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EXAMPLE 1: Sudden shock to L t . INITIAL VALUES are A = 1, y=k 0.5 , s = 0.20, δ = 0.05, n = 0.05, so ( δ +n) = 0.10. Let L t = 1 and K t = 4. Solution: 1. k steady-state = [(s*A)/( δ +n)] 1/(1- α ) = [0.20/0.10] 2 = 4 . 2. y = Ak α = 4 0.5 = 2 . 3. Actual investment per worker (which equals required investment per worker in a steady state) = sAk α = 0.2*1*4 0.5 = ( δ +n)k = (0.10)(4) = 0.4 4. c = y – i = 2 – 0.4 = 1.6 . 5. MPL = (1- α )Ak α = [0.50*1](4 0.5 ) = 1 6. MPK = ( α A)/(k 1- α ) = [0.50*1]/(4) 0.5 = 0.50/2= 0.25 . [[Recall that under constant returns-to-scale Y = MPL*L + MPK*K Æ put into per-worker terms to get y = MPL + MPK*k = 1 + 0.25*4 = 2.] Now let L t suddenly fall to 0.25. A = 1, y=k 0.5 , s = 0.20, δ = 0.05, n = 0.05, so ( δ +n) = 0.10. L t = 0.25 and K t = 4. Solution in SHORT RUN: 1. k actual = 4/0.25 = 16 (increase) 2. y actual = y = Ak α = 1* 16 0.5 = 4 (increase) 3. Actual investment per worker = sAk α = 0.20*1*16 0.5 = 0.80 (increase) 4. c actual = y actual – i actual = 4 - 0.80 = 3.2 . (increase) 5. MPL = (1- α )Ak α = [0.50*1](16 0.5 ) = 2 (increase) 6. MPK = ( α A)/(k 1- α ) = [0.50*1]/(16) 0.5 = 0.50/4= 0.125 . (decrease) Solution in LONG RUN (steady state solution). Since none of values in this formula (k steady-state = [(s*A)/( δ +n)] 1/(1- α ) ) have changed, the economy returns in the long run to the solution listed above under “Initial Values”. EXAMPLE 2: Sudden shock to K t . INITIAL VALUES are A = 1, y=k 0.5 , s = 0.20, δ = 0.05, n = 0.05, so ( δ +n) = 0.10. L t = 1 and K t = 4. Solution: 1. k steady-state = [(s*A)/( δ +n)] 1/(1- α ) = [0.20/0.10] 2 = 4 . 2. y = Ak α = 4 0.5 = 2 . 3. Actual investment per worker (which equals required investment per worker in a steady state) = sAk α = 0.2*1*4 0.5 = ( δ +n)k = (0.10)(4) = 0.4 4. c = y – i = 2 – 0.4 = 1.6 . 5. MPL = (1- α )Ak α = [0.50*1](4 0.5 ) = 1 6. MPK = ( α A)/(k 1- α ) = [0.50*1]/(4) 0.5 = 0.50/2= 0.25 . [Recall that Y = MPL*L + MPK*K Æ put into per-worker terms to get y = MPL + MPK*k = 1 + 0.25*4 = 2.] Now let K t suddenly fall to 1. A = 1, y=k 0.5 , s = 0.20, δ = 0.05, n = 0.05, so ( δ +n) = 0.10. L t = 1 and K t = 1. Solution in SHORT RUN: 1. k actual = 1/1 = 1 (decrease) 2. y actual = y = Ak α = 1* 1 0.5 = 1 (decrease) 3. Actual investment per worker = sAk α = 0.20*1*1 0.5 = 0.20 (decrease) 4. c actual = y actual – i actual = 1 - 0.20 = 0.8 .(decrease) 5. MPL = (1- α )Ak α = [0.50*1](1 0.5 ) = 0.50 (decrease) 6. MPK = ( α A)/(k 1- α ) = [0.50*1]/(1) 0.5 = 0.50 (increase) Solution in LONG RUN (steady state solution).

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