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All Chapters Covered
SOLUTIONS
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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
, @LECTJULIESOLUTIONSSTUVIA
Dividing by 8000 we get 176149/8000 = 22
years
, @LECTJULIESOLUTIONSSTUVIA
Fig 1.4.1 Set A
A figure of 22 years or even 15 years for any guarantee is very long indeed.
Company policy would have to be invoked – there are matters to consider in the
determination of guarantees other than the test data provided. These matters could
include corrosion, user abuse etc. Such factors are more likely to occur, the longer
the operating period. Questions need to be asked such as is there an industry
standard for such guarantees, what are competitors offering as guarantees, etc.
A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ
value is taken into account. The β value remains at 2 but the γ value is negative at
over 50 000 cycles! This implies that there is a probability of failure before entering
service. This data looks suspect and further tests should be done to confirm the
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reliability characteristics of DESIGN B.
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Fig 1.4.2 Set A with γ Calculation
ASSIGNMENT 1.5: Rolling Element Bearings
This assignment requires a full report as detailed in the text of the book and is therefore outside
the scope of this Instructor’s Manual
, @LECTJULIESOLUTIONSSTUVIA
ASSIGNMENT 1.6:
Case 1.1: Weibull Analysis at the New Era Fertilizer Plant
Here we are dealing with only three failures, but many
suspensions The data table appears as shown below:
Time to Failure Failure or Suspension Number of F or S Rank Mean Rank
Order =F%
250 hours F 1 1 1/25 = 4%
250 hours S 7
600 hours F 1 2.41 2.41/25=9.6%
600 hours S 7
2000 hours F 1 4.92 4.92/25=19.7%
2000 hours S 7
When using the software the β value comes out as Zero! Graphical methods indicate a
value of about 0.7 – certainly less than 1. If the mean ranks calculated above are
plotted as the F values on Weibull graph paper, against the three failure values, the β
value comes out as 0.7. Any value between 0 and 1 indicates a hyper-exponential
distribution ie a quality problem. We do not have enough information to be sure of
this as we only have three data points. We have only three failures, but the correlation
is very high – the three points lie almost exactly on a straight line.
Since values of β less than unity indicate a quality problem, then either the
manufacturer is selling poor quality bellows or they are being damaged on
installation.
Even without using Weibull analysis we can see from the failure pattern that
something is wrong here. One set of bellows lasted till 2000 hours until the first
failure. Perhaps those seven discarded bellows a 150 hours might also have lasted at
least 2000 hours. The same goes for the ones discarded at 600 hours.
Failure analysis like this is like detective work. We pick up clues and follow where
they lead This means we can now formulate a plan of action:
1. When the next failure occurs, only replace the failed item. We can build up set
of eight or so failures like this – five to add to the three that we have already.
2. When the next failure occurs, we must observe the installation to see if the
bellows is being damaged when fitted.
3. We must visit the manufacturer’s works and study his production and quality control
4. Also study past records from the company’s SCADA1 system to see whether
the bellows have been subjected to out-of-specification conditions, especially
high temperatures or high pressure. Reliability only applies to defined
operating conditions.
5. Check on the storage of the bellows – rubber components can exhibit shelf life
problems, perhaps leading to premature failures after installation. How long
, @LECTJULIESOLUTIONSSTUVIA
are the bellows stored and under what conditions? Rubber items should be
stored in a dark, cool room.
1 SCADA: Supervisory Control and Data Acquisition
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The object of this assignment is to demonstrate that we must use whatever data we
have in the interests of solving a problem. We will never have perfect data – the cost of
perfect information is infinite. As this example demonstrates, the little bit of data that
we have at least allows us to formulate a plan of investigation to establish the true
situation.
ASSIGNMENT 1.7:
Case 1.2: The Life History of a Hillman Vogue Sedan
1. The question is to find the following in the repair record:
Infant mortality failure
Incomplete repair
Life extension
Indication of failure
Retrofit
Visual Inspection
Preventive Measure, or Proxy Replacement ie replacing something
so that something else does not fail
Root Cause Analysis
The answers are given in bold face italics in the table below:
Kilometres Repair Action
42 300 New Clutch: This could well be an infant mortality failure as the clutch had a very
short life
89 500 Retrofit inline fuel filter Retrofit as stated
140 500 New head gasket, valve grind As regards the engine as a whole, this was an
incomplete repair as the head gasket failed again at 170 000 km
140 900 New clutch plate, engine rebuild Life Extension
170 000 Blown head gasket – car sold as scrap
2. The main issue in this case is that a Bathtub is present (Figure 1.17 in the case:
Major Repair Cost vs Years of Service). But this is not the failure rate vs time
bathtub of the literature. It is a cost bathtub. Many systems are withdrawn
from service when the costs maintain the system go to high, not because the
failure rate is increasing.
3. The difference between the two models was that the 1976 model was a “parts bin
special” – cars were being put together with what remained of component
production after manufacture of certain items had all but ceased. And quality
sometimes tails off at the end of a production run as workers and management loose
interest, and as special “one-off” parts might have to be made.
All Chapters Covered
SOLUTIONS
, @LECTJULIESOLUTIONSSTUVIA
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.
