Cooperation strategy of technology licensing based on evolutionary game

Abstract

This study analyzes the issue of technology-licensing cooperation between a firm with production patent technology and a firm with inferior production technology, and obtains the evolution trend of the technology-licensing deal and cooperation strategy under fixed-fee licensing and royalty licensing situation. We find that the probability of successful cooperation between the two firms increases when fixed technology-license fees and cost savings from technology licensing increase simultaneously, and change of fixed technology license fees and cost savings affects the willingness to cooperate of the firm with inferior production technology, and not the firm with production patent technology. Furthermore, modest royalty fees promote successful cooperation. In both licensing situations, for the firm with production patent technology, an increase in the market share of its products or non-licensing resource sharing cost-saving value reduces the cooperation probability. Meanwhile, for the firm with inferior production technology, an increase in the market share of its products promotes successful cooperation in the royalty licensing case, but requires conditions for fixed technology-licensing fees and cost savings lower than a certain value in the fixed-fee licensing case.

Appendix

Proof of Conclusion 1

If $\beta \Delta _A q_A -\alpha F<0$, namely if $q_A <\frac{\alpha F}{\beta \Delta _A }$ or $F>\frac{\beta \Delta _A q_A }{\alpha }$, we have $\frac{\beta \Delta _A q_A -\alpha F}{\beta \Delta _A q_A +F}<0$, there always is $y>\frac{\pi _{1nn} -\pi _{1yn} }{\pi _{1yy} -\pi _{1ny} +\pi _{1nn} -\pi _{1yn} }$. Here, $x=1$ is an ESS. Putting $p_A =a-q_B -bq_A$ into $\beta \Delta _A q_A -F<0$, we can obtain $F>\beta \Delta _A \frac{p_B -a+q_B }{\alpha b}$. Conclusion 1 is proved.

Proof of Conclusion 2

Under $0<\frac{\pi _{1nn} -\pi _{1yn} }{\pi _{1yy} -\pi _{1ny} +\pi _{1nn} -\pi _{1yn} }<1$ (that is $0<\frac{\beta \Delta _A q_A -\alpha F}{\beta \Delta _A q_A +F}<1)$, if $y>\frac{\beta \Delta _A q_A -\alpha F}{\beta \Delta _A q_A +F}$, ${F}'(x)|_{x=1} <0$ and ${F}'(x)|_{x=0} >0$, then $x=1$ is the stable point; if $y<\frac{\beta \Delta _A q_A -\alpha F}{\beta \Delta _A q_A +F}$, ${F}'(x)|_{x=1} >0$ and ${F}'(x)|_{x=0} <0$, then $x=0$ is the stable point. Conclusion 2 is proved.

Proof of Conclusion 3

If $x>\frac{\pi _{2nn} -\pi _{2ny} }{\pi _{2yy} -\pi _{2yn} +\pi _{2nn} -\pi _{2ny} }$, then $\frac{\alpha F-\beta \Delta _B q_B }{F-(\varepsilon q_B +\beta \Delta _B q_B )}<0$ and $(\varepsilon +\beta \Delta _B )q_B<F<\frac{\beta \Delta _B q_B }{\alpha }$. Substituting $q_B =\frac{a-q_A -p_A }{b}$ into $(\varepsilon +\beta \Delta _B )q_B<F<\frac{\beta \Delta _B q_B }{\alpha }$, we get $(\varepsilon +\beta \Delta _B )\frac{a-q_A -p_A }{b}<F<\frac{\beta \Delta _B (a-q_A -p_A )}{\alpha b}$. Therefore, when $(\varepsilon +\beta \Delta _B )q_B<F<\frac{\beta \Delta _B q_B }{\alpha }$ or $(\varepsilon +\beta \Delta _B )\frac{a-q_A -p_A }{b}<F<\frac{\beta \Delta _B (a-q_A -p_A )}{\alpha b}$, $y=1$ is an ESS. Conclusion 3 is proved.

