Industrial Organisation Class Notes uOttawa MCG 5171B
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Industrial Organisation...
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EMP5103 – Lecture 1
Thursday, January 10, 2013
EMP 5103 Reliability, Quality and Safety Engineering Dhillon, B, Reliability, Quality and Safety for Engineers, CRC Press, Boca Raton, Florida, 2004
Marks Report Mid Term Final Exam
30% 30% 40%
Mar 28 Feb 28 Apr 4
Report Format
1. 2. 3. 4.
Title Summary, it is not an introduction to the report. What is the main highlight of the document? What is contained in the report?
Introduction – familiarize reader on topic sections. Main Body – 2.1, 2.2, etc… Conclusions - future research and studies. References Books 1. Gorman, D, Human Reliability, Program Press, New York, 1986. Journal Articles 2. Williams, R, Engineering Design, Journal of Mechanical Engineering, Vol.10, 1982, pp 10-20. Conference Proceedings 3. Reiche, H., Reliability Management, Proceedings of the American Society for Quality Control Annual Conference, 1982, pp 10-30. Report 4. Gorman, S., Technical Management, Report No.1801, 1987, Available from the Department of Mechanical Engineering, University of Ottawa, Ontario, Canada.
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EMP5103 – Lecture 1
Thursday, January 10, 2013
Course Outline 1. 2. 3.
4.
5. 6. 7. 8. 9. 10. 11.
Reliability background and introductory mathematics. Failure data collection and Sources Static Reliability Evaluation Models Series system Parallel system K-out-n system, series-parallel system, parallel-series system Dynamic Reliability Evaluation Models Exponential distribution Weibull distribution Series, Parallel. K-out-n, series-parallel, parallel-series, Standby Redundancy- Related Topics Reliability Testing Reliability and Maintainability Management Safety Management Robot Reliability and Safety Quality Management Quality Control Charts
1. Reliability Background and Introductory mathematics Reliability Definition Reliability is the probability that an item will perform its function adequately for the desired period of time when operated according to specified conditions.
History of Reliability
Power system problems (1930’s) World War II (V1 and V2 rockets) U.S. Department of Defense (1945-1950) Advisory Group on the Reliability of Electronic Equipment (AGREE)—1950 AGREE report (1957) ( it took AGREE 7 years to report on the reliability)
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EMP5103 – Lecture 1
Thursday, January 10, 2013
History of Quality Control
C.N. Frazeo of Telephone Laboratories made use of statistical methods in inspection problems—1916 Walter A. Shewhart developed quality control charts—1924 Etc…
History of Safety
Pling, the Elder (AD 23-79) - Historia Naturalis –wearing of protective masks by workers to stop the inhalation of toxic substances Miner’s Safety Lamp-Humphrey Davy-18th century American Public Health Association—1885 - Occupational Health and Safety 1st Text book-Industrial Safety by H.W. Heinrich—1931
Publications Reliability Publications
IEEE Transactions on Reliability (1953) Microelectronics and Reliability (1962) Reliability Engineering and System Safety International Journal of Quality and Reliability Management Reliability Reviews Annual Reliability and Maintainability Symposium (1954)
Safety Publications
Journal of Safety Research (USA) National Safety Needs (USA) System Safety Conferences
Quality Control Publications American Society for Quality Control (ASQC) –Journals and Conferences
Design Reliability
Design specification (MTBF-mean time between failure ,MTTR-mean time to repair, Etc) Reliability allocation 2000 hrs
1000 hrs
1000 hrs
Allocate top reliability of Arm to subsystems right down to components’ level. 3
EMP5103 – Lecture 1
Thursday, January 10, 2013
Data collection and Analysis – writing information to databank and retrieving it FMEA-Failure Mode and Effect Analysis and FTA- Fault Tree Analysis (methodologies) Reliability growth – build a prototype and test it
R
t
Reliability demonstration – takes place at customer facility, performed at the system level and is typically set up as a success test Reliability warranty – determine what percentage of the failure population can be covered financially and estimating the time at which this portion of the population will fail Failure data feedback – citing the causes/reasons for failure occurrences
Reliability Areas and Applications: 1. 2. 3. 4. 5. 6. 7. 8.
Reliability general Mechanical reliability Software reliability Human reliability Reliability optimization Reliability growth modeling (see graph above) Power system reliability Life cycle costing (LCC) LCC = AC + OC with AC = Acquisition Cost and OC = Ownership Cost
9. Failure data collection and analysis 10. Maintainability and maintenance 11. Structural reliability 12. Robot reliability and safety 13. Etc… Safety Areas and Applications:
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EMP5103 – Lecture 1
Thursday, January 10, 2013
14. System safety 15. Safety management 16. Accidents 17. Etc… Quality Control Areas and Applications: 18. Statistical quality control 19. Food, textile, medical areas (health care), farm, etc…
Mathematics Probability Properties:
0 P( X ) 1 P( X 1 X 2 X 3 ... X n ) P ( X 1 ) P ( X 2 ) ... P ( X n ) (mutually exclusive events)
P( X 1 X 2 X 3 ... X n ) 1 (1 P( X i )) (independent events)
n
i 1
If n 2 : P ( X 1 X 2 ) P( X 1 ) P ( X 2 ) P( X 1 ) P( X 2 ) (independent events) P(S ) 1 S = sample space P( S ) 0
P( X 1 X 2 X 3 ... X n ) P ( X 1 ) P ( X 2 )...P ( X n ) (independent events)
Reliability Related Formulas: Cumulative Distribution Function: t
F (t )
f ( x ) dx
f(t): F(t):
failure density function cumulative distribution function
Failure density function: f (t )
dF (t ) dt
Total Area: F () 1
Reliability: R F 1
R is reliability and F is Failure
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EMP5103 – Lecture 1
Thursday, January 10, 2013 t
0
t
R (t ) 1 F (t ) 1 f ( x )dx f ( x) dx t
0
t
f ( x)dx f ( x)dx 1
t
t
f ( x) dx 1 f ( x )dx 0
Hazard rate: (t )
f (t ) f (t ) R (t ) 1 F (t )
Failure Time Distributions Exponential Distribution: f (t ) e t t: time λ: Constant failure rate
t
F (t ) ( )e t dt 1 e t
λ(t)
o
λ
R (t ) 1 F (t ) e t
(t )
f (t ) e t t R (t ) e
t
Example: 0.002 failures/year If t 10 F (t ) 1 e t 1 e 0.002*10 0.0198 R (t ) e 0.002*10 0.9802
Weibull Distribution: f (t )
b
(t )b1e
b (t )
for t ≥ α
and b,β,α > 0
b, β, and α are shape, scale, and location parameters, respectively
6 α=0
α=1
EMP5103 – Lecture 1
F (t ) 1 e
Thursday, January 10, 2013
b (t )
b = 1: exponential distribution
b = 2: Reyleigh distribution
b f (t ) (t ) R (t )
(t ) b 1 e
b 2; (t )
b
(t ) b 1
λ(t)
t b 1
b 1; (t )
b (t )
For α = 0: b
e
(t )
b (t )
b=1 1
2
t t
λ(t)
See page “16a” for function recapitulation.
t
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EMP5103 – Lecture 1
Thursday, January 10, 2013
Bathtub Hazard Rate Curve:
λ(t)
Burn-in period
Useful life period
Wear out period
t (t ) kct c1 (1 k )bt b1 e t
b
For b,c,β,λ > 0 , and 0 ≤ k≤ 1 t ≥ 0 And c = 0.5 and b = 1 to get the shape above for the bathtub b= 0.5 : bathtub curve b=1 : Extreme value distribution b,c = shape parameters β,λ = scale parameters t = time Mean Time To Failure (MTTF): E (t ) MTTF
tf (t )dt 0
MTTF
R(t )dt 0
MTTF lim R ( s ) s 0
Exponential Distribution:
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EMP5103 – Lecture 1
Thursday, January 10, 2013
f (t ) e t
E (t ) MTTF te t dt 0
F (t ) 1 e
1
t
R (t ) e t
MTTF e t dt 0
1
1 R( s) s MTTF lim s 0
1 1 s
2. Failure Data Collection and Sources 2.1 Purposes of Collecting Failure Data 1. To calculate hazard rate (failure rate) of an item. 2. To make decisions regarding the introduction of a redundant item. 3. To perform trade-off studies (cost-reliability studies). 4. To conduct the item’s replacement studies. 5. To perform preventive maintenance studies of the item. 6. To conduct effective design reviews. 7. To determine the maintenance needs of a new item. 8. To perform life cycle cost studies. 9. To predict the reliability and the availability of the item. 10. To recommend design changes for improving the item’s reliability. 11. Etc… 2.2 Data Collection Sources 1. 2. 3. 4. 5. 6. 7. 8.
