Designing free software samples: a game theoretic approach

Abstract

We develop a vertical differentiation game-theoretic model that addresses the issue of designing free software samples (shareware) for attaining follow-on sales. When shareware can be reinstalled, cannibalization of sales of the commercial product may ensue. We analyze the optimal design of free software according to two characteristics: the evaluation period allotted for sampling (potentially renewable) and the proportion of features included in the sample. We introduce a new software classification scheme based on the characteristics of the sample that aid consumer learning. We find that the optimal combination of features and trial time greatly depends on the category of software within the classification scheme. Under alternative learning scenarios, we show that the monopolist may be better off not suppressing potential shareware reinstallation.

Appendices

Appendix A: Basic model proofs

Below, we analyze the Subgame Perfect equilibria of the basic game. Using Proposition 1 we can rule out as an equilibrium the case where a monopolist would offer a free sample and consumers decline using the free sample and buy the product right away. One interesting insight from this proposition is that the payoff to the consumer from buying without receiving a sample or buying while declining to use the free sample must be the same.

Case 1

(A free sample is offered) The solution (${s^\ast,t^\ast}$) is such that ${s_{0}\leq s^\ast\leq 1}$ and ${t_{0}\leq t^\ast \leq 1}$.

By backward induction, the consumers who use the sample and then buy (B) must have their payoffs satisfy: (1) ${\delta{U}-\delta {P}+\delta ^{2}\theta {V}/(1-\delta) > \delta {U}+{d}UV\delta ^{2}/(1-\delta)}$ (Sampling and Buy > Sampling and Reinstall Sample). Condition (1) is equivalent to ${\theta > {(1-\delta)P}/{\delta {K}(s,t)}}$; where ${{K}(s,t)=[{1- ds({(1+t)}/{2}})]{V} > 0}$. The demand function is given by ${D(P,t,s)= {N}[1- \Psi ((1-\delta)P/\delta{K})]}$.

The monopolist’s profit is ${\Pi_{1}={D}(P,t,s)\times[P-c]-{F}-{N}\gamma c}$. The optimal solution for the price P ^*₁ satisfies the following first order condition:

$$ \frac{\partial \Pi}{\partial P}=\frac{\partial {D}}{\partial P} \left[{P-c} \right]+{D}=0 $$

(1)

Solving Eq. 1 leads to:

$$ P_1^\ast =\frac{c}{2}+\frac{\delta {K}^{\ast}\theta_{\rm H}} {2(1-\delta)} $$

(2)

To insure that any demand exists, we check that ${\theta_{H} > (1-\delta)P_1^\ast /\delta {K}^{\ast}}$, which is implied by ${\theta_{\rm H} > {((1-\delta)}/{\delta})c}$ when ${{K}^{\ast} > 1}$; where the variable K ^* represents the function ${K(\cdot}$) evaluated at the point (s ^*, t ^*). The monopolist’s optimal profit is ${\Pi _1^\ast=({{N}}/({\theta_{\rm H} -\theta_{\rm L}}))\times ({\delta {K}^{\ast}}/{4(1-\delta )})\times [{\theta_{\rm H}-(({(1-\delta )c})/{\delta {K}^{\ast}})}]^2-{F}- {N}\gamma c.}$

The optimal proportion of features s ^* is given by the following necessary first order conditions depending on whether the solution is a corner or interior one:

$$ \left\{ {{\begin{array}{l} {s^\ast =s_0}\\ {s^\ast =\bar {s}}\\ {s^\ast =1}\\ \end{array}}}\right.\frac{\partial \Pi}{\partial s}=\frac{\partial \mbox{D}}{\partial s}\left[ {P-c} \right]\left\{ {{\begin{array}{l} { < 0}\\ {=0}\\ { > 0}\\ \end{array} }} \right. \quad {\begin{array}{l} {(3\hbox{a})}\\ {(3\hbox{b})}\\ {(3\hbox{c})}\\ \end{array} } $$

