Homework: Week 4
Group 4: Maxine Catchlove, Sam Ridsdale, Stephanie Schott, Tim Hawley
August 25th 2015
Definitions and Assumptions:
- The aggregate production function is: \(Y_{t} = K_{t}^{\alpha}(A_{t}L_{t})^{1-\alpha}\)
(where \(Y\) is output, \(K\) is the capital stock, \(L\) is the labour force, and \(A\) is a Hicks-neutral technology.)
- The labour force grows according to \(L_{t+1}=L_{t}(1+n)\), and technology grows at \(A_{t+1}=A_{t}(1+g)\).
- The capital stock evolves according to \(K_{t+1}=(1 - \delta)K_{t}+I_{t}\), where \(\delta\) is the depreciation rate and \(I=sY\) is investment, with \(s\) being an exogenously determined savings rate.
- Firms in the economy maximise profits when: \(\Pi = K_{t}^{\alpha}(A_{t}L_{t})^{1-\alpha} - wL-rK\)
(with \(w\) being the prevailing wage rate and \(r\) being the cost of capital.)
Part 1: Working with the aggregate production function
Question 1:
- When firms are using the amount of capital that maximises profits, how much capital is used? What is the return on capital?
The profit maximising level of capital is obtained by finding \(\frac{\partial\Pi}{\partial K}\) and setting it to \(0\):
\(\frac{\partial \Pi}{\partial K_{t}} = \alpha K_{t}^{\alpha - 1}(A_{t}L_{t})^{1 - \alpha} - r = 0\)
\(\alpha K_{t}^{-1}K^{\alpha}(A_{t}L_{t})^{1 - \alpha} = r\)
\(\alpha \frac{Y_{t}}{K_{t}} = r\)
Above is the marginal product of capital. Rearranging this gives the profit maximising level:
\(K_{t}^{*} = \alpha\frac{Y_{t}}{r}\)
It can also be arranged to obtain the return on capital:
\(\frac{Y_{t}}{K_{t}} = \frac{r}{\alpha}\)
Question 2:
- Repeat the above, but for labour. If all income Y is either labour income wL or capital income rK, what is the relationship between the parameters of the production function and the capital/labour shares?
The profit maximising level of labour is obtained by finding \(\frac{\partial\Pi}{\partial L}\) and setting it to \(0\):
\(\frac{\partial \Pi}{\partial L_{t}} = (1 - \alpha) K_{t}^{\alpha}(A_{t}L_{t})^{1 - \alpha - 1} - w = 0\)
\((1 - \alpha) K_{t}^{\alpha}(A_{t}L_{t})^{1 - \alpha}(L_{t})^{-1} = w\)
\((1 - \alpha) \frac {Y_{t}}{L_{t}} = w\)
Above is the marginal product of labour. Rearranging this gives the profit maximising level:
\(L_{t}^{*} = (1 - \alpha)\frac{Y_{t}}{w}\)
It can also be arranged to obtain the return on labour:
\(\frac{Y_{t}}{L_{t}} = \frac{w}{(1 - \alpha)}\)
If all income is either labour income and capital income then the capital share tells us that it will either all be dedicated to labour or all dedicated to capital.
Question 3:
Another common production function is the “CES” production function, which takes the form \(Y = (\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}}\)
Derive the marginal products of capital and labour for this form.
Once again we can obtain MPK by finding \(\frac{\partial \Pi}{\partial K}\) and setting it to \(0\):
\(\frac{\partial Y}{\partial K} = \frac{\partial Y}{\partial u} . \frac{\partial u}{\partial K}\)
\(\frac{\partial Y}{\partial u} = \frac{1}{\rho}u^{\frac{1}{\rho}-1}\)
\(\frac{\partial u}{\partial K} = \alpha\rho K^{\rho - 1}\)
\(\frac{\partial Y}{\partial K} = \frac{1}{\rho}(\alpha\rho K^{\rho - 1})(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1}\)
\(\frac{\partial \Pi}{\partial K} = \frac{1}{\rho}(\alpha\rho K^{\rho - 1})(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1} - r = 0\)
\((\alpha\rho K^{\rho - 1})(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1} = r\rho\)
\((\alpha\rho K^{\rho - 1})Y^{1 - \rho} = r\rho\)
\(\alpha(\frac{Y}{K})^{1 - \rho} = r\)
And with the same process we obtain MPL:
\(\frac{\partial Y}{\partial L} = \frac{\partial Y}{\partial u} . \frac{\partial u}{\partial L}\)
\(\frac{\partial Y}{\partial u} = \frac{1}{\rho}u^{\frac{1}{\rho}-1}\)
\(\frac{\partial u}{\partial L} = (1-\alpha)\rho L^{\rho - 1}\)
\(\frac{\partial Y}{\partial L} = \frac{1}{\rho}(1-\alpha)\rho L^{\rho - 1}(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1}\)
\(\frac{\partial \Pi}{\partial L} = \frac{1}{\rho}(1-\alpha)\rho L^{\rho - 1}(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1} - w = 0\)
\((1-\alpha)\rho L^{\rho - 1}(\alpha K^{\rho} + (1 - \alpha) L^{\rho})^{\frac{1}{\rho}-1} = w\rho\)
\((1-\alpha)\rho L^{\rho - 1}Y^{1 - \rho} = w\rho\)
\((1-\alpha)(\frac{Y}{L})^{1 - \rho} = w\)
Part 2: Growth
Question 4
- Express the Cobb-Douglas version of the model in per-effective worker terms (both production function and capital accumulation equation). That is, divide both parts of the model by \(AL\). Call the resulting capital and output per effective units of labour \(k\) and \(l\).
