Chapter 4s: Reliability

9) Using the same information from question 8, what is the length of service time such that the percentages of units failing before this length is equal to 65%?

.65 = 1 - e^-T/50 .35 = e^-T/50 ln(.35) = -T/50 T = -50 ln(.35) T = 52.4911 ≈ 53 months

10) A calculator is expected to last for 45 months. If this can be determined using an exponential distribution: a. What is the probability that a calculator fails before 42 months? b. What is the probability that a calculator fails after 48 months? c. What length of service will result in a failure rate of 12%?

10) A calculator is expected to last for 45 months. If this can be determined using an exponential distribution: a. What is the probability that a calculator fails before 42 months? 1 - e^-42/45 = .6068 b. What is the probability that a calculator fails after 48 months? e^-48/45 = .3442 c. What length of service will result in a failure rate of 12%? .12 = 1 - e^ -T/45 .88 = e^-T/45 ln(.88) = -T/45 T = -45 ln(.88) T = 5.7525 ≈ 6 months

10. A copy machine has a service life that follows a normal distribution with a mean of 52 months and a standard deviation of 5 months. What is the percentages of failures for the following: Before 48 months Between 42 and 47 months Between 48 and 60 months Before 62 months

Before 48 months z= (48-52)/5=-0.8 P(z < -0.8) = .2119 Between 42 and 47 months z= (42-52)/5=-2.0 ≈ .0228 z= (47-52)/5=-1.0 ≈ .1587 P(-2.0 < z < -1.0) = .1587 - .0228 = .1359 Between 48 and 60 months z= (48-52)/5=-0.8 ≈ .2119 z= (60-52)/5=1.6 ≈ .9452 P(-0.8 < z < 1.6) = .9452 - .2119 = .7333 Before 62 months z= (62-52)/5=2.0 P(z < 2.0) = .9772

2) If a system has many components and each component has high reliability, the overall system reliability is still low. Why is this?

If a product is composed of a large number of parts, it conceivably can have a low reliability because its reliability is a function of the products of the individual reliabilities. For example, if a product has 20 parts, each with a reliability of .99, and all must operate, the overall product reliability will only be about .9920 = .8179. Also, keep in mind that the failure rate increases as the number of components increase.

3) How does redundancy help to improve the system's reliability?

Redundancy refers to backup parts or systems built into a product (or service). Their purpose is to increase reliability by taking over in the event that a primary part or system fails

1) Define the term reliability.

Reliability is a measure of the ability of a product or service to perform its intended function under a prescribed set of conditions.

6) A product has four parts, all of which must function. If one of the parts has a reliability of .875 and the other 3 parts have a reliability of .85, what is the overall reliability?

Rule 1: .875 × .85 × .85 × .85 = .5374

4) Suppose we have the following system: >>>>>> (.85) >>>>>>>> (.83) >>>>>>> What is the reliability? a. Of the overall system? b. The system where each component has a backup with the same reliability, and a switch that is 100% reliable. c. The system where each component has a backup with the same reliability, and a switch that is 95% reliable.

What is the reliability? a. Of the overall system? >>>>>> (.85) >>>>>>>> (.83) >>>>>>> Rule 1: .85 × .83 = .7055 b. The system where each component has a backup with the same reliability, and a switch that is 100% reliable. ...(.85)............................. (.83) ......^ .....................................^ ....(1) .....................................(1) ...... ^.......................................^ >>>>>> (.85) >>>>>>>> (.83) >>>>>>> First column: Rule 2 with switch: .85 + (1-.85)(1)(.85) = .9775 Second column: Rule 2 with switch: .83 + (1-.83)(1)(.83) = .9711 Overall (Rule 1): .9775 × .9711 = .9493 c. The system where each component has a backup with the same reliability, and a switch that is 95% reliable. ...(.85)............................. (.83) ......^ .....................................^ ....(.95) .............................( .95) ...... ^.......................................^ >>>>>> (.85) >>>>>>>> (.83) >>>>>>> First column: Rule 2 with switch: .85 + (1-.85)(.95)(.85) = .971125 Second column: Rule 2 with switch: .83 + (1-.83)(.95)(.83) = .964045 Overall (Rule 1): .971125 × .964045 = .9362

5) A product is comprised of six parts that must function in order for the product to work. If the product must have an overall reliability of 92.5%, what is the minimum reliability needed for each component? X >>> X >>> X >>> X >>> X >>> X >>> X

X * X * X * X * X * X = .925 X^6 = .925 X = 6^√(.925) = .925(1/6) = .987090462 = .9871

8) A computer game follows an exponential distribution with a mean of 50 months. Determine the following: a. The probability that a unit will last for at least 60 months b. The probability that a unit will last for at least 55 months c. The probability that a unit will fail before 50 months d. The probability that a unit will fail before 72 months e. The probability that a unit will fail between 40 and 65 months

a. The probability that a unit will last for at least 60 months e^-60/50 = .3012 b. The probability that a unit will last for at least 55 months e^-55/50 = .3329 c. The probability that a unit will fail before 50 months 1-e^-50/50 = .6321 d. The probability that a unit will fail before 72 months 1 - e^-72/50 = .7631 e. The probability that a unit will fail between 40 and 65 months e^-40/50 = .4493 e^-65/50 = .2725 .4493 - .2725 = .1768

7) An assembly robot has four parts that must function in order for the robot to do its job. The four parts have the following reliabilities: .97, .93, .91, and .90. ((DRAW PICTURE WITH SQUARES IF HELPFUL)) a. What is the overall reliability? b. If only one backup part, with the same reliability as the part it is backing up, can be used, where should the backup part be placed to get the highest overall reliability? c. If a backup part with a reliability of .95 is used, where should it be placed to get the highest overall reliability?

a. What is the overall reliability? Rule 1: .97 × .93 × .91 × .90 = .7388 b. If only one backup part, with the same reliability as the part it is backing up, can be used, where should the backup part be placed to get the highest overall reliability? Backup with Component 1: Rule 2: .97 + [(1 - .97) × .97] = .9991 Rule 1: .9991 × .93 × .91 × .90 = .7610 Backup with Component 2: Rule 2: .93 + [(1 - .93) × .93] = .9951 Rule 1: .97 × .9951 × .91 × .90 = .7905 Backup with Component 3: Rule 2: .91 + [(1 - .91) × .91] = .9919 Rule 1: .97 × .93 × .9936 × .90 = .8053 Backup with Component 4: Rule 2: .90 + [(1 - .90) × .90] = .99 Rule 1: .97 × .93 × .91 × .99 = .8127 Conclusion: Add the Backup to Component 4—this leads to the highest reliability of .8127. c. If a backup part with a reliability of .95 is used, where should it be placed to get the highest overall reliability? Backup with Component 1: Rule 2: .97 + [(1 - .97) × .95] = .9985 Rule 1: .9985 x .93 x .91 x .90 = .7605 Backup with Component 2: Rule 2: .93 + [(1 - .93) × .95] = .9965 Rule 1: .97 × .9965 × .91 x .90 = .7916 Backup with Component 3: Rule 2: .91 + [(1 - .91) × .95] = .9955 Rule 1: .97 × .93 × .9955 × .90 = .8082 Backup with Component 4: Rule 2: .90 + [(1 - .90) × .95] = .995 Rule 1: .97 × .93 × .91 × .995 = .8168 Conclusion: Add the Backup to Component 4—this leads to the highest reliability: .8168.