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Sebenta de Exercícios Resolvidos - O DGE - UBI

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44CAPÍTULO 5. MODELO IS-LM EM ECONOMIA FECHADA E PROCURA AGREGADA<br />

Temos então que <strong>de</strong>duzir o novo mo<strong>de</strong>lo IS-LM, sendo que a<br />

única expressão que se altera é a IS:<br />

Y = C + I + G + X − M ⇔<br />

⇔ Y = C + cY dp + I − bi + G ⇔<br />

⇔ Y = C + c ( a(Y − T ) + fY f) + I − bi + G ⇔<br />

⇔ Y = C + c ( a(Y − T ) + fgY ) + I − bi + G ⇔<br />

⇔ Y (1 − c(a + fg)) = C − caT + I − bi + G ⇔<br />

1 [ ]<br />

⇔ Y =<br />

C − caT + I − bi + G ⇔<br />

(1 − c(a + fg))<br />

1 [ ] b<br />

⇔ Y =<br />

C − caT + I + G −<br />

(1 − c(a + fg))<br />

(1 − c(a + fg)) i<br />

E recalcular<br />

⎧<br />

o equilíbrio:<br />

⎪⎨ i =<br />

⎪⎩ Y =<br />

⎧<br />

⎪⎨<br />

⇔<br />

⎪⎩ Y =<br />

1<br />

(1−c(a+fg))[C−caT +I+G]− 1 k M/P<br />

( h k + b<br />

(1−c(a+fg)))<br />

h/k<br />

(1−c(a+fg))[C−caT +I+G]− h/k<br />

k M/P<br />

( h k + b<br />

(1−c(a+fg)))<br />

i =<br />

+ 1 k M/P ⇔<br />

1<br />

(1−0.8(0.6+0.4×1.08)) [100−0.8×0.6×10+30+15]− 1 20 ×2500<br />

( 0.5<br />

20 + 350<br />

(1−0.8(0.6+0.4×1.08)))<br />

0.5/20<br />

(1−0.8(0.6+0.4×1.08))<br />

g) Isso quer dizer que<br />

0.5/20<br />

[100−0.8×0.6×10+30+15]− ×2500<br />

20<br />

(1−0.8(0.6+0.4×1.08)))<br />

( 0.5<br />

20 + 350<br />

C = C + cY dp<br />

I = I − bi<br />

G = G<br />

M/P = M/P<br />

L d = kY − hi<br />

Y dp = aY + f(Y f − T (1 + i))<br />

Y f = gY<br />

= 0.338<br />

+ 1 × 2500 = 125.01<br />

20<br />

em que<br />

C c I b G M/P k h a f g T<br />

100 0.8 30 350 15 2500 20 0.5 0.6 0.4 1.08 10

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