Systems Reliability Assignment Solutions - Professor Stupples, C125

This document presents a comprehensive solution to a Systems Reliability assignment. It begins with fault tree analysis, deriving logical expressions for gate outputs, minimal cut sets, and calculating upper and lower bounds for the top event's probability using the inclusion-exclusion theorem. The assignment then delves into geothermal power unit reliability, calculating unavailability, mean time to failure, and sketching cumulative failure probability. Further, the solution explores the exponential distribution, hazard rate estimation, and confidence intervals for the mean time to failure. It includes graphical analysis of cumulative failure probability and discusses model suitability. The assignment also addresses the preference for constant failure rates in reliability calculations, convolution functions, and preventive maintenance strategies, drawing conclusions from an article on the subject and suggesting modifications to reliability analysis. Finally, it provides detailed calculations, interpretations, and reasoned arguments for each question, demonstrating a thorough understanding of systems reliability principles.

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System Reliability
Name
Institution Affiliation

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1. The figure below gives a fault tree for a top event marked with the
probabilities for the basic events.
a. Write down the
logical expressions for the outputs from the gates A, B and C are as
follows and hence derive the minimal cuts sets for the system (Birolini,
2013).
P ( C ) =P ( 3 )+ P ( 4 ) +P ( 5 )−P ( 3 ) P ( 4 )−P ( 3 ) P ( 5 ) −P ( 4 ) P (5 )+ P ( 3 ) P ( 4 ) P ( 5 )
(1∈3 [ ¿ ] logic tree for success)
P ( B ) =P ( 2 ) P ( C ) =P(2 C) (1 in 2 [AND] logic tree for failure)
P ( B )=P ( 2 ) {P ( 3 ) +P ( 4 ) + P (5 )−P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 )−P ( 4 ) P ( 5 )+ P ( 3 ) P ( 4 ) P ( 5 ) }
P ( A ) = P(1 ∨ B) = P (1) + P ( B ) −P(1 ∧ B) (1 in 2 (OR)logic tree for success ¿
P ( A ) = P (1) +
P ( 2 ) { P ( 3 ) + P ( 4 ) + P ( 5 ) −P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 ) −P ( 4 ) P ( 5 ) + P ( 3 ) P ( 4 ) P ( 5 ) } −P (1) ∧ P ( 2 ) { P ( 3 ) + P ( 4 ) + P ( 5 ) −P ( 3 ) P
Minimal cuts.
A=1+ B

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B=¿2. C
C=3+ 4+5
b. Using the minimal cut set derived, use the inclusion –exclusion theorem
to calculate the upper and lower bounds for the probability of the top
event, T (Smith, 2011).
∑
i=1
N
P(C¿¿ i)−∑
i<1
P(Ci Λ C j )≤ P(T )≤∑
i=1
N
P(C¿ ¿i)¿ ¿
∑
i=1
N
P( C¿¿ i)= ( B1 ) + ( B2 X B3 ) + ( B2 X B4 ) + ( B2 X B5 ) ¿
( 0.4 ) + ( 0.2 X 0.3 ) + ( 0.2 X 0.2 )+ ( 0.2 X 0.4 )=0.58
∑
i< 1
P ( Ci Λ C j ) =0.58−P ( C1 Λ C2 ) +P (C2 Λ C3 ) + P ( C3 Λ C4 ) +P ( C1 Λ C2 ) +p ¿
C1= ( 1 ) −B1 ,1=0.4
C2= ( 2,3 ) −B1 , 2=0.2 , B2 , 2=0.3
C2= ( 2,3 ) −B1 , 2=0.2 , B2 , 2=0.3
C4= ( 2,5 ) −B1 ,4 =0.2 , B2 , 4=0.4
P ( C1 Λ C2 ) =B1,1 X ( B1,2 X B2,2 ) =0.4 X 0.2 X 0.3=0.024
P ( C1 Λ C3 ) =B1,1 X ( B1,3 X B2,3 ) =0.4 X 0.2 X 0.2=0.016
P ( C1 Λ C4 )=B1,1 X ( B1,4 X B2,4 )=0.4 X 0.2 X 0.4=0.032
P ( C2 Λ C3 ) = ( B1,2 X B2,2 ) X ( B1,3 X B2,3 )=0.2 X 0.3 X 0.2=0.012
P ( C2 Λ C4 )= ( B1,2 X B2,2 ) X ( B1,4 X B2,4 ) =0.2 X 0.3 X 0.4=0.024
P ( C3 Λ C4 )= ( B1,3 X B2,3 ) X ( B1,4 X B2,4 ) =0.2 X 0.2 X 0.4=0.016
∑
i< 1
P ( Ci Λ C j ) =0.024+ 0.016+0.032+0.012+0.024+ 0.016=0.124