with ${\frac{\partial {D}}{\partial s}={N}\times \frac{P}{{K}^{2}}\times \Psi'()\times\frac{\partial K}{\partial s}}$. Thus, Eqs. 3a, b, c become (given t ^*):

$$ \left\{{{\begin{array}{l} {s^\ast =s_0} \\ {s^\ast =\bar{s}}\\ {s^\ast =1}\\ \end{array}}} \right.\frac{\partial K}{\partial s}=-{d}\left( {\frac{1+t^\ast }{2}} \right)V^\ast +\left[{1-{d}s\frac{1+t^\ast}{2}} \right]\frac{\partial V}{\partial s}\left\{{{\begin{array}{l} { < 0} \\ {=0} \\ { > 0} \\ \end{array} }} \right. \quad {\begin{array}{l} {(4\hbox{a})}\\ {(4\hbox{b})}\\ {(4\hbox{c})}\\ \end{array}} $$

Similarly, the optimal trial time t ^* is given by the following first order conditions:

$$ \left\{{{\begin{array}{l} {t^\ast =t_0}\\ {t^\ast =\bar{t}}\\ {t^\ast =1} \\ \end{array}}} \right.\frac{\partial \Pi}{\partial t}=\frac{\partial {D}}{\partial t}\left[{P-c} \right]\left\{ {{\begin{array}{l} { < 0}\\ {=0}\\ { > 0}\\ \end{array}}} \right. \quad {\begin{array}{l} {(5\hbox{a})}\\ {(5\hbox{b})}\\ {(5\hbox{c})}\\ \end{array}} $$

Again, Eqs. 5a, b, c become (given s ^*):

$$ \left\{{{\begin{array}{l} {t^\ast =t_0}\\ {t^\ast =\bar{t}}\\ {t^\ast =1} \\ \end{array}}} \right.\frac{\partial K}{\partial t}=-\frac{{d}s^\ast }{2}V^\ast +\left[ {1-{d}s^\ast \frac{1+t}{2}} \right]\frac{\partial V}{\partial t}\left\{ {{\begin{array}{l} { < 0}\\ {=0}\\ { > 0}\\ \end{array}}} \right. {\begin{array}{l} {(6\hbox{a})}\\ {(6\hbox{b})}\\ {(6\hbox{c})}\\ \end{array} } $$

Since fixed costs are assumed independent of the sample’s actual values for s and t, our (local) optimal proportion of features and optimal trial time are determined at the margin by maximizing static demand. Maximizing demand in our framework translates into maximizing the function ${K(\cdot}$), which pins down the trade-off between the impact of learning on the desirability of purchasing the product versus the attractiveness of sample reinstallation.

Thus, our solutions are maximizing the function ${K(\cdot)}$ given the parameter values. Regarding interior solutions, a series of simple algebraic manipulations respectively using (4b) and (6b) leads to:

$$ \bar {s}(t^{\ast})=\frac{1}{d(1+t^{\ast})}- \frac{(1-\alpha)}{2\alpha}t^{\ast}-\frac{1}{2\alpha A} $$

(7)

$$ \bar {t}(s^{\ast})=\frac{1}{ds^{\ast}}-\frac{\alpha }{2(1-\alpha)}s^{\ast}-\frac{1+A(1-\alpha)}{2A(1-\alpha)} $$

(8)

It is important to note that the proportion of features s ^* and the trial time t ^* cannot jointly be interior solutions at the same time, since we can easily show that the Hessian is not negative semi-definite at the point (${\bar {s},\bar{t}}$). In fact, the pair (${\bar{s},\bar {t}}$) represents a saddle point solution in the feasible space.

Case 1.1

(s ^*, t ^*) = (1,1) will be true if (4c) and (6c) hold.

It is easy to see that (4c) is equivalent to ${1 < \bar{s}(1)}$ and (6c) is equivalent to ${1 < \bar {t}(1)}$. These two conditions will be satisfied when d is close to zero (reinstallation cost is infinite).

