Outline Solutions to Tutorial Sheet 1

Graham Crocker

Description

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 ...

Informations

• Publié par: Usso
• Nombre de lectures: 7
• Langue: English

Extrait

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.