# Parameter Estimation for a Class of Lifetime Models

Authors: Xinyang Ji; Shunhou Fan; Wei Fan
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Our purpose in this paper is to present a better method of parametric estimation for a bivariate nonlinear regression model, which takes the performance indicator of rubber aging as the dependent variable and time and temperature as the independent variables. We point out that the commonly used two-step method (TSM), which splits the model and estimate parameters separately, has limitation. Instead, we apply the Marquardt’s method (MM) to implement parametric estimation directly for the model and compare these two methods of parametric estimation by random simulation. Our results show that MM has better effect of data fitting, more reasonable parametric estimates, and smaller prediction error compared with TSM.


---

## Body

## 1. Introduction

In the study of fatigue lifetime evaluation of rubber materials, accelerated aging test is widely used as an effective procedure for obtaining the data on performance indicators(
y
), aging time (
t
), and aging temperature (
T
). In order to investigate the relationships among them, Dakin [1, 2] proposed the kinetic equation for aging; that is,
(1)
y
=
B
e
-
K
t
α
,
where y is the performance indicator of rubber, t is the aging time, K is an aging rate constant depending on the temperature T, B is a constant, and α is a constant in (
0,1
).Mott and Roland [3] and Wise et al. [4] interpreted that the K in (1) can be expressed in the Arrhenius form. In this paper, we also assume the convention that the K for rubber can be described by the Arrhenius type
(2)
K
=
A
e
-
C
/
T
,
where T is aging temperature and A and C are constants.By (2) and (1), we obtain the model
(3)
y
=
B
e
-
A
e
-
C
/
T
t
α
,
which is called the y
-
t
-
T bivariate nonlinear regression model in this paper. Here, B, A, C, and α are model parameters. In the past, one (see, e.g., [5–7]) usually split (3) into (1) and (2) to estimate the parameters in (3).The constantα is determined by successive approximation method, which is to minimize (4) to two decimal places:
(4)
I
=
∑
i
=
1
l
‍
∑
j
n
‍
(
y
i
j
-
y
i
j
^
)
2
,
where y
i
j and y
i
j
^ denote the experimental measurements and predicted values of the performance indicators of rubber when the aging temperature index is i and the experiment serial number is j, respectively. When α is assigned a value, (1) can be converted into the following linear form through logarithm transformation:
(5)
Y
=
a
+
b
X
,
where Y
=
ln
⁡
⁡
y, a
=
ln
⁡
⁡
B, b
=
-
K, and X
=
t
α.The values ofa and b are determined by the least squares method:
(6)
b
i
^
=
∑
‍
X
Y
-
(
∑
‍
X
∑
‍
Y
)
/
n
∑
‍
X
2
-
(
∑
‍
X
)
2
/
n
,
a
i
^
=
∑
‍
Y
n
-
b
i
∑
‍
X
n
.TheB is given by B
^
=
∑
‍
B
i
^
/
l, where l is the maximum index of aging temperature.TheK is given by K
i
^
=
-
b
i
^ and used as the known value in (2). Similarly, (2) can be converted into the following linear form through logarithm transformation:
(7)
W
=
D
+
E
Z
,
where W
=
ln
⁡
⁡
K, D
=
ln
⁡
⁡
A, E
=
-
C, and Z
=
T
-
1.The values ofD and E are also determined by the least squares method:
(8)
E
^
=
∑
‍
W
Z
-
(
∑
‍
W
∑
‍
Z
)
/
l
∑
‍
W
2
-
(
∑
‍
W
)
2
/
l
,
D
^
=
∑
‍
W
l
-
E
∑
‍
Z
l
.
The estimated values of A and C are given by A
^
=
e
D
^ and C
^
=
-
E
^.At last, the final estimates of the parameters are substituted into (3) to form the regression forecast model.However, the above-mentioned TSM has the following limitation.First, the estimates of the parameters in (1) and (2), obtained by the logarithm regression method, are generally not the least squares solution of the original variables [8].Second, substituting the estimates of the parametersK
i
^ in (1) into (2) may lead to large errors. This is because the estimates of the parameters A and C in (3) highly rely on the precision of K
i
^, if K
i
^ has a small change that will lead to considerable change of the values of A and C. Furthermore, TSM is a tedious calculation method.Finally, the parameterB in (3) is the average of B
i
^, whose goodness needs verifying.Regarding the limitation above, the purpose of this paper is to adopt MM to estimate the four parameters in (3).


