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=Fielded Systems=
#REDIRECT [[RGA Models for Repairable Systems Analysis]]
The previous chapters presented analysis methods for data obtained during developmental testing. However, data from systems in the field can also be analyzed in RGA. This type of data is called fielded systems data and is analogous to warranty data. Fielded systems can be categorized into two basic types: one-time or nonrepairable systems and reusable or repairable systems. In the latter case, under continuous operation, the system is repaired, but not replaced after each failure. For example, if a water pump in a vehicle fails, the water pump is replaced and the vehicle is repaired.
Two types of analysis are presented in this chapter. The first is repairable systems analysis where the reliability of a system can be tracked and quantified based on data from multiple systems in the field. The second is fleet analysis where data from multiple systems in the field can be collected and analyzed so that reliability metrics for the fleet as a whole can be quantified.
{{repairable systems analysis rga}}
 
{{parameter estimation fielded rga}}
 
{{fleet analysis rsa}}
 
==General Examples==
<br>
===Example 5 (fleet data)===
<br>
Eleven systems from the field were chosen for the purposes of a fleet analysis. Each system had at least one failure. All of the systems had a start time equal to zero and the last failure for each system corresponds to the end time. Group the data based on a fixed interval of 3000 hours and assume a fixed effectiveness factor equal to 0.4. Do the following:
<br>
<br>
1) Estimate the parameters of the Crow Extended model.
<br>
2) Based on the analysis does it appear that the systems were randomly ordered?
<br>
3) After the implementation of the delayed fixes, how many failures would you expect within the next 4000 hours of fleet operation.
 
<br>
 
{|style= align="center" border="1"
|-
|colspan="2" style="text-align:center"|Table 13.9 - Fleet data for Example 5
|-
!System
!Times-to-Failure
|-
|1|| 1137 BD1, 1268 BD2
|-
|2|| 682 BD3, 744 A, 1336 BD1
|-
|3|| 95 BD1, 1593 BD3
|-
|4|| 1421 A
|-
|5|| 1091 A, 1574 BD2
|-
|6|| 1415 BD4
|-
|7|| 598 BD4, 1290 BD1
|-
|8|| 1556 BD5
|-
|9|| 55 BD4
|-
|10|| 730 BD1, 1124 BD3
|-
|11|| 1400 BD4, 1568 A
|}
 
====Solution to Example 5=====
<br>
:1) Figure Repair1 shows the estimated Crow Extended parameters.
:2) Upon observing the estimated parameter  <math>\beta </math>  it does appear that the systems were randomly ordered since  <math>\beta =0.8569</math> . This value is close to 1. You can also verify that the confidence bounds on  <math>\beta </math>  include 1 by going to the QCP and calculating the parameter bounds or by viewing the Beta Bounds plot. However, you can also determine graphically if the systems were randomly ordered by using the System Operation plot as shown in Figure Repair2. Looking at the Cum. Time Line, it does not appear that the failures have a trend associated with them. Therefore, the systems can be assumed to be randomly ordered.
 
<math></math>
[[Image:rga13.8.png|thumb|center|300px|Estimated Crow Extended parameters.]]
<br>
<br>
[[Image:rga13.9.png|thumb|center|300px|System Operation plot.]]
<br>
 
===Example 6 (repairable system data)===
<br>
This case study is based on the data given in the article Graphical Analysis of Repair Data by Dr. Wayne Nelson [23]. The data in Table 13.10 represents repair data on an automatic transmission from a sample of 34 cars. For each car, the data set shows mileage at the time of each transmission repair, along with the latest mileage. The + indicates the latest mileage observed without failure. Car 1, for example, had a repair at 7068 miles and was observed until 26,744 miles. Do the following:
<br>
 
:1) Estimate the parameters of the Power Law model.
:2) Estimate the number of warranty claims for a 36,000 mile warranty policy for an estimated fleet of 35,000 vehicles.
 
<br>
 
{|style= align="center" border="1"
|-
|colspan="5" style="text-align:center"|Table 13.10 - Automatic transmission data
|-
!Car
!Mileage
!
!Car
!Mileage
|-
|1|| 7068, 26744+|| || 18|| 17955+
|-
|2|| 28, 13809+|| || 19|| 19507+
|-
|3|| 48, 1440, 29834+|| || 20|| 24177+
|-
|4|| 530, 25660+|| || 21|| 22854+
|-
|5|| 21762+|| || 22|| 17844+
|-
|6|| 14235+|| || 23|| 22637+
|-
|7|| 1388, 18228+|| || 24|| 375, 19607+
|-
|8|| 21401+|| || 25|| 19403+
|-
|9|| 21876+|| || 26|| 20997+
|-
|10|| 5094, 18228+|| || 27|| 19175+
|-
|11|| 21691+|| || 28|| 20425+
|-
|12|| 20890+|| || 29|| 22149+
|-
|13|| 22486+|| || 30|| 21144+
|-
|14|| 19321+|| || 31|| 21237+
|-
|15|| 21585+|| || 32|| 14281+
|-
|16|| 18676+|| || 33|| 8250, 21974+
|-
|17|| 23520+|| || 34|| 19250, 21888+
|}
 