Proof of Conclusion 4

Under $0<\frac{\alpha F-\beta \Delta _B q_B }{F-(\varepsilon +\beta \Delta _B )q_B }<1$, if $x>\frac{\alpha F-\beta \Delta _B q_B }{F-(\varepsilon +\beta \Delta _B )q_B }$, we have and $F^{\prime }(y)|_{y=0}>0$ so $y=1$ is the stable point at this time. if $x<\frac{\alpha F-\beta \Delta _B q_B }{F-(\varepsilon +\beta \Delta _B )q_B }$, then ${F}'(y)|_{y=0} <0$ and ${F}'(y)|_{y=1} >0$, so $y=0$ is the stable point. Conclusion 4 is proved.

Proof of Conclusion 7

According to $\frac{\pi _{2nn} -\pi _{2ny} }{\pi _{2yy} -\pi _{2yn} +\pi _{2nn} -\pi _{2ny} }=\frac{\gamma \alpha -\Delta _B }{\beta \Delta _B +\varepsilon -\gamma }$ and $\varepsilon >\gamma$, just when $\gamma <\frac{\Delta _B }{\alpha }$, then $\frac{\gamma \alpha -\Delta _B }{\beta \Delta _B +\varepsilon -\gamma }<0$. In this situation, there is always $x>\frac{\pi _{2nn} -\pi _{2ny} }{\pi _{2yy} -\pi _{2yn} +\pi _{2nn} -\pi _{2ny} }$. Therefore, $y=1$ is an ESS. Conclusion 7 is proved.

Proof of Conclusion 9

For $\frac{\partial s_F }{\partial F}=\frac{1}{2}(\frac{(1+\alpha )\beta \Delta _A q_A }{(F+\beta \Delta _A q_A )^{2}}+\frac{\alpha \varepsilon q_B +(\alpha -1)\beta \Delta _B q_B }{(F-(\varepsilon +\beta \Delta _B )q_B )^{2}})$, to make the expression more than zero, just to considering the sign of $\frac{\alpha \varepsilon q_B +(\alpha -1)\beta \Delta _B q_B }{(F-(\varepsilon +\beta \Delta _B )q_B )^{2}}$, obtain $\varepsilon >\frac{(1-\alpha )\beta \Delta _B }{\alpha }$, then $\frac{\partial s_F }{\partial F}>0$. From $\frac{\partial s_F }{\partial \varepsilon }=\frac{(\beta \Delta _B q_B -\alpha F)q_B }{2(F-(\varepsilon +\beta \Delta _B )q_B )^{2}}$,if $F<\frac{\beta \Delta _B q_B }{\alpha }$, then $\frac{\partial s_F }{\partial \varepsilon }>0$. From $\frac{\partial s_F }{\partial q_B }=\frac{((1-\alpha )\beta \Delta _B -\alpha \varepsilon )F}{2(F-(\varepsilon +\beta \Delta _B )q_B )^{2}}$, if $\varepsilon <\frac{(1-\alpha )\beta \Delta _B }{\alpha }$, then $\frac{\partial s_F }{\partial q_B }>0$. $\frac{\partial s_F }{\partial q_A }=-\frac{\beta \Delta _A F(1+\alpha )}{2(F+\beta \Delta _A q_A )^{2}}<0$. From $\frac{\partial s}{\partial \beta }=\frac{1}{2}(-\frac{(1+\alpha )F\Delta _A q_A }{(F+\beta \Delta _A q_A )^{2}}+\frac{(F-\alpha F-\varepsilon qr)\Delta _B q_B }{(F-(\varepsilon +\beta \Delta _B )q_B )^{2}})$, only the second fractions in the bracket $F-\alpha F-\varepsilon q_B <0$, that is $F<\frac{\varepsilon q_B }{1-\alpha }$, then $\frac{\partial s}{\partial \beta }<0$. From$\frac{\partial s_F }{\partial \alpha }=\frac{F(\beta \Delta _B q_B +\beta \Delta _A q_A +\varepsilon q_B )}{2(F+\beta \Delta _A q_A )(-F+(\varepsilon +\beta \Delta _B )q_B )}$, when the denominator $(\varepsilon +\beta \Delta _B )q_B -F>0$, that is $F<(\varepsilon +\beta \Delta _B )q_B$, then $\frac{\partial s_F }{\partial \alpha }>0$. From $\frac{\partial s_F }{\partial \Delta _B }=\frac{(F-\alpha F-\varepsilon q_B )\beta q_B }{2(F-(\varepsilon +\beta \Delta _B )q_B )^{2}}$, when $F>\frac{\varepsilon q_B }{1-\alpha }$, $\frac{\partial s_F }{\partial \Delta _B }>0$. $\frac{\partial s_F }{\partial \Delta _A }=-\frac{(1+\alpha )\beta Fq_A }{2(F+\beta \Delta _A q_A )^{2}}<0$. Conclusion 9 is proved.

