Applied Reliability Page 1 APPLIED RELIABILITY Techniques for Reliability Analysis with Applied Reliability Tools (ART) (an EXCEL Add-In) and JMP® Software AM216 Class 3 Notes Santa Clara University Copyright David C. Trindade, Ph.D. STAT-TECH ® Spring 2010 Applied Reliability Page 2 AM216 Class 3 Notes Weibull Distribution – Formulas – Specification Limits – Calculations – Rayleigh Distribution – Parameter Estimation Normal Distribution – Formulas – Standard Normal Variate – Normal Tables – Spreadsheet Functions for the Normal – Example Lognormal Distribution – Formulas – Relationship to Normal – Calculations – Parameter Estimation Applied Reliability Page 3 Life Distribution Models Parameters Distribution Parameters Interpretation Weibull c Characteristic Life m Shape Parameter Lognormal T50 Median Life Shape Parameter Applied Reliability Page 4 Life Distribution Models Applications WEIBULL: Extreme value distribution. Weakest link model. Modeling gate oxide failures is possible use, but lognormal more commonly used. LOGNORMAL: Model failure rates associated with multiplicative processes such as electromigration or corrosion. Applied Reliability Page 5 Weibull Life Distribution Probability Density Function (PDF) m m t m f (t) e(t /c) t c Weibull PDF Plot 5 4.5 4 m=10 3.5 3 2.5 m=0.5 2 m=4 1.5 Reciprocal Characteristic Life MultipleLife Characteristic Reciprocal 1 m=1 m=2 0.5 0 0 0.5 1 1.5 2 2.5 3 Multiples of Characteristic Life Applied Reliability Page 6 Weibull Distribution Cumulative Distribution Function (CDF) m F(t) 1 e(t/c) Weibull CDF Plot 1 m=4 m=10 m=2 0.9 m=1 0.8 m=0.5 0.7 0.6 0.5 CDF 0.4 0.3 0.2 0.1 0 0 0.5 1 1.5 2 2.5 3 Multiples of Characteristic Life Applied Reliability Page 7 Weibull Distribution Instantaneous Failure Rate (IFR) m1 m m t m t h(t) c c t c Weibull IFR Plot 10 9 8 m=10 7 m=4 6 5 m=2 4 3 Reciprocal Characteristic Life MultipleLife Characteristic Reciprocal m=.5 2 m=1 1 0 0 0.5 1 1.5 2 2.5 3 Multiples of Characteristic Life Applied Reliability Page 8 Weibull Distribution in ART In ART Add-In, select Weibull > PDF, CDF, HAZARD FUNCTION. Choose input type and specify value. Click OK. Output shown below for c = 100000, m = 0.5. Applied Reliability Page 9 Weibull Distribution Parameters The CDF is m F(t) 1 e(t /c) The parameters are : the characteristic life c the shape parameter m Note at t = c, m F(c) 1 ec/c 1 e1 0.6321 Thus, the characteristic life is the 63.2 percentile of the Weibull distribution for any shape parameter. Applied Reliability Page 10 Weibull Distribution Hazard Rate The hazard rate is m1 f (t) m t h(t) 1 F(t) c c Note for m = 1, the hazard rate is h(t)=1/c, that is, a constant failure rate. So for m = 1, the Weibull distribution becomes the exponential distribution with MTTF = c. For m > 1, the hazard rate increases in time t. For m < 1, the hazard rate decreases in time t. Applied Reliability Page 11 The Median of a Weibull Distribution The median is the 50th percentile. Set the CDF equal to 0.5 at t50 = median m t50 /c Ft50 1 e 0.5 Then m t /c e 50 0.5 m t50 / c ln0.5 So 1/m 1/m t50 cln2 c0.6931 Alternatively, the characteristic life c can be determined from the median using t c 50 0.69311/m Applied Reliability Page 12 Weibull Distribution Median Relationship of Median (t50) to Characteristic Life 1/ m 1/ m t50 cln2 c0.6931 1 0.9 0.8 0.7 0.6 0.5 Median 0.4 c = 1 0.3 0.2 0.1 0 0 1 2 3 4 5 6 7 8 9 10 11 Shape Parameter m The higher the shape parameter, the closer the median is to the characteristic life c. For decreasing hazard rate (m<1), t50 occurs early compared to c. Applied Reliability Page 13 Important Weibull Formulas To estimate the fraction failing F(t) at time t, given the characteristic life c and shape parameter m: m F(t) 1 e(t /c) To estimate the shape parameter m, given the characteristic life c and fraction failing at time t : ln ln1 F m lnt / c To estimate the time t to a specified fraction failing F(t), given the shape parameter m and characteristic life c: t c ln1 F1/m To estimate the characteristic life c, given the shape parameter m and fraction failing at time t : t c ln1 F1/m Applied Reliability Page 14 Weibull Calculations in ART In ART, select Weibull>Probability Calculations. Choose output value desired. Enter any