Case 1.2

(${s^\ast ,t^\ast )=(s_{0},t_{0})}$ will be true if (4a) and (6a) hold.

Both conditions are equivalent to having ${s_{0} > \bar {s}(t_0)}$ and ${t_{0} > \bar {t}(s_0)}$. These conditions will be satisfied even when d is large (reinstallation cost is low) but not arbitrarily close to 1; α is in a compact range (not too close to either 0 or 1); and some or all of the following hold: A, s ₀ and t ₀ are large. Moreover, when A is small then α must be large, and vice versa.

Case 1.3

${(s^\ast ,t^\ast)=(1,{t}_{0})}$ will be true if (4c) and (6a) hold.

These conditions are equivalent to ${1 < \bar {s}(t_0)}$ and ${t_{0} > \bar {t}(1)}$ as well as ${(1-\alpha)(1-t_{0}) < \alpha}$. These conditions are satisfied if ${\alpha}$ is large, t ₀ and d small; and both A and s ₀ are either large or small in tandem (d must be close to 0 when A and s ₀ are both small). The conditions A and s ₀ being either large or small together insure that ${{K}(1,t_{0})\geq \hbox{max}\{{K}(s^\ast ,t^\ast);s^\ast \in \{s_{0},\bar{s},1\}; t^\ast \in \{t_{0},\bar{t},1\}\}}$.

Case 1.4

${(s^\ast ,t^\ast)=(s_{0},1)}$ will be true if (4a) and (6c) hold.

These conditions are equivalent to ${{s}_{0} > \bar {s}(1)}$ and ${1 < \bar {t}(s_0)}$ implying ${2(1-\alpha) > \alpha s_{0}}$. These conditions are satisfied when α and s ₀ are small. A higher value for A will make the condition on α less binding.

Case 1.5

${(s^\ast ,t^\ast)=(\bar {s},t_{0})}$ will be true if (4b) and (6a) hold. The latter condition is equivalent to ${t_{0} > \bar {t}(\bar{s})}$ and both conditions imply that ${(1-\alpha)(1+t_{0}) < \alpha \bar{s}}$. This will be true when α and A are large and s₀ and d are small. The condition s₀ small is necessary otherwise, case 1.2 would dominate.

Case 1.6

${(s^\ast ,t^\ast)=(s_{0},\bar{t})}$ will be true if (4a) and (6b) hold. The first condition is equivalent to ${s_{0} > \bar {s}(\bar{t})}$ and both conditions imply ${(1-\alpha)(1+\bar{t}) > \alpha s_{0}}$. This is true when α is small, A and s₀ are large and d is large enough.

Case 1.7

(Not a solution) (s^*, t^*) = (${\bar {s},1)}$ will be true if (4b) and (6c) hold. This pair cannot be a solution. The latter condition is equivalent to ${\bar{t}(\bar{s}) > 1}$ and both conditions imply that ${(1-\alpha)(1+t_{0}) > \alpha \bar{s}}$ which will be true if α is small, A and t₀ large, with d not too small. This solution is dominated by (s₀, t^*) with ${t^\ast \in\{t_{0} ,\bar{t}, 1\}}$. The idea being that if A is large and α is small, reducing the sample features does not reduce its marketing power, while at the same time it prevents sample re-use.

Case 1.8

(Not a solution)${(s^\ast ,t^\ast)=(1,\bar{t})}$ will be true if (4c) and (6b) hold. Again, this pair cannot be a solution. The first condition is equivalent to ${\bar{s}(\bar{t}) > 1}$ and both conditions imply that ${(1-\alpha)(1+\bar{t}) < \alpha}$; which will be true if α and A are large and d small. This solution is dominated by picking (s^*, t₀) with ${s^\ast \in \{s_{0},\bar{s},1\}}$. The idea being that if α is large, trial time plays a minor role, and hence reducing it does not lower the attractiveness of the product since A is large.