## 2. Marquardt’s Method

The general form of the nonlinear regression model is(9)
Y
=
φ
(
x
1
,
x
2
,
…
,
x
p
;
β
1
,
β
2
,
…
,
β
r
)
+
ξ
,
where φ is a known nonlinear function, x
1
,
x
2
,
…
,
x
p are a set of p independent variables, β
1
,
β
2
,
…
,
β
r are a set of r unknown parameters to be estimated, and ξ is the random error. If y and x
1
,
x
2
,
…
,
x
p are observed for n times, n sets of observations (
(
x
i
1
,
x
i
2
,
…
,
x
i
p
,
y
i
), i
=
1,2
,
…
,
n
) can be obtained.Substituting thei
th set of observations of the independent variables into the model (9), we see that
(10)
φ
(
x
i
1
,
x
i
2
,
…
,
x
i
p
;
β
1
,
β
2
,
…
,
β
r
)
=
φ
(
x
i
,
β
)
.Sincex
i
1
,
x
i
2
,
…
,
x
i
p are known values, we deduce that φ
(
x
i
,
β
) is a function of β
1
,
β
2
,
…
,
β
r. For a given initial value β
(
0
)
=
(
β
1
(
0
)
,
β
2
(
0
)
,
…
,
β
r
(
0
)
), we expand φ
(
x
i
,
β
) using Taylor’s formula at β
(
0
) and omit the quadratic and above terms. The expansion is as follows:
(11)
φ
(
x
i
,
β
)
≈
φ
(
x
i
,
β
(
0
)
)
+
∂
φ
∂
β
1
|
β
=
β
(
0
)
(
β
1
-
β
1
(
0
)
)
+
∂
φ
∂
β
2
|
β
=
β
(
0
)
(
β
2
-
β
2
(
0
)
)
+
⋯
+
∂
φ
∂
β
r
|
β
=
β
(
0
)
(
β
r
-
β
r
(
0
)
)
.
All the numbers in (11) except the parameters β
1
,
β
2
,
…
,
β
r are known. It is clear that the right-hand side of (11) is a linear function of β
1
,
β
2
,
…
,
β
r. Thus, we apply the least squares method to (11) and set
(12)
Q
=
∑
i
=
1
n
‍
{
y
i
-
[
φ
(
x
i
,
β
(
0
)
)
+
∑
j
=
1
r
‍
∂
φ
∂
β
j
|
β
=
β
(
0
)
(
β
j
-
β
j
(
0
)
)
]
}
2
+
d
∑
j
=
1
r
‍
(
β
j
-
β
j
(
0
)
)
2
,
where d
≥
0 is called the damping factor. When d
=
0, this method of linearization becomes the Gauss-Newton method [9] which is a special case of MM. Even worse, the selection of initial values of iteration for the Gauss-Newton method is harder than that for MM.In order to minimizeQ, the first partial derivatives of Q with respect to β
1
,
β
2
,
…
,
β
r should be zero; that is,
(13)
0
=
∂
Q
∂
β
k
=
2
∑
i
=
1
n
‍
[
+
∑
j
=
1
r
‍
∂
φ
(
x
i
,
β
)
∂
β
j
|
β
=
β
(
0
)
(
β
j
-
β
j
(
0
)
)
]
y
i
-
φ
(
x
i
,
β
(
0
)
)
kkkkkkk
+
∑
j
=
1
r
‍
∂
φ
(
x
i
,
β
)
∂
β
j
|
β
=
β
(
0
)
(
β
j
-
β
j
(
0
)
)
]
k
k
k
k
×
∂
φ
(
x
i
,
β
)
∂
β
k
|
β
=
β
(
0
)
+
2
d
(
β
k
-
β
k
(
0
)
)
,
k
=
1,2
,
…
,
r
.
The equality (13) can be turned into the following form:
(14)
(
a
11
+
d
)
(
β
1
-
β
1
(
0
)
)
+
a
12
(
β
2
-
β
2
(
0
)
)
+
⋯
+
a
1
r
(
β
r
-
β
r
(
0
)
)
=
a
1
y
,
a
21
(
β
1
-
β
1
(
0
)
)
+
(
a
22
+
d
)
(
β
2
-
β
2
(
0
)
)
+
⋯
+
a
2
r
(
β
r
-
β
r
(
0
)
)
=
a
2
y
,
⋮
a
r
1
(
β
1
-
β
1
(
0
)
)
+
a
r
2
(
β
2
-
β
2
(
0
)
)
+
⋯
+
(
a
r
r
+
d
)
(
β
r
-
β
r
(
0
)
)
=
a
m
y
,
where
(15)
a
j
k
=
∑
i
=
1
n
‍
∂
φ
∂
β
j
|
β
=
β
(
0
)
·
∂
φ
∂
β
k
|
β
=
β
(
0
)
=
a
k
j
,
a
j
y
=
∑
i
=
1
n
‍
(
y
i
-
φ
(
x
i
-
β
(
0
)
)
)
·
∂
φ
∂
β
j
|
β
=
β
(
0
)
,
h
h
h
h
h
h
h
h
h
h
h
j
=
1,2
,
…
,
r
;
k
=
1,2
,
…
,
r
.
Thus
(16)
β
=
[
β
1
β
2
⋮
β
r
]
=
[
β
1
(
0
)
β
2
(
0
)
⋮
β
r
(
0
)
]
+
[
a
11
+
d
(
0
)
a
12
⋯
a
1
r
a
21
a
22
+
d
(
0
)
⋯
a
2
r
⋮
⋮
⋮
⋮
a
r
1
a
r
2
⋯
a
r
r
+
d
(
0
)
]
-
1
[
a
1
y
a
2
y
⋮
a
r
y
]
.Obviously, this solution depends on the initial valuesβ
1
(
0
)
,
β
2
(
0
)
,
…
,
β
r
(
0
), and d
(
0
). If all the absolute values |
β
j
-
β
j
(
0
)
| for i
=
1,2
,
…
,
r are quite small, the estimation will be considered successful. On the contrary, if each |
β
j
-
β
j
(
0
)
| is rather large, we will insert β
j calculated in the previous step into (14) as a new β
j
(
0
). Then we compute the updated values of β
j from (14) and insert them back to (14) as the new β
j
(
0
). Iterate this process until all the absolute values |
β
j
-
β
j
(
0
)
| can be ignored. Since a
1
y
,
a
2
y
,
…
,
a
r
y are fixed in (14), the larger the value of d is, the smaller the absolute values of β
1
-
β
1
(
0
)
,
β
2
-
β
2
(
0
)
,
…
,
β
r
-
β
r
(
0
) are. Therefore, the value of d should not be too large; otherwise the times of iteration will be increased. The boundary for selecting the value of d depends on whether the residual sum of squares is decreasing.


