Answer to Question #212367 in Macroeconomics for Kay

Question

Q.1 Consider the following information about a hypothetical economy:

1. Y = A (0.025K − 0.5N) N

2. A=2/3

3. K=2000

4. N^s=-18+(18/5)w

5. C=200+(2/3)(Y-T)-300r

6. T=-75+(1/4)Y

7. I =100−100r

8. G=100

9. L = 0.5Y − 200i

10. M=6300

11. π^e=0.10

Now using this information, answer the following: 

(c) Derive the equations of the IS, LM and AD curves. Determine the numerical values of the slopes and 

intercepts of these curves and interpret each? 

(d) Determine the equilibrium levels of all endogenous variables under the assumptions of the classical 

macroeconomic framework. 

(e) Beginning from the initial classical equilibrium, suppose that the central bank increases the money supply 

by 420 while price remains fixed at its initial long run equilibrium level. What will be the impact of this 

policy on all endogenous variables in short run and long run? 

(f) Compare the equilibrium positions in (d) and (e) in one graph indicating all points.

Answer 1

c)

IS equation,

$Y=C+I+G\\Y=200+\frac{2}{3}(y-T)-300r+100-100r+100\\Y=400+\frac{2}{3}(Y+75-\frac{1}{4}Y)-400r\\Y=400+\frac{2}{3}(\frac{3}{4}Y+75)-400r\\Y=400+\frac{1}{2}Y+50-400r\\\frac{1}{2}Y+400r=450\\\frac{1}{2}dY+400dr=0\\400dr=-\frac{1}{2}dY\\\frac{dr}{dY}=-\frac{1}{800}$

LM equation,

$\frac{M}{P}=L \space \space \space [r=i-\pi e]$

$\implies \frac{6300}{P}=0.5Y-200(r+\pi e)\\\implies\frac{6300}{P}=0.5Y-200r-20\\0.5Y-200r=\frac{6300}{P}+20\\0.5dY-200dr=0\\-200dr=-0.5dY\\\frac{dr}{dY}=\frac{0.5}{200}=\frac{1}{400}\\\frac{1}{2}Y+400r=450\\\frac{1}{2}Y-200r=\frac{6300}{P}+20$

$Y=\frac{\begin{vmatrix} 450 & 400 \\ \frac{6300}{P} & -200 \end{vmatrix}}{\begin{vmatrix} \frac{1}{2} & 400 \\ \frac{1}{2} & -200 \end{vmatrix}}$

$Y=\frac{-90000-\frac{2520000}{P}-8000}{-100-200}$

$Y=\frac{98000+\frac{2520000}{P}}{300}$

$Y=326.67+\frac{8400}{P}\\\frac{dY}{dP}=-{\frac{8400}{P^2}}$

d)

$Y=A(0.025K-0.5N)N\\Y=\frac{2}{3}(50-0.5N)N\\\frac{\delta Y}{\delta N}=\frac{2}{3}\times50-\frac{2}{3}\times\frac{1}{2}\times 2N\\\implies\frac{2}{3}(50-N)=w\\\implies Nd=50-\frac{3}{2}w$

$N^s=N^d\\\implies-18+\frac{18}{w}=50-\frac{3}{2}w\\\implies(\frac{18}{5}+\frac{3}{2})w=68\\\implies(4.6+1.5)w=68\\\implies w=\frac{68}{5.1}=13.33\\\therefore N=29.99 \approx30$

$r=\frac{\begin{vmatrix} \frac{1}{2} & 450\\ \frac{1}{2}& \frac{6300}{P}+20 \end{vmatrix}}{-300}$

$r=\frac{\frac{3150}{P}+10-225}{-300}$

$r=\frac{\frac{3150}{P}-215}{-300}$

$r=-\frac{10.5}{P}+0.716$

$Y=\frac{2}{3}(50-0.5\times30)30$

$Y=\frac{2}{3}(50-15)\times30\\=20\times35=700\\700=326.67+\frac{800}{P}\\\implies P=22.5\\r=0.25$

e)

$dN=420$

$22.5=0.5Y-200(r+0.1)\\\implies0.5Y-200r-20=298.67\\\implies \frac{1}{2}Y+400r=450(15)\\$

$Y=\frac{\begin{vmatrix} 450 & 400 \\ 318.67 & -200 \end{vmatrix}}{-300}$

$Y=724.89$

$r=\frac{\begin{vmatrix} \frac{1}{2} & 50 \\ \frac{1}{2} & 318.67 \end{vmatrix}}{-300}$

$r=\frac{144.94-225}{-300}$

f.

LM curve shifts rightward due to increase in money supply.