Question

Consider the following numerical example of the IS-LM model: C = 241 +0.62Y) | = 151 +0.12Y-8981 G = 204 T = 163 i = 0.05 Derive the IS relation. (Hint: You want an equation with Y on the left side of the equation and everything else on the right.) Y=||- | i. (Round your calculations of the intercept and slope terms to two decimal places.) The central bank sets an interest rate of 5%. In the equations given above, this decision is represented as i = and is known as the relation. Given this model's IS and LM equations, the equilibrium real output, Y, is . (Round your response to the nearest integer.) Use the expression given below along with the values of equilibrium real output and the interest rate to determine the level of the real money supply. (M) = 2.37–7,6611 = 2.3Y-7,6611 The real money supply is . (Round your response to the nearest integer.) Enter your answer in each of the answer boxes.

Answer 1

decreases. constant IS: 0.26 soy = C + I + G 2u1 + 0.62 14 163) + IS1 + 0.12Y - 89 8 il + zou 494.g - 898 i 1903.61 - 345.84 i i 5% = 0.05 IS relation Y= 1903.01 - 3453.84 x oos 17.30-91 - 2.3 Y - 7661 P - 2.3 x 1730.91 - 7661 - 3592.04 I Real m 0.05 .