, @LECTJULIESOLUTIONSSTUVIA
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
, @LECTJULIESOLUTIONSSTUVIA
Dividing by 8000 we get 176149/8000 = 22
years
, @LECTJULIESOLUTIONSSTUVIA
Fig 1.4.1 Set A
A figure of 22 years or even 15 years for any guarantee is very long indeed.
Company policy would have to be invoked – there are matters to consider in the
determination of guarantees other than the test data provided. These matters could
include corrosion, user abuse etc. Such factors are more likely to occur, the longer
the operating period. Questions need to be asked such as is there an industry
standard for such guarantees, what are competitors offering as guarantees, etc.
A further point to note is that DESIGN B exhibits very peculiar characteristics if the γ
value is taken into account. The β value remains at 2 but the γ value is negative at
over 50 000 cycles! This implies that there is a probability of failure before entering
service. This data looks suspect and further tests should be done to confirm the
, @LECTJULIESOLUTIONSSTUVIA
reliability characteristics of DESIGN B.
, @LECTJULIESOLUTIONSSTUVIA
Fig 1.4.2 Set A with γ Calculation
ASSIGNMENT 1.5: Rolling Element Bearings
This assignment requires a full report as detailed in the text of the book and is therefore outside
the scope of this Instructor’s Manual
, @LECTJULIESOLUTIONSSTUVIA
ASSIGNMENT 1.6:
Case 1.1: Weibull Analysis at the New Era Fertilizer Plant
Here we are dealing with only three failures, but many
suspensions The data table appears as shown below:
Time to Failure Failure or Suspension Number of F or S Rank Mean Rank
Order =F%
250 hours F 1 1 1/25 = 4%
250 hours S 7
600 hours F 1 2.41 2.41/25=9.6%
600 hours S 7
2000 hours F 1 4.92 4.92/25=19.7%
2000 hours S 7
When using the software the β value comes out as Zero! Graphical methods indicate a
value of about 0.7 – certainly less than 1. If the mean ranks calculated above are
plotted as the F values on Weibull graph paper, against the three failure values, the β
value comes out as 0.7. Any value between 0 and 1 indicates a hyper-exponential
distribution ie a quality problem. We do not have enough information to be sure of
this as we only have three data points. We have only three failures, but the correlation
is very high – the three points lie almost exactly on a straight line.
Since values of β less than unity indicate a quality problem, then either the
manufacturer is selling poor quality bellows or they are being damaged on
installation.
Even without using Weibull analysis we can see from the failure pattern that
something is wrong here. One set of bellows lasted till 2000 hours until the first
failure. Perhaps those seven discarded bellows a 150 hours might also have lasted at
least 2000 hours. The same goes for the ones discarded at 600 hours.
Failure analysis like this is like detective work. We pick up clues and follow where
they lead This means we can now formulate a plan of action:
1. When the next failure occurs, only replace the failed item. We can build up set
of eight or so failures like this – five to add to the three that we have already.
2. When the next failure occurs, we must observe the installation to see if the
bellows is being damaged when fitted.
3. We must visit the manufacturer’s works and study his production and quality control
4. Also study past records from the company’s SCADA1 system to see whether
the bellows have been subjected to out-of-specification conditions, especially
high temperatures or high pressure. Reliability only applies to defined
operating conditions.
5. Check on the storage of the bellows – rubber components can exhibit shelf life
problems, perhaps leading to premature failures after installation. How long
, @LECTJULIESOLUTIONSSTUVIA
are the bellows stored and under what conditions? Rubber items should be
stored in a dark, cool room.
1 SCADA: Supervisory Control and Data Acquisition
, @LECTJULIESOLUTIONSSTUVIA
The object of this assignment is to demonstrate that we must use whatever data we
have in the interests of solving a problem. We will never have perfect data – the cost of
perfect information is infinite. As this example demonstrates, the little bit of data that
we have at least allows us to formulate a plan of investigation to establish the true
situation.
ASSIGNMENT 1.7:
Case 1.2: The Life History of a Hillman Vogue Sedan
1. The question is to find the following in the repair record:
Infant mortality failure
Incomplete repair
Life extension
Indication of failure
Retrofit
Visual Inspection
Preventive Measure, or Proxy Replacement ie replacing something
so that something else does not fail
Root Cause Analysis
The answers are given in bold face italics in the table below:
Kilometres Repair Action
42 300 New Clutch: This could well be an infant mortality failure as the clutch had a very
short life
89 500 Retrofit inline fuel filter Retrofit as stated
140 500 New head gasket, valve grind As regards the engine as a whole, this was an
incomplete repair as the head gasket failed again at 170 000 km
140 900 New clutch plate, engine rebuild Life Extension
170 000 Blown head gasket – car sold as scrap
2. The main issue in this case is that a Bathtub is present (Figure 1.17 in the case:
Major Repair Cost vs Years of Service). But this is not the failure rate vs time
bathtub of the literature. It is a cost bathtub. Many systems are withdrawn
from service when the costs maintain the system go to high, not because the
failure rate is increasing.
3. The difference between the two models was that the 1976 model was a “parts bin
special” – cars were being put together with what remained of component
production after manufacture of certain items had all but ceased. And quality
sometimes tails off at the end of a production run as workers and management loose
interest, and as special “one-off” parts might have to be made.
