1. The Bathtub curve indicates failure probability, Which stage is NOT normally associated with the bathtub curve?_________
a. Normal-life where few failures occur
b. Wear-out where failure increases due to age
c. Infant-mortality where failures occur early
d. Pulling the plug where production is halted due to unacceptable level of failures
2. Failure occurs due to defective parts during the_________.
a. Wear out stage
b. Normal life stage
c. Early life stage
3. What refers to wear out failure_________.
a. Increasing failure rate
b. Decreasing failure rate
c. Depends upon type of the experiment
d. Depends upon the subject
4. What is /are the major purpose/s of using a bath tub curve?
a. To determine the capital maintenance in defense equipments
b. To compute lifts in the distillation column
c. To decide the maintenance of equipment
d. All of the above
5. What is the failure cost of a product possessing reliability R=1?
a. Zero
b. Unity
c. Infinity
d. None of the above
6. Which among the following exhibits inversely proportional relationship with the reliability?
a. Production cost
b. Design and development cost
c. Maintenance and repair cost
d. All of the above
7. Which among the below mentioned types of redundancy exhibits maximum failure rate?
a. Cold standby
b. Warm or Tepid
c. Hot or Active
d. All of the above
8. Which of the following is not a phase of bath tub curve of hardware reliability?
a. Useful life
b. wear out
c. Time
d. Burn-in
9. What is bath tub curve stands for?
a. maintenance schedule
b. failure rate
c. vibration chart
d. viscosity chart
10. Which one of the following relationships is incorrect?
ANS: f(t) = -dF(t)/ dt
11. Which of the following is not considered a reliability design method________.
a. Parts selection
b. Derating
c. Accessibility
d. Choice of technology
12. The probability density function of a Markov process is___________.
a. p(x1,x2,x3.......xn) = p(x1)p(x2/x1)p(x3/x2).......p(xn/xn-1)
b. p(x1,x2,x3.......xn) = p(x1)p(x1/x2)p(x2/x3).......p(xn-1/xn)
c. p(x1,x2,x3......xn) = p(x1)p(x2)p(x3).......p(xn)
d. p(x1,x2,x3......xn) = p(x1)p(x2 *x1)p(x3*x2)........p(xn*xn-1)
13. Markov analysis is a technique that deals with the probabilities of future occurrences by__________.
a. Using Bayes' theorem
b. Analyzing presently known probabilities
c. Time series forecasting
d. The maximal flow technique
14. How is reliability and failure intensity related to each other?
a. Direct relation
b. Inverse relation
c. No relation
d. None of the mentioned
15. What is MTTF ?
a. Maximum time to failure
b. Mean time to failure
c. Minimum time to failure
d. None of the mentioned
16. What is MTTR ?
a. Mean Time To Restore
b. Mean Time To Repair
c. Mean Time To Recovery
d. Mean Time to Restoration
17. Measure of reliability is given by ______ .
a. Mean Time between success
b. Mean reliable
c. Mean Time between failure (MTBF)
d. MTTR
18. Mean Time To Repair (MTTR) is the time needed to repair a failed equipment/component.
a. True
b. False
c. Can't say
d. None
19. Which one of the below is measured by MTBF?
a. Tolerance
b. Life time
c. Reliability
d. Quality
20. What is the area under a conditional Cumulative density function ?
a. 0
b. Infinity
c. 1
d. Changes with CDF
21. The components in the system below are exponentially distributed with the
indicated failure rates. Develop an expression for the reliability of the system. What is the
system reliability at time = 100 hours ?
Sol.) For the exponential distribution, the reliability R(t)
= e - λt
- The reliability of the system
․ R s (t) = ( e - 0.002t)( e - 0.002t)( e - 0.001t)( e - 0.003t) = e - 0.008t
- The hazard function
․ λ s = 0.002+0.002+0.001+0003 = 0.008
- At time = 100 hours, the reliability is
․ R s = e - 0.008(100) = 0.4493
SERIES System :
When components are in series and each component has a reliability RI If one component fails, the system fails.
1. The overall reliability of a series system shown above is :
2. R AB = R1 x R2 x R3
3. If R1 = R2 = R3 = 0.95
4. RAB = R1 x R2 x R3 = 0.95 x 0.95 x 0.95 = 0.86
5. R total is always < than R1 or R2 or R3
When components are in parallel and each component has a reliability Ri. If one component fails, the system does not fail.
1. RAB= 1 - probability (1 & 2 both fail)
2. The probability of 1 failing is = (1 - R1 )
3. The probability of 2 failing is = (1 - R 2 )
4. Overall reliability is R AB =1 - (1 - R 1 ) (1 - R 2)
5. If R1= 0.9 and R2 =0.8
6. R AB =1 - (1 - 0.9) (1 - 0.8) = 0.98
7. R Total ls always > than R1 or R2
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