### 2.1. Steps for Calculating the Parameters in (3)

There are two independent variables (aging timet and aging temperature T) and four unknown parameters (B, A, C, and α) in (3). The steps for solving the nonlinear equations for the four parameters are as follows.(a)
Calculate the partial derivatives ofy in (3) with respect to B, A, C, and α, respectively; and we obtain
(17)
∂
y
∂
B
=
e
-
A
e
-
C
/
T
t
α
,
∂
y
∂
A
=
-
B
e
-
A
e
-
C
/
T
t
α
+
e
-
C
/
T
t
α
,
∂
y
∂
C
=
B
A
2
t
α
T
e
-
A
e
-
C
/
T
t
α
-
C
/
T
,
∂
y
∂
α
=
-
B
A
t
α
ln
⁡
t
e
-
A
e
-
C
/
T
t
α
-
C
/
T
.
(b)
Select the initial iteration values of the parameters; that is,β
(
0
)
=
(
B
(
0
)
,
A
(
0
)
,
C
(
0
)
,
α
(
0
)
). Whether the selection of initial values is appropriate will determine the amount of calculation and the convergence of iteration process. This paper uses TSM to estimate the initial values of the parameters according to the aging data in the related paper [6]. These values are also considered the initial values of the parameters in the process of random simulation in Section 3.
(c)
Insert then sets of observations ((
y
i
,
t
i
,
T
i
), i
=
1,2
,
…
,
n) and the partial derivatives in (17) and β
(
0
) into (15), and obtain each of the coefficient values in (14). For the first iteration, set the initial value; that is, d
=
d
(
0
)
=
0.01
a
11, calculate from (14) the β in (16), and then insert the estimate of β into the original expression (12) to calculate the residual sum of squares: Q
(
0
)
=
∑
i
=
1
n
[
y
i
-
y
(
T
i
,
t
i
;
B
,
A
,
C
,
α
)
]
2. Obviously, the smaller the value of Q
(
0
) is, the better it is.
(d)
For the second iteration, setβ
(
0
)
=
β, and d
=
1
0
m
d
(
0
) for m
=
-
1,0
,
1,2
,
….
(i)
First, setm
=
-
1; that is, d
=
0.1
d
(
0
), and then obtain the new values β
=
(
B
(
1
)
,
A
(
1
)
,
C
(
1
)
,
α
(
1
)
). Next, calculate the new residual sum of squares: Q
(
1
)
=
∑
i
=
1
n
[
y
i
-
y
(
T
i
,
t
i
;
B
(
1
)
,
A
(
1
)
,
C
(
1
)
,
α
(
1
)
)
]
2.
(ii)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
0; that is, d
=
d
(
0
), recalculate β, and recalculate the residual sum of squares Q
(
1
).
(iii)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
1; that is, d
=
10
d
(
0
), and recalculate β and Q
(
1
).
(iv)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
2; that is, d
=
100
d
(
0
), and recalculate β and Q
(
1
). If Q
(
1
)
<
Q
(
0
), the second iteration is done. Keep on increasing the values of β until Q
(
1
)
<
Q
(
0
), and the fourth step is finished.
(e)
For the third iteration, take the terminal values at the second iteration ofd, β, and Q
(
1
) as the new d
(
0
), β
(
0
), and Q
(
0
), respectively. Repeat the whole process of the second iteration until a new Q
(
1
)
<
Q
(
0
).
(f)
Iterate the procedure as in processes (d) and (e) untilmax
⁡
1
≤
j
≤
r
⁡
|
β
j
-
β
j
(
0
)
|
≤
eps (tolerance) is satisfied. But we have to notice that the value of d should not be too large at this time; otherwise max
⁡
1
≤
j
≤
r
|
β
j
-
β
j
(
0
)
|
≤
eps would hold even though the actual iteration failed.


