Answer to Question #303348 in Finance for Lukas

Question

Assuming that Nations 1 and 2 are both large and starting from the equilibrium level of

national income and equilibrium in the trade balance in Nation 1 and given that 𝑀𝑃𝑆1 = 0.20; 𝑀𝑃𝑆2 = 0.15; 𝑀𝑃𝑀1 = 0.20 𝑎𝑛𝑑 𝑀𝑃𝑀2 = 0.10, find the change in the

equilibrium level of national income and the trade balance in Nation 1 for:

a) An autonomous increase in the exports of Nation 1 of 200 that replaces domestic

production in Nation 2

b) An autonomous increase in investment of 200 in Nation 1

c) An autonomous increase in investment of 200 in Nation 2.

Answer 1

A)"K^{"} = \\frac{1} { MPS1 + MPM1 + MPM2 (MPS1\/MPS2)}"

= $\frac{1}{ 0.20 +0.20+ 0.10 (0.20/0.15)}$

=$\frac{1}{0.533} = 1.88$

"\\Delta{YE} =(\\Delta{X})(K") = (200)\\times(1.88) = 376"

$\Delta{M} =(\Delta YE)(MPM_1)=(376)\times(0.20) = 75.2$

$\Delta S=(\Delta YE)(MPS_1)=376\times0.20$

$\Delta X = \Delta S + \Delta M = 75.2+ 75.2 = 150.4$ so that

$\Delta X-\Delta M$ = 75.2= Nation 1's trade surplus.

B)$K^{*} = \frac{1+ MPM2/MP52}{MPS1+ MPM1+ MPM2(MPS1/MPS)}$

= ($\frac{1+0.101}{0.15/0.533})$

=$\frac{1.667}{0.533}$ =3.13

$\Delta YE= (\Delta IX)= (200X3.13)$ =626

$\Delta M=(\Delta YE\times MPM_1)=(626\times0.20)$ =125.2

$\Delta S= (\Delta YE\times MPS1) =(626X0.20)$ =125.2

200+$\Delta X$ =125.2+125.2

and $\Delta X$ = 50.4

so that$\Delta X-\Delta M$ =50.4-125.2=74.8

c)$K^{*} = \frac{1+ MPM2/MP52}{MPS1+ MPM1+ MPM2(MPS1/MPS)}$

$(\frac{1+0.101}{0.20/0.533})$

$\frac{1.667}{0.106}=15.72$

$\Delta YE= (\Delta IX)= (200X15.72)=3144$

$\Delta M=(\Delta YEX\times MPM_1)=(626\times0.15)=93.9$

$\Delta S= (\Delta YE\times MPS1) =(626X0.15)=93.9$

200+$\Delta$ X=93.9+93.9=187.8

$\Delta X=$ 134.8

$\Delta X$ -$\Delta M$ =134.8-93.9=40.9