MREGC 5005 Quantitative Techniques for Asset Management Assignment 3

This assignment focuses on reliability analysis techniques used in asset management. It begins with an AMSAA model analysis of system failure times, calculating Beta and Lambda using maximum likelihood estimators and unbiased estimator equations, and calculating MTBF for both time-terminated and failure-terminated tests. The second part involves Weibull analysis, requiring the creation of a Weibull paper plot based on failure time data of components, and commenting on the Beta value. The final section involves estimating Weibull parameters (Beta and Eta) for a new asset using expert opinions and setting up and solving three inequalities based on failure probabilities over given time periods. The assignment aims to provide a comprehensive understanding of reliability modeling and its applications in asset management, providing a basis for decision-making regarding asset maintenance and replacement.

Assignment 3
The failure time of a system are noted. Perform a reliability growth analysis using the AMSAA model
(apply the Maximum Likelihood Estimators Equation and unbiased estimator equation to calculate Beta
& Lambda). Calculate MTBF for: Time terminated (Time – 1000 hours) and Failure terminated test. [5%]
Q1
Failure
No.
Failure
Time
(hours)
Cumulative
failure time
(hours)
Log(failure
time)
1 3 3 0.477121
2 10 13 1
3 13 26 1.113943
4 31 57 1.491362
5 57 114 1.755875
6 61 175 1.78533
7 80 255 1.90309
8 110 365 2.041393
9 125 490 2.09691
10 129 619 2.11059
11 144 763 2.158362
12 168 931 2.225309
13 229 1160 2.359835
14 297 1457 2.472756
15 321 1778 2.506505
16 328 2106 2.515874
17 366 2472 2.563481
18 397 2869 2.598791
19 421 3290 2.624282
20 438 3728 2.641474
21 501 4229 2.699838
22 620 4849 2.792392
Total 4849
hours
45.93451
^β= 22
22 ( 2.792392 ) −45.93451 = 1.419528
^λ= 22
10001.419528 = 1.2129 * 10-3
^p=1.419528∗1.2129∗10−3∗10001.419528 −1 = 0.03123

MTBF = 1/p(T) = 1/0.03123 = 32.022
Q2)
The failure time of the first 10 failures of a family of 40 identical components is recorded. Assuming that
the failure mode is simular, obtain the Weibull failure time distribution using: Weibull Papers. (you must
submit you Weibull Paper plot)
cycles Failure
median
rank
1/(1-median
rank)
ln(ln(1/
(1-
median
rank)))
ln(cycle
s)
180000 1 0.0673076
92
1.072164948 -
2.66384
12.1007
1
200000 2 0.1634615
38
1.195402299 -
1.72326
12.2060
7
380000 3 0.2596153
85
1.350649351 -
1.20202
12.8479
3
420000 4 0.3557692
31
1.552238806 -
0.82167
12.9480
1
460000 5 0.4519230
77
1.824561404 -0.5086 13.0389
8
500000 6 0.5480769
23
2.212765957 -
0.23037
13.1223
6
550000 7 0.6442307
69
2.810810811 0.03292
5
13.2176
7
600000 8 0.7403846
15
3.851851852 0.29903
3
13.3046
8
700000 9 0.8365384
62
6.117647059 0.59397
7
13.4588
4
800000 10 0.9326923
08
14.85714286 0.99268
9
13.5923
7

12 12.2 12.4 12.6 12.8 13 13.2 13.4 13.6 13.8
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
ln(cycles)
ln(ln(1/(1-median rank))
SUMMARY OUTPUT
Regression Statistics
Multiple R 0.974839
R Square 0.950312
Adjusted R
Square
0.944101
Standard
Error
0.263628
Observation
s
10
ANOVA
df SS MS F Significan
ce F
Regression 1 10.6337 10.6337 153.003
6
1.7E-06
Residual 8 0.555997 0.0695
Total 9 11.1897
Coefficien
ts
Standard
Error
t Stat P-value Lower
95%
Upper
95%
Lower
95.0%
Upper
95.0%
Intercept -29.1964 2.319571 -12.587 1.49E-
06
-34.5454 -
23.8475
-
34.5454
-
23.8475
ln(cycles) 2.208398 0.178536 12.3694
6
1.7E-06 1.796693 2.62010
3
1.79669
3
2.62010
3

a)
^β=¿ 2.208
Eta = ^β ln ԯ
Eta = 551625.24
b) Comment on the Beta Value of this analysis.
Since beta is more than one, the failure rate increases with time
c) What is the reliability at time = 800,000 cycles
F(t) = β
ԯ ( t
ԯ ) β−1
exp [ −( t
ԯ )
β
]
= 2.208
551625.24∗( 800000
551625.24 )2.208−1
∗exp [− ( 800000
551625.24 )2.208
]
= 6.4643*10-7
R(t) = 1 – f(t)
= 1 - 6.4643*10-7
= 0.9999994
d. If reliability of 90% is required, at what time should replacement occur? Calculate using the Formula
of Weibull.
Time = 1/4790000/3600 = 7.515 * 10-5 hours
MTBF = 7.515∗10−5
10 = 7.516 *10-6 hours
t = -[MTBF *ln(R)]
= -[7.516 *10-6 *ln0.9]= 7.9185 *10-7 hours
Q3)
A new designed asset is being implemented to a train. It is needed to estimate the Weibull parameters
for the asset to be used in a reliability model. A group of experts are gathered and provided answer to
the following questions:
• Out of 100 new assets, how many will fail before 7 years?
o 65% to 80 % will failure before 7 years
• Out of 100 new assets, how many will fail before 5 years?

o 40% to 60 % will failure before 5 years
• Out of 100 assets that are 7 years of age, how many will fail in the next 2 years?
o 70% to 90 % will failure in the next 2 years
Using the above answers set up the 3 inequalities equation and solve for the two Weibull Parameters
(Beta and Eta) [10%]
e
−7 β
ԯ ≤ 0.8 ………………1
e
−5 β
ԯ ≤ 0.6 ………………2
e
−7 β
2 ≤ 0.8 ………………3
2.7182
−7 β
2 ≤ 0.8
−7 β
2 log 2.8182≤ log 0.8
β=0.0617
Substituting
To equation 2
e
−5∗0.0617
ԯ ≤ 0.6
-0.3083/etalog2.7182 = log0.6
Eta = 0.604