Past experience with similar or identical material. Inspection data produced by quality control and manufacturing groups. Failure reporting system developed and used by customers. Reports generated by the repair facility. Data obtained during the breadboard or the development of the item. Tests: field demonstration, environmental qualification, field installation. Factory acceptance testing of equipment, modules, and assemblies. Warranty claims: Warranty Claims
Product failure related data
General data (product related) 9
EMP5103 – Lecture 1
Thursday, January 10, 2013
General Data: o Serial number and identification of model o Starting date of the warranty o Date of production o Name of the producing plant Product Failure Related Data: o Failed part o Failure: date, type, and severity o Cost of repair o Warranty coverage type o Age of the product at failure o Etc… Government Industry Data Exchange Program (GIDEP) Typical Deficiencies in Failure Data Sources: 1. 2. 3. 4. 5. 6. 7. 8.
Non-standardized data reporting. Poor description of the item in question. Time-consuming data feedback. Lack of accuracy in repair times required for many parts. Inconsistent and vague definitions and terms on reporting forms. Difficulty in pinpointing the cause of failure. Government security of company proprietary classification of the data. Etc…
Failure Data Forms: 1. Description of the item in question. 2. Location of the hardware. 3. Form number. 4. Serial number of the failed part. 5. Name of the manufacturer. 6. Failure description. 7. Date the failure occurred. 8. Date the failure was detected. 9. Date the failure was corrected. 10. Form completion date. 11. Serial number of the replaced part. 12. Manufacturer of the replaced part. 13. Time taken for repair. 14. Operating hours from previous failure. 15. Cost of repair. 16. Name of the repairperson. 17. Signatures. 18. Etc… See pp. 22 and 23.
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EMP5103 – Lecture 1
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MIL-HDBK-217: very important document that includes mathematical formulas (failure rates…). Best source for reliability (pp 23). Data Sources Information Documents: 1. Dhillon, B., Mechanical Reliability, American Institute of Aeronautics and Astronautics, Washington, D.C., 1988. Lists over 55 data sources. 2. Dhillon, B., Human Reliability with Human Factors, Pergamon Press, New York, 1986. Lists 24 human reliability related data sources. 3. Dhillon, B., Robot Reliability and Safety, Springer-Verlag, New York, 1991. List over 55 data sources. 4. Dhillon, B., Life Cycle Costing, Gordon and Breach Science Publisher. Lists over 20 data sources. See pp. 28. MIL-HDBK-217:
Example: Failure rate evaluation of an electronic component
ea b C exp kt λb: base failure rate t: Absolute temperature k: Boltzmann’s constant ea: activation energy C: constant
b Fq Fe ... Fq: the factor which takes into consideration the part quality level Fe: the factor which takes into consideration the influence of environment c Constant failure rate Example: Tunnel diode c b Fe Fq failures/106 hours Fq (quality level - Jan) = 5 note Jan is the unit Fe (environment – ground, Benign) = 1 note Benign is the unit λb (base failure rate) = 0.044 failures/106 hours c 0.044 1 5 0.22 failures/106 hours Note: failure rate is exponentially distributed.
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EMP5103 – Lecture 1
Thursday, January 10, 2013
Failure Rate Evaluation of Equipment: k
t q j ( g Fq ) j j 1
k = the number of different generic part classifications qj = the jth generic part quantity Fq = the jth generic part quality factor g = the constant failure rate of the jth component
3. Static Reliability Evaluation Models: Series Configuration:
1
2
3
k
E1
E2
E3
Ek
Ei: Successful event i
P ( E1 E 2 E 3 ...E k ) P ( E1 ) P ( E 2 )...P ( E k ) Let : Ri P ( E i ) k
RS R1 R2 R3 ...Rk Ri i 1
Example:
1
2
R1=0.8
R2=0.9
RS 0.8 0.9 0.72
IFF Ri 0.95 k
RS 1 Fi
(non-identical units)
i 1
Where Fi is the failure probability of unit i. For identical units: RS 1 kF
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EMP5103 – Lecture 1
Thursday, January 10, 2013
Parallel Configuration:
E1 1
E1 2
E1 3
E1 α
Fp P ( E1 E 2 E3 ...E ) P ( E1 ) P ( E 2 ) P ( E3 )...P ( E ) Let : Fi P ( Ei )
Fp Fi
(for non-identical units)
i 1
R p 1 Fi i 1
For identical units:
Rp 1 F Example: A parallel system is composed of four independent and active units. The unvreliability of units 1,2,3 and 4 is 0.4, 0.3, 0.2 and 0.1 respectively. Calculate the system reliability.
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EMP5103 – Lecture 1
Thursday, January 10, 2013
0.4
0.3
0.2
0.1
1
2
3
4
R p 1 F1 F2 F3 F4 1 0.4 0.3 0.2 0.1 0.9976 Rp 1 F As for identical units
1 0.95 Rp
α=3
0.90
α=2
0.85
α=1
0
0.1
0.2
0.3
F
14
0.4
EMP5103 – Lecture 1
Thursday, January 10, 2013 not mentioned
1 0.90 Rp
α=4
0.80 α=2
0.70
α=3 α=1
0.5
0.6
0.7
0.8
0.9
F K-out-of- α-unit System:
1
2
k
α
15
1
EMP5103 – Lecture 1
Thursday, January 10, 2013
1 R i R i i k i
Rk
where ! i i!( i )! n
(R
i
i 1
(identical units)
Fi ) 1
R F 3 R 3 3R 2 F 3RF 2 F 3 1 R2 R 3 3R 2 (1 R ).....F 1 R 3
1
α=3 Parallel System
0.80 System Reliability
0.60 2-ou-of-3 System
0.40 0.20
Series System
0
0.2
0.4
0.6 Unit Reliability
RS R 3 ( Series _ system) R p 1 (1 R 3 )( Parallel _ system) Rk R 2
3
( R F ) 3 R 3 3R 2 F 3RF 2 F 3
Stop here
16
0.8
1
EMP5103 – Lecture 1
Thursday, January 10, 2013
1 0.80 System Reliability
0.60
R = 0.9
0.40
R = 0.7
0.20 R = 0.5 0
1
2
3
4
5
6
k (requirement for the system to work) Example:
A parallel system is composed of three independent and active units. At least two units must be functioning successfully for system success. The reliability of units 1, 2 and 3 is 0.8, 0.9 and 0.7 respectively. Calculate the system’s reliability. (R1 + F1)(R2 + F2)(R3 + F3) = 1 Ri: is the unit’s reliability for i= 1, 2, 3 Fi: is the unit’s unreliability for i= 1, 2, 3 R1 R2 R3 + R 1 R2 F 3 + R 1 R3 F 2 + R 2 R3 F 1 + R 1 F 2 F 3 + R 2 F 1 F 3 + R 3 F 1 F 2 + F 1 F 2 F 3 = 1 R2/3 = R1 R2 R3 + R1 R2 F3 + R1 R3 F2 + R2 R3 F1 = R1 R2 R3 + R1 R2 (1-R3) + R1 R3 (1-R2) + R2 R3 (1R1) = 0.8*0.9*0.7+0.8*0.7*(1-0.9)+….. Series-Parallel configuration:
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EMP5103 – Lecture 1
Thursday, January 10, 2013 1
2
k
1
1
1
2
2
2
3
3
3
r
r
r
r
R pj 1 F ji
Fji: is the jth subsystem, ith unit’s unreliability
i 1
k
k
r
j 1
j 1
i 1
Rsp R pj (1 F ji )
For identical units:
Rsp 1 F r
k
Example: A system is composed of two independent, identical and active subsystems. Each subsystem consists of two identical units in parallel. The unit’s failure probability is 0.2. calculate the system’s reliability. F = 0.2 and r = 2 and k = 2
R sp 1 F r
k
1 0. 2 2
2
0.9216
0.2
0.2
0.2
0.2
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EMP5103 – Lecture 1
Thursday, January 10, 2013
r=3
1
F = 0.1
r=2
0.80 0.60
Rsp
r=3
0.40 0.20
r=2
F = 0.5
r=1
1
2
3
4
5
6
k
1
2
3
k 1
2
3
r
k
Rsj R ji i 1
r r k R ps 1 1 Rsj 1 1 R ji j 1 j 1 i 1 For identical units: r R ps 1 1 R k where R is the unit’s reliability Example:
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EMP5103 – Lecture 1
Thursday, January 10, 2013
k = r = 2, F = 0.2 0.2
0.2
0.2
0.2
R ps 1 1 R k 1 1 0.8 2 0.8704 R ps 1 1 R k r
2
r=3
1
R = 0.9
r=2
0.80 Rps
r
r=1
0.60 r=3
0.40 0.20
r=2
R = 0.5
r=1
1
2
3
4
5
6
k
1. Dynamic Reliability Evaluation Models. 1.1.General Reliability Function
t ( x ) dx R t e 0
t: is time and λ is units hazard rate or the instantaneous failure rate t
F (t )
f (t ) 1 dR (t ) (t ) R (t ) R (t ) dt t
t
0
0
f (t ) dt
0
dF (t ) f (t ) dt R (t ) F (t ) 1
1
(t )dt R(t ) dR(t )
R (t ) 1 F (t )
At t=0; R(0) = 1 20
EMP5103 – Lecture 1 R (t )
t
1 dR (t ) R (t )
(t )dt 0
Thursday, January 10, 2013
1
t
(t ) ln R (t ) 0
t
R (t ) e
( t ) dt 0
Example: (t ) t
R(t ) e
dt 0
e t
1.2.Mean Time To Failure (MTTF)
MTTF R (t )dt 0
MTTF lim R ( s) s 0
dont _ have _ to _ remember : t MTTF lim R (t ) dt t 0 t
Where __ f (t ) R (t ) dt 0
Laplace _ transform : lim f (t ) lim s f ( s ) t
s 0
R(s) s Then, _ MTTF lim R ( s ) but , _ f ( s )
s 0
Example:
21
EMP5103 – Lecture 1 (t )
Thursday, January 10, 2013
k (t k 1 )
where : k : shape _ parameter
: scale _ parameter t : time t
R (t ) e
( t ) dt 0
MTTF
e
e
1 tk
1
t
kt
dt
e
1 k
1 tk
dt
0
k 1
0
1
k
1
k
t
1 t
e dt
0
Example: k 1
1
where :
0.007 failure / hour t 50hours R(50) e (0.007 )(50) 0.7047
Series configuration:
1
1 1
2
For
1 2
1 3
3
k
1 k
, k
k
Rs (t ) R (t ) e i 1
1 t i
t
as we have:
i 1
MTTFs Rs (t ) dt
k
0
0
e i 1
1 t i
R (t ) e
dt e
k
i 1
0
1
i t
t
( t ) dt
dt
0
22
e
1 1 i 1 i k
1
1 dt 0
e
1 t 1
EMP5103 – Lecture 1
Thursday, January 10, 2013
1 1 t for ( t ) ≤ 0.05 i 1 i i Hazard rate: f (t ) 1 dRs (t ) s (t ) s Rs (t ) Rs (t ) dt 1 let : i i k
Rs (t ) 1
k
s (t )
i i 1
e
k
it
(e
k
it i 1
)
i 1
k
s i i 1
λ1 k
R (t ) e
λ2
λ3
λk
1 t i
i 1
1 1 t 0.05 t : true _ for : i 1 i i k
Rs (t ) 1
Series System Non-Constant Failure Rate:
1
2
3
k
j (t ) j j t
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EMP5103 – Lecture 1
Thursday, January 10, 2013
t
R (t ) e
( t ) dt 0
R j (t ) e
j t 1 jt 1
k
Rs (t ) e
j t 1 jt 1
i 1
Rs (t ) exp
k 1 t j k j 1 jt 1 j 1
tot 1 Rs (t ) exp 1 tot tot let :
k
tot i j 1 k
tot i j 1
ttot : gamma
Rs (t ) exp exp ttot e
k
it j 1
Mean Time To Failure for Various Simple Unit and Series Systems: System Structure
MTTF 1
Z(t)=λ
2k
Z(t)=kf
1 1 1 k m 1 (m 1) m 1 m 1
Z(t)=kfm
1 n
Z1=λ1
Z2=λ2
i
Z3=λ3
11
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EMP5103 – Lecture 1
Thursday, January 10, 2013
m
Z1 k1t Z 2 k 2t
Z n knt m
λα
R p (t ) 1 Fi (t ) i 1
i 1
for : i t 0.05 :
R p (t ) 1 i t i 1
For identical units
R p (t ) 1 1 e t
0.005 failures / hours
Eg: λ = 0.005 failures/hr
1
1
m 1
λ2
λ1
R p (t ) 1 1 e i t
n
ki 1 ( m 1) 11 m 1 m 1
m
Parallel Network:
Ri (t ) e i t
t = 100 hours
R p (100) 1 1 e ( 0.005)(100) 0.845
2 redundancy can be the best increasing. R p (t ) 1 (1 e t )
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EMP5103 – Lecture 1
Thursday, January 10, 2013
1 0.80 Rp(t)
0.60 0.40 α=4 α=3 α=2
0.20
α=1
0
0.5
1
1.5
2
2.5
λt
2
1
R p (t ) 2e t e 2t
0.80
R (t ) e t
0.60 Reliability 0.40
Rs (t ) 2e t
0.20
0
0.5
1
1.5 λt
Mean Time To Failure (MTTF): MTTF
R(t )dt 0
26
2
2.5
EMP5103 – Lecture 1
Thursday, January 10, 2013
λ1
λ2
2 t R p (t ) 1 1 e 1t 1 e
R p (t ) e 1t e 2 t e (1 2 )t MTTF p
e
1t
e 2 t e (1 2 )t dt
0
e 1t e 2 t e 1 2 t 1 2 2 1
0
1 1 1 1 2 1 2
k1tm
k2tm m 1 1 1 1 m 1 m 1 1 1 1 m 1 k1 m 1 k 2 m 1 k1 k 2 m 1
MTTF
( ) t 1e t dt 0
Reliability and MTTF Functions for Various Active Configurations for Exponentially Distributed Failure Times – Identical Components: Reliability Configuration
27
EMP5103 – Lecture 1 λ
Thursday, January 10, 2013 λ
λ
λ MTTF
1
2
1
1
λ
3
n
λ
2
λ
m
λ
MTTF
1
λ
1
λ
2
λ
2
λ
m
λ
m
λ
m
λ
2
1 m 1 i 1 i
λ
2
1
1 n
n
28
MTTF
1 n
n 1 (1) j 1 j 1 j i 1 i jm
EMP5103 – Lecture 1
Thursday, January 10, 2013
MTTF
11 12
1
2
MTTF
m 1
2
1
2
n
λ
λ
k
λ
m
λ
k _ out _ of _ m MTTF
System Reliability Plots for Various Configurations:
29
1 m 1 1 k i
1 m 1 n i 1 i
EMP5103 – Lecture 1
Thursday, January 10, 2013
R(t ) e t
1
R p (t ) 1 1 e t
0.80
3
R 2 (t ) 3e 2t 2e 3t 3
0.60 Reliability 0.40
Rs (t ) e 3t
0.20
0
0.5
1
1.5
2
2.5
λt Standby System: 1
2
3
m
Assumptions: 19. switch perfect¨ 20. standby units remain as good as new 21. unit failures are independent 22. identical units Rs (t ) e t MTTF
k 1
( t ) t j! j 0
Rs (t )dt 0
k
Approximate Reliability Equations for k-out-of-n type of Systems – Identical Components:
30
EMP5103 – Lecture 1
Situation 1 unit of 2 must be working for success 1 unit of 3 must be working for success 1 unit of 4 must be working for success 1 unit of n must be working for success 2 units of 3 must be working for success 3 units of 4 must be working for success (n-1) units of n must be working for success (n-2) units of n must be working for success n
Thursday, January 10, 2013
Formula R(t)
Approximation R(t)
2 e t e 2 t
1 t 2
3e t 3e 2t e 3t
1 t 3
4e t 6e 2t 4e 3t e 4t
1 t 4
n
n t e
1 1
1
1 t n
3e 2t 2e 3t
1 3 t 2
4e 3t 3e 4t
1 6 t 2
n t 2 2
ne ( n 1)t (n 1)e nt
1
n ( n 2 ) t n 1 n t e e (2n n 2 )e ( n 1)t 2 2
1
n t 3 3
n!
