# PS4Ans[1] - Economic Growth and Development EC 375 Problem...

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Economic Growth and Development Prof. Murphy EC 375 Problem Set 4 Answers Chapter 9 # 1, 4 (on pages 271-272). 1. The annual growth rate of productivity is given by the following equation: ˆ A = ˆ y + ! ˆ L . We are given a value of 1/3 for β and 0 for ˆ y , leaving the growth rate of the population, ˆ L , as the only unknown. To solve for ˆ L , we use the standard growth equation with the initial population as 4 million and the final population after 10,000 years as 170 million. The equation is: 4(1 + ˆ L ) 10,000 = 170 ˆ L = (170 / 4) (1/10,000) ! 1 = 0.000375. Now we substitute to find our growth rate of productivity over this period: ˆ A = 0 + (1/ 3)(0.000375) = 0.000125. That is, the growth rate of productivity over this period was roughly 0.0125 percent per year. 4. a. In any given year, the production of bread must equal the production of cheese in this economy. That is, Y b = Y c , always. Knowing that the productivity of each good is equal at this point in time, we can solve for the quantity of labor devoted to each sector as follows. Y b = Y c , A b L b = A c L c , L b = L c . Since L b + L c = L , L b = L c = L / 2. The labor force will be equally split between the two sectors. b. To calculate the growth rate of total output, we first calculate the growth rates of each sector by taking the natural log of both sides and differentiating with respect to time. For the bread sector:

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2 ln( Y b ) = ln( A b ) + ln( L b ), d dt ln( Y b ) = d dt ln( A b ) + ln( L b ), ˆ Y b = ˆ A b + ˆ L b . Similarly, for the cheese sector, we get, ˆ Y c = ˆ A c + ˆ L c . We know the value for the growth rate of productivity in both sectors. Furthermore, in our answer to Part (a), we found that labor is currently equally divided among the two sectors. Thus, the growth of labor in one sector must be offset by the growth of labor in the other ˆ L b = ( ! ˆ L c ). We substitute in these values and get, ˆ Y b = 2% + ˆ L b . And, ˆ Y c = 1% + ˆ L c = 1% !
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• Spring '10
• Jannett

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