A multi-release software reliability modeling for open source software incorporating dependent fault detection process

Abstract

The increasing dependence of our modern society on software systems has driven the development of software products become even more competitive and time-consuming. Single release software product no longer meets the increasing market requirements. Thereby it is important to release multiple version software products in order to add new features in the next release and fix remaining faults from previous release. In this paper, we develop a multi-release software reliability model with consideration of the remaining software faults from previous release and the new introduced-faults (from newly added features). Additionally, dependent fault detection process is taken into account in this research. In particular, the detection of a new fault for developing the next release depends on the detection of the remaining faults from previous release and the detection of the new introduced-faults. The proposed model is validated on the open source software project datasets with multiple releases.

Appendix

Proof of Lemma 1

We need to show that: (a) the function $g\left( t \right)$ in Eq. (7) is convex, and (b) the function $f\left( x \right)$ in Eq. (8) is concave.

• (a) Since d is non-negative and ${g}''(t)=d^{2}e^{dt+C_{0}}>0$, thus function g(t) is convex.

• (b) $f\left( x \right) =x^{\frac{1}{ab}}\left( {x-a} \right) ^{\frac{1}{a\left( {a-b} \right) }}\left( {x-b} \right) ^{-\frac{1}{b\left( {a-b} \right) }}$

  $$\begin{aligned} f'\left( x \right)= & {} \frac{1}{{ab}}{x^{\frac{1}{{ab}} - 1}}~{\left( {x - a} \right) ^{\frac{1}{{a\left( {a - b} \right) }}}}~{\left( {x - b} \right) ^{ - \frac{1}{{b\left( {a - b} \right) }}}} + \frac{1}{{a\left( {a - b} \right) }}{x^{\frac{1}{{ab}}}}~{\left( {x - a} \right) ^{\frac{1}{{a\left( {a - b} \right) }} - 1}}~{\left( {x - b} \right) ^{ - \frac{1}{{b\left( {a - b} \right) }}}} \\&-\frac{1}{{b\left( {a - b} \right) }}{x^{\frac{1}{{ab}}}}~{\left( {x - a} \right) ^{\frac{1}{{a\left( {a - b} \right) }}}}~{\left( {x - b} \right) ^{ - \frac{1}{{b\left( {a - b} \right) }} - 1}}\\= & {} {x^{\frac{1}{{ab}}}}~{\left( {x - a} \right) ^{\frac{1}{{a\left( {a - b} \right) }}}}~{\left( {x - b} \right) ^{ - \frac{1}{{b\left( {a - b} \right) }}}}\left( {\frac{1}{{abx}} + \frac{1}{{a\left( {a - b} \right) \left( {x - a} \right) }} - \frac{1}{{b\left( {a - b} \right) \left( {x - b} \right) }}} \right) \\= & {} {x^{\frac{1}{{ab}}}}~{\left( {x - a} \right) ^{\frac{1}{{a\left( {a - b} \right) }}}}~{\left( {x - b} \right) ^{ - \frac{1}{{b\left( {a - b} \right) }}}}\frac{{\left( {ab + a + b} \right) - \left( {a + b} \right) x}}{{abx\left( {x - a} \right) \left( {x - b} \right) }}\\= & {} f\left( x \right) \frac{{\left( {ab + a + b} \right) - \left( {a + b} \right) x}}{{abx\left( {x - a} \right) \left( {x - b} \right) }}\\= & {} {\left[ {f\left( x \right) } \right] ^{ - 1}}\frac{{\left( {ab + a + b} \right) - \left( {a + b} \right) x}}{{ab}}\\ f''\left( x \right)= & {} - {\left[ {f\left( x \right) } \right] ^{ - 2}}f'\left( x \right) \frac{{\left( {ab + a + b} \right) - \left( {a + b} \right) x}}{{ab}} - {\left[ {f\left( x \right) } \right] ^{ - 1}}\frac{{a + b}}{{ab}}\\= & {} - \frac{{f'\left( x \right) \left\{ {{{\left[ {f\left( x \right) } \right] }^{ - 1}}\frac{{\left( {ab + a + b} \right) - \left( {a + b} \right) x}}{{ab}}} \right\} }}{{f\left( x \right) }} - {\left[ {f\left( x \right) } \right] ^{ - 1}}\frac{{a + b}}{{ab}}\\= & {} - \frac{{{{\left[ {{f^{'\left( x \right) }}} \right] }^2}}}{{f\left( x \right) }} - \frac{{\frac{{a + b}}{{ab}}}}{{f\left( x \right) }} < 0 \end{aligned}$$

  Note that $f\left( x \right) =g\left( t \right) >0$ in Eq. (3), $a>0$ and $b>0$, therefore, function f(x) in Eq. (8) is concave. Since the function $g\left( t \right)$ is convex, and $f\left( x \right)$ is concave, the results in Lemma 1 follow accordingly. $\square$