Note: ( n i )!i! i Standby Systems: Situation 1 unit of 2 must be working for success 1 unit of n must be working for success
Formula R(t)
e t te t
e t te t
1 t 2 e t ... 1 t n 1 e t 2 (n 1)!
Fault Trees Analysis: Developed in Bell labs (minuteman launch control system) in 1962. 31
Approximation R(t) 1
1
t 2 2
t n n!
EMP5103 – Lecture 1
Thursday, January 10, 2013
Fault Tree Symbols: - AND Gate: output
inputs The AND gate denotes that an output event occurs if and only if all the input events occur. - OR Gate: output
inputs The OR gate denotes that an output event occurs if any one or more of the input events occur. - Resultant event:
A rectangle denotes an event which results from the combination of fault events through the input of a logic gate. - Basic fault event:
32
EMP5103 – Lecture 1
Thursday, January 10, 2013
A circle represents a basic fault event or the failure of an elementary component. The failure parameters such as unavailability probability, failure and repair rates of a fault event are obtained from the empirical data or other sources. Example: Build the constant a fault tree of a simple system concerning a room containing a switch and a light bulb. Assume the switch only fails to close. In addition, the top event is dark room.
Switch fails to close
?
Dark Room
?
Power off
Power failure
Bulb burnt out
Fuse failure
P ( A B ) P ( A) P ( B ) P ( A) P ( B ) n
P ( X 1 X 2 X 3 ... X n ) 1 1 P ( X i ) i 1
P ( X 1 X 2 X 3 ... X n )
n
P( X i )
i 1
output
inputs P ( A.B ) P ( A) P ( B )
Example:
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EMP5103 – Lecture 1
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A 1
2
5
6 D 7
3
4 B
8 C
E F
0.1421
0.1296
F
?
C
A
E
0.36
B
0.0144
0.36
1
2
3
4
0.2
0.2
0.2
0.2
D
34
7
8
0.2
0.2
0.36
5
6
0.2
0.2
EMP5103 – Lecture 1
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R A 0.8 0.8 0.64 F A 1 R A 1 0.64 0.36 FB 1 R B 1 0.64 0.36 RC 1 F A FB 1 0.36 0.36 0.8704 FD 1 R D 1 0.8 0.8 0.36 R E 1 0.36 0.36 0.2 0.9856 R N RC R E 0.8578 FN 1 R N 0.1421
2. Miscellaneous Redundancy Related Topics. 2.1.Determining Number of Parallel Units for Specified System Reliability System
R S 1 1 R n
1 R n
(1 RS )
n ln(1 R ) ln(1 RS ) n
ln(1 R S ) ln(1 R )
Example: RS=0.98
n
RS=0.85
ln(1 0.98) 2units ln(0.85)
RS 1 (1 0.85) 2 0.9775
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2.2.Triple Modular Redundancy (TMR)vs. Simplex System TMR: R R R Simplex System: R
R 2 3R 2 2 R 3 3
R F 3
R 3 3R 2 F 3RF 2 F 3 R 3 3R 2 F R 3 3R 2 (1 R) ... RSimplex R R 2 RSimplex 3
3R 2 2 R 3 R 3R 2 R 2 1 2 R 2 3R 1 0 b b 2 4ac 2a R1, 2 1or1 / 2 R1, 2
R (t ) e t 1 / 2 e t t 0.693
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RSimplex (t ) e t
1 Reliability 0.80
R 2 (t ) 3
0.60 0.50 0.40 0.20 λt=0.693 (mission time) 0
0.5
1
1.5
2
2.5
λt
2.3.Common-Cause Failures and Redundancy Definition: A common-cause failure is defined as any instance where multiple units or components fail due to a single cause. Some-causes: 1. Equipment design deficiency 2. Operations and maintenance errors 3. External normal environment (dust, temperature, humidity, vibrations, etc…) 4. External catastrophe – Fire, Flood, Earthquake, Tornado 5. Common Manufacturer 6. Common External Power Source 7. Functional Deficiency
2.4.Redundancy Equation Approximation (Effective Failure Rate or Average Failure Rate) With Repair: k-out-of-n system: 23. All units are active with equal unit failure rate and (n-q)-out-of –n required for success:
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1 μ 2
k
n
n!q 1
( n q )
(n q 1)! q 1 MTTFk n 1 MTTFk n
n
for : 0 :
( n q ) n
n
1 i inq
24. Two active on-line units with different failure repair rates. One out of two is required for success: A
λA,μA
B
λB,μB
A B 12 A B A B A B A B A B
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Unit A down Unit B up
λA
λB
μA Both Up
Both down λB μB
Unit B down Unit A up
λA
A B 0 A2B AB A 2 B 2 AB
12
Example: A system has five active units, each with a failure rate of 220 failures/10 6 hours and only three are required for successful operation. If one unit fails, it takes an average of three hours to repair it to an active state. What is the effective failure rate of this configuration? n5 q 53 2
1
35
3
5! 220 10 6
3 0.00575 failure / 10 6 hours
5 2 1! 13 2
Example: A ground radar system has 2-level-weather-channel with failure rate of 50 failures/10 6 hours and a 6level-weather-channel with a failure rate of 180 failures/10 6 hours. Although the 6 level weather channel provides more comprehensive coverage, the operation of either channel will result in acceptable system operation. What is effective failure rate of the two-level-weather-channel if one of two is required and mean time to repair is 1 hour? A 50 10 6 B 180 10 6 A B 1 12
50 10 6 180 10 6 1 1 50 10 6 180 10 6 1 1 1 1 50 10 6 180 10 6
12 0.018 failures / 10 6 hours
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Example: Determine the effective failure rate for 8 of 10 identical units required with no repair. The failure rate of a single unit is 60 failures / 106 hours. n 10 q2
60 10 6 (10 2 )
10
60 10 6 10
179 failures / 10 6 hours
1i
i 10 2
8units : 60 f / 10 6 hours ( failure _ rate 480 failures / 10 6 hours )
That means that by adding 2 (from 8 to 10) more units the failure rate will drop from 480 to 179 failures / 106 hours.
3. Reliability Testing. Reliability tests can be divided into three categories: 1. Reliability Development and Demonstration Testing 2. Qualification and Acceptance Testing 3. Operational Testing Standard used: MIL-STD-471
3.1.Reliability Development and Demonstration Testing The objectives of reliability and demonstration tests are: 1. To determine if the design must be improved to meet the reliability requirement 2. To indicate any design changes needed 3. To verify improvements in design reliability
3.2.Qualification and Acceptance Testing The objectives of qualification and acceptance tests are: 1. To determine if a part, assembly, or an end item should be accepted or rejected (either on an individual or lot basis) 2. To determine if a particular design should be considered as qualified for its intended application
3.3.Operational Testing Operational Testing:
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The objectives of operational testing:
To verify the reliability analysis performed during the project. To provide data indicating necessary modifications of operational procedures and policies, as they affect reliability and maintainability. To provide information to be used in later activities.