Case 2

(No Free Sample is Given) Assume that the monopolist strategy space is limited to picking a proportion of features s < s ₀. Since the proportion of features is below the minimum, then V = 1. Thus, the function ${{K}=1-{d}s({(({1+t})/{2})})}$ takes values in the interval [0,1]. It is easy to see that the demand ${D(P, t, s)}$ is decreasing in s and t. By not distributing a free sample, the monopolist is also not incurring the cost ${N\gamma c}$ of making and distributing copies of the sample. Thus, the optimal response for the monopolist is to set ${s^\ast}$ and ${t^\ast = 0}$. Moreover, the standard first order condition for selecting the optimal price ${P_2^\ast}$ leads to ${P_2^\ast =({c}/{2})+({\delta \theta_{\rm H}}/{2(1-\delta)})}$, and the profit is ${\Pi_2^\ast ={{N}}/({\theta_{\rm H}-\theta_{\rm L}})({\delta }/{4(1-\delta)})[ {\theta}_{\rm H} -({(1-\delta)}/{\delta}) c]^2}-F$. Lastly, for the demand to exist we need ${\theta_{\rm H} > ({(1-\delta )c}/{\delta})}$.

We need to check that selecting ${s^\ast}$ and ${t^\ast = 0}$ is optimal over the global strategy space. This will be the case if ${\Pi _2^\ast > \delta_{\rm m}\Pi _1^\ast}$ where ${\Pi _1^\ast}$ is the maximum profit generated in Case 1. It is easy to check that ${\Pi _2^\ast > \delta _{\rm m}\Pi _1^\ast}$ when γ large, δ _m small and ${K^{\ast} < 1}$. The latter will be true when the learning parameter A is small and/or s ₀ is close to 1, and/or t ₀ is close to 1, and/or the reinstallation cost is low (d large), and/or α is small. Otherwise, the solutions ${s_{0 }\le s^\ast \leq 1}$ and ${t_{0 }\le t^\ast \leq 1}$ explored in Case (1) are optimal over the global strategy space, i.e. including selecting s < s ₀ and t < t ₀, under contrary assumptions. □

Appendix B: Multi-period learning proofs

Case 1

(2-period learning) We assume that in addition to the strategies that were available before, consumers can also buy the product in period 2; but not beyond that period. Below, we analyze the equilibrium where two sequential market demands co-exit with one segment of consumers who prefer buying at the end of period 1, and another segment who will buy after two periods.

Subcase 1.1

The solution ${(s^\ast ,t^\ast)}$ is such that ${s_{0}\leq s^\ast \leq 1}$ and ${t_{0}\leq t^\ast \leq 1}$.

To keep the notations lighter we examine the consumers payoffs from the standpoint of present values at the end of period 1 (after sampling), and not the beginning of stage 1 as was done in the basic model. By backward induction, the consumers who will use the sample and then buy (B) at end of period 1 must have their payoffs satisfy the following conditions; given that learning occurs according to the learning function L ₁:

• (1) ${{U}-{P}+ \delta \theta {V}+\delta^{2}\theta \lambda {V}/(1-\delta) > {U}+\delta d {UV}+\delta ^{2}d {U} \lambda {V}/(1-\delta)}$ (Buy at end period 1 > Not-Buy)

• (2) ${{U}-{P}+ \delta \theta {V}+\delta ^{2}\theta \lambda {V}/(1-\delta) > {U}-\delta {P}+ \delta d {UV} +\delta^{2}\theta \lambda {V}/(1-\delta)}$ (Buy at end period 1 > Buy at end of period 2) . By backward induction, consumers who will use the sample and then buy (B) at end of period 2 must have their payoffs satisfy:

• (3) ${{U}-\delta {P}+ \delta d {UV} +\delta ^{2}\theta \lambda {V}/(1-\delta) > {U}+\delta d {UV}+\delta^{2}d {U} \lambda {V}/(1-\delta)}$ (Buy at end period 2 > Not-Buy)

Note that conditions (2) and (3) imply condition (1).

