Optimal know-how transfers in licensing contracts

Abstract

This paper studies optimal licensing contracts in the presence of moral hazard associated with costly provision of know-how by the licensor. In our setting, the target market is defined as the fraction of consumers that have a positive valuation for the product that is licensed. It is shown that, no matter how thin the target market is, know-how transfer always takes place. Consistent with actual practice, the optimal licensing contract includes a royalty on sales to attenuate the moral hazard problem. However, full know-how transfer will not occur for low enough maximum willingness to pay and high enough convexity of know-how cost. Finally, it is also shown that the effective (inclusive of the royalty) marginal cost exceeds the one when know-how transfer does not occur thus showing a potential malfunction of know-how transfer specially if the recipient is a developing country.

Appendix

1.1 Proofs

In order to prove the propositions, it is first important to present first the three different cases that arise when there is partial transfer of know-how. As presented in the text, there are two different intervals in the royalty rate that induce partial know-how transfers. The first one, for $[a(1-2m)-c]< r \le [a(1-2m)-c]\big (\frac{2\alpha a}{2\alpha a-c^{2}}\big )$, corresponds to $k(r) = k^{*}$ which implies en effective marginal cost of $\hat{c}= a(1-2m)$ and therefore all consumers in the relevant market buy the good, $q^*(\hat{c})=m$. Thus, the licensor maximizes the following function, $rm+ am^{2}-\frac{\alpha ( 1+( \frac{r-a(1-2m)}{c})^{2}}{2}$. The interior equilibrium royalty rate is $r^{*}=\frac{mc^{2}}{\alpha }+ [a(1-2m)-c]$, it is decreasing in m and leads to $k(r^{*}) = k^{*} =\frac{ mc}{\alpha }$ and $f^*=am^2$. Thus, for this region the licensor profits are equal to $\Pi ^{L*}_p= \frac{(mc)^{2}}{2\alpha }+ m(a(1-m)-c)$.

Consider now, $[a(1-2m)-c](\frac{2\alpha a}{2\alpha a-c^{2}}) < r \le \frac{ 2 \alpha a}{c}$. Noticing that $k(r) = k^{**}$ in this interval, and that $\hat{c}(r)= c(1- \frac{r c}{2 \alpha a})+ r$, so that not all consumers in the high-end market buy the good. The licensor maximizes the following function: $r ( \frac{ a- \hat{c}(r)}{2a} )+ \frac{ (a-\hat{c}(r))^{2}}{4a}- \frac{\alpha }{2}(\frac{r c}{2 \alpha a})^{2}$. Yielding an interior solution which equals, $r^{**}= \frac{2 \alpha a c^{2}(a-c)}{4 \alpha ^{2} a^{2}+2 \alpha a c^{2}-c^{4}}$. Given $r^{**}$, the equilibrium know-how, the fixed fee, licensor’s payoffs and technology payments are as follows: $k(r^{**})=\frac{c^{3}(a-c)}{4 \alpha ^{2} a^{2}+2 \alpha a c^{2}-c^{4}}$; $f^{**}= \frac{4 \alpha ^{4} a^{3} (a-c)^{2}}{(4 \alpha ^{2} a^{2}+2 \alpha a c^{2}-c^{4})^{2}}$; $\Pi ^{L**}_p= \frac{\alpha (2 \alpha a + c^{2})(a-c)^{2}}{ 2(4 \alpha ^{2} a^{2}+2 \alpha a c^{2}-c^{4})}$.

Next, it must be checked whether the above interior solutions $r^{*}$ and $r^{**}$ are consistent with the intervals at which they apply. That is, whether $r^{*} \in [a(1-2m)-c, (a(1-2m)-c)( \frac{2\alpha a}{2\alpha a-c^{2}})]$ and $r^{**} \in [(a(1-2m)-c)(\frac{ 2\alpha a}{2\alpha a-c^{2}}),\frac{2\alpha a}{c}]$. Considering first $r^{*}$, it is easy to show that it is always greater than the lower bound in the interval, but to satisfy the upper bound it is required that $m < \frac{\alpha (a-c)}{4 \alpha a-c^{2}}=m^{*}$. Then, we conclude that the equilibrium royalty is $r^{*}$ if $m< m^{*}$, while it is $\overline{r}$ if $m > m^{*}$ where $\overline{r}$ is precisely the upper bound in the interval, i.e. $\overline{ r} = (a(1-2m)-c)(\frac{2\alpha a}{2\alpha a-c^{2}})=(\frac{4\alpha a^2}{2\alpha a-c^{2}})(\overline{m}-m)$. Next consider $r^{**}$, it is easy to check that it is greater than the lower bound in the interval as long as $m > \frac{2 \alpha ^2 a(a - c)}{4 \alpha ^2 a^2+2 \alpha a c^2-c^4}= m^{**}$. Thus, if $m > m^{**}$ then the equilibrium royalty rate is $r^{**}$, being $\overline{r}$ otherwise. Finally, $r^{**}$ is smaller than the upper bound in the interval if and only if $\alpha > \frac{c ( \sqrt{4ac+c^2}-c)}{4 a}$.

