econ.studio
Solow–Swan Growth Model
Solutions

Solow–Swan Growth Model

Solutions

Exercise 1
(a) k∗=(0.25/0.08)3/2=(3.125)1.5≈5.525k^* = (0.25/0.08)^{3/2} = (3.125)^{1.5} \approx 5.525. y∗=(k∗)1/3≈1.768y^* = (k^*)^{1/3} \approx 1.768. c∗=0.75×1.768≈1.326c^* = 0.75 \times 1.768 \approx 1.326. (b) k∗/y∗=s/(n+g+δ)=0.25/0.08=3.125k^*/y^* = s/(n+g+\delta) = 0.25/0.08 = 3.125. (c) In steady state Y/(AL)Y/(AL) is constant, so Y/L=A⋅y∗Y/L = A \cdot y^* grows at rate g=0.02g = 0.02, i.e. 2% per year.
Exercise 2
(a) Golden Rule: 13(kGR∗)−2/3=0.08\frac{1}{3}(k^*_{GR})^{-2/3} = 0.08 gives (kGR∗)2/3=4.167(k^*_{GR})^{2/3} = 4.167, so kGR∗≈8.508k^*_{GR} \approx 8.508. (b) sGR=0.08×(kGR∗)2/3=0.08×4.167≈0.333=αs_{GR} = 0.08 \times (k^*_{GR})^{2/3} = 0.08 \times 4.167 \approx 0.333 = \alpha. (c) s=0.25<sGR=1/3s = 0.25 < s_{GR} = 1/3: below the Golden Rule (dynamically efficient). Raising ss would eventually raise c∗c^*, but the current generation bears a cost.
Exercise 3
(a) k∗∗=(0.36/0.08)3/2=(4.5)1.5≈9.545k^{**} = (0.36/0.08)^{3/2} = (4.5)^{1.5} \approx 9.545. y∗∗=(9.545)1/3≈2.121y^{**} = (9.545)^{1/3} \approx 2.121. (b) (2.121−1.768)/1.768≈20%(2.121 - 1.768)/1.768 \approx 20\%. (c) It remains g=2%g = 2\%. The savings rate increase raises the level of output per worker but not the long-run growth rate.
Exercise 4
(a) TFP=4.0−(0.35)(5.0)−(0.65)(2.0)=4.0−1.75−1.30=0.95%\text{TFP} = 4.0 - (0.35)(5.0) - (0.65)(2.0) = 4.0 - 1.75 - 1.30 = 0.95\%. (b) 0.95/4.0=23.75%0.95/4.0 = 23.75\%. (c) Capital deepening contribution: α×(K^−L^)=0.35×3%=1.05%\alpha \times (\hat{K} - \hat{L}) = 0.35 \times 3\% = 1.05\%. Fraction: 1.05/4.0=26.25%1.05/4.0 = 26.25\%.
Exercise 5
(a) ∣λ∣=(1−α)(n+g+δ)=(2/3)(0.08)≈5.33%|\lambda| = (1-\alpha)(n+g+\delta) = (2/3)(0.08) \approx 5.33\% per year. (b) From 0.10=e−∣λ∣t0.10 = e^{-|\lambda|t}: t=ln⁡(10)/0.0533≈43t = \ln(10)/0.0533 \approx 43 years.