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Fielded Systems

Repairable Systems Analysis

Using the Power Low to Analyze Complex Repairable Systems

General Examples Using Fielded Systems

Fielded Systems Data

Parameter Estimation for Repairable Systems

Suppose that the number of systems under study is K and the qth system is observed continuously from time Sq to time Tq, q = 1, 2, ... , K. During the period [Sq,Tq], let Nq be the number of failures experienced by the qth system and let Xi,q be the age of this system at the ith occurrence of failure, i = 1, 2, ... , Nq. It is also possible that the times Sq and Tq may be observed failure times for the qth system. If $X_{N_{q},q}=T_{q}$ then the data on the qth system is said to be failure terminated and Tq is a random variable with Nq fixed. If $X_{N_{q},q}<T_{q}$ then the data on the qth system is said to be time terminated with Nq a random variable. The maximum likelihood estimates of λ and β are values satisfying the Eqns. (4) and (5).

MATH (4)
MATH (5)

where 0ln0 is defined to be 0. In general, these equations cannot be solved explicitly for $\widehat{\lambda }$ and $\widehat{\beta },$ but must be solved by iterative procedures. Once $\widehat{\lambda }$ and $\widehat{\beta }$ have been estimated, the maximum likelihood estimate of the intensity function is given by: MATHIf S1 = S2 = ...= Sq = 0 and T1 = T2 =...= Tq (q = 1, 2, ... , K) then the maximum likelihood estimates $\widehat{\lambda }$ and $\widehat{\beta }$ are in closed form.

MATH (6)
MATH (7)

The following examples illustrate these estimation procedures.

Example 1

For the data in Table 13.1, the starting time for each system is equal to 0 and the ending time for each system is 2000 hours. Calculate the maximum likelihood estimates $\widehat{\lambda }$ and $\widehat{\beta }$.

Table 13.1 - Repairable system failure data

System 1 (Xi1)

System 2 (Xi2)

System 3 (Xi3)

1.2

1.4

0.3

55.6

35.0

32.6

72.7

46.8

33.4

111.9

65.9

241.7

121.9

181.1

396.2

303.6

712.6

444.4

326.9

1005.7

480.8

1568.4

1029.9

588.9

1913.5

1675.7

1043.9

 

1787.5

1136.1

 

1867.0

1288.1

 

 

1408.1

 

 

1439.4

 

 

1604.8

N1 = 9

N2 = 11

N3 = 14

Solution

Since the starting time for each system is equal to zero and each system has an equivalent ending time, the general Eqns. (4) and (5) reduce to the closed form Eqns. (6) and (7). The maximum likelihood estimates of $\hat{\beta}$ and $\hat{\lambda}$ are then calculated as follows: MATHMATH

The system failure intensity function is then estimated by:MATHFigure 13.2 is a plot of $\widehat{u}(t)$ over the period (0, 3000). Clearly, the estimated failure intensity function is most representative over the range of the data and any extrapolation should be viewed with the usual caution.

Figure

Figure 13.2: Instantaneous Failure Intensity vs. Time plot