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∑
i=1
N
P(C¿¿ i)−∑
i<1
P ( Ci Λ C j )=0.58−0.124=0.456 ¿
0.456 ≤ P(T ) ≤0.58
c. Calculate the exact value of the probability of the top event. Comment
on how closely this compares with limits found in part (b) (Nachlas,
2017).
P ( A ) = P (1) +
P ( 2 ) {P ( 3 ) + P ( 4 ) + P ( 5 )−P ( 3 ) P ( 4 ) −P ( 3 ) P ( 5 )−P ( 4 ) P ( 5 )+ P ( 3 ) P ( 4 ) P ( 5 ) }−P (1) ∧ P ( 2 ) {P ( 3 )+P ( 4 ) + P ( 5 )
¿ ( 0.4 )+ ( 0.2 ) { ( 0.3 ) + ( 0.2 ) + ( 0.4 )− ( 0.3 X 0.2 ) − ( 0.3 X 0.4 ) − ( 0.2 X 0.4 )−(0.3 X 0.2 X 0.4) }−(0.4 X 0.2) { ( 0.3 ) + ( 0.2 ) + ( 0
¿ 0.6 { ( 0.9 )− ( 0.06 )− ( 0.12 )−(0.08)− ( 0.024 ) }−(0.8) {0.9− ( 0.06 )− ( 0.12 ) −0.8+0.024 }
P ( A )=0.4797
The exact value of the probability is within therange obtained ∈ part ( b ) above
CITATION Lie 17 ¿ 1033(Lienig ,2017) .
2. The figure below shows the power output from a batch of experimental
geothermal power production units based in Northern Alaska. The
geothermal heat input is essentially constant, but the harsh conditions
have rendered the units somewhat unreliable, necessitating frequent
repair to return them to normal production

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a. From the figure, calculate the unavailability of a typical unit at 20
months, Q (20), 50 months, Q (50), and 60 months, Q (60).
At 20 months theunits 2,3 are unavailable ( 2Out of 8 )
Q ( 20 )= 2
8 =0.25
At 50 months theunits 1,4,7 are unavailable ( 3Out of 8 )
Q ( 20 )= 3
8 =0.375
At 60 months theunits 2,5,8 are unavailable ( 3Out of 8 )
Q ( 20 )= 3
8 =0.375
b. List the times to failure (TTF) that may be deduced from the graph and
hence deduce the mean time to failure, MTTF (Colombo, 2017).
Unit 1: TTF =16 months
Unit 2: TTF = 14 months
Unit 3: TTF =8 months

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Unit 4: TTF = 8 months
Unit 5: TTF = 24months
Unit 6: TTF =14 months
Unit 7: TTF =12 months
Unit 8: TTF =27 months
MTTF= 135
8 =16.88
c. Using your data sketch of the cumulative failure probability/unreliability,
for a typical system as a function of running time over range 0 to 35
months. Estimate the reliability of a system that has been running for
20months, R (20) (JA1012, 2014).
Time of failure (months) Failure number Failure probability
10 2 0.25
20 4 0.5
30 7 0.875
40 8 1

