Bounded learning-by-doing and sources of firm level productivity growth in colombian food manufacturing industry

Abstract

This paper models the bounded learning concept with the learning progress function characterized by the degree of efficiency and the specification of the learning progress as a logistic function capturing both the slow start-up and the limit in learning progress. We differentiate learning efficiency from the technical efficiency. The endogeneity corrected stochastic frontier model is then used to decompose the factor productivity growth into components associated with technological change, technical efficiency, scale, and learning. This productivity growth decomposition provides useful information and policy level insight in firm-level productivity analysis. Empirical results based on plant-level panel data on the Colombian food manufacturing industry for the period 1982–1998 suggest that productivity growth not only stems from technical progress, technical efficiency change, and scale but also from significant learning effect. The relative importance of the productivity growth components provides perspective for efficient resource allocation within the firm.

Appendices

Appendix 1

To solve the differential equation governing the learning progress for a general firm

$$\begin{array}{c}\frac{{dA}}{{dV}} = \alpha A - \frac{\alpha }{{\bar a}}{A^2} - \eta A\\ \frac{{dA}}{{dV}} = \frac{\alpha }{{\bar a}}A\left( {\left( {\alpha - \eta } \right){\textstyle{{\bar a} \over \alpha }} - A} \right)\end{array}$$

Using separation of variables we get,${\int} {\frac{{dA}}{{A\left( {\left( {\alpha - \eta } \right){\textstyle{{\bar a} \over \alpha }} - A} \right)}} = {\int} {\frac{\alpha }{{\bar a}}dV} } + c$

Partial fraction decomposition, ${\frac{1} {\left( \alpha - \eta \right) {\frac{\bar a}{\alpha}} }} {\int} \left({\frac{1}{A}}- {\frac{1}{\left(\alpha-\eta \right){\frac{\bar a}{\alpha}}-A}}\right)dA={\frac{\alpha}{\bar a}}V+c$

$$\begin{array}{l}\ln A - \ln \left( {(\alpha - \eta ){\textstyle{{\bar a} \over \alpha }} - A} \right) = (\alpha - \eta )V + c\\ \frac{A}{{(\alpha - \eta ){\textstyle{{\bar a} \over \alpha }} - A}} = k{e^{(\alpha - \eta )V}}\\ \end{array}$$

Initial condition, at V = 0, $A = {a_0}$ hence $k = \frac{{{a_0}}}{{(\alpha - \eta ){\textstyle{{\bar a} \over \alpha }} - {a_0}}}$

Putting the value of k we write, $(\alpha - \eta ){\textstyle{{\bar a} \over \alpha }}{a_0}{e^{(\alpha - \eta )V}} - {a_0}A{e^{(\alpha - \eta )V}} = A\left( {(\alpha - \eta ){\textstyle{{\bar a} \over \alpha }} - {a_0}} \right)$

$$A\left( V \right) = \frac{{\left( {\alpha - \eta } \right){\textstyle{{\bar a} \over \alpha }}}}{{1 + \frac{{\left( {\alpha - \eta } \right){\textstyle{{\bar a} \over \alpha }} - {a_0}}}{{{a_0}}}{\mathrm{exp}}\left\{ { - \left( {\alpha - \eta } \right)V} \right\}}}$$

Substituting η by ηα the differential equation becomes $\frac{{dA}}{{dV}} = \alpha A - \frac{\alpha }{{\bar a}}{A^2} - \eta \alpha A$ and the solution becomes $A\left( V_{t} \right) = \frac{{\left( {1 - \eta } \right)}{\bar \alpha} } {{1 + \frac{{\left( {1 - \eta } \right){\bar a} - {a_0} } } {a_0}}e^{-\alpha\left( l-\eta \right) V}}$

Appendix 2

Simulated plots of the learning progress function

3

Fig. 3

Simulated learning progress with different level of learning inefficiency