Transcript RELIABILITY IMPROVEMENT WARRANTIES

MODULE 3: CASE STUDIES
Professor D.N.P. Murthy
The University of Queensland
Brisbane, Australia
CASE STUDY - 1
Source: Warranty Cost Analysis
[Chapter 13]
Item:
Aircraft component
Problem:
 Item supplied without warranty
 Customer requests two-year warranty
 Select warranty terms
 Predict costs
Data and Analysis
 Operational data
88 failure times
65 service times
 Repairable item: Repaired back to new
 Special analysis required
 (“incomplete data”)
 Weibull distribution fits the data
(Increasing failure rate: Shape parameter > 1)
Data and Analysis [Cont.]
Summary
Failed items: MTTF = 2580 flight hours
Service times: Mean = 2081 flight hours
Estimate of overall MTTF: 3061 fl. hrs.
(Based on Weibull distribution.)
Policies Considered
1. Nonrenewing FRW, W = 5000 fl. hrs.
2. Nonrenewing FRW, W = 2 years,
calendar time
3. Rebate PRW, W = 5000
4. Rebate PRW, W = 2 years
(Average usage rate: 3061 flight hours per
year)
Results
Costs: cs = $9000 cb = $17500 cr = $5400
Policy
1
2
3
4
Estimated Cost
$15,669
18,978
13,300
16,098
CASE STUDY - 2
Product: Microwave Links
Major components
 Crystal Receiver
 Crystal Transmitter
 2Mb card
 2Mb PCM card
CASE STUDY - 2




Sold in lots (size varying from 1 - 100)
Sold with 3 year FRW policy
Failed items returned in batches
No information about
– the time at which the item was put in use
– the time at which the item failed
Manufacturing Cost Data
Customer
Number
Labour
Material
Overhead
Order
of Systems
Cost
Cost
Number
in Batch
($)
($)
($)
034-1605
2
1329
11474
034-1616
24
18155
034-1758
2
034-1809
Direct
Total
Average
Cost
Cost per
($)
($)
System ($)
929
0
13732
6866.00
131811
12355
1035
163356
6806.50
2665
11940
1865
0
16470
8235.00
38
30600
178243
21415
0
230258
6059.42
034-1899
2
1800
10250
1260
0
13310
6655.00
056-1976
4
2812
29448
1966
0
34226
8556.50
068-1838
4
3966
27528
2776
0
34270
8567.50
072-1955
2
1238
18782
867
0
20887
10443.50
099-1429
100
72900
529900
51100
7100
661000
6610.00
106-1682
16
14180
132379
9928
1386
157873
9867.06
Recovered Expenses
Repair Cost Data
Repair
Number
Labour
Material
Overhead
Job
of
Cost
Cost
Number
Failed Parts
($)
($)
($)
034-W348
4
230
435
034-W355
1
83
034-W364
2
034-W378
Direct
Total
Average
Cost
Cost per
($)
($)
System ($)
161
0
826
206.50
0
62
0
145
145.00
108
2
75
0
185
92.50
1
54
0
37
0
91
91.00
040-R215
1
63
0
44
0
107
107.00
040-W241
1
63
0
44
0
107
107.00
040-W252
3
111
2
78
0
191
63.67
Recovered Expenses
Failure Records
Repair
Failure
Customer
Despatch
Age
Serial No
Job #
Date
Job #
Date
(Days)
040-W285
9/16/94
040-1145
7/14/93
429
87
196-R344
5/3/95
196-1306
7/30/93
642
1376
040-W376
3/8/95
040-1168
9/20/93
534
1396
040-W376
3/8/95
040-1168
9/20/93
534
1397
040-W376
3/8/95
040-1168
9/20/93
534
1398
040-W376
3/8/95
040-1168
9/20/93
534
1300
040-W252
7/11/94
040-1359
10/29/93
255
1559
040-W252
7/11/94
040-1360
10/29/93
255
1498
Survival Records
Customer
Despatch
Number
Date of
Age (Days) at
Job
Date
of
Data
Date of Data
Systems
Collection
Collection
Number
196-1306
7/30/93
2
6/9/95
679
196-1322
9/11/93
2
6/9/95
636
355-1336
9/24/93
2
6/9/95
623
040-1358
10/29/93
2
6/9/95
588
040-1359
10/29/93
2
6/9/95
588
040-1360
10/29/93
2
6/9/95
588
040-1361
10/29/93
2
6/9/95
588
034-1393
11/12/93
10
6/9/95
574
Data used in MATLAB program for analysis
Batch
Batch Age of
Number of
Failure
Number
Size
Batch
Failures in Batch
Times
j
Nj
Tj
nj
xji
1
2
679
1
642
2
2
636
0
3
2
623
0
4
2
588
0
5
2
588
2
255
350
6
2
588
3
255
354
ESTIMATES BASED ON
DATA ANALYSIS
 Manufacturing Cost per Item, Cs =
$7316.18
 Repair Cost per Item, Cr = $143.94
 Weibull scale parameter,  = 0.43233
 Weibull shape parameter,  = 1.57479
1.40
Failure Rate r(t)
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0
1
2
3
4
Time (Years)
5
6
7
0.35
0.3
0.25
f(t)
0.2
0.15
0.1
0.05
0
0
1
2
3
4
Time (Years)
5
6
7
REPAIR COST PER ITEM VERSUS
NUMBER OF ITEMS RETURNED
400.00
Repair Cost per Item ($)
350.00
300.00
250.00
200.00
150.00
100.00
50.00
0.00
0
5
10
15
Number of Items
20
25
WARRANTY SERVICING COST FOR
DIFFERENT WARRANTY PERIODS.
Warranty
Expected
Warranty Warranty Servicing Cost
Period
Number of Failures Servicing
as a percentage of
(Years)
in Warranty Period
Cost ($)
W
M(W)
cr*M(W)
1
0.27
38.43
0.53
2
0.80
114.48
1.56
3
1.51
216.78
2.96
4
2.37
341.02
4.66
5
3.37
484.61
6.62
6
4.49
645.78
8.83
the manufacture cost
WARRANTY SERVICING COSTS FOR
DIFFERING REPAIR COSTS
Warranty
Repair
Period
Cost
Repair Cost
Repair Cost
Repair Cost
plus 1 Standard plus 2 Standard plus 3 Standard
Deviation
Deviations
Deviations
(Years)
$143.94
$238.87
$333.81
$428.75
1
38.43
63.78
89.12
114.47
2
114.48
189.99
265.49
341.00
3
216.78
359.77
502.76
645.75
4
341.02
565.95
790.89
1015.82
5
484.61
804.25
1123.90
1443.55
6
645.78
1071.73
1497.69
1923.64
WARRANTY SERVICING COSTS FOR
VARYING SCALE PARAMETER 
Warranty
99.9%
99%
95%
Point
95%
99%
99.9%
Period
Confidence
Confidence
Confidence
Estimate
Confidence
Confidence
Confidence
Interval for 
Interval for 
Interval for 
for
Interval for 
Interval for 
Interval for 
Lower Limit
Lower Limit
Lower Limit