Tests for the Validity of Assumed Failure Time Distributions:
Bartlett test (exponential distribution only) Kolmogorov-Smirnov test
Bartlett Test: Bartlett statistic is defined as:
Y k 2 12k 7k 1 ln X
S bk
X
1 k ti k i 1 k
Y ln t i i 1
where t i is the ith time to failure and k is the total number of failures in the sample. A sample of at least 20 failures is necessary for the test to discriminate effectively. If the failure times are exponentially distributed, then S bk is distributed as chi-square with k 1 degrees of freedom. Thus, a two-tailed chi-square approach is utilized.
S bk lower limit
upper limit
Example: A sample of 20 failure times (in days) of an air traffic control system is given in Table 1. Determine with the aid of the Bartlett test that the Table 1 data are representative of an exponential distribution with 90% confidence.
7 8
35 46
85 86
41
142 186
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20 19 34
X
45 63 64
111 112 141 Table 1
185 266 267
1 7 8 20 19 ... 267 96.10 20 20
Y ln t k 82.8311 k 1
82.8311 20 14.43 (7)(20) 1
ln(96.10)
S bk (12)(20) 2 Upper Limit:
, ( k 1) where 1 confidence level 1 0.9 0.1 2
2
Lower Limit:
2 1 , ( k 1) 2 0.1 , (20 1) 30.14 (from Table 12.4 in handout #1) 2
2 Upper limit:
Lower limit: 1 2
0.1 , ( 20 1) 10.12 (from Table 12.4 in handout #1) 2
The above result exhibits that there is no contradiction to the assumption of exponential distribution because S bk falls in between the upper limit and the lower limit. Confidence Limits for Mean Time Between Failures:
2 p, fd p : the quantity which is the function of the confidence coefficient fd : degrees of freedom f : the accumulated number of failures at time t* , where t* denotes the life test termination f * : the number of failures which were preassigned
m : the quantity of components which were put on test at zero time (t 0)
: the mean time between failures or mean life P : the acceptable error risk 1 p : the level of confidence
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To estimate confidence intervals, there are two cases to consider: (1) the test is terminated at a specified time,
t * , and (2) the test is terminated at a specified number of failures. t *:
Test Terminated at Time,
2
p , 2 f 2 2
,
2
2
lower limit
2 p 1 ,2 f 2
upper limit
Test Terminated at Specified Failures,
2 2 , p 2 p 2 ,2 f 1 ,2 f 2 2
lower limit
f *:
upper limit
The type of test determines the value of , for example, when a failure (failed units) is replaced or repaired (replacement test) the value of is given by:
mt * Similarly, for the non-replacement test, the value of gamma is given by: f
m f t * T j where T j denotes the j th failure j 1
Example: Fifteen components were put on test at time t=0, and testing was terminated when the 8 th failure occurred. That failure (8th) was observed at the 100th hour of testing. Calculate the components’ mean time between failures and mean life upper and lower confidence limits with 90% confidence level. Failed repaired back to its as good as new state:
(15)(100) 1500 hours 1500 187.5 hours 8 p 1 confidence level 1 0.9 0.1
2 f (2)(8) 16
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2 (2)(1500) 3000 2 376.9 hours 0.95,16 7.96 p Upper limit: (from handout #1) 2 1 ,2 f 2 2 (2)(1500) 3000 2 114 .07 hours 0.05,16 26.30 Lower limit: (from handout #1) 2 p ,2 f 2 Example: Twenty-five components were put on a test at time t=0. It was a non-replacement test and was terminated after 90 hours. During the period, five components failed at 20, 40, 30, 70, and 80 hours of operation respectively. Calculate the components’ mean time between failures and mean life upper and lower confidence units with 95% confidence level.
25 5 90 20 40 30 70 80 1800 240 2040 hours 2040 408 hours 5
2 Lower limit:
Upper limit:
p , 2 f 2 2
2
(2)( 2040) 4080 174.81 hours 23.34 (handout 1) 2 0.05 ,12 2
2 p 1 ,2 f 2 2
2
(2)(2040) 4080 1255.39 hours 3.25 0.05 (handout 1) 1 ,10 2
Economics of Testing:
Quality Control Testing
Cost of Testing:
CT P
NL n
Cost of Not-Testing:
I
C NT NR P I C R P W C R
W
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N : number of units under consideration R : total fraction defective previously observed or estimated P I : fraction defective which fails in plant P N : fraction defective which fails during the warranty period CR
W
: average cost of warranty repair
P : cost of test equipment n : number of units tested per hour L : labor and overhead rate per test hour cost of testing cost of not - testing C NT C T N
P
I
R P I C R PW C R
W
Ln
Sources of Product Unreliability: (excluding operation) Distribution of Causes (%) 20-40 40-65 15-20
Source Design and Development Quality of Components Quality of Workmanship
(t )
infant mortality failures
random failures
wear-out failures t
Causes of Infant Mortality Failures:
Inadequate test specifications Inadequate quality control Inadequate manufacturing processes Inadequate materials Improper handling Marginal components
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Overstressed components Etc…
Causes of Random Failures:
Insufficient design margin Misapplication – overstress Wrong use environment Cause “unknown” failures
Causes of Wear-Out Failures:
Material wear Aging Limited-life components Inadequate or improper preventive maintenance Etc…
7. Reliability and Maintainability Management To be continued next time …
Handout #1: Table 12.4 – Chi Square Distribution Degrees of Freedom 1 2
Probability 0.975 0.001 0.05
0.95 0.004 0.1
0.05 3.84 5.99
46
0.025 5.02 7.38
EMP5103 – Lecture 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
Thursday, January 10, 2013 0.22 0.48 0.83 1.24 1.69 2.18 2.7 3.25 3.82 4.4 5.01 5.63 6.26 6.91 7.56 8.23 8.91 9.59 12.40
0.35 0.71 1.15 1.64 2.17 2.73 3.33 3.94 4.58 5.23 5.89 6.57 7.26 7.96 8.67 9.39 10.12 10.85 13.85
7.82 9.49 11.07 12.59 14.07 15.51 16.92 18.31 19.68 21.92 22.36 23.69 25.00 26.30 27.59 28.87 30.14 31.41 36.42
9.35 11.14 12.83 14.45 16.01 17.54 19.02 20.48 21.92 23.34 24.74 26.12 27.49 28.85 30.19 31.53 32.85 34.17 39.36
Handout #2: Table 7.1 – Chi-Square Distribution Degrees of Freedom
Probability 0.975 0.05 0.48 1.24 2.18 3.25 4.40 5.63 6.91 8.23 9.59
2 4 6 8 10 12 14 16 18 20
0.95 0.10 0.71 1.64 2.73 3.94 5.23 6.57 7.96 9.39 10.85
0.90 0.21 1.06 2.20 3.49 4.87 6.30 7.79 9.31 10.87 12.44
0.05 5.99 9.45 12.59 15.51 18.31 21.03 23.69 26.30 28.87 31.41
0.025 7.38 11.14 14.45 17.53 20.48 23.34 26.12 28.85 31.53 34.17
0.01 9.21 13.28 16.81 20.09 23.21 26.22 29.14 32.00 34.81 37.57
7. Reliability and Maintainability Management 1. Dhillon, B., Engineering Reliability Management, IEEE Journal on Selected Areas in Communications, Vol. 4, 1986, pp. 1015-1025. 2. Dhillon, B., Reiche, H., Reliability & Maintainability Management, Vari Nostrand, Reinhold, New York, 1985. History:
1958 – USAF – Reliability Program Management Document, Exhibit 58-10
Steps to Improve Reliability of Engineering Products:
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Westinghouse Electric Corporation. Appoint a manager to look after reliability. Direct the manager to review present effort. Establish a high level reliability panel composed, of say, quality control, engineering, purchasing, and manufacturing managers. Define the panel’s responsibilities – have it report to top management. Establish an overall reliability and maintainability policy of the company. Provide training to people in reliability. Set reliability goals and refine them as necessary. Design products that are reliable. Conduct verification analysis of product reliability. Produce the products that are reliable. Audit reliability program on regular basis. Make use of services provided by the central reliability staff.
