All Chapters Covered
SOLUTIONS
, RELIABILITY ENGINEERING – A LIFE CYCLE APPROACH
INSTRUCTOR’S MANUAL
CHAPTER 1
The Monty Hall Problem
The truth is that one increases one’s probability of winning by changing one’s choice. The
easiest way to look at this from a probability point of view is to say that originally there is a
probability of ⅓ over every door. So there is a probability of ⅓ over the door originally chosen,
and a combined probability of ⅔ over the remaining two doors. Once one of those two doors
is opened, there remains a probability of ⅓ over the door originally chosen, and the other
unopened door now has the probability ⅔. Hence it increases one’s probability of winning the
car by changing one’s choice of door.
This does not mean that the car is not behind the door originally chosen, only that if one were
to repeat the exercise say 100 times, then the car would be behind the first door chosen about
33 times and behind the alternative choice about 66 times. Prove for yourself using Excel!
Another way to prove this result is to use Bayes Theorem, which the reader can source for
himself on the internet.
Assignment 1.2: Failure Free Operating Period
The FFOP (Failure Free Operating Period) is the time for which the device will run without
failure and therefore without the need for maintenance. It is the Gamma value for the
distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
probability of failure. It will be seen from the graph in the software with Beta = 2 that the
distribution is of almost perfect normal shape and that the distribution does not begin at the
origin. The gap is the 100 hours that the software calculates when asked.
When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the
three left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect
it at about 120 hours. This is the estimation of Gamma. In the days before software this was
always the most unreliable estimate of a Weibull parameter and the most difficult to obtain
graphically.
Assignment 1.3
When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that
the distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life.
Some items have failed before being put into service. This can apply in practice to rubber
components and paints, for example.
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,Assignment 1.4: The Choice between Two Designs of Spring
DESIGN A DESIGN B
Number Cycles to Failure Number Cycles to Failure
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
β=4
Correlation = 0.9943
F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A)
Hence R400k = 92%
For DESIGN B we get from the WEIBULL-DR software (not shown here)
Β=2
Correlation = 0.9867
F400k = 20%
Hence R400k = 80%
Hence DESIGN A is better
From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value
is 126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years
A conservative guarantee would therefore by 15 years.
NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
version of the software prints out)
DESIGN A
β=3
γ = 101 828.6 say 100 000
For F = 1% at 90% confidence, F = 176149
Dividing by 8000 we get 176149/8000 = 22 years
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, Fig f1.4.1 fSet fA
A ffigure fof f22 fyears for feven f15 fyears ffor fany fguarantee fis fvery flong findeed. fCompany
fpolicy fwould fhave fto fbe finvoked f– fthere fare fmatters fto fconsider fin fthe fdetermination fof
fguarantees fother fthan fthe ftest fdata fprovided. fThese fmatters fcould finclude fcorrosion, fuser
fabuse fetc. fSuch ffactors fare fmore flikely fto foccur, fthe flonger fthe foperating fperiod.
fQuestions fneed fto fbe fasked fsuch fas fis fthere fan findustry fstandard ffor fsuch fguarantees,
fwhat fare fcompetitors foffering fas fguarantees, fetc.
A ffurther fpoint fto fnote fis fthat fDESIGN fB fexhibits fvery fpeculiar fcharacteristics fif fthe fγ
fvalue fis ftaken finto faccount. fThe fβ fvalue fremains fat f2 fbut fthe fγ fvalue fis fnegative fat fover
f50 f000 fcycles! fThis fimplies fthat fthere fis fa fprobability fof ffailure fbefore fentering fservice.
fThis fdata flooks fsuspect fand ffurther ftests fshould fbe fdone fto fconfirm fthe freliability
fcharacteristics fof fDESIGN fB.
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SOLUTIONS
, RELIABILITY ENGINEERING – A LIFE CYCLE APPROACH
INSTRUCTOR’S MANUAL
CHAPTER 1
The Monty Hall Problem
The truth is that one increases one’s probability of winning by changing one’s choice. The
easiest way to look at this from a probability point of view is to say that originally there is a
probability of ⅓ over every door. So there is a probability of ⅓ over the door originally chosen,
and a combined probability of ⅔ over the remaining two doors. Once one of those two doors
is opened, there remains a probability of ⅓ over the door originally chosen, and the other
unopened door now has the probability ⅔. Hence it increases one’s probability of winning the
car by changing one’s choice of door.
This does not mean that the car is not behind the door originally chosen, only that if one were
to repeat the exercise say 100 times, then the car would be behind the first door chosen about
33 times and behind the alternative choice about 66 times. Prove for yourself using Excel!
Another way to prove this result is to use Bayes Theorem, which the reader can source for
himself on the internet.
Assignment 1.2: Failure Free Operating Period
The FFOP (Failure Free Operating Period) is the time for which the device will run without
failure and therefore without the need for maintenance. It is the Gamma value for the
distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
probability of failure. It will be seen from the graph in the software with Beta = 2 that the
distribution is of almost perfect normal shape and that the distribution does not begin at the
origin. The gap is the 100 hours that the software calculates when asked.
When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the
three left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect
it at about 120 hours. This is the estimation of Gamma. In the days before software this was
always the most unreliable estimate of a Weibull parameter and the most difficult to obtain
graphically.
Assignment 1.3
When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that
the distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life.
Some items have failed before being put into service. This can apply in practice to rubber
components and paints, for example.
1
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isis
mmiciicsis
oolala
titoionn
,Assignment 1.4: The Choice between Two Designs of Spring
DESIGN A DESIGN B
Number Cycles to Failure Number Cycles to Failure
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
β=4
Correlation = 0.9943
F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A)
Hence R400k = 92%
For DESIGN B we get from the WEIBULL-DR software (not shown here)
Β=2
Correlation = 0.9867
F400k = 20%
Hence R400k = 80%
Hence DESIGN A is better
From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value
is 126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years
A conservative guarantee would therefore by 15 years.
NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
version of the software prints out)
DESIGN A
β=3
γ = 101 828.6 say 100 000
For F = 1% at 90% confidence, F = 176149
Dividing by 8000 we get 176149/8000 = 22 years
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, Fig f1.4.1 fSet fA
A ffigure fof f22 fyears for feven f15 fyears ffor fany fguarantee fis fvery flong findeed. fCompany
fpolicy fwould fhave fto fbe finvoked f– fthere fare fmatters fto fconsider fin fthe fdetermination fof
fguarantees fother fthan fthe ftest fdata fprovided. fThese fmatters fcould finclude fcorrosion, fuser
fabuse fetc. fSuch ffactors fare fmore flikely fto foccur, fthe flonger fthe foperating fperiod.
fQuestions fneed fto fbe fasked fsuch fas fis fthere fan findustry fstandard ffor fsuch fguarantees,
fwhat fare fcompetitors foffering fas fguarantees, fetc.
A ffurther fpoint fto fnote fis fthat fDESIGN fB fexhibits fvery fpeculiar fcharacteristics fif fthe fγ
fvalue fis ftaken finto faccount. fThe fβ fvalue fremains fat f2 fbut fthe fγ fvalue fis fnegative fat fover
f50 f000 fcycles! fThis fimplies fthat fthere fis fa fprobability fof ffailure fbefore fentering fservice.
fThis fdata flooks fsuspect fand ffurther ftests fshould fbe fdone fto fconfirm fthe freliability
fcharacteristics fof fDESIGN fB.
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