Upper Limit
Upper Limit
Upper Limit
(Years)
0.42295
0.42499
0.42674
0.43233
0.43793
0.43968
0.44172
1
37.12
37.41
37.65
38.43
39.21
39.46
39.75
2
110.59
111.43
112.15
114.48
116.82
117.56
118.42
3
209.42
211.01
212.38
216.78
221.21
222.61
224.24
4
329.43
331.93
334.10
341.02
347.99
350.19
352.75
5
468.14
471.70
474.77
484.61
494.52
497.64
501.28
6
623.84
628.58
632.68
645.78
658.98
663.15
668.00
WARRANTY SERVICING COSTS FOR
VARYING SHAPE PARAMETER 
Warranty
99.9%
99%
95%
Point
95%
99%
99.9%
Period
Confidence
Confidence
Confidence
Estimate
Confidence
Confidence
Confidence
Interval for 
Interval for 
Interval for 
for
Interval for 
Interval for 
Interval for 
Lower Limit
Lower Limit
Lower Limit

Upper Limit
Upper Limit
Upper Limit
(Years)
1.56540
1.56744
1.56920
1.57479
1.58038
1.58214
1.58418
1
38.73
38.67
38.61
38.43
38.25
38.19
38.13
2
114.63
114.60
114.57
114.48
114.38
114.35
114.32
3
216.25
216.37
216.47
216.78
217.10
217.20
217.31
4
339.27
339.65
339.98
341.02
342.06
342.39
342.78
5
481.11
481.87
482.52
484.61
486.70
487.36
488.13
6
640.03
641.27
642.35
645.78
649.23
650.32
651.58
CASE STUDY:
PHOTOCOPIER
[Service Agent Perspective]
DATA FOR MODELLING
Count
60152
60152
60152
132079
132079
132079
220832
220832
220832
220832
252491
252491
252491
252491
365075
365075
Day
29
29
29
128
128
128
227
227
227
227
276
276
276
276
397
397
Component
Cleaning Web
Toner Filter
Feed Rollers
Cleaning Web
Drum Cleaning Blade
Toner Guide
Toner Filter
Cleaning Blade
Dust Filter
Drum Claws
Drum Cleaning Blade
Cleaning Blade
Drum
Toner Guide
Cleaning Web
Toner Filter
 Supplied by the
service agent
 Single machine:
Failures over a 5 year
period
 Part of the data is
shown on the left
side
MODELLING
 One can either use number of copies
(count) or time (age) as the variable in
modelling at both component and
system level
 The count and time between failures are
correlated (correlation coefficient 0.753)
COMPONENT FAILURES
Failed Component Frequency
Cleaning web
15
Toner filter
6
Feed rollers
11
Drum blade
2
Toner guide
7
Cleaning blade
7
Dust filter
6
Drum claws
5
Crum
6
Ozone filter
8
Upper fuser roller
5
Upper roller claws
5
TS block front
2
Charging wire
6
Lower roller
2
Optics PS felt
3
Drive gear D
2
 Photocopier has
several components
 Frequency
distribution of
component failures
is given on the left
COMPONENT FAILURES
SYSTEM LEVEL MODELLING
SERVICE CALLS
 Service calls modelled as a point
process through rate of occurrence
of failure (ROCOF) which defines
probability of service call in a short
interval as a function of age (time)
 ROCOF: Weibull intensity function
  : Scale parameter
(  1)

