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Problem_Set_1_Solution

Problem_Set_1_Solution - Economic Growth and Development...

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Economic Growth and Development MS&E 249 Professor Olivier de La Grandville Fall 2008 Problem Set 1 Solutions 1. (20 points) For Leontief production, Y = F ( K, L ) = min( K a , L b ). Therefore by equation (6) in Solow paper, ˙ r = sF ( r, 1) - n 3 r = s min( r a , 1 b ) - n 3 r . As r 0 < a b , r 0 a < 1 b , so initially equation (6) becomes ˙ r = sr a - n 3 r = ( s a - n 3 ) r . Hence by the solution to this first-order differential equation, r ( t ) = r 0 e ( s a - n 3 ) t (1) As by assumption n 3 < s a , n 3 r < sr a . Thus r increases until it reaches the value a b , say at time t 1 . Note that t 1 satisfies r ( t 1 ) = r 0 e ( s a - n 3 ) t 1 = a b . From t = t 1 onward, min( r a , 1 b ) = 1 b . Thus, for t t 1 , equation (6) equals ˙ r = s b - n 3 r resulting in r ( t ) = ( a b - s bn 3 ) e - n 3 ( t - t 1 ) + s bn 3 (2) Since n 3 r < n 3 a b < s a a b = s b , ˙ r > 0. Thus, r increases further asymptotically towards s bn 3 as t goes to . One example of r ( t ) is shown in Figure 1. Figure 1: r ( t ) when n 3 < s a and r 0 < a b 1
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2. (20 points) Let’s start with equation (6) of Solow paper, ˙ r = sF ( r, 1) - nr = sr α - nr . Applying the change of variable x = r 1 - α , we get dx = (1 - α ) r - α dr and ˙ x = (1 - α ) r - α ( sr α - nr ) (3) = (1 - α )( s - nr 1 - α ) (4) = (1 - α ) s - (1 - α ) nx (5) By solving this first-order differential equation, we get x ( t ) = ( r 1 - α 0 - s n ) e - (1 - α ) nt + s n , which gives r ( t ) = ( r 1 - α 0 - s n ) e - (1 - α ) nt + s n 1 1 - α (6) Note that at equilibrium, ˙ r = 0, and therefore sr * α - nr * = 0, which yields r * =
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