required information. Click OK. Output shown below is time to 10% CDF for c = 10000 and m = 1.5. Applied Reliability Page 15 Class Project Weibull Distribution Calculations Given that the population distribution is Weibull, find: a) The characteristic life c necessary for 10% failures at 168 hours, given shape parameter of 2.0. b) The expected percent fallout F(t) at 1,000 hrs, given a characteristic life of 1,000,000 hrs and shape parameter of 0.5. c) The time t to reach 20% failures, given characteristic life of 50,000 hrs and shape parameter of 10. d) The shape parameter m for 5% failures in 2,000 hrs, given a characteristic life of 30,000 hrs. Applied Reliability Page 16 Class Project Weibull Distribution and Spec Limits A manufacturer produces fireworks that have a median time to explode following ignition of 5 seconds. The population distribution is Weibull with a shape parameter of 8. If the specification limits for acceptable fireworks is 3.5 to 6.2 seconds, what fraction of the units are rejectable? Applied Reliability Page 17 Class Project Weibull Distribution and Spec Limits A manufacturer produces fireworks that have a median time to explode following ignition of 5 seconds. The population distribution is Weibull with a shape parameter of 8. If the specification limits for acceptable fireworks is 3.5 to 6.2 seconds, what fraction of the units are rejectable? t 5 5 c 50 5.23 0.69311/m 0.69311/8 0.9552 8 F(3.5) 1 e(3.5/ 5.23) 1 e0.04 0.0392 8 1 F (6.2) e ( 6.2 / 5.23) e 3.87 0.0208 Total rejectable is 0.0392 + 0.0208 = 0.060 or 6.0 %. Applied Reliability Page 18 Class Project Solving Via ART Under Weibull>Calculations, first find characteristic life c from median: Applied Reliability Page 19 Class Project Solving Via ART Then, find areas outside specifications. Applied Reliability Page 20 Class Project: Fireworks Problem: Weibull CDF and PDF Fireworks: Weibull CDF 1 0.9 0.8 0.7 0.6 0.5 F(t) 0.4 0.3 0.2 0.1 0 1 2 3 4 5 6 7 t Fireworks: Weibull PDF 0.6 0.5 0.4 0.3 f(t) 0.2 0.1 0 1 2 3 4 5 6 7 t Applied Reliability Page 21 Alternate Forms of the Weibull Distribution In the literature, the CDF may be expressed as m F(t) 1 e(t/c) or F(t) 1 e(t/) or m F(t) 1 et /c or F(t) 1 et / We will use the first form. Use caution when interpreting the parameters in various references. Applied Reliability Page 22 The Rayleigh Distribution Special Case of Weibull with Shape Parameter m= 2 The CDF is : 2 F(t) 1 e(t/c) The failure rate increases linearly with time : 2 h(t) t c 2 Radial Error Consider a robotic arm trying to position itself at a point in a plane. Assume the error in each x and y coordinate is normally distributed with mean zero and the same standard deviation . Then the distance from the target location (total radial error) is R X 2 Y 2 The CDF for R can be derived and shown to be 2 2 F(r) 1 er /2 which is a Rayleigh distribution with c 2 Applied Reliability Page 23 Radial Calculations in ART Under Weibull, select Rayleigh Radial Error. Enter information needed. Click OK. Applied Reliability Page 24 Class Project Weibull Distribution and Radial Error An expert dart thrower claims he can throw darts at a target and end up within 0.5 cm of the origin of the target, in any direction (radial error). The bullseye is centered at the origin and has a width of 0.7 cm. If he throws a dart at the target 50 times, what's his expected number of bullseyes? What assumptions are you making? What if the distribution were uniform? accuracy radius = 0.5 cm bullseye width 0.7 cm Hint: The standard deviation is approximately 2(0.5)/6 = 0.166 cm Applied Reliability Page 25 Weibull Parameter Estimation To estimate the parameters c and m for a Weibull distribution based on complete or censored data, the following methods are available: 1) Maximum Likelihood 2) Linear Rectification and Least Squares Regression ML estimates generally have many desirable properties but usually require special computer programs such as ART for solution. Linear rectification is easy to set up, and least squares regression analysis is readily available on computers. However, these analytical or graphical estimates do not have desirable properties which would permit, for example, calculation of confidence bounds on the parameters.