# Default Detection Rate-Dependent Software Reliability Model with Imperfect Debugging
## 4 FDR-Related Reliability Model  
Here, based on the two unified types of imperfect debugging framework models obtained above, the analysis is carried out according to the \(b(t)\) proposed in the reference [32; 33; 34; 35]. \(b(t)\) is mainly set in the following five forms (\(b1(t)\), \(b2(t)\), \(b3(t)\), \(b4(t)\) and \(b5(t)\)). They are specific \(b(t)\) functions, representing different failure detection rates. These \(b(t)\) functions composed of different parameters depict the changing trend of fault detection rate under different test environments. Then, cumulative fault detection functions \(m(t)\) are obtained in the following five cases:  
(i) If \(b_{1}(t)=b\alpha\beta e^{-\beta t}\), \(m(t)\) can be obtained as follows:  
\[m_{ID\_1}(t)=e^{-\int_{0}^{t}b\alpha\beta e^{-\beta t}pu}\int_{0}^{t}b\alpha \beta e^{-\beta\gamma}ae^{\int_{0}^{\alpha}b\alpha\beta e^{-\beta\gamma}}ds \tag{14}\]  
\[m_{ID\_1}(t)=a\int_{0}^{t}b\alpha\beta e^{-\beta t}e^{-\int_{0}^{u}p(1-r( \tau))b\alpha\beta e^{-\beta\gamma}d\tau}du \tag{15}\]\(m_{IDI\_1}(t)\) and \(m_{IDI\_1}(t)\) represent the specific cumulative fault detection function \(m(t)\) obtained by two kinds of imperfect debugging framework models with the participation of the same \(b(t)\). IDI indicates imperfect debugging type one, and IDII means imperfect debugging type two. The following situation is the same.  
At this point, as the test continues, \(b_{1}(t\rightarrow\infty)=0\).  
(ii) If \(b_{2}(t)=\frac{b}{1+\beta e^{-bt}},\)\(m(t)\) can be obtained as follows:  
\[m_{IDI\_2}(t)=e^{-\int_{0}^{t}\frac{b}{1+\beta e^{-bt}}pdu}\int_{0}^{t}\frac{ b}{1+\beta e^{-bt}}ae^{\int_{0}^{\tau}\frac{b}{1+\beta e^{-bt}}pds}dv \tag{16}\]  
\[m_{IDII\_2}(t)=a\int_{0}^{t}\frac{b}{1+\beta e^{-bt}}e^{-\int_{0}^{\tau}p(1-r( \tau))\frac{b}{1+\beta e^{-bt}}d\tau}du \tag{17}\]  
At this point, as the test continues, \(b_{2}(t\rightarrow\infty)\to b\).  
(iii) If \(b_{3}(t)=b\alpha\beta te^{-\beta t^{2}/2}\), the corresponding \(m(t)\) is:  
\[m_{IDI\_3}(t)=e^{-\int_{0}^{t}ba\beta te^{-\beta t^{2}/2}pdu}\int_{0}^{t}ba \beta ve^{-\beta t^{2}/2}ae^{\int_{0}^{\tau}ba\beta se^{-\beta t^{2}/2}pds}dv \tag{18}\]  
\[m_{IDII\_3}(t)=a\int_{0}^{t}ba\beta te^{-\beta t^{2}/2}e^{-\int_{0}^{u}p(1-r( \tau))b\alpha\beta te^{-\beta t^{2}/2}d\tau}du \tag{19}\]  
At this point, as the test continues, \(b_{3}(t\rightarrow\infty)=0\).  
(iv) If \(b_{4}(t)=b^{2}t/(1+bt),\)\(m(t)\) can be obtained as follows:  
\[m_{IDI\_4}(t)=e^{-\int_{0}^{t}b^{2}u/(1+bu)pdu}\int_{0}^{t}b^{2}u/(1+bu)ae^{ \int_{0}^{\tau}b^{2}s/(1+bs)pds}dv \tag{20}\]  
\[m_{IDII\_4}(t)=a\int_{0}^{t}b^{2}u/(1+bu)e^{-\int_{0}^{\tau}p(1-r(\tau))b^{2} u/(1+bu)d\tau}du \tag{21}\]  
At this point, as the test continues, \(b_{4}(t\rightarrow\infty)\to b\).  
(v) If \(b_{5}(t)=\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)t}}\), the corresponding \(m(t)\) is:  
\[m_{IDI\_5}(t)=e^{-\int_{0}^{t}\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)t}} pdu}\int_{0}^{t}\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)v}}ae^{\int_{0}^{ \tau}\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)v}}pds}dv \tag{22}\]  
\[m_{IDII\_5}(t)=a\int_{0}^{t}\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)u}}e^{ -\int_{0}^{u}p(1-r(\tau))\frac{b(1+\sigma)}{1+\sigma e^{-b(1+\sigma)v}}d\tau}du \tag{23}\]  
At this point, as the test continues, \(b_{5}(t\rightarrow\infty)\to b(1+\sigma)\).