Condition (3) is equivalent to ${\theta > \bar {\theta}_1={((1-\delta)P)}/{\lambda \delta {K}(s,t)}}$; where $K(\cdot)$ is defined as before in the basic model. Condition (2) is equivalent to ${\theta > \bar {\theta}_2 ={(1-\delta )P}/{\delta {K}(s,t)}}$. It is easy to show that ${\bar{\theta}_2 > \bar {\theta}_1}$ since ${\lambda > 1}$. More importantly, the segment of consumers that satisfy ${\theta > \bar {\theta}_2}$ will buy at the end of period 1, and the segment of consumers satisfying ${\bar{\theta}_2 > \theta > \bar{\theta}_1}$, will buy in the second period. The demand from consumers buying in period 1 is ${{D}_{1}({P}, t,s)={N}[1-\Psi(\bar{\theta}_2)]}$. The residual demand from consumers buying in period 2 is ${{D}_{2}({P}, t,s)={N}[\Psi(\bar \theta_2)-\Psi(\bar{\theta}_1)]}$. The monopolist’s profit (valued at end of period 1) is ${\Pi = [{D}_{1}+\delta_{\rm m}{D}_{2}]\times [{P}-c]-F-N\gamma c}$; which is the present value of profits obtained from the total demand. Let us denote the total demand by ${{D}=[{D}_{1}+\delta_{m}{D}_{2}]}$. The optimal solution for the price ${P^\ast}$ is found using the following first order condition:

$$ \frac{\partial \Pi}{\partial P}=\frac{\partial {D}}{\partial P}\left[{P-c}\right]+{D}=0 $$

(9)

Solving Eq. 9 leads to:

$$ P^\ast =\frac{c}{2}+\frac{\delta {K}^{\ast}{\theta }_{\rm H}}{2(1-\delta)}\times \frac{\lambda}{\lambda-\delta_m (\lambda -1)} $$

(10)

Given the expression for the total demand and the price above, the monopolist’s optimal profit is given by ${\Pi ^\ast=({{N}}/({\theta_{\rm H}-\theta_{\rm L} }))\times ({\delta {K}^{\ast}}/{(4(1-\delta))})\times ({\lambda}/({\lambda -\delta_m (\lambda -1)}))\times [{\theta_{\rm H}-({\lambda-\delta_m (\lambda -1)})/{\lambda}\times ({(1-\delta )c})/{\delta {K}^{\ast}}}]^2-{F}- {N}\gamma c.}$

Just as was done in Appendix A, it is straightforward to show that maximizing the profit function is equivalent to maximizing the function ${K(\cdot)}$ with respect to trial time and proportion of features. Therefore, the solutions (${s^\ast ,t^\ast}$) are the same as the ones described in Appendix A.

Subcase 1.2

The solution (${s^\ast ,t^\ast}$) is such that ${s^\ast < s_{0}}$ and ${t^\ast < t_{0}}$. This sub-case is the same as in Appendix A.

Case 2

(One-period learning; no reinstallation possible) Since the sample is essentially non-durable it is straightforward to show that the optimal solution (${s^\ast,t^\ast) =(1,T}$). The consumers that sample and buy must satisfy ${U-P+\delta^{\rm T}\lambda ^{\rm T-1}{V}_{\rm max}\theta /(1-\delta^{\rm T}) > U}$; since no reinstallation is possible; and given that learning occurs according to the learning function L ₂: where ${{V}_{\rm max} =A(\alpha +(1-\alpha)T)+1}$. Maximizing profit with respect to price gives:

$$ P_3^\ast=\frac{c}{2}+\frac{\lambda^{T-1}\delta ^TV_{\rm max} \theta_{\rm H}}{2(1-\delta^T)}, $$

(11)

