Econ 154a Problem Set 5

# Econ 154a Problem Set 5 - I. Problem 1: 1. Part 1 a) Graph...

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I. Problem 1: 1. Part 1 a) Graph of actual investment per worker sAf(k) and (n+d)k: Investment Per Worker 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0 0.5 1 1.5 2 2.5 k Investment sAf(k) (n+d)k b) Let k 0 be the present level of capital per worker. If P is the total present worker population, then the total present capital is k 0 P. If the growth rate of the worker population is n, then the worker population in the next period will be (1+n)P. Additionally, the present capital will depreciate and be worth (1–d)k 0 P in the future. Thus, the value of future capital will be (1–d)k 0 P+I and the value of future capital per worker will be k f = ([1– d]k 0 P+I)/([1+n]P). Performing some algebra: k f (1+n)P = (1– d)k 0 P + I; I = k f (1+n)P+(d–1)k 0 P. In the steady state, k f = k 0 , so I = k 0 (1+n)P+(d–1)k 0 P = k 0 P(1+n+d–1) = k 0 P(n+d). Dividing by P gives I/P = (n+d)k. Thus, it can be seen that (n+d)k is equal to I/P (investment per worker) if k is constant from one period to the next. 2. Part 2 a) As was demonstrated in Part 1, k is constant if investment per worker equals (n+d)k. It follows that if investment is greater than (n+d)k, k will be increasing; if investment is less than (n+d)k, k will be decreasing; and if investment equals (n+d)k, k will be constant (a steady state). For the first half of the curve, let sAk α = (n+d)k; (0.1)(1)k 0.3 = (0.05+0.01)k; 0.15k – 0.1k 0.3 = 0 = k(0.15 – 0.1k -0.7 ). So, k = 0 and 0.15 – 0.1k -0.7 = 0 b k -0.7 = 1.5 b 0.5603. For the second half of the curve, let s λ Ak α = (n+d)k; (0.1)(2)(1)k 0.3 = (0.05+0.1)k; 0.15k – 0.2k 0.3 = 0 = k(0.15 – 0.2k -0.7 ). So, k = 0 and 0.15 – 0.2k -0.7 = 0 b 0.75 = k -0.7 b k = 1.5083. Thus, the steady states occur at k = 0, 0.5603 and 1.5083; k is increasing on the range (0,0.5603)U[1,1.5083); and k is decreasing on the range (0.5603,1)U(1.5083,2]. 3. Part 3

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Evolution of capital per worker through 25 time intervals with k 0 A = 1.1 (provided graphically to save space): Capital Per Worker Over Time 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 5 10 15 20 25 30 Time Capital Per Worker b) From the graph in part 1 and the conclusions in part 2, it is apparent that the capital per worker will increase from the initial value of 1.1 until it reaches the steady state level at k = 1.5083. This is supported by the graph of capital per worker over time, which seems asymptotic to 1.5083. 4. Part 4 a) Evolution of capital per worker through 25 time intervals with k 0 = 0.9 (provided graphically to save space): Capital Per Worker Over Time 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 30 Time From the graph in part 1 and the conclusions in part 2, it is apparent that the capital per worker will decrease from the initial value of 0.9 until it reaches the steady state level at k = 0.5603. which seems asymptotic to 0.5603.
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## This note was uploaded on 07/19/2008 for the course ECON 154 taught by Professor Bjoernbruegemann during the Fall '07 term at Yale.

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Econ 154a Problem Set 5 - I. Problem 1: 1. Part 1 a) Graph...

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