Reliability Engineering Department Responsibilities:
Establishing reliability policy, plan, and procedures. Reliability allocation. Reliability prediction (MIL-HDBK-217). Specification and design reviews with respect to reliability. Reliability growth monitoring. Providing reliability related inputs to design specification and proposals. Reliability demonstration (MIL-STD-471). Training reliability manpower and performing reliability-related research and development work. Monitoring subcontractors’, if any, reliability activities. Auditing the reliability activities. Failure data collection and reporting. Failure data analysis. Consulting. Etc…
Tasks of a Reliability Engineer: To be continued next class… Tasks of a Reliability Engineer: 25. Performing analysis of a proposed design. 26. Analyzing customer complaints with reliability. 27. Investigating field failures. 28. Running tests on the system, sub-system and parts. 29. Developing tests on the system, subsystem and components. 30. Budgeting the tolerable system failure down to the component level. 31. Developing a reliability program plan. 32. Determining reliability of alternative designs. 33. Providing information to designers or management concerning reliability. 34. Monitoring sub-contractor’s reliability performance. 35. Participating in evaluating requests for proposals. 36. Developing reliability models and techniques. 37. Participating in design reviews.
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38. Etc… Reliability Cost Categories: Reliability cost = PC+ AC + IFC + EFC PC: Prevention Cost AC: Appraisal Cost IFC: Internal Failure Cost EFC: External Failure Cost Prevention Cost: 39. Redundancy 40. Parts 41. Hourly cost and overhead rates for design engineers, reliability engineers, etc… Appraisal Cost: 42. Hourly cost and overhead rates for evaluation, reliability qualification, reliability demonstration, life-testing, etc… 43. Vendor assurance cost for new component qualification, inspection, etc… 44. Etc… Internal Failure Cost: 45. Hourly cost and overhead rates for troubleshooting and repair, retesting, failure analysis, etc… 46. Replaced part’s cost. 47. Spare parts inventory. 48. Etc… External Failure Cost: 49. Cost to failure or repair. 50. Replaced parts cost. 51. Cost of failure analysis. 52. Warranty administration and reporting cost. 53. Liability insurance. 54. Etc… Reliability Program Cost Estimation: RAM (Reliability and Maintainability) Program Plan (in man hours): = 2.73 (NOT)2 Where NOT = number of MIL-STD-785 tasks required. Min – 4 Max – 22 Reliability Modeling and Allocation: = 4.05 (MAC)2(NOV) Where: MAC: Modelling and Allocation Complexity Series system = 1 Simple redundancy = 2 Very complex redundancy = 3 NOV: Number of items in allocation process
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Min – 7 Max – 445 Reliability Prediction: = 4.54(LOD)2(RF)2(POC) Where: LOD: Level of detail 1 – Prediction exists 2 – Prediction made using similar system data 3 – Full MIL- HDBK-217 stress prediction RF: Report formality 1 – Internal report 2 – Formal report POC: Percentage Commercial, Hardware used 4 – 025% 3 – 2650% 2 – 5175% 1 – 76100% FMEA (Failure Modes and Effect Analysis): = 17.79(NOI) Where: NOI: number of equipment for equipment level FMEA NOI – min 3 – max 206 Reliability Testing: = 182.07(HC) Where: HC: Hardware complexity 1 if < 15000 parts 2 if between 15000 and 25000 parts 3 if > 25000
8. Safety Management In year 2000, in the USA: 55. 5,200 deaths 56. 2.9 million disabling injuries 57. It costs 131 billion dollars to the nation History: 58. Pliny the Elder (23-79 AD): grinding – wear masks 59. 1893 in the USA: Rail Safety Act 60. 1938 in the USA: Food, Drugs and Cosmetic Act 61. 1970: Occupational Safety and Health Act 62. Nuclear regulatory commission 63. Consumer product safety 64. Commission 65. National Transportation Safety Board 66. Federal Aviation Agency (FAA) 67. Etc…
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Safety Periodicals: 68. Journal of Occupational Accident 69. Professional Safety 70. Concern 71. Journal of Safety Research 72. Protection 73. National Safety News 74. Nuclear Safety 75. Accident Prevention 76. Accident Facts 77. Safety Management Journal 78. Etc… Selective Texts on Safety: 1. Heinrich, H.W., Industrial Accident Prevention, McGraw Hill, New York, 1831 2. Handley, Industrial Safety Handbook, McGraw Hill, New York, 1969 3. Gloss, D.S, Introduction to Safety Engineering, Wiley, New York, 1994 4. Hammer, W., Product Safety and Engineering, Prentice Hall, New Jersey, 1980 5. Dhillon, B.S, Safety Assessment, A Quantitative Approach, Lewis Publish, New York. 1994 Safety-Related Data Sources: 79. Government Industry Data Exchange Program (GIDEP), Fleet Missile Systems Analysis, and Evaluation Group, U.S. Navy, Corona, California. 80. International Occupational Safety and Health Information Center, Bureau International du Travail, Geneva, Switzerland. 81. Loss Management Information System (LOMIS), Gulf Canada 800 Bay Street, Toronto. Factors for Developing Safety Requirements: 82. Occupational Safety and Health Act regulations. 83. Environmental Protection Agency regulations. 84. Nuclear Regulatory Commission regulations. 85. Company safety policy with respect to plant and administrative procedures. 86. State and local government requirements. 87. Etc… Safety-Related Activities for the top management: 88. Safety training. 89. Safety Inspections. 90. Safety problem diagnosis ns solutions. 91. Accident investigations. 92. Employee participation in safety programs. Safety Engineer’s Responsibilities: 93. Accident prevention and analysis. 94. Management of safety training. 95. Acting as a consultant to management on safety-related matters. 96. Design layout of equipment with respect to safety. 97. Study human factors (ergonomics) with respect to safety. 98. Focus on the interface between the workplace and environment. 99. Keep abreast of new literature on safety. 100. Etc…
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System Safety Analysis Methods: 101. Fault tree analysis. 102. Failure Mode and Effect Analysis (FMEA). 103. Event tree analysis. 104. Job safety analysis. 105. Single point failure analysis. 106. Etc… Fault tree analysis in Safety Studies: ?
Operator fails to wear safety glasses
Chip in eye (Grinding)
Person without safety glasses other than operator s close to operation
?
?
Machine operating
?
People in the area (motive to go into area by non operator)
Safety glasses not worn
52
Operator fails to stop oeration
EMP5103 – Lecture 1
Thursday, January 10, 2013
?