When the equilibrium royalty is $\overline{r}$ then $k(\overline{r})=(\frac{2 a c}{2\alpha a-c^{2}})(m'-m), \hat{c}(\overline{r})=a(1-2m)$, also $q(\hat{c}(\overline{r}))=m$, the fixed fee equals $am^2$ and the licensor’s profits are $\frac{8 \alpha ^2 a^2 m (a(1-m)-c)-c^2(\alpha (a-c)^2+2am^2(2 \alpha a-c^2)) }{2(2\alpha a-c^2)^2}$.

Finally, once the licensor knows how payoffs depend on m for each possible level of know-how transfer, it must decide which one to implement. In order to properly undertake the analysis, a first step must be done which is to rank the four thresholds on m that determine the licensor’s payoffs.

• (i) First, it is easy to find that $m^*<m^{**}<m'$ just assuming $\alpha > \frac{c^2}{2a}$.

• (ii) However, the position of $\hat{m}$ depends on the relationship between maximum willingness to pay, initial marginal cost of production and efficiency in transmitting know-how as follows:

  • (a) $\hat{m} < m^*$ if either $c<a<2c$ and $\frac{c^2}{2a}<\alpha <\frac{c}{2}$ or $2c<a$ and $\frac{c}{4}<\alpha <\frac{c}{2}$.

  • (b) $\hat{m} < m^{**}$ if $2c<a$ and $\frac{c(\sqrt{4ac+c^2}-c)}{4a}<\alpha < \frac{c}{4}$.

  • (c) $\hat{m}$ is smaller than $\overline{m}$ for all $c<a$ and all $\frac{ c^2}{2a}<\alpha <\frac{c}{2}$.

Then four different situations are identified depending on the size of both a and $\alpha$:

• (Case A) Small maximum willingness to pay case: $c<a<2c$, and $\frac{c}{4}< \frac{c}{4a}(\sqrt{4ac+c^2}-c)<\frac{c^2}{2a}<\alpha <\frac{c}{2}$, implying the following threshold ordering: $0< \hat{m} < m^*<m^{**}<m'<\frac{1}{2}$.

• (Case B) Large maximum willingness to pay case: $2c<a$.

• (Case B.1) $\frac{ c^2}{2a}< \frac{c}{4a}(\sqrt{4ac+c^2}-c)<\frac{c}{4}<\alpha <\frac{c}{2}$, which implies the following threshold ordering: $0< \hat{m} < m^*<m^{**}<m'<\frac{1}{2}$.

• (Case B.2) $\frac{c^2}{2a}< \frac{c}{4a}(\sqrt{4ac+c^2}-c)<\alpha <\frac{c}{4} <\frac{c}{2}$, which implies the following threshold ordering: $0< m^*<\hat{m} <m^{**}<m'<\frac{1}{2}$.

• (Case B.3) $\frac{c^2}{2a}< \alpha <\frac{c}{4a}(\sqrt{4ac+c^2}-c)<\frac{c}{4} <\frac{c}{2}$, which implies the following threshold ordering: $0< m^*< m^{**}<\hat{m} <m'<\frac{1}{2}$.

Proof of Proposition 1

Comparing $\Pi _{p}^{L}(m)$ with $\Pi _{n}^{L}(m)$ we find that:

• (a) $a(1-m)m-cm + \frac{m^2 c^2}{2 \alpha }$ is greater than $\Pi _{n}^{L}(m)=a(1-m)m-cm$ for $0<m<m^*$, just by inspection.

• (b) Consider now, $m^*<m<m^{**}$, $\frac{8 \alpha ^2 a^2 m (a(1-m)-c)-c^2(\alpha (a-c)^2+2am^2(2 \alpha a-c^2))}{2(2\alpha a-c^2)^2}$ is greater than $\Pi _{n}^{L}(m)=a(1-m)m-cm$ iff $m \in \big [ \frac{(a-c) \alpha }{2(3 a \alpha - c^2) }, \frac{a-c}{2a}\big ]$. But notice that $\frac{(a-c) \alpha }{2(3 a \alpha - c^2) }<m^*<m^{**}<\frac{a-c}{2a}$ so that the result holds for the relevant interval in m,

• (c) Next, for $m^{**}<m<m'$, $\frac{\alpha (2 \alpha a+ c^2) (a-c)^2}{2(4 \alpha ^2 a^2+ 2\alpha a c^2-c^4)}$ is greater than $\Pi _{n}^{L}(m)=a(1-m)m-cm$ for all $m \in [0, m']$. Just notice that $\Pi _{n}^{L}(m)$ is increasing in m and when evaluated at $m'$ the expression obtained is always smaller than $\frac{ \alpha (2 \alpha a+ c^2) (a-c)^2}{2(4 \alpha ^2 a^2+ 2\alpha a c^2-c^4)}$.

