Economics homework help.

Econ 100B: Economic Analysis – Macroeconomics Problem Set #3

Due Date: September 25, 2020

General Instructions:

• Please upload a PDF of your problem set to Gradescope by 11:00 pm. • Late homework will not be accepted. • Please put your name and student ID at the upper right corner of the front page.

1. Write out the steps need to show that the last equation on slide 21 of lecture 5 follows from the first equation on the same slide. Hint 1: A review of the first 2 pages of the memo1 has some helpful hints. Hint 2: the expression for κ can be written as K = κEL.

2. The variable κ is capital per worker per unit of worker efficiency, or capital per worker modified to reflect the difference in efficiency among workers. Consider two countries, A and B, with labor efficiencies EA = 1.5 and EB = 3, respectively.

(a) If the size of the labor force is the same in each country and κ∗ is the same in each country, which country will have a larger aggregate capital K in steady state?

(b) Given your answer to question 2a above, briefly explain why κ provide a better measure of the capital per worker needed in an economy.

3. By how much does the half-life shown on slide 10 of lecture 6 change if gE doubles?

4. Briefly explain the reason each of the bullet points below for the shock in the saving rate shown on slide 12 of lecture 6.

• At t = 0:

– s up, so κ∗ up.

– κ(t) unchanged (unch).

– Negative gap created.

– t1/2 unch.

– Balanced growth up.

– Y (t)/L(t) unch.

• For t > 0:

– κ(t) evolves to κ∗.

– Gap relaxes to zero.

– gκ = 0 for t < 0 and gκ > 0 for t > 0.

– gY/L > gE during relaxation.

– Long-term gY/L in unchanged from its value for t < 0.

1R.J. Hawkins (2020) The Solow-Swan and Romer Models.

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Due Date: September 25, 2020

General Instructions:

• Please upload a PDF of your problem set to Gradescope by 11:00 pm. • Late homework will not be accepted. • Please put your name and student ID at the upper right corner of the front page.

1. Write out the steps need to show that the last equation on slide 21 of lecture 5 follows from the first equation on the same slide. Hint 1: A review of the first 2 pages of the memo1 has some helpful hints. Hint 2: the expression for κ can be written as K = κEL.

2. The variable κ is capital per worker per unit of worker efficiency, or capital per worker modified to reflect the difference in efficiency among workers. Consider two countries, A and B, with labor efficiencies EA = 1.5 and EB = 3, respectively.

(a) If the size of the labor force is the same in each country and κ∗ is the same in each country, which country will have a larger aggregate capital K in steady state?

(b) Given your answer to question 2a above, briefly explain why κ provide a better measure of the capital per worker needed in an economy.

3. By how much does the half-life shown on slide 10 of lecture 6 change if gE doubles?

4. Briefly explain the reason each of the bullet points below for the shock in the saving rate shown on slide 12 of lecture 6.

• At t = 0:

– s up, so κ∗ up.

– κ(t) unchanged (unch).

– Negative gap created.

– t1/2 unch.

– Balanced growth up.

– Y (t)/L(t) unch.

• For t > 0:

– κ(t) evolves to κ∗.

– Gap relaxes to zero.

– gκ = 0 for t < 0 and gκ > 0 for t > 0.

– gY/L > gE during relaxation.

– Long-term gY/L in unchanged from its value for t < 0.

1R.J. Hawkins (2020) The Solow-Swan and Romer Models.

1

5. We will now build a Solow-Swan calculator using the Euler algorithm discussed in lecture. An example of my Excel version of this calculator is shown below. You can use your preferred platform (e.g., Python, Matlab, Octave, Mathematica, etc.), but the steps presented below will given in terms of Excel.

(a) In column A, create a time range running from -10 to 50 in steps of 1.

(b) In column B, create the saving rate, s, as a function of time:

i. To include a shock at t = 0, put the pre-shock value for s of 0.1 in cell B24.

ii. Set cell B26 equal to cell B24.

iii. For cells B27 to B86 set the value in each row to be equal to the value in the previous row (e.g., the equation in cell B27 is = B26).

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iv. We introduce a shock at t = 0 by (i) adding a shock multiplier (the value of 2 in this case) in cell B23 and then (ii) multiplying the value in cell B36 by it: i.e., the equation in cell B36 is = B35 ∗ B23.

