0.2Y 5r = 120 – 5r = 0.2Y 120 – r = Y 24 – Hence, LM equation: r = Y 24 – (3) Equilibrium level of income and interest rate: Substituting the value of r in the IS equation we have Y = 1100 25r – Y = 1100 25(Y 24) – – Y = 1100 (1Y - 600) – Y = 1100 Y + 600 – 2Y = 1700 Y = 850 Thus the equilibrium of income is 850. To obtain the equilibrium rate of interest we substitute the value of Y in LM equation and we get r = Y 24 – r = × 850 24 – r = 34 24 = 10 – Thus equilibrium rate of interest is 10 per cent. Consider the following economy: 20. Problem: C = 100 + 0.8Yd I = 50 25r – G = T = 50 = 200 Md = Y 25r – 1. Calculate the IS and LM curves. 2. Calculate the equilibrium levels of output (national income) and interest.
0.2Y – 5r = 120 5r = 0.2Y – 120 r = Y – 24 Hence, LM equation: r = Y – 24 (3) Equilibrium level of income and interest rate: Substituting the value of r in the IS equation we have Y = 1100 – 25r Y = 1100 – 25(Y – 24) Y = 1100 – (1Y - 600) Y = 1100 – Y + 600 2Y = 1700 Y = 850 Thus the equilibrium of income is 850. To obtain the equilibrium rate of interest we substitute the value of Y in LM equation and we get r = Y – 24 r = × 850 – 24 r = 34 – 24 = 10 Thus equilibrium rate of interest is 10 per cent. 20. Problem: Consider the following economy: C = 100 + 0.8Yd I = 50 – 25r G = T = 50 = 200 Md = Y – 25r 1. Calculate the IS and LM curves. 2. Calculate the equilibrium levels of output (national income) and interest.
chapter-_5_is-_lm_model_econ-_102_2.docx