• (d) Finally for $m'<m$, $\frac{\alpha (2 \alpha a+ c^2) (a-c)^2}{2(4 \alpha ^2 a^2+ 2\alpha a c^2-c^4)}$ is greater than $\Pi _{n}^{L}(m)=\frac{(a-c)^2}{4a}$, since the same argument as in (c) works.

Therefore $\Pi _{p}^{L}(m) > \Pi _{n}^{L}(m)$ for all $m>0$ and $\alpha \in \big [\frac{c^2}{2a},\frac{c}{2}\big ]$. $\square$

Proof of Proposition 2

Proposition 2 gives the conditions for partial equilibrium to arise at equilibrium.

• (a) Consider both case (A) and (B.1) as defined above, which results in the following ordering: $0< \hat{m} < m^*<m^{**}< m'<\frac{1}{2}$; and also implies that $m^* <\frac{\alpha }{c}$, thus partial know-how transfer never collapses to the full know-how transfer case.

• (a.i) First, notice that for $0< m<\hat{m}$ the relevant comparison between the profits corresponding to partial know-how transfer and those corresponding to full is whether $(a(1-m)-c)m + \frac{m^2 c^2}{2 \alpha }$ is greater than $am(1-m)-\frac{\alpha }{2}$. This is always true since the difference is decreasing in m for $m<\frac{\alpha }{c}$ and positive for $m=\hat{m}$.

• (a.ii) Next for $\hat{ m}< m< m^*$ the relevant comparison is whether $(a(1-m)-c)m + \frac{m^2 c^2}{ 2 \alpha }$ is greater than $\frac{a}{4}(1-(\frac{2\alpha }{c})^{2}) -\frac{ \alpha }{2}$ and this occurs when $m^2(\frac{c^2}{2 \alpha }-a)+(a-c)m+\frac{\alpha }{2}-\frac{a}{4}+\frac{a\alpha ^2}{c^2}>0$. This happens for $m \in [m^-,m^+]$. But $m^+=\frac{2 c^2 (a-c) \alpha +\sqrt{2} \sqrt{ a c^2 \alpha \left( 8 a \alpha ^3-4 c^3 \alpha +c^4 \right) }}{2 c^2 \left( 2 a \alpha -c^2\right) }>m'$, therefore we need $m>m^-=\frac{2 c^2 (a-c) \alpha -\sqrt{2} \sqrt{ a c^2 \alpha \left( 8 a \alpha ^3-4 c^3 \alpha +c^4 \right) }}{2 c^2 \left( 2 a \alpha -c^2\right) }.$ Since $m^-$ is smaller than $\hat{ m}$ for all $\alpha >\frac{c^2}{2a},$ then the claim is always true.

• (a.iii) Next for $m^*< m< m^{**}$ the relevant comparison is whether the profits corresponding to partial know-how transfer, $\frac{8 \alpha ^2 a^2 m (a(1-m)-c)-c^2(\alpha (a-c)^2+2am^2(2 \alpha a-c^2))}{2(2\alpha a-c^2)^2},$ are smaller than those for full transfer, $\frac{a}{4}(1-(\frac{2\alpha }{c})^{2}) -\frac{\alpha }{2}$. The former are more conveniently written as: $\frac{a[8 \alpha ^2 a^2 m m'-2 a \alpha c^2 m'^2-(4 a^2 \alpha ^2+2 a \alpha c^2-c^4)m^2)}{(2\alpha a-c^2)^2},$ which is increasing in m for all $m<m^{**}$. The latter can also be written as $a(\hat{m}+\frac{2\alpha }{c})\hat{m}-\frac{\alpha }{2},$ which are increasing in $\hat{m}.$ The difference between the latter and the former is a convex quadratic function in m which is negative iff m is between the following roots and positive otherwise. The roots are $m_{iii}^{+,-}= m^{**} \pm \frac{(2 a \alpha -c^2)(4a \alpha ^2+2c^2 \alpha -c^3)}{2c(4 a^2 \alpha ^2+2 a \alpha c^2-c^4)}$, or more precisely, $m_{iii}^{-}= \hat{m}$ and $m_{iii}^{+}= 2m^{**}-\hat{m}$. Then, the larger is always greater than $m^{**}$ and the smaller root is smaller than $m^{*}$, that is $\hat{m}<m^{*}$ if $\frac{\frac{a}{c}(2 a \alpha -c^2) (4 a \alpha -c^2)(4a \alpha ^2+2c^2 \alpha -c^3)-(2a \alpha (a-c)(2 a \alpha -c^2)^2 ) }{2a(4 a \alpha -c^2)(4 a^2 \alpha ^2+2 a \alpha c^2-c^4)} >0$ or equivalently iff $(4 a \alpha -c^2)(4a \alpha ^2+2c^2 \alpha -c^3)>2c \alpha (a-c)(2 a \alpha -c^2)$, which is true for $\alpha \ge \frac{c}{4}$. Note that under the conditions of (Case A), $c<a<2c$, and $\frac{c}{4}< \frac{c}{4a}(\sqrt{4ac+c^2}-c)<\frac{c^2}{2a}<\alpha <\frac{c}{2}$, the $\alpha \ge \frac{c}{4}$ is satisfied for all the range of $\alpha$, while for (Case B.1) the condition $\frac{c}{4}\le \alpha < \frac{c}{2}$ is required for partial know-how transfer to be preferred rather than full transfer.

