Planned Obsolescence with Network Effects

Proof of Lemma 1

Equation (0) is a first-order ODE of the form y′x=cy/x+d, where y is the dependent variable, x the independent variable, and c and d are positive constants (in particular, c≡1−Q,d≡Q/R+1). To solve Eq. (0) we just have to take into account:

1. Differential equations of the form y′x=Fy/x have the following solution: lnx=∫duFu−u+const. In our case u = y/x, which leads toyx=Const⋅c−1xc−dc−1x.

2. The “initial condition” for the differential equation is m(n = I₀ + E₀) = I₀.

   Imposing this condition we obtain m(n) in Eq. (1).□

Proof of Proposition 1

1. The existence of R^* ensures that the entrant can obtain a market-share above 50% (numerical examples are also provided along the paper). By taking m(N) = N/2 in Eq. (1), simple manipulations lead to R*=N1−I0+E0NQ/N/2−I0NI0+E01−Q−1, the quality ratio for which the entrant ties the game (so R^* gives the threshold quality ratio such that the two firms obtain a market share of 50%). For superior levels of R, the entrant wins the battle. Simple manipulations also give us the alternative expression for R^*.

2. By taking derivatives of R^* with respect to Q, I₀, N we check that R^* decreases with Q and N, and increases with I₀:

   For Q: From R*=12i0+e01−Q−e0/12i0+e01−Q−i0, it is easy to check that ∂R^∗ /∂Q is negative by taking into account that ∂/∂Qi0+e0Q−1<0 (because i₀ + e₀ < 1) and that i₀ > e₀ (by assumption).

   For N: R^∗ can be expressed as NQ−D1NQ−D2<0, with D₁ ≡ D × E₀, D₂ ≡ D × I₀ and D≡2I0+E0Q−1.

   It is immediate to check that sign∂R*/∂N=signqNQ−1D1−D2, which is negative since D₁ < D₂.

   For I₀: R*=NQ−D1NQ−D2=1+D2−D1NQ−D2.

   D₂ − D₁ increases with I₀ : ∂D2−D1∂I0=2I0+E0Q−11+Q−1I0−E0I0+E0, which is positive. And N^Q − D₂ decreases with I₀ (because D₂ increases with I₀). So D2−D1NQ−D2 increases with I₀.

   The inexistence of R^* under some parameter configurations implies that the entrant cannot win the battle in some cases. It is not difficult to find parameter configurations for which N/2−I0NI0+E01−Q≤0, which implies that R^* does not exist. By (b), those configurations will be favored as Q lowers, N lowers, I₀ increases.

1. According to Eq. (0), m′n=1−Qmn+Q1−RR+1: the variation in the number of consumers who opt for the incumbent’s product equals the probability that an arbitrary consumer chooses I’s product. As m(n) is concave (see Eq. (1)) m′(n) decreases with n, and consequently the probability associated to choosing the incumbent’s product also decreases with n. So, after the moment at which 1−Qmn+Q1−RR+1=12, the probability associated to buying the I’s product will remain below ½1/2. Solving 1−Qmn+q1−RR+1=12 leads to n1=AC−B1/Q. Finally, solving m/n = 1/2 leads to n2=A12−B1/Q. For more details see the proof of Proposition 2.

To show that the moment at which the entrant’s market-share reaches 50%, when n = n₁, occurs after the probability associated to choosing the E’s product reaches ½1/2, when n = n₂, we just have to show that 1−Qmn+Q1−RR+1=12 necessarily implies that m/n is above 1/2. As 1−Qmn+Q1−RR+1=12, given that coefficients 1 − Q and Q sums 1 and that 1R+1is below 1/2, m/n must be above ½1/2 (alternatively: compare AC−B1/Q to A12−B1/Q and observe that C > 1/2).□

Proof of Proposition 2

Assume that the entrant wins the battle (we have provided scenarios for that to happen in Proposition 1). In such a case, when all consumers have made their respective choices, inequality m(N) < N/2 necessarily holds. So there must be a moment in time at which m/n = 1/2 because ratio m/n is a decreasing function of n (take into account that m(n) is concave, see Eq. (1)). So the probability associated to network effects, m/n, will be below ½1/2 after the moment at which m/n reaches ½1/2 and until all consumers have made their purchases. On the other hand, 1−RR+1 is below ½1/2 at any time (because R > 1). A convex combination of numbers below ½1/2 is necessarily below ½1/2 (a convex combination satisfies that the sum of coefficients is one). Then both effects will favor future consumers to choose the entrant’s product, so network effects enhance the quality effect from the moment at which n = n₂ until the end (n = N). The time at which both effects operate jointly starts when m(n) = n/2, which operating leads to n2=A12−B1/Q.□

Proof of Proposition 3

The first order condition for the problem (P1) is −QAX−Q−1+K/X2=0, from which we obtain the optimal market size for the incumbent, X*=KQA11−Q, valid whenever I₀ + E₀ < X^* < N; otherwise a corner solution arises, and then the incumbent chooses either X^* = I₀ + E₀ or X^* = N.

X^* increases with K (directly observed); it decreases with Q (since the objective function, AX^−Q + B − K/X², decreases with Q); and also decreases with R (since X^* decreases with A, which in turn increases with R).

The equilibrium market share of the incumbent is obtained by substituting X*=KQA11−Q in m(X)/X. Condition m(X^*)/X^* > 1/2 is equivalent to B+AKQA−Q1−Q>12. It is easy to find situations where this condition is violated, which means that the incumbent can maximize its objective function getting a lower market-share than the entrant. For example, consider N = 1000, Q = 0.7, R =3, I₀ = 200, E₀ = 100, K = 120. Under this parameter configuration the solution is interior. The incumbent, trying to maximize (m − K)/X, fixes a market size of X^* = 860 consumers, 386 of whom opt for the incumbent’s product (45% of market share).□

Proof of Proposition 4

We identify social welfare with the entrant’s market share, 1 − m(X^*)/X^*, which is equal to 1−B−AKQA−Q1−Q when (P1) admits an interior solution. This expression increases with K (directly observed).

Social welfare has not a monotonic dependence with respect to Q and R. Let us see it. If N = 1000, R = 3, I₀ = 200, E₀ = 100, K = 120, the entrant’s market share is inversely U-shaped with respect to Q: if Q = 0.6, 0.7, 0.8 the entrant’s market share (identified with social welfare in our setup) is 0.544, 0.551, 0.549 respectively (in all cases X^* has an interior solution). With other parameter configurations, however, the entrant’s market share can increase with Q. If R = 2 (ceteris paribus) then market share of the entrant increases with Q reaching 2/3 when Q = 1. This conclusion is intuitive (the entrant’s market share rises as Q rises) but it does not hold with generality, as we have seen when R = 3 (the incumbent can strategically respond to the high average quality-orientation of the population Q by cutting off the market sooner).

If instead we fix Q and let R vary, efficiency also can be inversely U-shaped: consider N=1000, I₀ = 200, E₀ = 100, K = 120, Q = 0.7; if R = 1.5, 2.75, 4; then the entrant’s market share is 0.555, 0.559, 0.509, respectively (in the first case X^* = 1000, a corner solution).