Person enters to bring item to the area
Person enters to carry away items
Person enters area for other reason
0.08012
0.00012
0.08
0.1
0.06
0.02
0.01
0.1
0.2
0.03
Formulas / Models Related to Safety: 107. American National Standard Institute (ANSI), Z-1601, 1985. Title: Method of Recording and Work Injury Experience. Formula / Index: Estimating the disabling-injury frequency rate (DIFR)
DIFR
NDI (100000) EHE
EHE = the Employee-Hours of Exposure. NDI = Number of Disabling Injuries. Formula II:
DISR
D(100000) EHE
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DISR = Disabling Injury Severity Rate D = Total number of days charged Shortcomings: 108. Not sensitive enough to serve to serve as accurate indicator of safety effectiveness. 109. The smaller the work force, the less reliable the frequency rate, severity rate. 110. Many accidents, particularly the less severe ones, are never reported. 111. Etc…
9. Robot Reliability and Safety 112. 5000 – Egyptians built water-powered clocks. Chinese and Greeks built water and steam powered toys. 113. Aristotle: “If every instrument could accomplish its own work, obeying or anticipating the will of others…” 114. 1818 – Many Shelley wrote the science fiction “Frankenstein” “Machine Monster” 115. 1920 – Karl Capek wrote the science fiction “Rossums Universal Robots”. 116. 1942 – Asimov – 3 laws: o A robot may not injure a person nor, through inaction allow a person to come to harm. o A robot must always obey orders from people except in circumstances in which such orders conflict with the above (first) law. o A robot must protect its own existence except in circumstances in which it is in conflict with above two laws. 117. 1959 – Planet Corporation commercialized the first robot in USA. 118. 1967 – Japan got their fist robot. 119. 1970 – Conference about robot. 120. 1971 – Japanese Industrial Robot Association. 121. 1975 – Robot Institution of America. History of Robot Reliability and Safety: 122. 1985 – Japanese Industrial Safety and Health Association. 123. 1986 – American Nation Standard “Industrial Robots and Systems – Safety Requirements”. 124. 1986 – Robot Safety, edited by Bonney and Yong. Robot Population Worldwide: 125. 1981 – US Automotive industry o 30% of labor cost 126. Japan (135 companies concerned with robots). 127. US (100 companies concerned with robots). 128. UK (30 companies concerned with robots). 129. Today around 1 million 130. 2010 – should be about 5 million robots. Japan: 131. 132. 133. 134. 135.
Automobile (36%) Electric machinery (30%) Plastic molding products (10%) Metal working Steel making
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EMP5103 – Lecture 1 136. 137. 138. 139. USA: 140. 141. 142. 143. 144. 145. In 1990, the
Thursday, January 10, 2013
Textile Chemical Ship building Etc… Assembly (35 – 40%) Arc welding (15 – 20 % ) Material handling (30 – 35%) Paint spraying (5%) Spot welding (3 – 5%) Other areas (7 – 10%) US had about 100,000 robots
Robot Accident Examples: 146. A repair person climbed over a safety fence without shutting off the power to the robot and worked in its area while it was temporarily stopped. When the robot recommenced movement, it pushed the repair person into a grinding machine and, consequently, the person died. 147. A worker switched on a welding robot meanwhile another person was still in its working area, consequently, that person was pushed into a positioning fixture by the robot and died later. Common Reason for Deaths: 148. A human entered the “danger zone” to rectify fault. 149. The human was either pushed into another machine or crushed against something else by the robot. The robot itself did not kill the human. 150. The human struck from behind by the robot in such a situation, the concerned human was not aware that the robot was moving until it was too late. 151. Even though the human was experienced, but through appropriate training should have been alerted of existing dangers. Robot Accident Around the World: 152. Japan 153. Sweden 154. U.K. 155. U.S.A. Japan Accidents: 156. 4 persons died and several cases of injuries. 157. 1978 – 1982 (190 plants – 4341 robots) o 2 deaths o 2 cases of lost time (3 days, 7 days) Sweden Accidents: Survey data (Jan 1976 – June 1978) involving a total of 270 robots: 158. 7 accidents per year. Survey data (1979 –1983) involving a total of 36 robots: 159. 8 accidents per year. Breakdown by industry: 160. Foundries (4 accidents) 161. Plastics (4 accidents) 162. Automotive (9 accidents)
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EMP5103 – Lecture 1 163. 164.
Thursday, January 10, 2013
Metal working (16 accidents) Others (3 accidents)
U.K. Accidents: 165. 2623 robots 166. Survey 37 robot systems different designs 167. Approximately 22,000 robot production hours. 168. 73 accident occurrence 169. 1 led to human injury 170. 57 damage to machinery 171. 15 no damage accidents 172. Approximately 25% of system production time lost due to accidents. U.S.A. Accidents: - 13,000 robots (1984). - 1 death - General Motors Corporation (used robots over 23 year) - One serious injury. - One minor injury. U.S.A. – Canada Accidents: - 17 accidents: o 1 death. o 3 serious injuries. o 5 minor injuries. o 8 mal-functions. Causes of Robot Accidents: - Japanese study-causes of 18 near accidents. - Incorrect action by the robot during manual operation. - Incorrect movement of peripheral equipment during teaching or testing. - Erroneous movement of the robot during teaching or testing. - Sudden entry of the human to the robot area. - Incorrect movement of peripheral equipment during normal operation. - Etc… Possible Sources of Robot Accidents: - Engineering Factors: o Control panel failure o Robot arm’s high speed (the speed is the factor) o Poor software design o Poor control panel design o Etc… - Usability (user-friendliness) Engineering Factors. - Organizational factors: o Inadequate robot training programs-repairman, operators, programmers, etc… o Incorrect procedures for initial robots start-up. o Operator carrying robot adjustments (they know how to operate but not how to maintain). o Etc…
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Solutions to Safety-Related Problems: Improving robot reliability. Improving design of mechanical hardware components. Providing appropriate safety training to people concerned with maintenance, operation, and testing. Developing an effective sensory detection capability of the robots. Paying proper attention to human factors during the design of human workstation layout. Paying more attention to human factor during the design of the robot systems. Etc…
Robot Reliability: - Expected life: 40,000 hours. - MTBF at least 400 hours. - MTTR of 8 hours or less. - Maximum MTBF: 2500 hours - Cost of maintenance: approximately 11% of the procurement cost - Availability: 0.98 (98%) (for power station 0.9999) Publications: 1. Engelberger, J.F., Three Million Hours of Robot Field Experience, The Industrial Robot, 1974, pp. 164-168. 2. Pollard, B.W., RAM for Robots: Reliability, Availability, and Maintainability, Robotics Today, 1981, pp. 209-220. 3. Bonney, M.C., Yong, Y.F., Editors. Robot Safety, Springer, New York, 1985. 4. Dhillon, B., Survey On Robot Reliability and Safety, Microelectronics and Reliability, Vol. 27, 1987, pp.105-118. 5. Dhillon, 1991 (publications).
