## 6 A Method to Calculate \(\boldsymbol{k_{b/m}}\)
### The Three Inequations to Be Fulfilled by \(k_{b/m}\)  
We first rewrite the three inequations to be satisfied by \(k_{b/m}\): Equations (34), (42) and (45). Besides, we will rename the thresholds of the three inequations as \(\alpha_{0}\), \(\alpha_{1}\) and \(\alpha_{2}\), respectively. \[k_{b/m}>\frac{-MOS_{m}\cdot J_{b}}{MOS_{b}\cdot J_{m}} =\alpha_{0}\] \[k_{b/m}\geq\frac{MOS_{m}\cdot J_{b}\cdot(\hat{n}_{m}-2)}{MOS_{b }\cdot J_{m}\cdot\hat{n}_{b}} =\alpha_{1}\] \[k_{b/m}\geq\frac{MOS_{m}\cdot J_{b}\cdot\hat{n}_{m}}{MOS_{b} \cdot J_{m}\cdot(\hat{n}_{b}-2)} =\alpha_{2}\]  
We need to find a value for \(k_{b/m}\) that satisfies the three inequations. First of all, the range of solutions for \(k_{b/m}\) is \(\left]K_{b/m},+\infty\right)\), where \(K_{b/m}\) will be the maximum value among \(\alpha_{0}\), \(\alpha_{1}\) and \(\alpha_{2}\). The probability \(p_{i}^{*}(k_{b/m})\) of sending (I+P) frames through the best path is depicted in Figure 7. The limit of \(p_{i}^{*}(k_{b/m})\) when \(k_{b/m}\rightarrow\infty\) (horizontal asymptote) can be obtained from Equation (26) and it has the following value:  
\[\lim_{k_{b/m}\rightarrow\infty}p_{i}^{*}(k_{b/m})=\frac{\hat{n}_{b}}{2} \tag{47}\]  
The vertical asymptote occurs at \(k_{b/m}\)-value that makes the denominator zero (_i.e._, \(k_{b/m}=\alpha_{0}\)). We should find a value for \(k_{b/m}\) in the range \(\left]K_{b/m},+\infty\right)\) with which \(p_{i}^{*}(k_{b/m})\) changes softly throughout time. This way, the transition in the selection between the best and the medium-quality paths will be smooth producing a more stable system. For that, we calculate the first derivative \(\frac{\partial p_{i}^{*}(k_{b/m})}{\partial k_{b/m}}\), which represents the slope value for each \(k_{b/m}>K_{b/m}\) (_i.e._, Zone of interest). \[\frac{\partial p_{i}^{*}(k_{b/m})}{\partial k_{b/m}}=\frac{\left( \frac{\hat{n}_{b}\cdot MOS_{b}}{J_{b}}\right)\cdot\left(\frac{2\cdot k_{b/m} \cdot MOS_{b}}{J_{b}}+\frac{2\cdot MOS_{m}}{J_{m}}\right)}{4\cdot\left(\frac{ k_{b/m}\cdot MOS_{b}}{J_{b}}+\frac{MOS_{m}}{J_{m}}\right)^{2}}\\ -\frac{\left(\frac{2\cdot MOS_{b}}{J_{b}}\right)\cdot\left(\frac {k_{b/m}\cdot MOS_{b}\cdot\hat{n}_{b}}{J_{b}}+\frac{MOS_{m}\cdot(2-\hat{n}_{ m})}{J_{m}}\right)}{4\cdot\left(\frac{k_{b/m}\cdot MOS_{b}}{J_{b}}+\frac{MOS_{m}}{J_{m}} \right)^{2}} \tag{48}\]  
After simplifying the equation, we obtain:  
\[\frac{\partial p_{i}^{*}}{\partial k_{b/m}}=\frac{\frac{MOS_{b}}{J_{b}}\cdot \frac{MOS_{m}}{J_{m}}\cdot(\hat{n}_{b}+\hat{n}_{m}-2)}{2\cdot\left(\frac{k_{b/ m}\cdot MOS_{b}}{J_{b}}+\frac{MOS_{m}}{J_{m}}\right)^{2}} \tag{49}\]  
Now, we isolate \(k_{b/m}\) from Equation (49) in terms of \(\frac{\partial p_{i}^{*}(k_{b/m})}{\partial k_{b/m}}\) and we get the following expression:\[k_{b/m}=\frac{\pm\sqrt{\frac{\frac{MOS_{b}}{J_{b}}\frac{MOS_{m}}{J_{m}}\cdot( \hat{n}_{b}+\hat{n}_{m}-2)}{2^{\frac{\partial p_{i}}{\partial k_{b/m}}}}-\frac{ MOS_{m}}{J_{m}}}}{\frac{MOS_{b}}{J_{b}}} \tag{50}\]  
Here, we can see that \((\hat{n}_{b}+\hat{n}_{m}-2)\leq 0\), since \(\hat{n}_{b}=\frac{n_{rb}}{n_{s}}=\frac{n_{rb}}{n_{sb}+n_{sm}}\leq 1\), and \(\hat{n}_{m}=\frac{n_{rm}}{n_{s}}=\frac{n_{rm}}{n_{sb}+n_{sm}}\leq 1\). Look at Equation (40) to see the easy justification for both expressions. Consequently, we need that \(\frac{\partial p_{i}^{*}}{\partial k_{b/m}}\leq 0\) to compute a proper \(k_{b/m}\) value.