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在前两期咱們简略论述了竞争计谋和批發竞合计谋两制造商的最優產量,最優利润函数和影响各自最優解的身分。這一期咱們看第三種计谋—專利允许竞合计谋。
In the first two issues, we briefly elaborated on the optimal output, optimal profit function, and factors that affect the optimal solutions of two manufacturers: competitive strategy and wholesale competition strategy. In this issue, we will look at the third strategy - patent licensing competition strategy.
在專利允许竞合计谋下,新能源汽車制造商之間以授权動力電池技能專利的情势举行互助,即新能源汽車制造商1将動力電池技能專利授权给竞争敌手,并向後者收取必定的專利利用费,但在新能源汽車终端市場互相竞争。文章對這類竞合计谋也有必定的文献支持。
Under the strategy of patent licensing competition韓國保濕棒,, new energy vehicle manufacturers 邱大睿,cooperate in the form of authorizing power battery technology patents, that is, new energy vehicle manufacturer 1 grants power battery technology patents to competitors and charges them DIY手鍊,a certain patent usage fee, but competes with each other in the newDIVIN 葡萄酒櫃, energy vehicle terminal market. The article also has some literature support for this competitive strategy.
當两新能源汽車制造商施行專利允许竞應時,两邊起首针對動力電池的專利利用费(包含入門费和提成费)開展會商。此時的决议计划次序為:起首新能源汽車制造商1和2經由過程會商肯定動力電池技能的專利的入門费和提成费;其次两家新能源汽車制造商同時决议本身的產量,以實現本身长處的最大化。
When two new energy vehicle manufacturers implement patent licensing competition, the two sides first negotiate the patent usage fee (including entry fee and co妹妹ission fee) for the power battery. The decision sequence at this point is: Firstly, new energy vehicle manufacturers 1 and 2 negotiate to determine the entry fee and co妹妹ission fee for the patent of power battery technology; Secondly, both new energy vehicle manufacturers simultaneously determine their own production to maximize their own interests.
新能源汽車制造商1和2的利润函数别離為
The profit functions of new energy vehicle manufacturers 1 and 2 are respectively
给定入門费和提成费,經由過程構建古诺竞争博弈模子,求得定理4。
Given the entry fee and co妹妹ission fee, by constructing a Cournot competitive game model, Theorem 4 is obtained.
定理4 在專利允许竞合计谋下,新能源汽車制造商1和2的最優產量别離為
Theorem 4: Under the patent licensing competition strategy, the maximum production of new energy vehicle manufacturers 1 and 2 are:
在Mathematica长進行代码复現,获得成果為
Reproduce the code on Mathematica and the result is |
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