• (a.iv) Finally, for $m^{**}\!<\!m\!<\!m'$, the relevant comparison is whether $\frac{\alpha (2 \alpha a+ c^2) (a-c)^2}{2(4 \alpha ^2 a^2+ 2\alpha a c^2-c^4)}$ is greater than $\frac{a}{4}(1-(\frac{2\alpha }{c})^{2}) -\frac{\alpha }{2}$. It happens that the claim is always true. Therefore, we conclude that partial transfer of know-how always yields greater licensor’s profits than full transfer under the conditions stated in parts (i) and (ii.a) in the proposition.

• (b) Consider now, (Case B.2) which implies the following threshold ranking: $0< m^*<\hat{m} <m^{**}<m'<\frac{1}{2}$; and also implies that $\frac{\alpha }{c}<m^*$, thus partial know-how transfer collapses to the full know-how transfer case for $\frac{\alpha }{c}<m<\hat{m}$.

• (b.i) Take, first $0< m<\frac{\alpha }{c}<m^*<\hat{m}$. The relevant comparison is that in (a.i) and then partial transfer is preferred.

• (b.ii) Consider now $\frac{\alpha }{c}<m<\hat{m}$, as just noted both partial and full know-how transfer are equivalent with $k=1$ in both cases, then full transfer of know-how is implemented.

• (b.iii) Take $\hat{m}<m<m^{**}$. The relevant difference in profits between full and partial know-how transfer is the one in (a.iii). Since now $\alpha \in [ \frac{c}{4a}(\sqrt{4ac+c^2}-c),\frac{c}{4}]$, we already know that $m^*<\hat{m}<m^{**}$, therefore partial know-how is implemented if $\hat{m}<m<m^{**}$, which is the case.

• (b.iv) Take $m^{**}<m<m'$. The relevant comparison is that in (a.iv) and then partial transfer is preferred. Therefore, we conclude that partial transfer of know-how yields greater licensor’s profits than full transfer for either $0< m<\frac{\alpha }{c}$ or $\hat{m}<m$ under the conditions stated in part (ii.b) in the proposition.

• (c) Consider (Case B.3) which implies the following threshold ranking: $0< m^*< m^{**}<\hat{m}<m'<\frac{1}{2}$, and since $m^{**}<\hat{m}$, partial know-how transfer collapses to the full know-how transfer case for $\frac{\alpha }{c}<m$.

• (c.i) Take, first $0< m<\frac{\alpha }{c}$. The relevant comparison is that in (a.i) and then partial transfer is preferred.

• (c.ii) For $\frac{\alpha }{c} <m$ both partial and full know-how transfer are equivalent with $k=1$ in all cases.

Therefore, we conclude that partial transfer of know-how yields greater licensor’s profits than full transfer when $0< m<\frac{\alpha }{c}$ under the conditions stated in part (ii.c) in the proposition. $\square$

Proof of Result 2

We first compute the effective marginal cost for any possible situation, If there is full transfer $\hat{c}=\hat{r}=\frac{2 \alpha a}{c}$. And $\hat{c}>c$ if $\alpha >\frac{c^2}{2a}$ which is the case.

In case of partial know-how transfer we have three possible cases depending on the value of m.

• If $0<m<m^*<m'$ then $\hat{c}=c+a(1-2m)$, once we have substituted for the precise expressions for $r^*$ and $k^*$. It is, then, easy to find that $\hat{c}>c$ iff $m<\frac{1}{2}$, which is true since $m^*<\frac{1}{2}$.

• If $m^*<m<m^{**}<m'$ then $\hat{c}=a(1-2m)$. Which is larger than c iff $m<m'$.

• If $m^{**}<m$ then $\hat{c}=r^{**}(1-\frac{c^2}{2\alpha a})+c$. And therefore $\hat{c}>c$ iff $\alpha >\frac{c^2}{2a}$ which is the case. $\square$