318-595 Statistics & Reliability Basic Statistics Review Example:

318-595 Statistics & Reliability Basic Statistics Review Example:

318-595 Statistics & Reliability Basic Statistics Review Example: The following data represents the amount of time it takes 7 people to do a 355 exam problem. X i= 2, 6, 5, 2 ,10, 8, 7 in min. n = 7 where Xi = index notation for each individual. where n = 7 people n Calculate the mean(average): n = 7 people m= X = Mean: Xi n Step 1 Step 22 Equation X = (2+6+5+2+10+8+7)/7 = 5.7 minutes n Calculate the standard deviation: Xi = Sum of the individual times where X and m = Average or Mean where i=1

( Xi - X ) = s= Step 4 53.43 (Xi - X) (Xi - X) Square each one =s= 7-1 2-5.7= -3.7 13.69 6-5.7= .3 .09 Then Add All 5-5.7 = -.7 .49 Standard Deviation: 2-5.7 = -3.7 13.69 = s = 2.98 minutes 10-5.7 = 4.3 18.49 2 8-5.7 = 2.3 5.29 7-5.7 = 1.3 1.69Step 3 (Xi - X) = 53.43 i =1 n-1 2 where = s = Standard Deviation Sum or Variance

Equation Definition: Range = Max - Min Median = Middle number when arranged low to high Mode = Most common number This Example: Range = 10 - 2 = 8 minutes Median = 6 minutes Mode = 2 minutes Std Deviation is a measure of how spreadout the data is 318-595 Statistics & Reliability Bar Chart or Histogram 25 22 20 18 15 12 10 8 5 2 0 1.238 1.240 1.242 1.244 Provides

Providesaavisual visualdisplay displayof ofdata datadistribution distribution Shape Shapeof ofDistribution DistributionMay Maybe beKey Keyto toIssues Issues 1. 2. 3. 4. 5. Normal (Bell Shaped) Uniform (Flat) Bimodal (Mix of 2 Normal Distributions) Skewed left or right Total number of bins is flexible but usually no more than 10 By using an infinite number of bins, resultant curve is a distribution Use T-Test to Compare Means, F-Test to Compare Variances 318-595 Statistics & Reliability Target Target Specification Limit Specification Limit Histogram

Specification Some Chance of Failure 1s 1s The Thehigher higherthe thenumber number (Z) (Z)in infront frontof ofthe the sigma sigmasymbol symbolthe thelower lower the thechance chanceof of producing producingaadefect defect 3s = z 66807ppm

Much Less Chance of Failure 1s1s 6s = z 3.4ppm 318-595 Statistics & Reliability Target Target Specification Limit Specification Limit Sigma vs Specification Some Chance of Failure 1s 1s The Thehigher higherthe thenumber number (Z) (Z)in infront frontof ofthe the sigma

sigmasymbol symbolthe thelower lower the thechance chanceof of producing producingaadefect defect 3s = z 66807ppm Much Less Chance of Failure 1s1s 6s = z 3.4ppm 318-595 Statistics & Reliability PPM (with 1.5 shift) 1,000,000 IRS - Tax Advice (phone-in) (140,000 PPM)

100,000 Restaurant Bills Doctor Prescription Writing 10,000 Average Average Company Company 1,000 Payroll Processing Airline Baggage Handling 100 10 Best-in-Class Best-in-Class 1 2 3 4 5 6

7 Domestic Airline Flight Fatality Rate Examples of Fault/Failure Rates on The Sigma Scale (0.43 PPM) 318-595 Statistics & Reliability MANUFACTURABILITY ASPECT Critical Measurement in Production: The desired state reality.... Unless you designed for it 318-595 Statistics & Reliability Over time, a typical process will shift and drift by ~ 1.5 Inherent Capability of the Process . . . also called short-term capability . . . reflects within group variation Shift and Drift Time 1 Time 2 Time 3 Time 4 Sustained Capability of the Process . . . also called long-term capability . . . reflects total process variation LSL

