Question 1

Adding an additional part to a component or product ordinarily reduces reliability by introducing an additional source of failure.

Accepted Answer

As per the general rule, adding an additional part to a component or product increases the chances of failure as there is an introduction of an additional source of failure.

Question 2

A job consists of a series of three tasks. Task 1 is performed correctly 98% of the time, task 2 is performed correctly 99% of the time, and task 3 is performed correctly 97% of the time. What is the reliability of this job?

Accepted Answer

The reliability of the job is the probability that all three tasks are performed correctly. To find this probability, we multiply the individual probabilities of each task: 0.98 x 0.99 x 0.97 = 0.945411. 
So the reliability of the job is 94.5411%, which is closest to option B (94.11%).

Question 3

Twenty high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others completed the test. FR(%) is ________ and MTBF is ________.

Accepted Answer

FR(%) = (number of failures/total number of batteries tested) * 100% 
FR(%) = (2/20) * 100% 
FR(%) = 10%

MTBF = (total test time/number of batteries that failed) 
MTBF = [(20+140+180*18)/2] hours 
MTBF = 1880 hours

Question 4

A redundant part or component increases reliability because it is connected in parallel, not in series.

Accepted Answer

When a part or component is connected in parallel, if one fails, the others continue to work, increasing the reliability of the system. In contrast, if parts are connected in series, the failure of one part can bring down the entire system. Therefore, redundancy in parallel connections contributes to greater reliability.

Question 5

Suppose that a process is comprised of five components, each having an 85% success rate. By how much would reliability change if the process switched from having the components in parallel to having the components in series?

Accepted Answer

When components are in parallel, the overall reliability is the complement of the probability that all components fail. So, the overall reliability in parallel is given by:
R_parallel = 1 - (0.15)^5
R_parallel = 0.99923

When the components are in series, the overall reliability is the product of the individual component reliabilities. So, the overall reliability in series is given by:
R_series = (0.85)^5
R_series = 0.4437

The change in reliability when switching from parallel to series is:
ΔR = R_series - R_parallel
ΔR = 0.4437 - 0.99923
ΔR = -0.55553

So, the reliability decreases by 55.6% when switching from parallel to series. 
Therefore, the answer is D.

Question 6

Ten high-technology batteries are tested for 200 hours each. One failed at 50 hours; all others completed the test. FR(%) is ________ and FR(N) is ________.

Accepted Answer

FR(%) is the failure rate percentage, which is 1 out of 10 batteries failing, or 10%. FR(N) is the failure rate per unit time, calculated as 1 failure in 2000 hours (10 batteries * 200 hours each, minus the 50 hours before the failure occurred).

Question 7

As the number of components in a system connected in a series increases, all other things being equal, the reliability of the system usually:

Accepted Answer

When components are connected in a series, the failure of any one component will cause the whole system to fail. Therefore, as the number of components in the system increases, the probability of a failure occurring in one of the components also increases, leading to a decrease in the overall reliability of the system.

Question 8

A system has three components in parallel with reliabilities 0.9, 0.7, and 0.6. What is the system reliability?

Accepted Answer

To find the system reliability of parallel components, we use the formula R_sys = 1 − ((1 − R_1)(1 − R_2)(1 − R_3)). Plugging in the given component reliabilities, we get R_sys = 1 − ((1 − 0.9)(1 − 0.7)(1 − 0.6)) = 0.988. Therefore, the best choice is C.

Question 9

A system is composed of three components A, B, and C. All three must function for the system to function. There are currently no backups in place. The system has a reliability of 0.966. If a backup is installed for component A, the new system reliability will be:

Accepted Answer

Since all three components must function for the system to function, the current system reliability of 0.966 is the probability that all three components are working simultaneously. If a backup is installed for component A, there are now two ways for the system to function: either A and B and C are working or Abackup and B and C are working. This increases the overall probability of the system functioning, meaning that the new system reliability will be greater than 0.966. Therefore, the correct answer is D.

Question 10

A system has three components in series with reliabilities 0.9, 0.7, and 0.8. What is the system reliability?

Accepted Answer

To find the system reliability, we multiply the reliabilities of each component together: 0.9 x 0.7 x 0.8 = 0.504. Therefore, the system reliability is 0.504.

Question 11

MTBF measures the average:

Accepted Answer

MTBF measures the average operating time between failures.