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5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
Time of failure(Months)
Failure Probability
R ( 20 ) =1−Q(20)
¿ the graph Q ( 20 )=0.5
R ( 20 )=1−0.5 = 0.5
d. Lists the times to repair (TTR) that may be deduced from the graph and
use this list to calculate the mean time to repair (MTTR).
Unit 1: TTR =10 months
Unit 2: TTR = 18 months
Unit 3: TTR =16 months
Unit 4: TTR = 14 months
Unit 5: TTR = 16 months
Unit 6: TTR =12 months

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Unit 7: TTR =12 months
Unit 8: TTR =12 months
MTTR=110
8 =13.75
e. Assuming that both the rate of failure and the rate of repair at constant,
i. Calculate the conditional failure intensity, λ ,and use this values to
calculate the expected Unit reliability after 20 months, R (20).Comment
on any difference between this figures and that found under part (c)
above (Dahoo, 2016).
λ= 1
MTTF = 1
16.88 =0.05924
R (20) =
Reliability = 1- unreliability
R (t) = e-ʎt = e-0.05924×20 = 0.0.3058
ii. Calculate the steady–state unavailability, Q( ∞).Would you expect this
figure to be similar to Q(50), calculate above, and if so, why?
Q (∞) = 1- R (∞) =1 - e-ʎt = 1 – e (∞) = 1 – 0 = 1
Comparing this value with Q (50)
Q (50) =
Q (50) = 1- R (50) =1 - e-ʎt =
= 1 - e-0.05924×50 = 0.9483
Comparing the unavailability at 50 months and that at infinite time, we can see
that the unavailability is tending towards 1. This means that as time progresses,

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the system losses its reliability and will become totally unreliable at an infinite
time (Dhillon, 2017).
3. (a) (i) Write down the equation defining the reliability of a system with
exponential failure distribution.
R ( t )=e− λt
(ii) Recast your expression for the reliability R (t), of a system with an
exponential distribution into a form where the hazard rate, λ ,may be found
from a straight-line graph (Lee, 2014).
Hazard rate ʎ = (ℎ 𝑡) = f (t)
R (t) = f (t )
1−F (t)
Where f (t) - is the probability density function of the T.T.F
F (t) - is the failure probability
R (t) - is the reliability probability
(b) 20 system are subjected to an endurance test.The table below gives failure
data for the first 8 system to fail (Gouriveau, 2016).
Failure
number
Time of failure(months) Failure probability
1 1.4 0.125
2 3.2 0.25
3 3.4 0.375
4 3.9 0.5
5 5.6 0.625
6 7.1 0.75
7 8.4 0.875
8 8.8 1
i. Assuming that the distribution is exponential, use a graphical method to
estimate the hazard rate, λ , and hence the mean time to failure, MTTF
=θ, for the system under test (Mobley, 2012).

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λ = Gradient of the curve = 0.875−o.5
8.4−3.9 = 0.375
4.5 =0.0833
MTTF= 1
λ = 1
0.0833 =12 months
ii. Use the χ2 table provided at the back of this question paper to calculate
the 90% confidence interval for the mean time to failure, MTTF.
MTTF=12 months C=0.90 Number of failures(r) =8
Degree of freedom = (2r) =16
C=1−2 α 0.9=1−2 α α =0.05 And 1−α =0.95
χ2
0.05 ,16 And χ2
0.95 ,16
¿ the table
χ2
0.05 ,16 =¿7.96
χ2
0.95 ,16 =¿26.30

