Software Engineering | Schick-Wolverton software reliability model
Prerequisite – Jelinski Moranda software reliability model
The Schick-Wolverton (S-W) model is a modification to the J-M model. It is similar to the J-M model except that it further assumes that the failure rate at the ith time interval increases with time ti since the last debugging. In the model, the program failure rate function between the (i-1)th and the ith failure can be expressed as
![$$\lambda (t_i)=\phi [N-(i-1)]t_i$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-180aed9134cb9969845e03297e4bed89_l3.png "Rendered by QuickLaTeX.com")
where 
and N are the same as that defined in the J-M model and ti is the test time since the (i-1)th failure.
The pdf(probability distribution function) of 
can be obtained as follows:
![$$f(t_i)=\phi [N-(i-1)]t_ie^{\frac{[N-(i-1)]t_i^2}{2}} \; \; for \;i=1, 2, ..., N$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-654f631a11f277dfcff42af66fa77f70_l3.png "Rendered by QuickLaTeX.com")
Hence, the software reliability function is
![$$R(t_i)=e^{- \int_0^{t_i} \lambda (t_i)dt_i }$$ $$=e^{-\frac{\phi [N-(i-1)]t_i^2}{2}}$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-ee292d79d7b19e870793b9b273d0e7ef_l3.png "Rendered by QuickLaTeX.com")
We now wish to estimate N assuming that is given. Using the MLE method, the log likelihood function is given by,
![$ln L(N) &=ln\left \{ \prod_{i=1}^nf(t_i)\right \}\\$ $$&=ln\left \{ \prod_{i=1}^n\left [ \phi [N-(i-1)]t_ie^{\frac{[N-(i-1)]t_i^2}{2}\right ]\right \}\\$$ $$&=nln\phi + \sum_{i=1}^nln[N-(i-1)]+\sum_{i=1}^nlnt_i-\sum_{i=1}^n\phi [N-(i-1)]\frac{t_i^2}{2}\\$$ $$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-40e7ff95a06538b6d032132f37115732_l3.png "Rendered by QuickLaTeX.com")
Taking the first derivative with respect to N, we have,
![$$ \frac{\partial}{\partial N}[ln L(N)]=\sum_{i=1}^n \frac{1}{N-(i-1)}-\phi \sum_{i=1}^n \frac{t_i^2}{2}=0 $$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-f39b0a76063ee0f0eeb91fd2827f21fc_l3.png "Rendered by QuickLaTeX.com")
Therefore, the MLE of N can be obtained by solving the following equation:
Next, we assume that both N and ? are unknown. Hence we obtain,
![$$\frac{\partial}{\partial \phi}[ln L(N, \phi )]=\sum_{i=1}^n \frac{1}{N-(i-1)}-\phi \sum_{i=1}^n \frac{t_i^2}{2}=0$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-a96d7975d21e95c049a6f9303cf03637_l3.png "Rendered by QuickLaTeX.com")
and
![$$\frac{\partial}{\partial \phi}[ln L(N, \phi )]=\frac{n}{\phi }-\sum_{i=1}^n [N-(i-1)]\frac{t_i^2}{2}=0$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-e3b3ec336b85ff416707e43fb2a0ffab_l3.png "Rendered by QuickLaTeX.com")
Therefore, the MLEs of N and can be found by solving the two equations simultaneously as follows:
![$$\phi = 2\sum_{i=1}^n \frac{1}{[N-(i-1)]T}$$ $$N=\frac{2n}{\phi T}+\frac{\sum_{i=1}^n (i-1)t_i^2}{T}$$](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-693215b16babe9b46a725edc84b24c9a_l3.png "Rendered by QuickLaTeX.com")
where
Please Login to comment...