At this point your values for the saving rate should look like those in the figure above. To verify that this is working, if you (i) change the value in cell B23 to 1 then all of the saving rates should be 0.1 and (ii) if you change the value in cell B3 to 3 then all saving rates before t = 0 should be 0.1 and the rest should be 0.3.

(c) Copy cells B23 to B86 into columns C, D, E, and F, and change the pre-shock values in row 24 and the column headings in row 25 as shown in the figure above.

(d) Populate cells G26 to G86 with values of κ∗

κ∗ =

( s

gE + gL + δ

) 1 1−α

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using the value of the variables in each row. Note how the shock to s changes κ∗

at t = 0

(e) To initialize and evolve κ(t):

i. Set the initial condition κ(t = −10) = κ∗ in cell H26 and in cell I26 put the result of

dκ

dt = sκα − (gE + gL + δ)κ

with the time derivative calculated using values at t = −10. ii. Evolve κ(t) with the Euler algorithm

κ(t+ ∆t) = κ(t) + dκ

dt × dt

with the equation in cell H27 of

= H26 + I26 ∗ 1.

for our time step of 1 year.

iii. Calculate dκ/dt in cell I27 in the same way that you calculated the value in cell I26.

iv. Complete the evolution for all times by copying the equations in cells H27 and I27 into all cells below to cells H86 and I86.

(f) Calculate the gap in column J using the values for κ∗, κ(t), and α in columns G, H, and F. Note how the economic is initially in steady state (gap = 0), has a negative gap in response to the shock at t = 0, and how the gap relaxes after the shock as a result of κ(t) moving toward κ∗ for t > 0.

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(g) Calculate E(t) in column K in a manner similar to the way that you set up the saving rate. First enter the shock, column heading and initial value in cells K23 to K26. Then model the evolution of E(t) in cells K27 though K86 using:

E(−9) = E(−10)egEt

with, for example, the value in cell K27 given by

= K26 ∗ EXP(C26 ∗ 1)

and similarly for the remaining cells in column K.

(h) Add the ability to incorporate the shock to E(t) at t = 0 with the equation =K35*EXP(C35*1)*K23 in cell K36.

(i) The shock to E(t) at t = 0 will also impact κ(t). In addition, a shock to K(t)/L(t) at t = 0 will also impact κ(t). We will incorporate the shocks to E(0) and K(0)/L(0) through the use of factors that we will refer to as fE and fK/L, respec- tively. At t = 0 we have that

κ(t) = K(t)

E(t)L(t) =

( 1

E(t)

)( K(t)

L(t)

) into which we can introduce the shock factors as follows

κ(t) −→ (

1

fE × E(t)

)( fK/L ×

K(t)

L(t)

) =

K(t)

E(t)L(t) × ( fK/L fE

) or

κ(t) −→ κ(t)× ( fK/L fE

) .

with the equation =(H35+I35*1)*(H23/K23) in cell H36.

(j) Add the Balanced and Actual growth paths for Y (t)/L(t) in columns L and M using

Y (t)/L(t) =

κ ∗αE(t) for Balanced.

κ(t)αE(t) for Actual.

(k) Create graphs as shown.

Validate your calculator by replicating the shock shown in the figure above. Your solution for this question is a screenshot of your simulation with a gL shock equal to 4 (i.e., the post-shock values of gL should be 0.08). This should be the only shock; all other shock multipliers, including that of the saving rate, should be equal to 1.

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6. Use your calculator to model the the German economy shown on slide 16 of lecture 6:

(a) Obtain the data for the German economy from the Maddison Project Database. There are two series for GDP and while a result for either is acceptable, let’s use cgdppc.

(b) Add another column of dates with 1946 cell N36 and prior and subsequent dates listed in cells N26 through N86. We will approximate the shocks in 1945 and 1946 by a single shock in 1946.

(c) Add the Maddison data for Y (t)/L(t) in Germany in column O.

(d) Add the Maddison data to your graph.

(e) Use the base case values of s = 0.118, gE = 0.023, gL = 0.03, δ = 0.0638 and α = 0.29. Vary the initial value of E until you get a good fit by eye.

Your solution for this question is (i) a screenshot of your graph and (ii) your estimate of the shock to K(t)/L(t).

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