All Chapters Covered
dg dg
SOLUTIONS
, RELIABILITY ENGINEERING – A LIFE CYCLE APPROACH dg dg dg dg dg dg
INSTRUCTOR’S d g MANUAL
CHAPTER 1 d g
The Monty Hall Problem
dg dg dg
The truth is that one increases one’s probability of winning by changing one’s choice. The
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
easiest way to look at this from a probability point of view is to say that originally there is a
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
probability of ⅓ over every door. So there is a probability of ⅓ over the door originally chosen, and
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
a combined probability of ⅔ over the remaining two doors. Once one of those two doors is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
opened, there remains a probability of ⅓ over the door originally chosen, and the other
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
unopened door now has the probability ⅔. Hence it increases one’s probability of winning the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
car by changing one’s choice of door.
dg dg dg dg dg dg dg
This does not mean that the car is not behind the door originally chosen, only that if one were to
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
repeat the exercise say 100 times, then the car would be behind the first door chosen about 33
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
times and behind the alternative choice about 66 times. Prove for yourself using Excel!
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Another way to prove this result is to use Bayes Theorem, which the reader can source for
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
himself on the internet.
dg dg dg dg
Assignment 1.2: Failure Free Operating Period d g d g d g d g d g
The FFOP (Failure Free Operating Period) is the time for which the device will run without
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
failure and therefore without the need for maintenance. It is the Gamma value for the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
probability of failure. It will be seen from the graph in the software with Beta = 2 that the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution is of almost perfect normal shape and that the distribution does not begin at the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
origin. The gap is the 100 hours that the software calculates when asked.
dg dg dg dg dg dg dg dg dg dg dg dg dg
When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the three
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect it at about
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
120 hours. This is the estimation of Gamma. In the days before software this was always the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
most unreliable estimate of a Weibull parameter and the most difficult to obtain graphically.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Assignment 1.3 d g
When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Some items have failed before being put into service. This can apply in practice to rubber
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
components and paints, for example.
dg dg dg dg dg
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,Assignment 1.4: The Choice between Two Designs of Spring dg dg dg dg dg dg dg dg
DESIGN A d g DESIGN B dg
Number Cycles to Failure dg dg Number Cycles to Failure dg dg
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
dg dg dg dg dg dg dg dg dg
β=4 dg dg
Correlation = 0.9943 dg dg
F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A) Hence
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
R400k = 92%
dg dg dg
For DESIGN B we get from the WEIBULL-DR software (not shown here)
dg dg dg dg dg dg dg dg dg dg dg
Β=2 dg dg
Correlation = 0.9867 dg dg
F400k = 20% dg dg
Hence R400k = 80% dg dg dg
Hence DESIGN A is better dg dg dg dg
From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years A
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg d g
conservative guarantee would therefore by 15 years.
dg dg dg dg dg dg dg
NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
version of the software prints out)
dg dg dg dg dg dg
DESIGN A dg
β=3 dg dg
γ = 101 828.6 say 100 000
dg dg dg dg dg dg
For F = 1% at 90% confidence, F = 176149
dg dg dg dg dg dg dg dg dg
Dividing by 8000 we get 176149/8000 = 22 years
dg dg dg dg dg dg dg dg dg
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, Fig 1.4.1 Set A dg dg dg
A figure of 22 years or even 15 years for any guarantee is very long indeed. Company policy
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
would have to be invoked – there are matters to consider in the determination of guarantees
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
other than the test data provided. These matters could include corrosion, user abuse etc. Such
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
factors are more likely to occur, the longer the operating period. Questions need to be asked
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
such as is there an industry standard for such guarantees, what are competitors offering as
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
guarantees, etc.
dg dg
A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ value is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
taken into account. The β value remains at 2 but the γ value is negative at over 50 000 cycles!
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
This implies that there is a probability of failure before entering service. This data looks suspect
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
and further tests should be done to confirm the reliability characteristics of DESIGN B.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
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dg dg
SOLUTIONS
, RELIABILITY ENGINEERING – A LIFE CYCLE APPROACH dg dg dg dg dg dg
INSTRUCTOR’S d g MANUAL
CHAPTER 1 d g
The Monty Hall Problem
dg dg dg
The truth is that one increases one’s probability of winning by changing one’s choice. The
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
easiest way to look at this from a probability point of view is to say that originally there is a
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
probability of ⅓ over every door. So there is a probability of ⅓ over the door originally chosen, and
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
a combined probability of ⅔ over the remaining two doors. Once one of those two doors is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
opened, there remains a probability of ⅓ over the door originally chosen, and the other
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
unopened door now has the probability ⅔. Hence it increases one’s probability of winning the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
car by changing one’s choice of door.
dg dg dg dg dg dg dg
This does not mean that the car is not behind the door originally chosen, only that if one were to
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
repeat the exercise say 100 times, then the car would be behind the first door chosen about 33
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
times and behind the alternative choice about 66 times. Prove for yourself using Excel!
