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Published byLilian Bottum Modified about 1 year ago

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Up to Now – we covered Basic Model: –failure intensity --- by number of failures (only linear relation model) Logarithmic Poisson Model: – failure intensity ---- by number of failures Basic Model: – # of Failures ---- by execution time Logarithmic Poisson Model: –# of Failures ---- by execution time

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Two Failure Intensity Models in terms of Execution Time (t) Basic Model : – f(t) = f0 e -[(f0*t)/v] where f0 = initial failure intensity t = time v = estimated total number of failures Logarithmic Poisson Model: – f(t) = f0 / [(f0*k*t) + 1] where f0 = initial failure intensity t = time k = decay parameter

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Failure Intensity as a function of Execution Time (Graphical Curves) Logarithmic Poisson model Basic model execution time, (t) failure-intensity, (f) We need to lengthen the x-axis now that we have changed it from # of failures to execution time. This is because the “time” between failures lengthens as more failures are encountered and fixed the ‘real’ curves would be much smoother!

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Examples Basic model: – for f0 = 10 failures/cpu-time; v= 100 total failures; and t = 10 cpu- time – F (10) = f0 e -[(f0*t)/v] = 10 e -[(10*10)/100] = 10 e -1 = 10 *.368 = 3.68 failures/cpu-time Logarithmic Poisson model: –for f0= 10 failures/cpu-time ; k =.02 decay factor ; and t = 10 cpu- time – F (10) = f0 / [(f0*k*t) + 1] = 10 / [(10*.02*10)+1] = 10 / [2 + 1] = 3.33 failures/cpu-time

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Quick Comparison of Models (reminder) BasicLogarithmic Poisson Initial intensity Estimated decay Estimated total failures f0 V --- k

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Scenarios of Basic Model where F0 ≡ initial failure intensity F0a > F0b and same vF0a F0b v Same F0, but va > vb vavb F0 X-axis = # of failures experiencedX-axis = execution time F0a F0b F0a > F0b and same v F0 t for vb t for va t for v Same F0, but va > vb

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Scenarios of Logarithmic Poisson Model F0a > F0b and same kF0a F0b Same F0, but ka > kb F0 X-axis = # of failures experiencedX-axis = execution time F0a F0b F0a > F0b and same k F0 v Same F0, but ka > kb ka kb ka

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Using “derived” value for projection Can we use these model equations for some projections? Assume we started with f0 failure intensity and reached a failure intensity of f1. We want to get to a lower failure intensity of f2 as the objective of our “quality plan”. Then can we project the # of more failures (or ∆v) that must be found to get to f2? f0 v f1 f2 v2v1 Basic model Since : f1 = f0 - [(f0 *v1)/v] (from Basic intensity model formula) f2 = f0 - [(f0* v2)/v] f1 – f2 = (f0*v2)/v - (f0*v1)/v ; subtracting out f0 v ( f1- f2 ) = f0 (v2 –v1) [ v*(f1- f2) ] / f0 = v2 - v1 = ∆ v ∆v = v/f0 * (f1 – f2) for Basic Model ∆v Similarly, if we go through the formula manipulation We will get ∆v = (1/k) * ln(f1/f2) for Logarithmic Poisson Model

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Example Suppose, for Basic Model we have: –f0 = 10 and estimated v = 100 –Assume we are at f1 failure intensity = 3.68 –Assume that we want to get to f2 =.000454 –Then using the formula: ∆v = 100/10 (3.68 -.000454) ≡ 10 * 3.67 ≡ 36.7 more failures need to be found.

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“Derived” ∆ for execution time Basic model: – ∆t = (v/f0) * ln( f1/f2 ) where f0 = initial failure intensity f1 = present failure intensity f2 = desired failure intensity Logarithmic Poisson model: – ∆t = 1/k (1/f2 – 1/f1 )

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Example Suppose, for Basic Model we have same assumption as before: –f0 = 10 and estimated v = 100 –Assume we at f1 failure intensity = 3.68 –Assume that out objective is to get to f2 =.000454 Question: how much more execution time units are required to get to that objective? –Then using the formula: ∆t = 100/10 [ln (3.68 /.000454)] ≡ 10 * ln (8106) ≡ 10 * 9 ≡ 90 more execution time units

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