Up to Now – we covered

1. 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

2. 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

3. Failure Intensity as a function of Execution Time(Graphical Curves) Logarithmic Poisson model failure-intensity, (f) the ‘real’ curves would be much smoother! Basic model execution time, (t) 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

4. 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

5. Quick Comparison of Models (reminder) Basic Logarithmic Poisson Initial intensity f0 f0 V --- Estimated total failures --- k Estimated decay

6. Scenarios of Basic Modelwhere F0 ≡ initial failure intensity F0a F0a > F0b and same v F0a F0a > F0b and same v F0b F0b t for v v Same F0, but va > vb Same F0, but va > vb F0 F0 vb va t for vb t for va X-axis = # of failures experienced X-axis = execution time

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

8. 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 morefailures (or ∆v) that must be found to get to f2? 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 f0 Basic model f1 f2 Similarly, if we go through the formula manipulation We will get ∆v = (1/k) * ln(f1/f2) for Logarithmic Poisson Model v v1 v2 ∆v

9. 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.

10. “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 )

11. 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