T Two Challenges: USL Two Challenges: Center the Process and Eliminate Variation! 318-595 Statistics & Reliability Other Methods to Present/Analyze Data 318-595 Statistics & Reliability Plot or Scatter Plot Linear Correlation of Input to Output 1.4 1.2 1 Output 0.8 mArms 0.6 Vrms 0.4 0.2 0 0 0.5

1 1.5 Input Used Usedto toIllustrate IllustrateCorrelation Correlationor orRelationships Relationships 318-595 Statistics & Reliability Pareto Chart Root Cause Failures Example Failure Root Causes 70 60 50 OW 40 # IW 30 20 10 0 Dehumidifier Pow er Supply

No Defect Found Workmanship Other Root Cause Used Usedto toIllustrate IllustrateContributions Contributionsof ofMultiple MultipleSources Sources Excellent Excellentwhen whendata datais isabundant abundant 318-595 Statistics & Reliability Fishbone Diagram Ambient Ambient Temp Temp Load Load Res Res Line Line Voltage Voltage Effect: Temp Of Amp For example Line

Line Frequency Frequency Volume Volume Input Input Amplitude Amplitude Illustrates Cause & Effect Relationship 318-595 Statistics & Reliability Year to Date Summary Replacement Parts Example Warranty Replacements 35 Units 40% Rate 35% 30 30% 25 25% > Warranty Rate Total Units 40

20 15 10 20% 15% 10% 5 5% 0 0% J F M A M J J A S O N

D 318-595 Statistics & Reliability Intro to Reliability Evaluation Basic Series Reli Method of an Electronic System: Component 1 Component 2 1 2 Component i Component N i N Each component has an associated reliability The System Reli ss is the sum of all the component ss = i Reli is expressed in FITs failure units x FIT = x Failures/109 hours Note: 109 hours = 1 Billion Hours 318-595 Statistics & Reliability Definitions FIT = Failures per 109 hours MTBF = 1/MTBF = 1/Mean time between failure in hours MTBF (years) = 1x109 / ( FIT * 8766 hours /year ) R(t) = (Reliability at Time t) = Probability that a system will not fail for a time period t, assuming constant failure rate, R(t) = e-t, Note: MTBF in hr-1 and t in hr

Bathtub curve- typical reliability vs. time behavior 318-595 Statistics & Reliability Example An electronics assembly product has an MTBF of 20000 hours; constant failure rate What is the probability that a given unit will work continuously for one year? Reliability R(t) = e-t = 1/MTBF = 1/20000 hr = 0.00005 hr-1 (Failure rate) t = 8766 hours (1 year) R(1yr) =e-(8766/20,000 = 0.65 = 65% In other words, the Mean Time Between failures is 20,000 hours or about 2.3 years but 35% of the units would likely fail in the first year of operation. 318-595 Statistics & Reliability Example An electronics product team has a goal of warranty cost which requires that a Minimum reliability after 1 year be 99% or higher, R(1yr) >= 0.99 What MTBF should the team work towards to meet the goal? Equations: R = e -t and MTBF = 1/ Solve for MTBF: MTBF = 1/ = 1/ {(-1/t) * ln R }, R = 0.99, t = 8766 hrs MTBF >= 872,000 hours (99.5 yrs) ! 318-595 Statistics & Reliability Some Typical Stresses Environmental: Temp, Humid, Pressure, Wind, Sun, Rain Mechanical: Shock, Vibration, Rotation, Abrasion Electrical: Power Cycle, Voltage Tolerance, Load, Noise ElectroMagnetic: ESD, E-Field, B-Field, Power Loss Radiation: Xray (non-ionizing), Gamma Ray (ionizing) Biological: Mold, Algae, Bacteria, Dust Chemical: Alchohol, Ph, TSP, Ionic 318-595 Statistics & Reliability Intro to Reliability Evaluation

