Question #303348

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.



1
Expert's answer
2022-02-28T11:34:10-0500

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

= 10.20+0.20+0.10(0.20/0.15)\frac{1}{ 0.20 +0.20+ 0.10 (0.20/0.15)}

=10.533=1.88\frac{1}{0.533} = 1.88

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

ΔM=(ΔYE)(MPM1)=(376)×(0.20)=75.2\Delta{M} =(\Delta YE)(MPM_1)=(376)\times(0.20) = 75.2

ΔS=(ΔYE)(MPS1)=376×0.20\Delta S=(\Delta YE)(MPS_1)=376\times0.20

ΔX=ΔS+ΔM=75.2+75.2=150.4\Delta X = \Delta S + \Delta M = 75.2+ 75.2 = 150.4 so that

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

B)K=1+MPM2/MP52MPS1+MPM1+MPM2(MPS1/MPS)K^{*} = \frac{1+ MPM2/MP52}{MPS1+ MPM1+ MPM2(MPS1/MPS)}

= (1+0.1010.15/0.533)\frac{1+0.101}{0.15/0.533})

=1.6670.533\frac{1.667}{0.533} =3.13

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

ΔM=(ΔYE×MPM1)=(626×0.20)\Delta M=(\Delta YE\times MPM_1)=(626\times0.20) =125.2

ΔS=(ΔYE×MPS1)=(626X0.20)\Delta S= (\Delta YE\times MPS1) =(626X0.20) =125.2

200+ΔX\Delta X =125.2+125.2

and ΔX\Delta X = 50.4

so thatΔXΔM\Delta X-\Delta M =50.4-125.2=74.8

c)K=1+MPM2/MP52MPS1+MPM1+MPM2(MPS1/MPS)K^{*} = \frac{1+ MPM2/MP52}{MPS1+ MPM1+ MPM2(MPS1/MPS)}

(1+0.1010.20/0.533)(\frac{1+0.101}{0.20/0.533})

1.6670.106=15.72\frac{1.667}{0.106}=15.72

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

ΔM=(ΔYEX×MPM1)=(626×0.15)=93.9\Delta M=(\Delta YEX\times MPM_1)=(626\times0.15)=93.9

ΔS=(ΔYE×MPS1)=(626X0.15)=93.9\Delta S= (\Delta YE\times MPS1) =(626X0.15)=93.9

200+Δ\Delta X=93.9+93.9=187.8

ΔX=\Delta X= 134.8

ΔX\Delta X -ΔM\Delta M =134.8-93.9=40.9


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