 (t )  ( t
/ )
  : Shape parameter
SERVICE CALLS
 The shape parameter  > 1 implies that
service call frequency increases (due to
reliability decreasing) with time (age)
 Data indicates that this is indeed the
case. The next slide verifies this where
TTF denotes the time between service
calls.
Trend Analysis for TTF-System
Linear Trend Model
Yt = 75.6950 - 1.70911*t
Actual
Fits
Actual
Fits
TTF-System
100
50
MAPE:
MAD:
MSD:
0
0
10
20
Time
30
40
44.989
14.479
344.096
MODELLING ROCOF
40
30
20
10
250
500
750
1000
1250
1500
MODELLING ROCOF
 Time used as the variable in the
modelling
  = 157.5 days,  = 1.55
 Estimated average number of service
calls per year:
Year
Estimated Service Calls
1.





3.7
7.1
9.4
11.3
13.0





14.6 16.0 17.3 18.5 19.7
COMPONENT LEVEL MODELLING
COMPONENT: CLEANING WEB
MODELLING
 Failed components replaced by new
ones
 Time to failure modelled by a failure
distribution function F(t)
 The form of the distribution function
determined using the failure data
available (black-box modelling)
MODELLING





Several distribution function were
examined for modelling at the
component level. Some of them were:
2- and 3-parameter (delayed) Weibull
Mixture Weibull
Competing risk Weibull
Multiplicative Weibull
Sectional Weibull
COMPONENT LEVEL
 A list of the different distributions
considered can in found in the following
book: Murthy, D.N.P., Xie, M. and Jiang,
R. (2003), Weibull Models, Wiley, New
York
 We consider modelling based on both
“counts” and “age”
Cleaning Web
Counter
HISTOGRAM
(COUNTS)
5
Frequency
4
3
2
1
0
0
40000
80000 120000 160000 200000 240000
cleaning web counter
Cleaning Web Day
HISTOGRAM
(AGE)
6
Frequency
5
4
3
2
1
0
50
100
150
200
cleaning web day
250
WPP PLOT
 WPP plot allows one to decide if one of
the Weibull models is appropriate for
modelling a given data set
 For 2-parameter Weibull: WPP is a
straight line
 For more on WPP plot, see Weibull
Models by Murthy et al (cited earlier)
NOTATION






Two sub-populations
Scale parameters 1, 2
Shape parameters 1, 2
Location parameter

Mixing parameter p
error: (square of the error between model
and data on the WPP)
MODEL FIT
 




t0
p
error
Weibull(2)
1.060
*
80035
*
*
*
*
0.543
Delayed Weibull
1.060
*
80035
*
0
*
*
0.543
Mixture
0.851
5.230
79377
67923
*
*
0.674
0.092
Multiplicative
11.013
1.060
390693
80090
*
*
*
0.543
Competing Risk
0.630
1.266
1046923
96110
*
*
*
0.502
Sectional
0.926
1.199
107232
84078
3483.4
11832.9
*
0.484
Model
MIXTURE MODEL (COUNT)
SPARES NEEDED
 The average number of spares needed
each year can be obtained by solving the
renewal integral equation. See, the book
on Reliability by Blischke and Murthy
(cited earlier) for details. It is as follows:
Year
Estimated Cleaning Webs










2.83
3.04
3.04
3.04
3.04
3.04
3.04
3.04
3.04
3.04
REFERENCE
 For further details of this case study, see,
Bulmer M. and Eccleston J.E. (1992),
Photocopier Reliability Modeling Using
Evolutionary Algorithms, Chapter 18 in
Case Studies in Reliability and
Maintenance , Blischke, W.R. and Murthy,
D.N.P. (eds) (1992), Wiley, New York
Thank you