4. Consider the following numerical example of the IS-LM model C 0.8(Y T); I 1520 240i;...
1. Consider the following numerical example of the IS-LM model: C = 100 + 0.3YD I = 150 + 0.2Y - 1000i T = 100 G = 200 i = .01 (M/P)s = 1200 (M/P)d = 2Y - 4000i a. Find the equation for aggregate demand (Y). b. Derive the IS relation. c. Derive the LM relation if the central bank sets an interest rate of 1%. d. Solve for the equilibrium values of output, interest rate, C and I....
please help me Consider the following numerical example of the IS-LM model: C = 100 + 0.3YD I = 150 + 0.2Y - 1000i T = 100 G = 200 i = .01 (M/P)s = 1200 (M/P)d = 2Y - 4000i Find the equation for aggregate demand (Y). Derive the IS relation. Derive the LM relation if the central bank sets an interest rate of 1%. Solve for the equilibrium values of output, interest rate, C and I. Expansionary monetary...
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,...
Consider the following numerical example of the IS-LM model: C = 182 +0.5Y) 1 = 157 +0.12 Y- 890i G = 219 T = 175 i = 0.04 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=O-Di (Round your calculations of the intercept and slope terms to two decimal places.)
d 21. Consider the following IS-LM model: C = co +61 (Y – T) I = bo + b Y – bai M d¡Y – dzi Р M P Р a. where (b+c) <1 b. Derive IS equation. Derive and determine its sign. [5 points]- di di c. b. Derive LM equation. Derive and determine its sign. [5 points]- d. c. Assume that LM curve is ** dY t dY () M P =dY Solve for the equilibrium output and...
Stacked An economy is initially described by the following equations: C = 60+ 0.8(Y-T) I = 120-5 M/P = Y-25r G = 200 T = 200 M = 3000 P = 3 a. Derive and graph the IS and LM curves. Use the accompanying diagram to graph the IS and LM curves by placing the endpoints at the correct location, then place point A at the equilibrium interest rate and level of income. IS: Y= LM: Y= IS: Y= LM:...
4. Consider the IS-LM model: Y =C+I+G C = co + C(Y - T) - Car T = to +tįY I = io ti Y - ir M = m;Y + mo - mar, where the endogenous variables of the system are Y and r. The simultaneous solution of the first four equations defines the set of values of Y and r that establishes equilibrium in the goods market. While the fifth equation defines the values of Y and r...
Consider the following IS-LM model: C= 300+ 0.5YD, I=200+0.3Y-2000i, G=500, T=300 (a) Derive the IS relation. (The relationship of Y and i). (b) The central bank sets an interest rate of 10 %. How is that decision represented in the equations? (LM relation) (c) What is the level of real money supply when the interest rate is 10 %? Use the expression: (M/P) = 1.5.Y − 4000.i (d) Solve for the equilibrium values of C and I. (e) Suppose that...
Just e) f) and g) if possible please Question 5: The IS-LM model Consider the following IS-LM model: Consumption: C = 200 +0.25YD Investment: I=150 + 0.25Y - 10001 Government spending: G=250 Taxes: T=200 Money demand: L(i,Y)-2Y - 8000 Money supply: Ms /P=1600 (a) Derive the equation for the IS curve. (Hint: You want an equation with Y on the lefthand side and all else on the right) (b) Derive the equation for the LM curve. (Hint: It will be...
I need help with the last part of this questions with the new curve etc A= C + + G + X - M C = 0.75Y = 1200 - 50 G = 100 X = 300 M= 200 Y = A L = 0.25Y - 25i (M/P) = 250 L = (M/P) Derive the IS equation from the above model. Derive the LM equation from the above model. Derive the equilibrium levels of Income Y and Interest Rate i....