\(\frac{Y_{t}}{A_{t}L_{t}} = \frac{K_{t}^{\alpha}(A_{t}L_{t})^{1-\alpha}}{A_{t}L_{t}}\)
\(= K_{t}^{\alpha}(A_{t}L_{t})^{-\alpha}\)
\(= (\frac{K_{t}}{A_{t}L_{t}})^{\alpha}\)
\(= k_{t}^{\alpha}\)
\(\frac{K_{t+1}}{A_{t+1}L_{t+1}} = \frac{(1+\delta)K_{t}+sY_{t}}{A_{t+1}L_{t+1}}\)
\(\frac{K_{t+1}}{A_{t+1}L_{t+1}} = \frac{(1+\delta)K_{t}+sY_{t}}{(1+n+g)A_{t}L_{t}}\)
\(k_{t+1} = \frac{(1-\delta)}{(1+n+g)}k_{t} + \frac{1}{(1+n+g)}sY_{t}\)
Question 5
- Under balanced growth, the capital-output ratio is constant. Express the balanced growth path of \(y^{*}\) and \(k^{*}\) in terms of the exogenous variables. How quickly are capital and output growing during balanced growth?
Under balanced growth, \(k_{t+1} = k_{t} = k^{*}\)
\(k = \frac{(1-\delta)k+sk^{\alpha}}{(1+n+g)}\)
\((\delta+n+g)k = sk^{\alpha}\)
\(k^{\alpha-1}(\delta+n+g) = s\)
\(k^{*} = (\frac{s}{\delta+n+g})^{\frac{1}{1-\alpha}}\)
\(y = k^{\alpha}\)
\(y^{*} = (\frac{s}{\delta+n+g})^{\frac{\alpha}{1-\alpha}}\)
During balanced growth output and capital grow at the same constant rate, given by the exogenous variables.
Question 6
- We don’t always assume that economies are at the equilibrium point; more, it is an attracting point. During WWII, much of Germany’s capital stock was destroyed, though technology was not. Illustrate what happened to Germany during the post-war years on a Solow-Swan diagram. Tell a story!
In the Solow-Swan model, a reduction in capital stock results in a movement along the ‘investment’ curve. This implies an immediate impact of lower output per worker, lower investment and lower capital per worker. (In the diagram this is the movement from the steady state levels, \(k^{*}\) to the new levels of \(k'\).)
However, since at this point investment more than offsets the depreciation of capital stock, there is an incentive to make higher levels of investment. Therefore, over time, the economy finds itself back at the steady state level, assuming no change in technology or the savings rate.
Of course, in the case of Germany’s economic recovery, this was not the only mechanism at play. Intervention by foreign governments also prevented the ability of markets to correct and there were other political factors such as the political incentive for the domestic government to increase the standard of living from the poor state which it had been left in, partly by borrowing massive sums of money from foreign countries such as the United States.
They also had the benefit of fairly significant improvements in technology during this time which meant that they would eventually overtake the previous steady state of the economy and once again become one of the strongest economies in the world, which is something to celebrate:
Question 7
Finally, we’re not interested in just finding the balanced growth path; we want to find the one that maximises some measure of wellbeing. One measure of wellbeing is the dollar value of material goods that we purchase.
Plot the savings rate (x axis) against the amount of consumption in the economy (y axis). What is the shape? If a country wanted to maximise its consumption, what rate would you recommend?
The above diagram shows that the maximum level of consumption will correspond with a certain savings rate, known the the ‘Golden Rule’ savings rate. This is where the slope of the consumption function is zero.
We also know that the maximum level of consumption has a corresponding level on the Solow-Swan diagram, since it is where the slope of the ‘output’ function is equal to the slope of the ‘depreciation’ function.
Therefore the value of the ‘Golden Rule’ savings rate can be found mathematically in the following way:
\(C = y - sy\)
\(C = k^{\alpha} - (d+n+g)k\)
\(\frac{\partial C}{\partial k} = \alpha k^{\alpha - 1 } - (\delta+n+g) = 0\)
\(\alpha k^{\alpha} = (\delta+n+g)k\)
\(\alpha k^{\alpha} = sk^{\alpha}\)
\(s = \alpha\)