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Another way to prove this result is to use Bayes Theorem, which the reader can source for
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
himself on the internet.
dg dg dg dg
Assignment 1.2: Failure Free Operating Period d g d g d g d g d g
The FFOP (Failure Free Operating Period) is the time for which the device will run without
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
failure and therefore without the need for maintenance. It is the Gamma value for the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution. From the list of failure times 150, 190, 220, 275, 300, 350, 425, 475, the Offset is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
calculated as 97.42 hours – say 100 hours. This is the time for which there should be no
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
probability of failure. It will be seen from the graph in the software with Beta = 2 that the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution is of almost perfect normal shape and that the distribution does not begin at the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
origin. The gap is the 100 hours that the software calculates when asked.
dg dg dg dg dg dg dg dg dg dg dg dg dg
When the graph is studied for Beta = 2 it will be seen that there is a downward trajectory in the three
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
left hand points. If this trajectory is taken down to the horizontal axis it is seen to intersect it at about
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
120 hours. This is the estimation of Gamma. In the days before software this was always the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
most unreliable estimate of a Weibull parameter and the most difficult to obtain graphically.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Assignment 1.3 d g
When the offset is calculated it is seen to be negative at – 185.59 (say 180). This indicates that the
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
distribution starts before zero on the horizontal axis. This is the phenomenon of shelf life.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
Some items have failed before being put into service. This can apply in practice to rubber
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
components and paints, for example.
dg dg dg dg dg
10
@@SSeeisis
mmiciicsisoo
lala
titoionn
,Assignment 1.4: The Choice between Two Designs of Spring dg dg dg dg dg dg dg dg
DESIGN A d g DESIGN B dg
Number Cycles to Failure dg dg Number Cycles to Failure dg dg
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
dg dg dg dg dg dg dg dg dg
β=4 dg dg
Correlation = 0.9943 dg dg
F400k = 8% (measured from the graph in the Weibull printout below Fig M1.4 Set A) Hence
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
R400k = 92%
dg dg dg
For DESIGN B we get from the WEIBULL-DR software (not shown here)
dg dg dg dg dg dg dg dg dg dg dg
Β=2 dg dg
Correlation = 0.9867 dg dg
F400k = 20% dg dg
Hence R400k = 80% dg dg dg
Hence DESIGN A is better dg dg dg dg
From Fig 1.4.1 Set A we can read in the table that for F = 1% at 90% confidence, the R value is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
126922 cycles. For an average use of 8000 cycles per year we get 126922/8000 = 15.86 years A
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg d g
conservative guarantee would therefore by 15 years.
dg dg dg dg dg dg dg
NOTE: The above calculations ignore the γ value. If this is calculated, the following figures
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
emerge as shown in Fig 1.4.2 (the obscuration of some of the figures is the way the current
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
version of the software prints out)
dg dg dg dg dg dg
DESIGN A dg
β=3 dg dg
γ = 101 828.6 say 100 000
dg dg dg dg dg dg
For F = 1% at 90% confidence, F = 176149
dg dg dg dg dg dg dg dg dg
Dividing by 8000 we get 176149/8000 = 22 years
dg dg dg dg dg dg dg dg dg
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mmiciicsisoo
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titoionn
, Fig 1.4.1 Set A dg dg dg
A figure of 22 years or even 15 years for any guarantee is very long indeed. Company policy
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
would have to be invoked – there are matters to consider in the determination of guarantees
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
other than the test data provided. These matters could include corrosion, user abuse etc. Such
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
factors are more likely to occur, the longer the operating period. Questions need to be asked
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
such as is there an industry standard for such guarantees, what are competitors offering as
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
guarantees, etc.
dg dg
A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ value is
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
taken into account. The β value remains at 2 but the γ value is negative at over 50 000 cycles!
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
This implies that there is a probability of failure before entering service. This data looks suspect
dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg dg
and further tests should be done to confirm the reliability characteristics of DESIGN B.
dg dg dg dg dg dg dg dg dg dg dg dg dg dg
10
@@SSeeisis
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