## 2.1. Steps for Calculating the Parameters in (3)

There are two independent variables (aging timet and aging temperature T) and four unknown parameters (B, A, C, and α) in (3). The steps for solving the nonlinear equations for the four parameters are as follows.(a)
Calculate the partial derivatives ofy in (3) with respect to B, A, C, and α, respectively; and we obtain
(17)
∂
y
∂
B
=
e
-
A
e
-
C
/
T
t
α
,
∂
y
∂
A
=
-
B
e
-
A
e
-
C
/
T
t
α
+
e
-
C
/
T
t
α
,
∂
y
∂
C
=
B
A
2
t
α
T
e
-
A
e
-
C
/
T
t
α
-
C
/
T
,
∂
y
∂
α
=
-
B
A
t
α
ln
⁡
t
e
-
A
e
-
C
/
T
t
α
-
C
/
T
.
(b)
Select the initial iteration values of the parameters; that is,β
(
0
)
=
(
B
(
0
)
,
A
(
0
)
,
C
(
0
)
,
α
(
0
)
). Whether the selection of initial values is appropriate will determine the amount of calculation and the convergence of iteration process. This paper uses TSM to estimate the initial values of the parameters according to the aging data in the related paper [6]. These values are also considered the initial values of the parameters in the process of random simulation in Section 3.
(c)
Insert then sets of observations ((
y
i
,
t
i
,
T
i
), i
=
1,2
,
…
,
n) and the partial derivatives in (17) and β
(
0
) into (15), and obtain each of the coefficient values in (14). For the first iteration, set the initial value; that is, d
=
d
(
0
)
=
0.01
a
11, calculate from (14) the β in (16), and then insert the estimate of β into the original expression (12) to calculate the residual sum of squares: Q
(
0
)
=
∑
i
=
1
n
[
y
i
-
y
(
T
i
,
t
i
;
B
,
A
,
C
,
α
)
]
2. Obviously, the smaller the value of Q
(
0
) is, the better it is.
(d)
For the second iteration, setβ
(
0
)
=
β, and d
=
1
0
m
d
(
0
) for m
=
-
1,0
,
1,2
,
….
(i)
First, setm
=
-
1; that is, d
=
0.1
d
(
0
), and then obtain the new values β
=
(
B
(
1
)
,
A
(
1
)
,
C
(
1
)
,
α
(
1
)
). Next, calculate the new residual sum of squares: Q
(
1
)
=
∑
i
=
1
n
[
y
i
-
y
(
T
i
,
t
i
;
B
(
1
)
,
A
(
1
)
,
C
(
1
)
,
α
(
1
)
)
]
2.
(ii)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
0; that is, d
=
d
(
0
), recalculate β, and recalculate the residual sum of squares Q
(
1
).
(iii)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
1; that is, d
=
10
d
(
0
), and recalculate β and Q
(
1
).
(iv)
IfQ
(
1
)
<
Q
(
0
), the second iteration is done. But if Q
(
1
)
≥
Q
(
0
), set m
=
2; that is, d
=
100
d
(
0
), and recalculate β and Q
(
1
). If Q
(
1
)
<
Q
(
0
), the second iteration is done. Keep on increasing the values of β until Q
(
1
)
<
Q
(
0
), and the fourth step is finished.
(e)
For the third iteration, take the terminal values at the second iteration ofd, β, and Q
(
1
) as the new d
(
0
), β
(
0
), and Q
(
0
), respectively. Repeat the whole process of the second iteration until a new Q
(
1
)
<
Q
(
0
).
(f)
Iterate the procedure as in processes (d) and (e) untilmax
⁡
1
≤
j
≤
r
⁡
|
β
j
-
β
j
(
0
)
|
≤
eps (tolerance) is satisfied. But we have to notice that the value of d should not be too large at this time; otherwise max
⁡
1
≤
j
≤
r
|
β
j
-
β
j
(
0
)
|
≤
eps would hold even though the actual iteration failed.