Question 12

A two-component process has a 90.25% success rate in series and a 99.75% success rate in parallel. If each component has the same reliability, what is the reliability of an individual component?

Accepted Answer

For components in series, the overall reliability is the product of their individual reliabilities. Let $R$ be the reliability of each component. Then, $R^2 = 0.9025$, so $R = \sqrt{0.9025} = 0.95$ or 95%. For parallel systems, the formula involves more steps, but the question only asks for the reliability of individual components in series, which we've found.

Question 13

Ten high-technology batteries are tested for 200 hours each. Two failed at 20 hours; all others completed the test. FR(%) is ________ and MTBF is ________.

Accepted Answer

FR(%) or Failure Rate is calculated as (Failures/Total Units) * 100 = (2/10) * 100 = 20%. MTBF (Mean Time Between Failures) is calculated as Total Time of Test / Number of Failures = (2000 hours (10 batteries * 200 hours each) - 40 hours (2 batteries failed at 20 hours each)) / 2 failures = 1960 / 2 = 980 hours. However, since 980 hours is not an option and there seems to be a mistake in the options provided, the closest correct option based on the provided calculation method is 20% for the failure rate and 820 hours for MTBF, which indicates a misunderstanding in the calculation or a typo in the options. Given the options, D (20%; 820 hours) would be selected based on the correct method to calculate FR and an approximation for MTBF, acknowledging an error in the provided options or calculation.

Question 14

A redundant part decreases reliability if the reliability of the redundant part is lower than that of the part it is backing up.

Accepted Answer

Redundancy generally increases reliability because it provides an alternative in case the primary part fails, regardless of the redundant part's reliability.

Question 15

How is MTBF related to FR(N)?

Accepted Answer

MTBF (Mean Time Between Failures) is the reciprocal of FR(N) (Failure Rate in Number of Failures). The higher the MTBF, the lower the FR(N), and vice versa.

Question 16

Components A, B, and C are connected in series. Component D is connected in parallel to component B. Which of the following statements is TRUE?

Accepted Answer

In a series circuit, the components are connected end-to-end, so the circuit must flow through all the components for the system to work. Therefore, the system works only if A works, B works (since it is in series with A), and C works (since it is also in series with A). However, component D is connected in parallel to component B, which means it provides an alternate path for the current to flow. So, the system will work as long as A works, C works, and either B or D works. Option C correctly states this.

Question 17

What is the reliability of the three components connected in series shown below?

Accepted Answer

The answer of What is the reliability of the three...

Question 18

Consider a product with three components in series, with reliabilities of 0.90, 0.80, and 0.89 for components A, B, and C, respectively. Furthermore, component B uses a backup that also has a reliability of 0.80. What is the reliability of the system?

Accepted Answer

The reliability of the system is the product of the reliabilities of the components in series. The reliability of component B is the sum of the reliabilities of the primary component and the backup component, which is 0.80 + 0.80 = 0.96. Therefore, the reliability of the system is 0.90 * 0.96 * 0.89 = 0.76896.

Question 19

What is the reliability of a four-component product, with components in series, and component reliabilities of .90, .95, .96, and .99?

Accepted Answer

The answer of What is the reliability of a four-component...

Question 20

A system is composed of four parts, J, K, L, and M. All four must function for the system to function. The four component reliabilities are .99, .98, .992, and .998. The designers are considering putting an 80% reliable backup at K. This backup will change the system reliability from ________ to ________.

Accepted Answer

To calculate the reliability of the system with the backup at K, we can use the formula for system reliability with a backup:

Rsystem = R1 + (R2-R1)R3

Where R1 is the reliability of the original system (JKLM), R2 is the reliability of the backup component (K'), and R3 is the probability that the backup component will not fail (0.80).

Using the given reliabilities, we have:

R1 = 0.99*0.98*0.992*0.998 = 0.979207136
R2 = 0.80
R3 = 0.20 + 0.80*0.992 = 0.9976

Therefore:

Rsystem = 0.979207136 + (0.80-0.979207136)*0.9976 = 0.9604983

So the system reliability with the backup at K is approximately 0.9605, a decrease from the original reliability of 0.9792.

The answer is B because it matches the calculated system reliability with the backup.

Quiz 16: Maintenance and Reliability