Each may be impacted by other factors or stresses, : Some commonly used factors T = Temperature Stress Factor V = Electrical Stress Factor E = Environmental Factor Q = Quality Factor Overall Component = B * T * V * E * Q Where B = Base Failure Rate for Component 318-595 Statistics & Reliability Reliability Prediction Methods/Standards Bellcore (TR-TSY-000332): Developed by Bell Communications Research for general use in electronics industry although geared to telecom. Highest Stress Factor is Electrical Stress Data based upon field results, lab testing, analysis, device mfg data and US Military Std 217 Stress Factors include environment, quality, electrical, thermal US Military Handbook 217E: Developed by the US Department of Defense as well as other agencies for use by electronic manufacturers supplying to the military Describes both a parts count method as well as a parts stress method Data is based upon lab testing including highly accelerated life testing (HALT) or highly accelerated stress testing (HAST) Stress factors include environment and quality 318-595 Statistics & Reliability Reliability Prediction Methods/Standards HRD4 (Hdbk of Reliability Data for Comp, Issue 4):

Developed by the British Telecom Materials and Components Center for use by designers and manufacturers of telecom equipment Stress factors include thermal as well as environment, quality with quality being dominant Standard describes generic failure rates based upon a 60% confidence interval around data collected via telecom equipment field performance in the UK CNET: Developed by the French National Center of Telecommunications Similar to HRD4, stress factors include thermal as well as environment and dominant quality Data is based upon field experience of French commercial and military telecom equipment 318-595 Statistics & Reliability Reliability Prediction Methods/Standards Siemens AG (SN29500): Developed by Siemens for internal uniform reliability predictions Stress factors include thermal and electrical however thermal dominates Standard describes failures rates based upon applications data, lab testing as well as US Mil Std 217 Components are classified into technology groups each with tuned reliability model 318-595 Statistics & Reliability Reliability Prediction Basic Series Reli Method of an Electronic System: Component 1 Component 2 1 2

Component i i Component N N Above Reliability Prediction Model is flawed because; Components may not have constant reliability rates ss = i Component applications, stresses, etc may not be well matched by the method used to model reliability Not all component failures may lead to a system failure Example: A bypass capacitor fails as an open circuit 318-595 Statistics & Reliability 595 Standard Failure Rates in FIT (Data is not accurate in all cases) Component Type Method A Method B Method C Method D Method E BJT/FET 5.0 3.8 3.2 7.6

4.0 Switch 5.0 44.0 30.0 1.0 20.0 Metal Film Res 0.7 2.5 0.05 0.05 0.2 18.2 2.7 1.0 1.1 2.6 Varistor, tc Res

6.0 1.0 10.0 1.0 10.0 Electrolytic Cap 210 22.0 120 16.0 120 Polyester Cap 8.5 2.0 3.0 0.5 7.0 Ceramic Cap 2.0

1.0 0.5 0.25 1.2 Si PN, Shottkey, PIN Diode 2.4 1.6 3.6 1.6 2.4 Zener Diode 3.2 13.6 17.4 18.8 70.0 LED 9.0 15.0

280 65.0 1.0 BJT Dig IC <100 Gates 20.0 138 2.3 1.0 6.7 BJT Dig IC < 1000 Gates 30.0 150 4.0 1.5 10.0 MOS Dig IC < 1000 Gates 27.3 301 9.0

1.0 13.3 MOS Dig IC => 1000 Gates 55.0 550 16.0 2.2 31.0 EM Coil Relay 385 302 220 715 44.0 SSR, Optocoupler 120 105 47.0 190

12.0 BJT Linear IC < 1000 Transistors 14.0 27.0 4.3 1.0 50.0 MOS Linear IC < 1000 Transistors 19.0 54.0 9.0 1.0 13.3 Transformer < 1VA 33.0 90.0 70.0 60.0 50.0

