10,000 but less than $200,000, or mitigatible environmental damage without violation of law or regulation where restoration activities can be accomplished.

Negligible — Could result in injury or illness not resulting in a lost work day, loss exceeding $2,000 but less than $10,000, or minimal environmental damage not violating law or regulation.

Catastrophic uses the term severe, Critical uses reversible, Marginal uses mitigatible, and Negligible uses minimal. Major is less than severe and has more impact than minimal. It’s a judgment call whether major corresponds with Critical or Marginal.

(B) I and V only

Inherent availability is the ideal-state minimal repair time (active corrective maintenance time); it is sometimes referred to as the steady-state availability or potential availability.

Capability, dependability, and effectiveness are terms used in quality, reliability, and systems engineering not directly concerning a measure of availability.

Achieved availability includes preventative and corrective maintenance time, thus is a bit more realistic then the inherent availability.

Operation availability is what actually occurs. It includes all delays and all maintenance time necessary to restore a system to operation.

(D) II, III, and IV only

Risk assessment in general is working to identify hazards that include loss of life, accidents, or injuries; equipment, property, or environmental damage. Although financial riskmay be considered in a program, it not part of a safety risk assessment.

(A) promoting employee safety through education and training

Employee safety, although important, is not an aspect of product safety.

(A) discovery

By definition, discovery is

“the entire efforts of a party to a lawsuit and his/her/its attorneys to obtain information before trial through demands for production of documents, depositions of parties and potential witnesses, written interrogatories (questions and answers written under oath), written requests for admissions of fact, examination of the scene and the petitions and motions employed to enforce discovery rights. The theory of broad rights of discovery is that all parties will go to trial with as much knowledge as possible and that neither party should be able to keep secrets from the other (except for constitutional protection against self-incrimination). Often much of the fight between the two sides in a suit takes place during the discovery period.” [Directly from http://dictionary.law.com/default.aspx?selected=530 on May 31, 2016]

(B) I and III only

II

is not generally true given the additional equipment and design and programming needed to automatically collect and report data.