## 3. Random Simulation and Result Analysis

### 3.1. Random Simulation and Data Processing

Random simulation is a method which uses random numbers to conduct computer simulation. The sample observations obtained by random sampling were utilized to estimate the parameters of the models (see, e.g., [10, 11]).This paper uses MatLab programming to simulate the data. Follow the steps below.(1)
Determine the model and initial values of the parameters.
(2)
Select a constantT randomly to compute T
i
=
T
+
10
i
(
i
=
1,2
,
…
,
n
).
(3)
Assume thatt follows the uniform distribution on the interval (1,100) and generates the random numbers from (1,100) as t. The number of times of simulations is N.
(4)
After obtaining all the simulated values, insert them into the model and add toy by a random number following the uniform distribution on (−0.1, 0.1). We can simulate n sets of subsamples eventually.In this paper,T
=
370, n
=
5, and N
=
20. The initial values of the parameters are B
=
1, A
=
8000, C
=
5000, and α
=
0.6. Then we can simulate 5 sets of data (each with 20 numbers) in Table 1.Table 1
Data fory-t-Tby random simulation.
T
1
=
380
T
2
=
390
T
3
=
400
T
4
=
410
T
5
=
420
t
1
y
1
t
2
y
2
t
3
y
3
t
4
y
4
t
5
y
5
2
0.90164
2
1.06520
2
0.97163
14
0.74929
6
0.95249
5
0.96176
7
0.86762
6
0.90061
15
0.77756
12
0.70445
7
0.87269
14
0.88055
11
0.94456
21
0.85007
14
0.81271
9
0.84975
19
0.86093
16
0.89346
25
0.74324
15
0.8594
15
0.83473
28
0.90947
23
0.81057
27
0.65854
27
0.59827
21
0.84506
31
0.84498
23
0.81645
28
0.81902
38
0.61045
22
0.90798
35
0.84552
30
0.87126
29
0.68834
43
0.52327
38
0.90154
41
0.85828
32
0.81683
37
0.66011
46
0.62483
46
0.85955
47
0.76119
41
0.68749
43
0.62437
49
0.53331
49
0.87239
51
0.89269
49
0.67741
52
0.66685
52
0.47344
58
0.80548
60
0.72179
60
0.63145
52
0.66147
55
0.48434
61
0.87837
65
0.76753
64
0.70384
56
0.70915
56
0.48452
65
0.75141
66
0.75146
65
0.76225
59
0.57080
63
0.55642
73
0.82056
67
0.76682
73
0.62422
63
0.57548
65
0.49338
74
0.91427
77
0.65527
75
0.57640
66
0.62424
78
0.39203
79
0.77127
87
0.82239
76
0.59947
72
0.67121
79
0.51148
80
0.80846
92
0.63481
80
0.65660
76
0.58539
80
0.54383
93
0.85508
93
0.72213
84
0.72900
78
0.60945
82
0.38676
98
0.74497
94
0.75969
87
0.72015
79
0.60864
85
0.43803
99
0.82579
95
0.74208
88
0.74480
89
0.64069
98
0.40408According to the simulated data, we first use MatLab programming to figure out that the approximate value ofα is 0.80 and then calculate the values of B, A, C, and α by TSM using SPSS software. The results are displayed in Table 2.Table 2
Analysis for two-step method.
Statistics
T
1
T
2
T
3
T
4
T
5
R
2
0.335
0.643
0.671
0.564
0.809
Sig.
0.008
0.000
0.000
0.000
0.000
B
i
^
0.906
0.973
0.955
0.831
0.917
K
i
^
0.003
0.008
0.011
0.01
0.022
Residual sum of squares
0.056
0.098
0.149
0.107
0.231
B
^
=
(
∑
‍
B
i
^
)
/
5
=
0.916
=
0.916, A
^
=
193493.655, and C
^
=
6746.318.
The total residuals sum of squares = 0.641.Then we estimate the parameters in the bivariate nonlinear model (3) by MM using SPSS software (the initial values of the parameters here are the same as those used in random simulation). The results are displayed in Table 3.Table 3
Analysis for Marquardt's method.
Sum of squares
DF
Mean square
Regression
55.235
4
13.809
Residuals
0.325
96
0.003
Total
55.5602
100
R
2 = 1 − (residual sum of squares)/(total) = 0.843.
B
^
=
0.973, A
^
=
35616.923, C
^
=
5689.481, and α
^
=
0.633.