Transformer > 1VA 3.0 9.0 7.0 6.0 5.0 Plastic Shell Connector, Plug, Jack 100.0 55.0 150.0 120.0 105.0 Metal Shell Connector, Plug, Jack 33.0 18.0 57.0 40.0 35.0 7.0

1.0 50.0 8.0 22.0 Carbon Res Pb, NCd, Li, Lio, NmH Battery 318-595 Statistics & Reliability 595 Standard Stress Factors Factor Definitions (may not represent standard models) T = Temperature Stress Factor = e[Ta/(Tr-Ta)] 0.4 Where Ta = Actual Max Operating Temp, Tr = Rated Max Op Temp, Tr>Ta V = Cap/Res/Transistor Electrical Stress Factor = e[(Va)/Vr-Va]-2.0 Where Va = Actual Max Operating Voltage, Vr = Abs Max Rated Voltage, Vr>Va E = Environmental (Overall) Factor >>> Indoor Stationary = 1.0 Indoor Mobile = 1.5 Outdoor Stationary = 2.0 Outdoor Mobile = 2.5 Automotive = 3.0 Q = Quality Factor (Parts and Assembly)

Mil Spec/Range Parts = 0.75 100 Hr Powered Burn In = 0.75 Commercial Parts Mfg Direct = 1.0 Commerical Parts Distributor = 1.25 Hand Assembly Part = 3.0 318-595 Statistics & Reliability Stress Factors Drive Simple: 595 Standard Deratings Resistors, Potentiometers <= 50% maximum power Caps/Res <= 60% maximum working voltage Transistors <= 50% maximum working voltage Note: Most discrete devices as well as linear ICs have parameters which will vary with temperature which is expressed as Tc (temp coefficient). Typically a delta or percent of change per deg C from ambient. 318-595 Statistics & Reliability MTBF Data Input Sheet for e-Reliability.com COST: $500 per report System / Equipment Name: Assembly Name: Quantity of this assembly: Parts List Number: Environment: Select One Of : GB, GF, GM, NS, NU, AIC, AIF, AUC, AUF, ARW, SF, MF, ML, or CL Parts Quality:

Select Either: Mil-Spec or Commercial/Bellcore Quantity Description ---------Bipolar Integrated Circuits IC / Bipolar, Digital 1-100 Gates IC / Bipolar, Digital 101-1000 Gates IC / Bipolar, Digital 1001-3000 Gates IC / Bipolar, Digital 3001-10000 Gates IC / Bipolar, Digital 10001-30000 Gates IC / Bipolar, Digital 30001-60000 Gates IC / Bipolar, Linear 1-100 Transistors IC / Bipolar, Linear 101-300 Transistors IC / Bipolar, Linear 301-1K Transistors IC / Bipolar, Linear 1001-10K Transistors , etc. EXAMPLE: Actual Reli Tool Input List of components, their number, Environment conditions, components quality 318-595 Statistics & Reliability Example Reliability calculation using actual MIL-HDBK-217F Failure rate of a Metal Oxide Semiconductor (MOS) can be expressed as p (C1 T C2 E ) Q L failures/1 06 hours Parameters are listed in MIL Data base. Temperature factor is modeled using Arrhenius type Eqn T 0.1 exp[ Ea / 8.617e 5(1 / Taccel 1 / Tworking )] where Ea is the activation energy. For MOS Ea 0.35 eV. For many components Ea is listed in MIL data base. 318-595 Statistics & Reliability Example Reliability report --------------------------------------------------------------------------------------| |

| | | Failure Rate in | | | | | | Parts Per Million Hours | | Description/ | Specification/ | Quantity | Quality |-------------------------| | Generic Part Type | Quality Level | | Factor | | | | | | | (Pi Q) | Generic | Total | | | | | | | | |=====================|================|==========|=========|============|============| | Integrated Circuit/ | Mil-M-38510/ | 16 | 1.00 | 0.07500 | 1.20000 |

