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Chapter 4 Supplement

Chapter 4 Supplement. Reliability. Reliability. Reliability : The ability of a product, part, or system to perform its intended function under a prescribed set of conditions Failure : Situation in which a product, part, or system does not perform as intended MTTF: Mean time to failure

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Chapter 4 Supplement

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  1. Chapter 4 Supplement Reliability

  2. Reliability Reliability: The ability of a product, part, or system to perform its intended function under a prescribed set of conditions Failure: Situation in which a product, part, or system does not perform as intended MTTF: Mean time to failure MTBF: Mean time between failures MTBF: Mean time before failure

  3. Probability that the product or system will function as expected Instantaneous reliability Function when activated Continuous reliability Function for a given length of time Independent events Reliability is a Probability

  4. Reliability: Rule 1 Things in series (e.g., All must work for the system to work) as follows: If two or more events are independent and “success” is defined as the occurrence of all of the events, then the reliability (probability of success) rt is equal to the product of the reliability probabilities of the events (components), i.e., rt = r1 × r2 ×…rn OR In this series system: multiply individual probabilities together to get the overall reliability probability rt= r1 x r2 =.9 x .8 = .72

  5. Reliability (series) Determine the reliability of the system shown .95 .90 .88 Ma Bell Friend’s Phone My Phone rtele= rmy x rBell x rfriend =.95 x .90 x .88 = .75 All must work to place a call!

  6. Sample Question • A semiconductor has three components in series. Component 1 has a reliability of 0.98; component 2, 0.95; component 3; 0.99. What is the reliability of the system? rt= r1 x r2 x r3 =.98 x .95 x .99 = .92

  7. Reliability: Rule Two – Objects in Parallel Reliability of Objects in Parallel (Redundant, Backups) is calculated by either: Rt=reliability1 + (failure1)(reliability2) … Rt=.90 +(1-.90)(.88) = .988 OR .90 Rt=1 – (failure1 x failure2) … Rt=1 - (1-.90) x (1-.88) = .988 .88 My Phones One of the two phones must work for me to make a call!

  8. .90 .92 .95 .90 .88 Example S-1: Reliability Determine the reliability of the system shown Friend’s phones Ma Bell My Phones

  9. .90 .92 .95 .90 .88 How Reliable is My Call? The system can be reduced to a series of three components Rfriend =.95+(1-.95)(.92)=.996 Rbell =.90 Rme = .88 + (1-.88)(.90) = .988 Friend’s phones My Phones Ma Bell OR .996 .90 .988 Rfriend =1-(1-.95)(1-.92) =.996 Rbell =.90 Rme = 1-(1-.88)(1-.90) = .988 Friend’s phones My Phones Ma Bell Rtotal = Rfriend Rbell Rme =.996 x .90 x .988 =.866 .866 Rtotal

  10. How would you attack this? 2 4 5 1 3 6 5 7 8 5

  11. Failure Rate

  12. Exponential Distribution Reliability = e -T/MTTF p(it died before T) =1- e -T/MTTF T Time Represents the cumulative reliability where failures are random (not infant mortality or wear-out periods) mean time to failure (MTTF) The average length of time before failure of a product or component. If a computer hard drive has a MTTF of 300,000 hours (34 years). What is the probability it will not fail before 3 years?

  13. Example: Class Computer So far this term I have had in the classroom the computer fail about once every 2 days. Assume an exponential distribution Assuming the computer is working at the beginning of the day, what is the probability it will not crash today? P(no failure) = e –T/MTBF = e-1/2 = .6065

  14. An electronic chess game has a useful life that is exponentially distributed with a mean of 30 months. Determine the following: • a) The probability that any given unit will operate for at least (1) 39 months, (2) 48 months, (3) 60 months. • b) The probability that any given unit will fail sooner than (1) 33 months, (2) 15 months, (3) 6 months. • c) The length of service time after which the percentage of failed units will approximately equal: (1) 85 %, (2) 50 %, (3) 95 %, (4) 99%.

  15. Answer to chess game question

  16. Normal Distribution – Wear out Remember this? Things that wear out like breaks can be modelled using the normal distribution

  17. # 17 A major television manufacturer has determined that its 42-inch colour TVs have a mean service life that can be modeled by a normal distribution with a mean of six years and a standard deviation of one-half years. • A. What probability can you assign to services lives of at least: 1) Five years? (2) Six years? (3) Seven and one-half years? • B. If the manufacturer offers service contracts of four years on these TVs, what percentage can be expected to fail from wear-out during the service period? • C. If the cost of replacement is $500 each what should they charge for a 5 year warranty?

  18. Answer to TV question

  19. Availability • The fraction of time a piece of equipment is expected to be available for operation MTBF = mean time between failures MTTR = mean time to repair On Average: If you computer runs for 4 hours before it crashes and it takes you 30 min to reboot to get back to where you were before the crash, what is the machine availability?

  20. ImprovingReliability • Component design • Production/assembly techniques • Testing α, β, RC • Redundancy/backups • Preventive maintenance procedures • User education • System design: modular

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