### 3.2. Result Analysis

(1) In regression analysis, the coefficient of determinationR
2 = 1 − (residual sum of squares)/(total sum of squares of deviations) is a statistic that measures the goodness of fit of the model under consideration. Specifically, the coefficient of determination is a statistical measure of how well the regression line fits the real data points. The closer to 1 the R
2 is, the closer the points of practical observations to the sample line and the better the goodness of fit of the model are. From Tables 2 and 3, it can be seen that the R
2 of MM is larger than that of TSM, which indicates that the prediction model of MM is more suitable for fitting the simulated data.(2) Comparing the estimates of the parameters obtained by the two methods, we can easily find out that the estimates of the parameters of MM are closer to the initial values, which indicates that using MM to estimate the parameters in they
-
t
-
T model is more reasonable.(3) We compare their residual sum of squares. The residual sum of squares for MM is 0.325, and that for TSM is 0.641. The former is only half of the latter. Obviously, the prediction error of they
-
t
-
T model resulting from using MM is smaller, and the precision of its fitted equation is higher.


## 3.1. Random Simulation and Data Processing

Random simulation is a method which uses random numbers to conduct computer simulation. The sample observations obtained by random sampling were utilized to estimate the parameters of the models (see, e.g., [10, 11]).This paper uses MatLab programming to simulate the data. Follow the steps below.(1)
Determine the model and initial values of the parameters.
(2)
Select a constantT randomly to compute T
i
=
T
+
10
i
(
i
=
1,2
,
…
,
n
).
(3)
Assume thatt follows the uniform distribution on the interval (1,100) and generates the random numbers from (1,100) as t. The number of times of simulations is N.
(4)
After obtaining all the simulated values, insert them into the model and add toy by a random number following the uniform distribution on (−0.1, 0.1). We can simulate n sets of subsamples eventually.In this paper,T
=
370, n
=
5, and N
=
20. The initial values of the parameters are B
=
1, A
=
8000, C
=
5000, and α
=
0.6. Then we can simulate 5 sets of data (each with 20 numbers) in Table 1.Table 1
Data fory-t-Tby random simulation.
T
1
=
380
T
2
=
390
T
3
=
400
T
4
=
410
T
5
=
420
t
1
y
1
t
2
y
2
t
3
y
3
t
4
y
4
t
5
y
5
2
0.90164
2
1.06520
2
0.97163
14
0.74929
6
0.95249
5
0.96176
7
0.86762
6
0.90061
15
0.77756
12
0.70445
7
0.87269
14
0.88055
11
0.94456
21
0.85007
14
0.81271
9
0.84975
19
0.86093
16
0.89346
25
0.74324
15
0.8594
15
0.83473
28
0.90947
23
0.81057
27
0.65854
27
0.59827
21
0.84506
31
0.84498
23
0.81645
28
0.81902
38
0.61045
22
0.90798
35
0.84552
30
0.87126
29
0.68834
43
0.52327
38
0.90154
41
0.85828
32
0.81683
37
0.66011
46
0.62483
46
0.85955
47
0.76119
41
0.68749
43
0.62437
49
0.53331
49
0.87239
51
0.89269
49
0.67741
52
0.66685
52
0.47344
58
0.80548
60
0.72179
60
0.63145
52
0.66147
55
0.48434
61
0.87837
65
0.76753
64
0.70384
56
0.70915
56
0.48452
65
0.75141
66
0.75146
65
0.76225
59
0.57080
63
0.55642
73
0.82056
67
0.76682
73
0.62422
63
0.57548
65
0.49338
74
0.91427
77
0.65527
75
0.57640
66
0.62424
78
0.39203
79
0.77127
87
0.82239
76
0.59947
72
0.67121
79
0.51148
80
0.80846
92
0.63481
80
0.65660
76