| Bipolar, Digital | B | | | | | | 30001-60000 Gates | | | | | | | | | | | | | | Integrated Circuit/ | Mil-M-38510/ | 8 | 1.00 | 0.01700 | 0.13600 | | Bipolar, Linear | B | | | | | | 101-300 Transistors | | |

| | | | | | | | | | | Diode/ | Mil-S-19500/ | 2 | 2.40 | 0.00047 | 0.00226 | | Switching | JAN | | | | | | | | | | | | | | | | | |

| | Diode/ | Mil-S-19500/ | 4 | 2.40 | 0.00160 | 0.01536 | | Voltage Ref./Reg. | JAN | | | | | | (Avalanche & Zener) | | | | | | | | | | | | | | Transistor/ | Mil-S-19500/ | 4 | 2.40 | 0.00007 | 0.00067 | | NPN/PNP

| JAN | | | | 318-595 Statistics & Reliability Reliability Prediction Drawbacks Prediction Methods not always effective in representing future reality of a product. Tend to be pessimistic Best utilized for design comparison, only if the same method is used Single Stress Factors must be employed to represent a composite average or worst case of the population. Difficult to predict average stress levels, peak stress levels Methods give an overall average failure rate, one dimensional Time to failure distributions (Weibull) are two dimensional describing infantile failures as well as end of life failures 318-595 Statistics & Reliability Reliability Growth Methods: HALT HALT Strategy: Highly Accelerated Life Testing One or more stresses used at accelerated amplitudes from what the product would see during application Stress level is gradually increased until failure is detected Failure is then autopsied to fundamental root cause Corrective/Preventive action taken to remove chance of recurring failure Test is then restarted Must be prepared to destroy prototypes Failure must be detectable, identifiable 318-595 Statistics & Reliability Example of Accelerated Life Test (595 Team Project): Rotating Bicycle Apparatus Potential reliability stress is the periodic g-load (start-stop cycles). This causes fatigue

failure mode (cracks in ceramic material, creep of plastics, adhesives, solder electrical contacts failure). Estimation of the test protocol, plan and execution time. The start-stop requirements for cycle: 12 s to accelerate from 0 to 5 rev/sec max rotational speed (60 mph) 5 s to decelerate from 5 rev/sec to 0. 35 starts-stops cycles per day One cycle time (from start to stop) is going to be: T = 12+5+5=20s, where 5 s is added as a lag time to accommodate the transition from stopping back to starting Assuming the throughput 35 start-stops/day for 365 days/year the total number of rotation cycles for 1 year is 35*365=12775 cycles /year (=12775 start-stops). Assuming 20% overhead the total number of cycles is going to be 1.2*12775=15330 cycles/year. Test time worth of 1 year of the number of cycles is going to be 15330*20/(3600*24)=3.5 days life time, years 1 3 5 10 test time, days 3.5 10.5 17.5 35 318-595 Statistics & Reliability Reliability Growth Methods: HAST HAST Strategy: Highly Accelerated Stress Testing One or more stresses used at accelerated amplitudes from what the product would see during application Stress level is constant, time to failure is primary measurement Failure may also be autopsied to fundamental root cause Corrective/Preventive action NOT necessarily taken Test is then restarted using higher or lower stress amplitude to get additional data points Used to find empirical relationship between stress level

and time to failure (life) 318-595 Statistics & Reliability Reliability Growth Methods: HASS HASS Strategy: Highly Accelerated Stress Screening Used in production to accelerate infantile failures and keep them from shipping to customers Must have HAST data to understand how much life is expended with stress screen One or more stresses used at slightly accelerated amplitudes from what the product would see during application Common application is powered burn-in time during which electronics are powered and thermal cycled. (Example MIL-STD-883) Assemblies tested during or after burn-in for failure inducements 318-595 Statistics & Reliability Failures/Time Reliability Bathtub Curve infant mortality constant failure rate wearout Time Infant mortality- often due to manufacturing defects .. Can be screened out In electronics systems, prediction models assume constant failure rates (Bellcore model, MIL-HDBK-217F, others) Understanding wearout requires knowledge of the particular device failure physics - Semiconductor devices should not show wearout except at long times - Discrete devices which wearout: Relays, EL caps, fans, connectors, solder

