*****INSTANT DOWNLOAD***PDF****Solutions Manual to Reliability Engineering (2nd Edition) by Bradley

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SOLUTIONS

, RELIABILITYENGINEERING –ALIFECYCLE APPROACH v v v v v v




INSTRUCTOR’S MANUAL v




CHAPTER 1 v




The Monty Hall Problem
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The truth is that one increases one’s probability of winning by changing one’s choice. The easiest
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way to look at this from a probability point of view is to say that originally there is a probability of
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⅓ over every door. So there is a probability of ⅓ over the door originally chosen, and a combined
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probability of ⅔ over the remaining two doors. Once one of those two doors is opened, there
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remains a probability of ⅓ over the door originally chosen, and the other unopened door now has
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the probability ⅔. Hence it increases one’s probability of winning the car by changing one’s
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choice of door.
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This does not mean that the car is not behind the door originally chosen, only that if one were to
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repeat the exercise say 100 times, then the car would be behind the first door chosen about 33
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times and behind the alternative choice about 66 times. Prove for yourself using Excel!
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Another way to prove this result is to use Bayes Theorem, which the reader can source for himself
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on the internet.
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Assignment 1.2: Failure Free Operating Period v v v v v




The FFOP (Failure Free Operating Period) is the time for which the device will run without
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failure and therefore without the need for maintenance. It is the Gamma value for the
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distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
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calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
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probability of failure. It will be seen from the graph in the software with Beta = 2 that the
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distribution is of almost perfect normal shape and that the distribution does not begin at the
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origin. The gap is the 100 hours that the software calculates when asked.
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When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the three
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left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect it at about
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120 hours. This is the estimation of Gamma. In the days before software this was always the most
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unreliable estimate of a Weibull parameter and the most difficult to obtain graphically.
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Assignment 1.3 v




When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that the
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distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life. Some
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items have failed before being put into service. This can apply in practice to rubber components
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and paints, for example.
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,Assignment 1.4: The Choice between Two Designs of Spring v v v v v v v v




DESIGN A v DESIGN B v




Number Cycles to Failure v v Number Cycles to Failure v v




1 726044 1 529082

2 615432 2 729000

3 807863 3 650000

4 755000 4 445834

5 508000 5 343280

6 848953 6 959900

7 384558 7 730049

8 666600 8 973224

9 555201 9 258006

10 483337 10 730008



Using the WEIBULL-DR software for DESIGN A above we get
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β=4 v v




Correlation = 0.9943 v v




F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A) Hence
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R400k = 92%
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For DESIGN B we get from the WEIBULL-DR software (not shown here)
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Β=2 v v




Correlation = 0.9867 v v




F400k = 20% v v




Hence R400k = 80% v v v




Hence DESIGN A is better v v v v




From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value is
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126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years A
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conservative guarantee would therefore by 15 years.
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NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
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emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
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version of the software prints out)
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DESIGN A v




β=3 v v




γ = 101 828.6 say 100 000
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For F = 1% at 90% confidence, F = 176149 Dividing
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by 8000 we get 176149/8000 = 22 years
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, Fig 1.4.1 Set A
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A figure of 22 years or even 15 years for any guarantee is very long indeed. Company policy
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would have to be invoked – there are matters to consider in the determination of guarantees other
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than the test data provided. These matters could include corrosion, user abuse etc. Such factors
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are more likely to occur, the longer the operating period. Questions need to be asked such as is
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there an industry standard for such guarantees, what are competitors offering as guarantees, etc.
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A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ value is taken
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into account. The β value remains at 2 but the γ value is negative at over 50 000 cycles! This implies
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that there is a probability of failure before entering service. This data looks suspect and further tests
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should be done to confirm the reliability characteristics of DESIGN B.
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