Outline Solutions to Tutorial Sheet 1

Outline Solutions to Tutorial Sheet 1

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Outline Solutions to Tutorial Sheet 81. The exponential probability plot is :Exponential Probability Plot for Hrs to Fail99 ML EstimatesM ean: 1.41989795908070605030100 5 10Data1The plot gives the estimate of the mean as 1.41. Thus, since the mean is , the estimate of λ isλ1= 0.7091.41−0.709×1(i) p(X < 1) = F(1) = 1− e = 1− 0.4921 = 0.5079−0.709×2(ii) p(X > 2) = R(2) = e = 0.2422(iii) p(1.5 < X < 2.5) = F(2.5) − F(1.5) or−0.709×1.5 −0.709×2.5R(1.5) − R(2.5) = e − e = 0.3452 − 0.1699 = 0.1753The hazard function is h(t) = λ = 0.709 (i.e. constant) and the MTBF is 1.41 hours2. The Weibull plot shows a good fit. Estimates of α (scale) and β (shape) are given as 0.735 and2.61 respectively.For this value of β, (>2) we expect a hazard function shapePercentThis shows positive ageing. Failure rates are increasing at an increasing rate.β 2 .61 t t − − α 0 .735 R ( t ) = e = e (in units of 1 million hours)2.610.75 − 0.735 −1.054 Then R(0.75) = e = e = 0.3485 1 1 MTBF = αΓ1+ = 0.735Γ 1+ = 0.735Γ(1.38) = 0.888537 from tables of the gamma β 2.61 function. Thus the MTBF is approximately 889,000 hours.Weibull Probability Plot for Hrs(millions99 ML Estimates95 Shape: 2.608719080 Scale: 0.73496570605040302010 5 3 2 10.1 1.0Data3. The Distribution Overview Plot for the Part B’s exponential lifetimes is:Overview Plot for Part BNo censoringProbability Density Function ...

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Outline Solutions to Tutorial Sheet 8
1.
The exponential probability plot is :
The plot gives the estimate of the mean as 1.41. Thus, since the mean is
λ
1
, the estimate of
λ
is
709
.
0
41
.
1
1
=
(i)
5079
.
0
4921
.
0
1
1
)
1
(
)
1
(
1
709
.
0
=
=
=
=
<
×
e
F
X
p
(ii)
2422
.
0
)
2
(
)
2
(
2
709
.
0
=
=
=
>
×
e
R
X
p
(iii)
)
5
.
1
(
)
5
.
2
(
)
5
.
2
5
.
1
(
F
F
X
p
=
<
<
or
1753
.
0
1699
.
0
3452
.
0
)
5
.
2
(
)
5
.
1
(
5
.
2
709
.
0
5
.
1
709
.
0
=
=
=
×
×
e
e
R
R
The hazard function is
709
.
0
)
(
=
=
λ
t
h
(i.e. constant) and the MTBF is 1.41 hours
2.
The Weibull plot shows a good fit. Estimates of
α
(scale) and
β
(shape) are given as 0.735 and
2.61 respectively.
For this value of
β
, (>2) we expect a hazard function shape
10
5
0
99
98
97
95
90
80
70
60
50
30
10
Data
Percent
Exponential Probability Plot for Hrs to Fail
ML Estimates
Mean:
1.41
This shows positive ageing. Failure rates are
increasing
at an
increasing
rate.
61
.
2
735
.
0
)
(
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
=
=
t
t
e
e
t
R
β
α
(in units of 1 million hours)
Then
3485
.
0
)
75
.
0
(
054
.
1
735
.
0
75
.
0
61
.
2
=
=
=
÷
ø
ö
ç
è
æ
e
e
R
888537
.
0
)
38
.
1
(
735
.
0
61
.
2
1
1
735
.
0
1
1
=
Γ
=
÷
ø
ö
ç
è
æ
+
Γ
=
÷
÷
ø
ö
ç
ç
è
æ
+
Γ
=
β
α
MTBF
from tables of the gamma
function. Thus the MTBF is approximately 889,000 hours.
3.
The
Distribution Overview Plot
for the
Part B
’s exponential lifetimes is:
In particular, notice the reverse-J shape of the pdf and the constant hazard function (constant failure
rate). Also note the rapid decline of the survival function as so many components fail early.
1.0
0.1
99
95
90
80
70
60
50
40
30
20
10
5
3
2
1
Data
Percent
Weibull Probability Plot for Hrs(millions
ML Estimates
Shape:
Scale:
2.60871
0.734965
6000
5000
4000
3000
2000
1000
0
99
95
90
80
70
60
50
40
30
20
10
5
1
Exponential Probability
Percent
6000
5000
4000
3000
2000
1000
0
1.0
0.5
0.0
Survival Function
Probability
6000
5000
4000
3000
2000
1000
0
0.00120
0.00115
0.00110
Hazard Function
Rate
6000
5000
4000
3000
2000
1000
0
0.0010
0.0005
0.0000
Probability Density Function
Overview Plot for Part B
No censoring
Exponential
ML Estimates
Mean:
861.142
Fail. Rate:
1.16E-03
MTBF:
861.142
For the Weibull lifetimes of
Part D,
the plot is:
Note the following.
(i)
The distribution is positively skew
(ii)
The parameter estimates for
α
and
β
. The fact that the shape parameter
β
is greater than 2
suggests that the hazard function should increase at an increasing rate (as the plot shows).
(iii)
Compared to the exponential survival function above, more components are surviving in
the early stages.
Calculating the MTBF from the estimates of
α
and
β
gives:
MTBF =
3
.
833
)
886
.
0
)(
548
.
940
(
)
43
.
1
(
)
548
.
940
(
317
.
2
1
1
548
.
940
1
1
=
=
Γ
=
÷
ø
ö
ç
è
æ
+
Γ
×
=
÷
÷
ø
ö
ç
ç
è
æ
+
Γ
β
α
hours
which is the same as the estimate on the plot.
Finally, for the two types of
Part F
,
normal distributions
provide good fits. The combined overview
plot is then:
1000
100
99
95
90
80
70
60
50
40
30
20
10
5
1
Weibull Probability
Percent
2000
1000
0
1.0
0.5
0.0
Survival Function
Probability
2000
1000
0
0.008
0.007
0.006
0.005
0.004
0.003
0.002
0.001
0.000
Hazard Function
Rate
2000
1000
0
0.0010
0.0005
0.0000
Probability Density Function
Overview Plot for Part D
No censoring
Weibull
ML Estimates
Shape:
2.317
Scale: 940.548
MTBF:
833.319
1
2
4000
3000
2000
99
95
90
80
70
60
50
40
30
20
10
5
1
Normal Probability
Percent
4000
3000
2000
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Survival Function
Probability
1
2
4000
3000
2000
0.015
0.010
0.005
0.000
Hazard Function
Rate
4000
3000
2000
1000
0.002
0.001
0.000
Probability Density Function
Overview Plot for Part F
No censoring
The pdf shows that Type 2 components have a higher average but more variable lifetime than Type 1.
As a result of this, the survival plot shows that, at a given time, a higher proportion of Type 2
components will survive. The hazard functions (increasing at an increasing rate) show that, at a given
time, Type 2 components have a lower risk of instantaneous failure. The fact that the hazard plot for
Type 1 rises more steeply than that for Type 2 is a reflection of the fact that type 1 lifetimes are less
variable – i.e. cover a smaller range.