318-595 Statistics & Reliability Reliability of Electronic Circuits Causes of Electronic Systems Failure Failures can generally be divided between intrinsic or extrinsic failures Intrinsic failures- Inherent in the component technology Electromigration (semiconductors, substrates) Contact wear (relays, connectors, etc) Contamination effects- e.g. channeling, corrosion, leakage CTE mismatch and other Interconnection joint fatigue Extrinsic failures- External stress to the components ESD or Electrostatic discharge energy Electrical overstress (over voltage, overload, overheat) Shock (Sudden Mechanical Impact) Vibration (Periodic Mechanical G force) Humidity or condensable water Package Mishandling, Bending, Shear, Tensile Many random and infantile failures of components are due to extrinsic factors Wearout failures are usually due to intrinsic failures 318-595 Statistics & Reliability Physics of Failure: Accumulated Fatigue Damage (AFD) is related to the number of stress cycles N, and mechanical stress, S, using Miners rule AFD N * S Exponent B comes from the S-N diagram. It is typically between 6 and 25 Example: Solder Joint Shear voids Effective cross-sectional Effective crossForce Area: D/2 F

sectional Area: D F Applied stress: S D Let = 10, then AFD(no voids) N * S 10 Applied stress: S 2* F 2 * S D AFD 1024 * N * S10 1024 * AFD(no voids) AFD with voids will age about 1000x faster than AFD with no voids Voids in solder joints 318-595 Statistics & Reliability Physics of failure: Thermal Fatigue Models Coefficients for Coffin - Manson Mechanical Fatigue Model The Coffin-Manson model is most often used to model mechanical failures caused by thermal cycling in mechanical parts or electronics. (Most electronic failures are mechanical in nature) Ncycles a Tb N cycles = number of cycles to failure at reference condition

b = typical value for a given failure mechanism, a = prop constant The values of the coefficient b for various failure mechanisms and materials (derived or taken from empirical data) General Failure Mechanism b Ductile Metal Fatigue 1 to 2 Commonly Used IC Metal Alloys and 3 to 5 Intermetallics Brittle Fracture 6 to 8 Reference: EIA Engineering Bulletin: Acceleration Factors, SSB 1.003, Electronics - Industries Alliance, Government Electronics and Information Engineering Department, 1999. Technology Association Failure Mechanism/Material b 316 Stainless Steel 1.5 4340 Steel 1.8 Solder (97Pb/03Sn) T > 30C 1.9 Solder (37Pb/63Sn) T < 30C 1.2 Solder (37Pb/63Sn) T > 30C 2.7 Solder (37Pb/03Ag & 91Sn/09Zn) 2.4 Aluminum Wire Bond

3.5 Au4Al fracture in wire bonds 4.0 PQFP Delamination / Bond Failure 4.2 ASTM 2024 Aluminum Alloy 4.2 Copper 5.0 Au Wire Bond Heel Crack 5.1 ASTM 6061 Aluminum Alloy 6.7 Alumina Fracture 5.5 Interlayer Dielectric Cracking 4.8-6.2 Silicon Fracture 5.5 Silicon Fracture (cratering) 7.1 Thin Film Cracking 8.4 318-595 Statistics & Reliability Normal operating conditions cycling 15C to 60C (T=45C) Plan for N Stress (Accelerated) cycles 40 to 125 C (T=165C) Find Mean life at stress level MTTF=4570 hrs=0.5 yrs Calculated acceleration factor and MTTF (and B10) @ normal stress: AF = Nuse / Nstress = (b(165/45)2.7 = 33.4 N cycles a T b MTTF (use)=MTTF(stress)*AF = 4570*33.4 = 152638hrs = 17.4 yrs B(10)use * AF * 0.1051 / 4768 *15.3 * 0.1051 / 12 6.9 yrs