Proof of Conclusion 10

According to $\frac{\partial s_R }{\partial \gamma }=\frac{\Delta _B -\alpha \beta \Delta _B -\alpha \varepsilon }{(\beta \Delta _B +\varepsilon -\gamma )^{2}}+\frac{(1+\alpha )\beta \Delta _A q_A q_B }{(\beta \Delta _A q_A +q_B \gamma )}$, we find that as long as the first fraction of the molecule is greater than zero, namely $\varepsilon <\frac{\Delta _B -\alpha \beta \Delta _B }{\alpha }$, $\frac{\partial s_R }{\partial \gamma }$ is constant higher than zero; according to $\frac{\partial s_R }{\partial \varepsilon }=\frac{\alpha \gamma -\Delta _B }{(\beta \Delta _B +\varepsilon -\gamma )^{2}}>0$, just when $\gamma >\frac{\Delta _B }{\alpha }$, $\frac{\partial s_R}{\partial \varepsilon }$ is constant higher than zero; $\frac{\partial s_R }{\partial q_B }=\frac{(1+\alpha )\beta \Delta _A q_A \gamma }{(\beta \Delta _A q_A +q_B \gamma )^{2}}>0$; $\frac{\partial s_R }{\partial q_A }=-\frac{(1+\alpha )\beta \Delta _A q_B \gamma }{(\beta \Delta _A q_A +q_B \gamma )^{2}}<0$; from $\frac{\partial s_R }{\partial \alpha }=\frac{q_B \gamma }{\beta \Delta _A q_A +q_B \gamma }-\frac{\gamma }{\beta \Delta _B +\varepsilon -\gamma }$, it is easy to get that when $\gamma <\frac{\varepsilon q_B +\beta \Delta _B q_B -\beta \Delta _A q_A }{2q_B }$, $\frac{\partial s_R }{\partial \alpha }>0$; for $\frac{\partial s_R }{\partial \beta }=\frac{\Delta _B (\alpha \gamma -\Delta _B )}{(\beta \Delta _B +\varepsilon -\gamma )^{2}}-\frac{(1+\alpha )\gamma \Delta _A q_A q_B }{(\beta \Delta _A q_A +q_B \gamma )}$, as long as the front fraction of the molecule is less than zero, that is when $\gamma <\frac{\Delta _B }{\alpha }$, it can guarantee that $\frac{\partial s_R}{\partial \beta }$ is less than zero constantly; we also obtain that $\frac{\partial s_R }{\partial \Delta _B }=\frac{\varepsilon +(-1+\alpha \beta )\gamma }{(\beta \Delta _B +\varepsilon -\gamma )^{2}}>0$ and $\frac{\partial s_R }{\partial \Delta _A }=\frac{-(1+\alpha )\gamma \beta q_A q_B }{(\beta \Delta _A q_A +q_B \gamma )^{2}}<0$. Conclusion 10 is proved.