Dynamic control of a multiproduct monopolist firm’s product and process innovation

Abstract

In this paper, we develop a dynamic control model of a multiproduct monopolist’s product and process innovation with knowledge accumulation resulting from learning by doing. Our work main features have four aspects: (i) the multiproduct monopolist’s instantaneous cost functions of product and process innovation depend on both the corresponding investments and the accumulations of knowledge through learning by doing in product and process innovation activities, respectively; (ii) the change rates of knowledge accumulations of product and process innovation are state variables; (iii) product innovation is understood as a reduction in product substitutability and process innovation as a lowering of marginal production cost; and (iv) monopolist’s inverse demand function depends jointly on the output level for product varieties and product substitutability. Our main objective is to investigate the optimal decision behavior of a multiproduct monopolist investing in product and process innovation with knowledge accumulation resulting from learning by doing under the monopolist optimum and social planner optimum. Our results show that: (i) there exists a unique saddle stable steady-state equilibrium under the monopolist optimum and social planner optimum, respectively; (ii) the learning rates of product and process innovation affect the monopolist’s investments in product and process innovation under the monopolist optimum and social planner optimum; and (iii) both investments in product and process innovation are lower under the monopolist optimum than that under the social planner optimum. Our results are valuable complements and development to the results drawn from the standard product and process innovation model.

Appendices

Appendix A

Proof of Lemma 1

According to expression (3.12) as well as differential dynamic Eqs. (2.1) and (2.2), one obtains:

$$ \frac{{\partial \lambda_{1} (t)}}{\partial k(t)} = \frac{{\partial \lambda_{1} (t)}}{\partial s(t)}\frac{\partial s(t)}{\partial k(t)} = \frac{n[a - c(t)]}{{4(r - \sigma )[1 + (n - 1)s(t)]^{3} }}[1 - e^{(r - \sigma )(t - T)} ]\frac{\partial s(t)}{\partial k(t)} $$

(A.1a)

$$ \frac{{\partial \lambda_{1} (t)}}{\partial h(t)} = \frac{{\partial \lambda_{1} (t)}}{\partial c(t)}\frac{\partial c(t)}{\partial h(t)} = \frac{n[a - c(t)]}{{2(r - \sigma )[1 + (n - 1)s(t)]^{2} }}[1 - e^{(r - \sigma )(t - T)} ]\frac{\partial c(t)}{\partial h(t)} $$

(A.1b)

$$ \frac{\partial s(t)}{\partial k(t)} = \frac{{\partial \int_{0}^{t} {\dot{s}(\tau )d\tau } }}{\partial k(t)} = - t $$

(A.2a)

$$ \frac{\partial c(t)}{\partial h(t)} = \frac{{\partial \int_{0}^{t} {\dot{c}(\tau )d\tau } }}{\partial h(t)} = - t $$

(A.2b)

Substituting expressions (A.2a), (A.2b) into (A.1a) and (A.1b), respectively, we have

$$ \frac{{\partial \lambda_{1} (t)}}{\partial k(t)} = \frac{{\partial \lambda_{1} (t)}}{\partial s(t)}\frac{\partial s(t)}{\partial k(t)} = - \frac{n[a - c(t)]t}{{4(r - \sigma )[1 + (n - 1)s(t)]^{3} }}[1 - e^{(r - \sigma )(t - T)} ] < 0 $$

(A.3a)

$$ \frac{{\partial \lambda_{1} (t)}}{\partial h(t)} = \frac{{\partial \lambda_{1} (t)}}{\partial c(t)}\frac{\partial c(t)}{\partial h(t)} = - \frac{n[a - c(t)]t}{{2(r - \sigma )[1 + (n - 1)s(t)]^{2} }}[1 - e^{(r - \sigma )(t - T)} ] < 0 $$

(A.3b)

Similar to the case of analyzing expression (3.12), according to expression (3.13) as well as differential dynamic Eqs. (2.1) and (2.2), one gets

$$ \frac{{\partial \lambda_{2} (t)}}{\partial k(t)} = \frac{{\partial \lambda_{2} (t)}}{\partial s(t)}\frac{\partial s(t)}{\partial k(t)} = - \frac{t}{{4(r - \delta )[1 + (n - 1)s(t)\}^{2} }}[1 - e^{(r - \delta )(t - T)} ] < 0 $$