====Solution to Example 6====
<br>
:1) The estimated Power Law parameters are shown in Figure Repair3.
:2) The expected number of failures at 36,000 miles can be estimated using the QCP as shown in Figure Repair4. The model predicts that 0.3559 failures per system will occur by 36,000 miles. This means that for a fleet of 35,000 vehicles, the expected warranty claims are 0.3559 * 35,000 = 12,456.
 
<math></math>
[[Image:rga13.10.png|thumb|center|300px|Entered transmission data and the estimated Power Law parameters.]]
 
<math></math>
[[Image:rga13.11.png|thumb|center|300px|Cumulative number of failures at 36,000 miles.]]
 
<br>
 
===Example 7 (repairable system data)===
<br>
Field data have been collected for a system that begins its wearout phase at time zero. The start time for each system is equal to zero and the end time for each system is 10,000 miles. Each system is scheduled to undergo an overhaul after a certain number of miles. It has been determined that the cost of an overhaul is four times more expensive than a repair. Table 13.11 presents the data. Do the following:
<br>
:1) Estimate the parameters of the Power Law model.
:2) Determine the optimum overhaul interval.
:3) If  <math>\beta <1</math> , would it be cost-effective to implement an overhaul policy?
 
<br>
 
{|style= align="center" border="1"
|-
|colspan="3" style="text-align:center"|Table 13.11 - Field data
|-
!System 1
!System 2
!System 3
|-
|1006.3|| 722.7|| 619.1
|-
|2261.2|| 1950.9|| 1519.1
|-
|2367|| 3259.6|| 2956.6
|-
|2615.5|| 4733.9|| 3114.8
|-
|2848.1|| 5105.1|| 3657.9
|-
|4073|| 5624.1|| 4268.9
|-
|5708.1|| 5806.3|| 6690.2
|-
|6464.1|| 5855.6|| 6803.1
|-
|6519.7|| 6325.2|| 7323.9
|-
|6799.1 ||6999.4|| 7501.4
|-
|7342.9 ||7084.4|| 7641.2
|-
|7736 ||7105.9|| 7851.6
|-
|8246.1|| 7290.9|| 8147.6
|-
| || 7614.2|| 8221.9
|-
| || 8332.1|| 9560.5
|-
| || 8368.5|| 9575.4
|-
| || 8947.9||
|-
| || 9012.3 ||
|-
| || 9135.9 ||
|-
| || 9147.5 ||
|-
| || 9601 ||
|}
 
====Solution to Example 7====
:1) Figure Repair5 shows the estimated Power Law parameters.
:2) The QCP can be used to calculate the optimum overhaul interval as shown in Figure Repair6.
:3) Since  <math>\beta <1</math>  then the systems are not wearing out and it would not be cost-effective to implement an overhaul policy. An overhaul policy makes sense only if the systems are wearing out. Otherwise, an overhauled unit would have the same probability of failing as a unit that was not overhauled.
 
<math></math>
[[Image:rga13.12.png|thumb|center|300px|Entered data and the estimated Power Law parameters.]]
<br>
<br>
[[Image:rga13.13.png|thumb|center|300px|The optimum overhaul interval.]]
 
===Example 8 (repairable system data)===
<br>
Failures and fixes of two repairable systems in the field are recorded. Both systems start from time 0. System 1 ends at time = 504 and system 2 ends at time = 541. All the BD modes are fixed at the end of the test. A fixed effectiveness factor equal to 0.6 is used. Answer the following questions:
<br>
:1) Estimate the parameters of the Crow Extended model.
:2) Calculate the projected MTBF after the delayed fixes.
:3) What is the expected number of failures at time 1,000, if no fixes were performed for the future failures?
 
====Solution to Example 8====
:1) Figure CrowExtendedRepair shows the estimated Crow Extended parameters.
:2) Figure CrowExtendedMTBF shows the projected MTBF at time = 541 (i.e. the age of the oldest system).
:3) Figure CrowExtendedNumofFailure shows the expected number of failures at time = 1,000.
 
<math></math>
[[Image:rga13.14.png|thumb|center|300px|Crow Extended model for repairable systems.]]
<br>
<br>
[[Image:rga13.15.png|thumb|center|300px|MTBF's from Crow Extended model.]]
<br>
<br>
[[Image:rga13.16.png|thumb|center|300px|Cumulative number of failures at time = 1,000.]]
 
<br>

Latest revision as of 06:11, 23 August 2012

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