318-595 Statistics & Reliability Appendices 318-595 Statistics & Reliability Reliability Distributions are non-Normal, require 2 parameters Intro: Weibull Distribution - ( t / ) F(t) = 1 - e beta, - slope/shape parameter ln ln (1 / (1 F(t))) = ln(t) ln() F(t) = Cumulative fraction of parts that have failed at time t Y=X+a eta, characteristic life or scale parameter Knowing the distribution Function allows to address the following problem (anticipated future failure): when t = F(t) = 63.2% What is the probability, P , that the failure will occur for the period of time T if it did not occur yet for the period of time t ? (T>t) T t P={F(T)-F(t)}/[1-F(t)]= 1 exp{ [( ) ( ) ]} 318-595 Statistics & Reliability

Physical Significance of Weibull Parameters When Weibull distribution parameters are defined, B10 and MTTF can be computed. 99 MTTF = mean time to failure (non-repairable) Cumulative Failure (%) F(t) = ( 1 + 1/ ) When = 1.0, MTTF = When = 0.5, MTTF = 2 MTBF = mean time between failure (repairable) (MTBSC) = total time on all systems / # of failures Slope = 10 When there is no suspension data, MTBF = MTTF 1 B10 10 Time to Failure (t) 100 The slope parameter, Beta (), indicates failure type <1 =1 > 1 rate of failure is decreasing rate of failure is constant rate of failure is increasing

infantile (early) failure random failure wear out failure 318-595 Statistics & Reliability Estimating Reliability from Test Data In testing electronics assemblies or parts, there are frequently few (or no) failures How do you estimate the reliability in this case? Use the chi-squared distribution and the following equation: MTBF = 2 * Number of hours on test * Acceleration factor / 2 In this equation, 2 is a function of two variables n, the degrees of freedom, defined as n= 2 * number of failures + 2 and F, the confidence level of the results (e.g. 90%, 95%, 99%) 318-595 Statistics & Reliability Example The following test was conducted: A new design was qualified by testing 20 boards for 1000 hours The test was conducted at elevated temperatures, where the test would accelerate failures by 10X the usage rate One board failed at 500 hours, the other 19 passed for the full 1000 hours What is the MTBF of the board design at 90% confidence? Solution: First, determine n = 2 * number of failure + 2 = 4; so 2 = 7.78 (at 90% confidence) Second, determine number of hours = 19 samples * 1000 + 1 * 500 = 19, 500 hours So, the answer is: MTBF = 2 * 19500 (total hours) * 10 (acceleration factor) / 7.78 = 50, 128 hours 318-595 Statistics & Reliability The calculations are based on the Binomial Distribution and the following formula: Pass/Fail Test Sample Sizes? where: Confidence Level CL =

n = sample size p = proportion defective r = number defective = probability of k or fewer failures occurring in a test of n units Example: Suppose that 3 failed parts have been observed in the test equivalent to 1 year life, what minimum sample size is needed to be 95% confident that the product is no more than 10% defective? Inputs in the formula are: p =0.1(10%), r = 3, CL = 0.95(95%), P(r

3 Equal Allocation System B(10), years System MTTF, years 1 5 10 Subsystem MTTF years 9.5 28.5 47.5 142.4 94.9 284.7 MTTF~=10 years (B10=1 year) results in failure rate 1-F=1-exp(-1/10*10)=0.63, i.e. 63% of units on average will fail for 10 years MTTF= 47.5 years (B10=5 years) results in failure rate 1-F=1-exp(-1/47.5*10)=0.19, i.e. 19% of units on average will fail for 10 years. System Reliability Target Must be Allocated

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