(A.4a)

$$ \frac{{\partial \lambda_{2} (t)}}{\partial h(t)} = \frac{{\partial \lambda_{2} (t)}}{\partial c(t)}\frac{\partial c(t)}{\partial h(t)} = - \frac{t}{2(r - \delta )[1 + (n - 1)s(t)]}[1 - e^{(r - \delta )(t - T)} ] < 0 $$

(A.4b)

Finishing the proof. □

Appendix B

Proof of Proposition 1

• (i) According to differential dynamic Eqs. (3.20) and (3.21), one can immediately derive

  $$ \frac{{\partial \dot{k}(t)}}{{\partial b_{1} }} = - \frac{\mu }{2\alpha (r + \gamma )}[r - \sigma (1 - e^{(r + \gamma )(t - T)} )] < 0;\;\frac{{\partial \dot{h}(t)}}{{\partial b_{2} }} = - \frac{\xi }{2\beta (r + \theta )}[r - \delta (1 - e^{(r + \theta )(t - T)} )] < 0. $$

• (ii) The reaction of product innovation investment rate to a change in process innovation investment is given by

  $$ \frac{{\partial \dot{k}(t)}}{\partial h(t)} = \frac{{\partial \dot{k}(t)}}{\partial c(t)}\frac{\partial c(t)}{\partial h(t)} $$

  (B.1)

  Using differential dynamic Eq. (2.2), one obtains:

  $$ \frac{\partial c(t)}{\partial h(t)} = \frac{{\partial \int_{0}^{t} {\dot{c}(s)ds} }}{\partial h(t)} = - t < 0 $$

  (B.2)

  From expression (3.20), one gets $\frac{{\partial \dot{k}(t)}}{\partial c(t)} = \frac{n[a - c(t)]}{{4\alpha \{ [1 + (n - 1)s(t)]\}^{2} }} > 0$, so we have $\frac{{\partial \dot{k}(t)}}{\partial h(t)} < 0$.

  Likewise, we can write:

  $$ \frac{{\partial \dot{h}(t)}}{\partial k(t)} = \frac{{\partial \dot{h}(t)}}{\partial s(t)}\frac{\partial s(t)}{\partial k(t)} $$

  (B.3)

  From differential dynamic Eq. (2.1), one obtains:

  $$ \frac{\partial s(t)}{\partial k(t)} = \frac{{\partial \int_{0}^{t} {\dot{s}(\tau )d\tau } }}{\partial k(t)} = - t < 0 $$

  (B.4)

  From expression (3.21), one gets $\frac{{\partial \dot{h}(t)}}{\partial s(t)} = \frac{n(n - 1)[a - c(t)]}{{4\beta [1 + (n - 1)s(t)]^{2} }} > 0$, so we have $\frac{{\partial \dot{h}(t)}}{\partial k(t)} < 0.$

• (iii) Differentiating expressions (3.20) and (3.21) with respect to c(t) and s(t), respectively, one obtains:

  $$ \frac{{\partial \dot{k}(t)}}{\partial c(t)} = \frac{n(n - 1)[a - c(t)]}{{4\alpha [1 + (n - 1)s(t)]^{2} }} > 0,\quad \frac{{\partial \dot{h}(t)}}{\partial s(t)} = \frac{n(n - 1)[a - c(t)]}{{4\beta [1 + (n - 1)s(t)]^{2} }} > 0. $$

  Finishing the proof. □

Appendix C

Proof of Proposition 3

• (i) According to differential equation system (4.12), one can derive

  $$ \frac{{\partial \dot{k}(t)}}{{\partial b_{1} }} = - \frac{\mu }{2\alpha (r + \gamma )}[r - \sigma (1 - e^{(r + \gamma )(t - T)} )] < 0;\quad \frac{{\partial \dot{h}(t)}}{{\partial b_{2} }} = - \frac{\xi }{2\beta (r + \theta )}[r - \delta (1 - e^{(r + \theta )(t - T)} )] < 0 $$

• (ii) The proof is similar to Proposition 1–(ii).

• (iii) Differentiating $\dot{k}(t)$ and $\dot{h}(t)$ differential equation system (4.12) with respect to c(t) and s(t), respectively, one obtains:

  $$ \frac{{\partial \dot{k}(t)}}{\partial c(t)} = \frac{(2n + 1)(n - 1)[a - c(t)]}{{4\alpha [1 + (n - 1)s(t)]^{2} }} > 0,\quad \frac{{\partial \dot{h}(t)}}{\partial s(t)} = \frac{n(n - 1)[a - c(t)]}{{4\beta [1 + (n - 1)s(t)]^{2} }} > 0. $$

  Finishing the proof. □