0.58539
80
0.54383
93
0.85508
93
0.72213
84
0.72900
78
0.60945
82
0.38676
98
0.74497
94
0.75969
87
0.72015
79
0.60864
85
0.43803
99
0.82579
95
0.74208
88
0.74480
89
0.64069
98
0.40408According to the simulated data, we first use MatLab programming to figure out that the approximate value ofα is 0.80 and then calculate the values of B, A, C, and α by TSM using SPSS software. The results are displayed in Table 2.Table 2
Analysis for two-step method.
Statistics
T
1
T
2
T
3
T
4
T
5
R
2
0.335
0.643
0.671
0.564
0.809
Sig.
0.008
0.000
0.000
0.000
0.000
B
i
^
0.906
0.973
0.955
0.831
0.917
K
i
^
0.003
0.008
0.011
0.01
0.022
Residual sum of squares
0.056
0.098
0.149
0.107
0.231
B
^
=
(
∑
‍
B
i
^
)
/
5
=
0.916
=
0.916, A
^
=
193493.655, and C
^
=
6746.318.
The total residuals sum of squares = 0.641.Then we estimate the parameters in the bivariate nonlinear model (3) by MM using SPSS software (the initial values of the parameters here are the same as those used in random simulation). The results are displayed in Table 3.Table 3
Analysis for Marquardt's method.
Sum of squares
DF
Mean square
Regression
55.235
4
13.809
Residuals
0.325
96
0.003
Total
55.5602
100
R
2 = 1 − (residual sum of squares)/(total) = 0.843.
B
^
=
0.973, A
^
=
35616.923, C
^
=
5689.481, and α
^
=
0.633.


## 3.2. Result Analysis

(1) In regression analysis, the coefficient of determinationR
2 = 1 − (residual sum of squares)/(total sum of squares of deviations) is a statistic that measures the goodness of fit of the model under consideration. Specifically, the coefficient of determination is a statistical measure of how well the regression line fits the real data points. The closer to 1 the R
2 is, the closer the points of practical observations to the sample line and the better the goodness of fit of the model are. From Tables 2 and 3, it can be seen that the R
2 of MM is larger than that of TSM, which indicates that the prediction model of MM is more suitable for fitting the simulated data.(2) Comparing the estimates of the parameters obtained by the two methods, we can easily find out that the estimates of the parameters of MM are closer to the initial values, which indicates that using MM to estimate the parameters in they
-
t
-
T model is more reasonable.(3) We compare their residual sum of squares. The residual sum of squares for MM is 0.325, and that for TSM is 0.641. The former is only half of the latter. Obviously, the prediction error of they
-
t
-
T model resulting from using MM is smaller, and the precision of its fitted equation is higher.


## 4. Conclusion

In this paper, we demonstrate that the MM is more suitable for estimating the parameters of the aging lifetime model by the theoretical analysis and random simulation. Our method not only avoids a plenty of tedious calculation in TSM but also adds the damping factor, which loosens the limitation of selecting the initial values. Furthermore, compared with TSM, MM greatly decreases the fitting error between the predicted values and the practical observed values, and we obtain the best-fit parameters. In addition, the model estimated by MM has higher fitting precision than that by TSM.We note that the parametric estimation in this paper can also be used in the prediction of lifetime of other materials, such as composite materials (see [12]).



---

*Source: 950401-2014-06-01.xml*