6. If initial K = 750, d = .1, and I = 150, what is the...

6. If initial K = 750, d = .1, and I = 150, what is the value of K in period t+1? _______

7. Using the data in problem 6, what is K in period t+5? ______

8. In the steady state, if Y = 500 and s=.1, C is? ________

9-10. (2 points). Draw the graph of the Solow model for a decrease in the investment rate, showing

the move to the new steady state.

Expert Solution

6) Capital stock is 750. Depreciation is 0.1 or 10% which is 75. Investment is 150. Hence capital in next period (t + 1) = 750 + 150 - 75 = 825

7) K t + 5 is 1057.10.

This is done in the table below. For example, K in t + 3 will be K in t + 2 = 892.50, added with investment +

150 and subtracted depreciation 0.1*(K t) = 0.1*825 = 82.5. This gives K in t + 3 = 953.3.

Time	Capital stock	Investment	Depreciation	Addition to capital
0	750	150.0	75.0	75.0
1	825.0	150.0	82.5	67.5
2	892.5	150.0	89.3	60.8
3	953.3	150.0	95.3	54.7
4	1007.9	150.0	100.8	49.2
5	1057.1	150.0	105.7	44

8) Y is 500 and saving rate is s = 0.1. This implies savings are 0.1 x 500 = 50. Hence C = Y - S = 500- 50 = 450

9) The graph is shown below. As the investment rate (saving rate) declines, the investment function shifts down. The new steady state is lower than the original steady state of capital per worker.

Rahul Sunny answered 1 year ago

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