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θ0.95 =2 x 8 x 12
7.96 =24.12 months
θ0.95 =2 x 8 x 12
26.30 =7.3 months
7.3 ≤ MTTF ≤24.12
(c) You are now shown further data on the final 12 system to fail.
Failure
number
Time to
failure(months)
Minimum Estimate Maximu
m
Central
9 8.9 0.45 0.41 0.15 0.38
10 9.5 0.5 0.42 0.23 0.40
11 9.8 0.55 0.50 0.25 0.44
12 11.3 0.6 0.57 0.3 0.5
13 14.5 0.65 0.64 0.37 0.58
14 16.2 0.7 0.68 0.39 0.62
15 17.9 0.75 0.7 0.42 0.66
16 20.7 0.8 0.72 0.46 0.69
17 23.2 0.85 0.80 0.50 0.72
18 26.4 0.9 0.84 0.55 0.8
19 30.5 0.95 0.88 0.59 0.84
20 39.2 1.0 0.92 0.64 0.88
i. Plot the cumulative failure probability, F (t), for times up to 40
months. Show :
1. The estimate of F (t) from the data given

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5 10 15 20 25 30 35 40 45
0
0.2
0.4
0.6
0.8
1
1.2
Minimum Estimate Central Maximum
Time of failure (Months)
F (t)
15 months
2. The central estimate based on an exponential failure
distribution.
¿ 13 months
3. The lower limit for F (t) based on an exponential failure
distribution.
¿ 10 months
4. The upper limit for F (t) based on an exponential failure
distribution.

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¿ 23 months
ii. How well does the exponential distribution model their failure
distribution for this system? What other model might you use?
For this system, the exponential distribution does not fully exhibit the
behavior of the system because of the limit nature of the data. Therefore,
other model like Weibull could be used. Though Weibull tends to be
cumbersome, it would be the best to truly depict the failure probability of
this system and hence properly analyze the reliability of the system.
4.
(i) Why do engineers prefer to use to ‘constant failure rates’ for
reliability calculations? Consider also the ‘convolution functions’
Constant rates of failure are the reasonable assumptions in practical gears and
yields mathematical expressions that are controllable and give precise feel of
complex repair and failure process. The assumption made in the process includes;
a. Prime of life of a component (either new or repaired and maintained).
b. Component with many subcomponents with different failure rates or ages.
c. Data are limited thus, more elegant mathematical handling are unjustified.
constant failure rates analysis, simplifies the constant hazard rate h (t) and
conditional failure intensity, λ ( t )=h ( t )=λ where λis used for constant value

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instead of “h”.Simplification for unreliability F(t),reliability R(t) and failure
probability density F(t) which are given by the following formulas.
F ( t ) =1−e−λt
R ( t )=e− λt
F ( t )=λ e−λt
F ( t ) Is the exponential distribution and failures are random and not concerted
around the end of life limit. The MTTF are obtained near the result.
MTTF= 1
λ
MTTF is obtained as arithmetic mean of time to fail and allows calculation of λ
and other parameters of reliability. MTTF analysis helps to it gain a reasonable
assumption on failed component being repaired. Combined process between
failure to repair and then to failure again can be analysed by using constant
failure rate and the constant repair rate by applying Laplace Transform as shown
below
w ( t ) =f ( t ) + ∫
u=0
u=t
f ( t−u ) v ( u ) du
v ( t ) = ∫
u=0
u=t
g ( t−u ) w ( u ) du
The transforms are useful in change of convolutions to products.
         thLthLduuhuthL
t
21
0
21 




 ∫

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           
        
      
      






















s
eLeLtgL
s
eLeLtfL
twLtgLtvL
tvLtfLtfLtwL
tt
tt
:rate,repairand,rate,failureconstantFor
Finding the inverse of the transform results to unconditional failure density v (t)
and the repair intensity v (t)
   
   t
t
etv
etw






















2
Failure and repair is dynamic initially and very stable to settle out a steady state
very fast with constant values for reliable parameters. Steady state unavailability
Q(∞) and the availailability A (∞ ¿ is given by;
 
  MTTRMTTF
MTTF
A
MTTRMTTF
MTTR
Q
















 




 
11
1
11
1
And the time taken to reach the steady state is given by dividing Q (t) by Q(∞)
Q (t) which is derived from Markov analysis concerned to one state to the
other.by probability of transition which is dependent on a system state on
time.Markov analysis is applied to failure rate and repair rate. The probability of