Again, given the assumption that completely preventing the unauthorized reinstallation of free samples is costless for the monopolist (independent of d), the monopolist’s optimal profit must given by ${\Pi_3^\ast =({{N}}/({\theta_{\rm H} -\theta_{\rm L}}))\times ({\lambda^{T-1}\delta ^T{V}_{\rm max}}/({4(1-\delta ^T)}))\times [{\theta_{\rm H}-({(1-\delta ^T)c})/{\lambda^{T-1}\delta^T{V}_{\rm max}}}]^2-F-N \gamma c}$. Comparing Case 1 with Case 2, we surmise that Case 1 will be favored by the monopolist when ${\delta_{m}\Pi^\ast > \delta ^{T}_{m}\Pi_3^\ast}$. This is true when the following sufficient condition holds:

$$ \frac{\delta {K}^{\ast}}{(1-\delta)}\times \frac{\lambda }{\lambda -\delta_m (\lambda -1)} > \frac{\lambda^{T-1}\delta ^T{V}_{\rm max}}{(1-\delta ^T)} $$

(12)

Let ${\bar{T}(A, d, \alpha , t_0,\delta , \delta_{m}, \lambda)}$ be such that:

$$ \frac{\lambda^{\bar{T}-1}\delta^{\bar{T}}\bar{V}_{\rm max}} {(1-\delta^{\bar{T}})}= \frac{\delta {K}^{\ast}} {(1-\delta)}\times \frac{\lambda}{\lambda -\delta_m(\lambda-1)} $$

(13)

A solution ${\bar{T}}$ exists if the RHS of (12) is monotonic in T, and under other conditions on the parameters defined below. Thus (12) is equivalent to:

$$ \frac{\lambda^{\bar{T}-1}\delta^{\bar{T}}{\bar {{V}}}_{\rm max}} {(1-\delta^{\bar T})} > \frac{\lambda^{T-1}\delta^T{V}_{\rm max}} {(1-\delta^T)} $$

(14)

From Eq. 14, we can see that Case 1 will be favored when ${T\geq \bar {T}({A}, d, \alpha ,t_0, \delta , \delta _{m}, \lambda)}$ as long as ${{\partial({\lambda^{T-1}\delta^T{V}_{\rm max}}/{(1-\delta^T)})}/{\partial T} < 0}$. It is easy to check that this latter derivative being negative is implied by λ close to 1. Moreover, these conditions with δ _m small and d large are also sufficient for the value of ${({\delta{K}^{\ast}}/{(1-\delta)})\times({\lambda}/({\lambda -\delta_m(\lambda -1))})}$ to be contained in the interval of minimum and maximum values for the function ${{\lambda^{T-1}\delta ^T{V}_{\rm max}}/{(1-\delta ^T)}}$. Furthermore, basic comparative statics on Eq. 13 reveals that ${\bar{T}}$ must be increasing in A, ${\alpha, s_0, t_0}$; since d is assumed large. However, the effect of raising the discount rate ${\delta}$ on ${\bar{T}}$ is ambiguous.

On the other hand, ${{\partial({\lambda ^{T-1}\delta^T{V}_{\rm max}}/{(1-\delta^T)})}/{\partial T} > 0}$, implies that Case 1 will be favored when ${\bar{T}({A}, d, \alpha, t_0, \delta,\delta_{m},\lambda) \quad \ge T}$. It is easy to check that this latter derivative being positive is implied by ${\lambda}$ large. Moreover, these conditions with ${\delta_{m}}$ small and d small are also sufficient for the value of ${({\delta{K}^{\ast}}/{(1-\delta)})\times ({\lambda}/{(\lambda -\delta_m (\lambda -1))})}$ to be contained in the interval of minimum and maximum values for the function ${{\lambda^{T-1}\delta^T{V}_{\rm max}}/{(1-\delta^T)}}$. The comparative statics from (13) follows from the same argument as before. It shows that ${\bar {T}}$ must be decreasing in A, ${\alpha, s_0}$ and ${t_0}$, and increasing in ${\delta}$; especially when d is assumed small. □