Consider the following example of the IS-LM model:
C = 300 + 0.5(Y–T)
I = 400 – 1000(r + x)
G = 250
T =
200
x=0.03
πe = 0.02

as given that T = 200
We know that real interest rate if nominal interest rate less expected inflation. So,
πe
πe = 0.04 - 0.02 = 0.02


IS Equation : It is a relation between interest rate and income such that goods market is in equilibrium.








This is the IS equation.
When i = 0.04, equilibrium value of Y is given by:




Therefore equilibrium value of
A) Now the risk premium x has increased to 0.1 and Fed wants to maintain equilibrium value of income at 1600, so the required real interest rate is:









So, the real interest rate required to achieve equilibrium income of 1600 is -5 %.
B) Zero lower bound constraint means that interest rate cannot fall below zero i.e. interest rate cannot reach negative values. It means that there is lower limit of 0% on interest rates.
Here, the real interest rate in part (a) does not satisfy zero lower bound constraint as it falls below zero to maintain the equilibrium income.
Consider the following example of the IS-LM model: C = 300 + 0.5(Y–T) I = 400...
Consider the following IS-LM model, in which the central bank targets an interest rate of i C 11500.3YD 1 = 2000 + 0.3Y-8000i G 2000 T 1500 = 4Y-16,000i -= 0.02 c) Solve for value of the real money supply that the central bank must set to achieve its interest rate target, and the equilibrium level of output, by combining the IS and LM relations. Compute the values of equilibrium consumption, investment, and money demand. If the GDP deflator is...
19) Consider the following IS LM model: C= 300 + 25YD I = 150 + 25 Y - 1,000 i G = 350 T= 100 i*=.02 1. Derive the Irelation. Solve for equilibrium real output. 2. Solve for equilibrium values on C and I, and verify the value you obtained for Y by adding C, I, and G. 3. Now suppose that the central bank increases the interest rate to 8%. How does this change the LM curve? Solve for...
Consider IS- LM Model Real Sector: Y C+IG C ab (1-t) Y I d-e t-income tax rate i-rate of interest G Go Money Market: Md Ms Md kY - Ms Mo Mo - exogenous stock of money 1) Setup the system of solutions in general form, with variables vector in the following order: Y, C, I, i; (6 points) 2) Now, suppose we have the following values of parameters: a 10; b 0.7;t= 0.2; d 25; k 0.25;1 0.04; e...
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...
4. Points = 18. Consider IS-LM Model: Real Sector: Y=C+I+G C = a +b (1-t) Y I=d-ei G=Go t-income tax rate i-rate of interest Money Market: Ma=M Ma= kY-li Mg = Mo Mo - exogenous stock of money 1) Setup the system of solutions in general form, with variables vector in the following order: Y, C, I, i; (6 points) 2) Now, suppose we have the following values of parameters: a = 10; b = 0.7; t = 0.2; d...
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....
Consider IS-LM Model: Real Sector: Y=C+I+G C=a+b (1-t) Y I=d-ei G=GO t-income tax rate i- rate of interest Money Market: Ma=M Ma=ky-li M = Mo Mo - exogenous stock of money 1) Setup the system of solutions in general form, with variables vector in the following order: Y, C, I, i; (6 points) 2) Now, suppose we have the following values of parameters: a= 10; b = 0.7; t = 0.2; d = 25; k = 0.25; 1 = 0.04;...
4. Consider the following numerical example of the IS-LM model C 0.8(Y T); I 1520 240i; T 150 0.25Y; G 200; (M/P)s 1800 (M/P)D 300 0.75Y 300i a. Derive the IS and LM relation. (10%) b. Solove for the equilibrium values of output, interest rate, disposable income.(10%) 400 and T becomes T 350 0.25Y c. Suppose that G rises by 200 to G = Simultaneously, the central bank decreases money supply to 1500. Calculate what will happen to Y* and...
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...
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...