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failing state at time is unavailability Q (t) and probability of normal state is
availability A (t) =1- Q (t).
(ii) What conclusions do you draw from the article for preventative
maintenance? From your analysis and understanding of the course, how
would you modify reliability analysis to comply with the article or would you
accept the limitations?
Most machines failures as a result of poor design and use of substandard materials
in the transmission system maintenance. Mechanical failure such as one that
occurs in gearbox causes longest failure rates. The rates system failure rates
increases as the design differs from expected design to be established. The main
cause of failure is wearing out ( hazard rate) that increases during the last stage of
component design life .The failure rates of a subsystem such as wind turbine
represents about 4% of the total number of failures.
In electrical machines major cause of failure are related to winding and bearing.
The size of the electrical machine influences the components to fail in a given
power requirement. Rotor winding is difficult in small generator. Banding,
conductor and stator winding failure are related to maintenance and contamination
issues. Stator winding, failure of bearing and rotor winding failure underwrite to
80 % of the total failure in induction machine. The translation failure distribution
includes; Stator (37%), bearing (41%), rotor (10%) and the other faults to about
(12%).Preventive maintenance can be explained as the measures taken in order to
avert the probability of system failure. This maintenance is performed without
considering the actual condition of the system either perfectly running or in a poor

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state. It is very important to undertake these maintenance procedure so as to
evade any downtime that can be caused when the system fails and also to escape
the fatalities that can be caused to the operator of the system or to the system itself
(Amruthnath, 2018).
The analysis of residuals are known as errors and they define the actual
observation and the one predicted from the model.
ei= yi− y⏞ i
, i =1, 2…..n
Where ei Is the residual /error yi is the actual observation and y⏞ i
,I is the predicted
model observation. Plotting the residual shows the best fit and deviation from
previous linear regression assumptions. The output of the machine is followed by
nacelle temperature that affects the stator winding temperature. The expected
generated output causes more current flow via the windings and heat generated is
proportional to current squared. Nacelle temperature is the ambient temperature
that is produced by the generator. Insulators are used in the design to reduce the
ambient temperature. Temperature increase to more than 40 oC results to
shutdown of the generator thus, maintaining the temperature limit for insulation.
The temperature increase is obtained from the differences between hottest
temperature of insulation and the ambient temperature. To stator overheating
blocked air filter are used for effective cooling. Linear regression model meets
accurate criteria’s for the modelling of relationships between variables. The F –
test shows that the regression method is significant. The SR deficiency of data
distribution that is caused by nonlinear behavior .Aerodynamics rotor blade are

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changed rapidly preventing excess power generation and loadings of machines
(wind turbine).
.
References
Amruthnath, N. &. (2018). Fault class prediction in unsupervised learning using model-based
clustering approach. London: International Conference on Information and Computer
Technologies .
Birolini, A. (2013). Reliability engineering: theory and practice. London: Springer Science &
Business Media.
Colombo, A. G. (2017). Systems Reliability Assessment. London : Elsevier .
Dahoo, P. R. (2016). .Analysis of Failures, Predictive Reliability. Delhi: Longhorns .
Dhillon, B. S. (2017). Engineering systems reliability, safety, and maintenance . New York:
CRC Press.
Gouriveau, R. M. (2016). From prognostics and health systems management to predictive
maintenance. Mumbai: John Wiley & Sons.
JA1012, S. A. (2014). A guide to the reliability-centered maintenance (RCM) standard. London:
SAE International.
Lee, J. W. (2014). Mechanical systems and signal processing. Prognostics and health
management design for rotary machinery systems, 314-334.
Lienig, J. &. (2017). Reliability analysis. . Fundamentals of electronic systems design , 45-73.

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Mobley, R. K. (2012). An introduction to predictive maintenance. London: Elsevier .
Nachlas, J. A. (2017). Reliability engineering: probabilistic models and maintenance methods.
New York: CRC Press.
Smith, D. D. (2011). Reliability, Maintainability and Risk. New York : Springer.