Robot Reliability Measures: Mean Time to Robot Failure:
Rr ( s ) Rr (t )dt slim 0
MMTFR
0 t
Rr (t ) e
r (t ) dt
MTTFR
0
PHR DTDTRF NRF
where : PHR is the production hours of the robot. DTDTRF is the downtime due to robot failure expressed in hours. NRF is the number of robot failures. Example: PHR=15,500 hours, DTDRF=200 hours, and NRF=10 MTTFR = (15500-200)/10 = 1530 hours Mean Time to Robot-Related Problems:
57
EMP5103 – Lecture 1 MTRP
Thursday, January 10, 2013
PHR DTDTRP NRP
PHR is the production hours of the robot. NRP is the number of robot-related problems. DTDTRP is the downtime due to robot-related problems expressed in hours. Example: MTRP = (20000-280)/20 = 986 hours Robot Reliability: t
Rr (t ) e
r (t ) dt 0
r (t ) r t
Rr (t ) e
r (t ) dt 0
e r t
Pareto Principle – Robot Quality Assurance: Alfredo Pareto (1848-1923) His principle in relation to quality control work simply states that there are always a few kinds of defects in the hardware manufacture which loom in occurrence frequency and severity. Robot Model with Preventive Maintenance:
μp Robot down for preventive
Robot Operating
maintenance
p
λf
λp
Robot Failed
o
μf
f
Andrei Markov (1856-1922): - Assumptions: 1. The probability of more than one transition in time interval Δt form one state to the next step is negligible. 2. The occurrences are independent. 3. The transitional probability from one state to the next state in the time interval Δt is given by λΔt, where λ is the constant failure rate associated with Markov states. 4. (λΔt) (λΔt)→ 0 i denotes the ith sate of the robot: i = O (operating) i = p (preventive maintenance), i = f (failed) Pi(t) is the probability that the robot is in state i at time t. λp is the robot preventive maintenance constant rate λf is the robot constant failure rate. μp is the robot constant repair rate due to preventive maintenance. μf is the robot constant repair rate. Po (t t ) Po (t )(1 f t )(1 p t ) P f (t ) f t Pp (t ) p t P f (t t ) P f (t )(1 f t ) Po (t ) f t Pp (t t ) Pp (t )(1 p t ) Po (t ) p t
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Po (t t ) Po (t ) ( f p ) Po (t )t P f (t ) f t Pp (t ) p t P (t t ) Po (t ) lim o ( f p ) Po (t ) P f (t ) f Pp (t ) p t t 0 dPo (t ) ( f p ) Po (t ) P f (t ) f Pp (t ) p dt dPp (t ) dt dP f (t )
p Pp (t ) Po (t ) p
f P f (t ) Po (t ) f dt At _ time _ t 0, Po (0) 1, Pp (0) P f (0) 0 Laplace _ transform : p( s)
e
st
p (t ) dt
0
p (t ) e
at
dp (t ) dt
p (s ) 1 sa sp ( s ) p (0)
f (t ) lim sf ( s ) Final value Theorem: tlim 0 s 0
sPo ( s) Po ( s) ( f p ) Po ( s) Pp ( s ) p P f ( s ) f sPo ( s) 1 ( f p ) Po ( s ) Pp ( s) p P f ( s ) f Po ( s)
( s f )( s p )
s s 2 s( f p p f ) f p p f f p
A s 2 s( f p p f ) f p p f f p f (s p ) Pp ( s ) sA p (s f ) Pf (s) sA f p Ass Po lim sPo ( s ) f p p f f p s 0
B f p p f f p Pf Pf
f p B p f B
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EMP5103 – Lecture 1
Po (t ) Pp (t ) P f (t )
f p k1k 2
p f k1k 2
f p k1k 2
Thursday, January 10, 2013
(k1 p )(k1 f ) k t (k 2 p )(k 2 f ) k t e 1 e 2 k ( k k ) k ( k k ) 1 1 2 2 1 2 p k1 p f k t f k 2 k t e 1 e 2 k ( k k ) k ( k k ) 1 1 2 2 1 2 f k1 f p k t f k 2 k t e 1 e 2 k ( k k ) k ( k k ) 1 1 2 2 1 2
( p f f p ) ( f p p f ) 2 4( f p p f f p ) k1 , k 2 2 Maximizing Income of a Robot System Subject to Failure and Repair:
λ Robot Up 0
Robot Down 1
μ dP0 (t ) P0 (t ) P1 (t ) dt dP1 (t ) P1 (t ) P0 (t ) dt At _ time _ t 0, P0 (0) 1, P1 (0) 0
e ( )t availability U (t ) P1 (t ) e ( )t unavailability A(t ) P0 (t )
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1
A(t)
1
0 t
Ass
1 MTTR
1 1 MTTF MTTR
MTTF uptime MTTR MTTF uptime downtime
Periodic Cost of the robot system maintenance crew is given by:
MC k
k MTTR
where: k is the robot system maintenance cost (constant) depending on the nature of the robot system. The expected periodic income from the robot system output:
EI I AV
where: I is the periodic income from the robot system output, if the robot system worked full time.
MTTF MTTF MTTR
EI I r AV I r
Thus the net income, NI of the robot system is:
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NI EI MC I r MTTF k NI MTTF MTTR MTTR dNI I MTTF k 0 2 dMTTR MTTF MTTR MTTR 2 MTTF MTTR* 1/ 2 I MTTF 1 k
10. Quality Management. Quality Control: This is a management function whereby control of the quality of manufactured item and raw materials is expected to prevent the manufacture of defective items. History: - 1916: C.N. Frazee - 1924: Quality control charts developed by Walter A. Shewhart. - 1946: ASQC (American Society for Quality Control) Journals (publications): - Quality Progress - Journal of Quality Technology - IEEE Transactions on Reliability - Annual Quality Congress transactions of the ASQC Some -
Elements of Quality Discipline: Statistical Quality Control Procurement Quality Control Quality Costs Applied Quality Control Quality Circles Etc…
Present Quality Trends Facing Industry: - 7 to 10% (of the total cost of production?) - The cost of quality has risen to a very high level - Customers’ quality requirements have been rising at a alarming rate - Because of the above factors, present methods and practices associated with quality are rapidly becoming outmoded. Functions of the Quality Control Engineering: - Process quality control (manufacturing aspect) - Quality control in new design - Quality control of incoming material - Inventory evaluation quality control - Special studies concerning quality control Quality Control Manual: - Benefits:
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It becomes useful when making quality-related decisions. It serves as a reference document. It can be used as a textbook when training quality personnel. It helps in continuity of operations of the quality control organization despite the personnel turnover. Information: o Responsibilities. o Statistical methodology. o Personnel. o Organization charts. o Quality policies and procedures. o Vendor quality control procedure. o Quality costs and inspection procedures. o Measuring equipment. o Defect prevention. o Promote quality o Etc… o o o o
-
Quality Costs: - Post delivery failure costs: o Warranty charges o Complain adjustment o Returned material o Etc… - Prevention costs: o Quality planning o Design review o Supplier evaluation o Process control o Training o Equipment calibration o Etc… - Costs of internal failures: o Repair and rework o Scrap o Re-inspection o Downtime of facilities because of defects o Etc… - Evaluation Costs: o Incoming material inspection o In process inspection o Assembly inspection o Review and recording data o Auditing the quality system o Etc… Indexes: - Vendor Rating Program Index:
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EMP5103 – Lecture 1
QI
Thursday, January 10, 2013
Cvq C p Cp
QI : Value _ of _ the _ quality _ cos t _ performance _ index Cvq : The _ vendor _ quality _ cos t C p : The _ purchase _ cos t QI = 1.1 or more indicates that there is an immediate need for corrective measures. Example: Vendor quality cost = $2,000 Purchase cost = $50,000 QI = (2000+50000)/50000 = 1.04 1.000 1.009: Excellent performance 1.01 1.03: Good performance -
Evaluating Quality Costs:
a 100 100 b : Quality _ Index a : Quality _ cos t b : The _ value _ of _ the _ output In real life situation, a value of θ = 105 can readily be achieved. θ = 110130: quality costs are ignored - Determining Accuracy and waste of inspector: Formula I: 100 where : θ: the percent of defects correctly identified by the regular inspector. β: is the number of defects missed by the regular inspector as reveled by the check inspector. λ: is the number of defects discovered by the regular inspector. μ: is the number of units without defects rejected by the regular inspector as revealed by the check inspector. Example: A regular inspector inspected a number of units in a lot and found 60 defects. All units (i.e.: good plus defective) of the lot were reexamined by the check inspector. Thus according to the findings of the check inspector, the values of μ and β were 10 and 15.
100 50100 76.92% 15 50 15 60 10 50
Formula II:
100 m
σ: the percent of good units rejected by the regular inspector. m: the total number of units inspected.
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Quality Control Charts: 1. The p-chart. 2. The X -chart 3. The R-chart 4. The c-chart The p-chart: Components with defects versus components free of defects. Upper control limit: UCLp = μ + 3σ Lower control limit: LCLp = μ - 3σ μ: the mean of the binomial distribution σ: the standard deviation of the binomial distribution
3σ
3σ
99.7%
N m Where : N: is the total number of defectives in classification
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EMP5103 – Lecture 1
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m: is the size of the sample β: is the number of samples
1 m
1
2
Example: 8 samples were taken from a production line. Each sample consists of 40 mechanical parts. After inspection, it was found that samples 1, 2, 3, 4, 5, 6, 7, and 8 contained 5, 6, 4, 2, 8, 10, 12, and 9 defectives respectively. Develop the p-chart. The fraction of defectives in sample 1:
5 0.1250 40
Similarly, the fraction of defectives in samples 2, 3, 4, 5, 6, 7, and 8 are0.15, 0.10, 0.05, 0.20, 0.25, 0.30, and 0.225 respectively. 56 0.175 40 8
0.1751 0.175 2 0.06 40 UCL p 3 0.3552 1
UCL p 3 0 : because _ negative _ value
0.40
UCLp=0.352
0.35 0.30 0.25 0.20
μ=0.175
0.15 0.10 0.05 0
1
2
3
4
5
6
7
Sample number
66
8
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