IS-LM Model Example ex2

# IS-LM Model Example ex2 - IS-LM Model Example Given the...

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IS-LM Model Example Given the following information, derive the equations for the IS and LM curves. C = 200 + 0.25 Y D (A) I = 150 + 0.25 Y – 1000 r (B) G = 250 (C) T = 200 (D) Y D = Y – T (E) (M/P) d = 2 Y – 8000 r (F) M/P = 1,600 (G) You can use equations (A) through (E) to determine the equation for the IS curve. Please note that this is the long way to do it. Many steps can be skipped here, but it is important to understand that this is the same process used in developing the simple Keynesian model, which was introduced in chapter 10. E = C + I + G E = 200 + 0.25 Y D + 150 + 0.25 Y – 1000 r + G E = 200 + 0.25 (Y – T) + 150 + 0.25 Y – 1000 r + G Recall that in equilibrium E = Y, therefore rewrite above equation: Y = 200 + 0.25 (Y – T) + 150 + 0.25 Y – 1000 r + G Y = 200 + 0.25Y – 0.25 T + 150 + 0.25 Y – 1000 r + G Group all the Y’s on the left side of the equation. Y – 0.25 Y – 0.25 Y = 200 – 0.25 T + 150 – 1000 r + G 0.5 Y = 200 – 0.25 T + 150 – 1000 r + G

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