Question

4. Assume the following set of equations characterize a small open economy E R o nd (1) Y = 10,000 (2) Y=C+I+G + NX (3) C = 0.75(Y-T) D (4) I = 3.000 - 100r (5) NX = 500 - 500€ (6) CF = -100r Elo (7) CF = NXPO ato hone (8) G = 2.5500 TO FOOD (9) T = 1,800. (10)r=r* = 8.5% mbo TO is net exports, CF is net capital outflow, and is the real exchange rate. Solve these equations for the equilibrium values of C, I, NX, CF,r, and .

Answer 1

sol: Set of Equations are characterize a small. open economy. ( y = 10,000 (2) y=C+I+G+NX Here given Data : (3) C=0.75 (Y-T). → Nx is Net Exports 4) 1 - 3,000 -loor CF is net capital out flow (5) NX = 500-500 € & is the real Exchange (6) CP =1000 rate. CF =NX (8) G = 21550 Then we have to solving O) T-1800 these Equations for the (10) rar*- 8.5% Equilibrum values of c, I, NN, CF r and e. Net Exports: These are the Difference between a Countries total value of exports and toral value of imports. Depending on whether a country imports more goods. Net capital but Flow It is the net Flow of Funds bend invested abroad by a country during a certain penod of time. Real Exchange Rate & It is the ratio of a foreign price Level and the domestic price tenele multiplied by the nominal Exchange rate. - The Real Exchange Rate measures the relative price of goods.

CF=NX Or) -10r = 500-500 € @ =) -1007 +500E = 500 con) 100 - 500 E2-500 7=C+I+G + NX we know these values from the given data 10,000 = 0.75 C10000 - 1800) + 3000 -100r + 2500 +500-500e 10,000 = 12200 -100r-500 e 100r + 5001 = 2200 > 0 Solving © & © Te 8.51€ = 2.7] . Nx = 500-5000 = 500-500x2.7 2.850 [CF = NX=-850 C=0.75(Y-T) -0.75 (10000 -1800) = 6150 ] I : 3000 -1001 = 3000 - 100 (8.5) II = 2150