Worked-Out Numerical Exercise for the Open-Economy IS-LM Model

To help you work through the algebra needed for numerical problems in this chapter, here is a worked-out numerical exercise as an example for solving the open-economy IS-LM model.

Consider an economy that is described by the following equations:

C^d = 140 + 0.8( Y - T) - 200r

T = 400 + 0.1Y

I^d = 1000 - 700r

NX = 50 - 0.07Y + 0.15Y_For - 100r + 400r_For

L = 0.5Y - 1000i

Y = 5000; G = 400; M = 489,900; π^e = 0.17; Y_For = 4000; r_For = 0.2; P_sr = 213.

Find the short-run and long-run equilibrium values of Y, P, r, T, C, I, NX, and i. To solve this problem, we follow these steps:

Step 1: Find the equation for the IS curve by using the goods market equilibrium con­dition. The goods market equilibrium condition is Y = C^d + I^d + G + NX. Substitute the equation for T into the equation for C^d from above. Then substi­tute the resulting equation and the equations for I^d and NX from above along with the values of G, Y_For,and r_For into Y = C^d + I^d + G + NX to obtain:

Y = {140 + 0.8[Y - (400 + 0.1Y)] - 200r} + [ 1000 - 700r] + 400 + [50

- 0.07Y + (0.15 ? 4000) - 100r + (400 ? 0.2)].

Rearrange this equation as an equation for r in terms of Y:

Y = {140 + 0.8Y - [(0.8 ? 400) + (0.8 ? 0.1Y)] - 200r} + [ 1000 - 700r]

+ 400 + [50 - 0.07Y + 600 - 100r + 80],

so

Y = [ 140 + 0.8Y - 320 - 0.08Y - 200r] + [ 1000 - 700r]

+ 400 + [50 - 0.07Y + 600 - 100r + 80],

so

0.35Y = [ 140 - 320 + 1000 + 400 + 50 + 600 + 80] - (200 + 700 + 100) r,

so

1000r = 1950 - 0.35Y.

Therefore, r = 1950/1000 - (0.35/1000)Y,so

r = 1.95 - 0.00035Y.

This equation is the IS curve.

Step 2: Find the equation for the LM curve by using the asset market equilibrium condition.

a. First find the equation for the LM curve with an unspecified value of

the price level. The asset market equilibrium condition equates real money supply to real money demand. Real money demand is given by L = 0.5Y - 1000i = 0.5Y - 1000(r + π^e) = 0.5Y - 1000(r + 0.17), and real money supply is M∣P = 489,900/P. In asset market equilibrium, 489,900/P = 0.5Y - 1000(r + 0.17), so 1000r = 0.5Y - 170 - 489,900/P. Therefore, r = (0.5/1000)Y - (170/1000) - (489,900∕1000)∕P, so

r = 0.0005Y - 0.17 - 489.9/P. This equation is the equation of the LM curve for an unspecified value of P.

b. Then find the equation for the LM curve when P = P_sr. Set P = P_sr = 213 in the LM curve to obtain r = 0.0005Y - 0.17 - 489.9 / 213, so r = 0.0005Y - 0.17 - 2.30. Therefore, r = 0.0005Y - 2.47. This equa­tion is the equation of the LM curve when P = P_sr.

Step 3: Find the short-run equilibrium.

a. Find the intersection of the IS and LM curves to find the short-run equilib­rium values of Y and r. We have written the equations of the IS and LM curves so that the left side of each equation is simply r. Setting the right side of the IS curve equal to the right side of the LM curve yields 1.95 - 0.00035Y = 0.0005Y - 2.47, so 4.42 = 0.00085Y. Therefore, Y = 4.42/0.00085 = 5200. Now use the value of Y in either the IS or the LM curve. In the IS curve: r = 1.95 - 0.00035Y = 1.95 -(0.00035 ? 5200) = 1.95 - 1.82 = 0.13. In the LM curve: r = 0.0005Y - 2.47 = (0.0005 ? 5200) - 2.47 = 2.60 - 2.47 = 0.13.

b. Plug these equilibrium values of Y and r into other equations to find equilibrium values for T, C, I, NX, and i.

T = 400 + 0.1Y = 400 + (0.1 ? 5200), so T = 920.

C = 140 + 0.8(Y - T) - 200r = 140 + 0.8(5200 - 920) - (200 ?

0.13) = 140 + 3424 - 26 = 3538.

I = 1000 - 700r = 1000 - (700 ? 0.13) = 1000 - 91 = 909.

NX = 50 - 0.07Y + 0.15Y_For - 100r + 400r_For = 50 - (0.07 ? 5200) + (0.15 ? 4000) - (100 ? 0.13) + (400 ? 0.2) = 50 - 364 + 600 - 13 + 80 = 353.

(Note that C + I + G + NX = 3538 + 909 + 400 + 353 = 5200, which equals Y.)

i = r + π^e = 0.13 + 0.17 = 0.3.

Step 4: Find the long-run equilibrium.

a. Use the fact that in long-run equilibrium, Y = Y. Plug the equilibrium level of output into the IS equation to find the equilibrium real interest rate. Use Y = 5000 and the IS equation, r = 1.95 - 0.00035Y, to obtain r = 1.95 - (0.00035 ? 5000) = 1.95 - 1.75 = 0.20.

b. Plug the equilibrium values of Y and r into other equations to find equi­librium values for T, C, I, NX, and i.

T = 400 + 0.1Y = 400 + (0.1 ? 5000), so T = 900.

C = 140 + 0.8(Y - T) - 200r = 140 + 0.8(5000 - 900) - (200

?0.2) = 140 + 3280 - 40 = 3380.

I = 1000 - 700r = 1000 -(700 ? 0.2) = 1000 - 140 = 860.

NX = 50 - 0.07Y + 0.15Y_For - 100r + 400r_For = 50 - (0.07 ? 5000) +

(0.15 ? 4000) -(100 ? 0.2) + (400 ? 0.2) = 50 - 350 + 600

-20 + 80 = 360.

(Note that C + I + G + NX = 3380 + 860 + 400 + 360 = 5000, which equals Y.)

i = r + π^e = 0.20 + 0.17 = 0.37.

c. Plug the equilibrium values of Y and i into the money demand equation to obtain the value of real money demand, L. Then find the value of P that equates real money supply, M/P, with real money demand, L.

The money demand curve is L = 0.5Y - 1000i = (0.5 ? 5000) - (1000 ? 0.37) = 2500 - 370 = 2130. Setting real money supply equal to real money demand gives 489,900/P = 2130, so P = 489,900/2130 = 230.

Here is an additional problem for calculating the real exchange rate and the nominal exchange rate in the economy above:

Problem: Suppose the real exchange rate e is given by e = 67 - 0.005Y + 0.002Y_For + 100r - 300r_For and the foreign price level is P_For = 115 units of foreign currency per foreign good. Calculate the values of the real exchange rate and the nominal exchange rate for the long-run equilibrium given this information.

Solution: Plugging in the long-run equilibrium values of Y and r and the values of Y_For and r_For yields e = 67 - (0.005 ? 5000) + (0.002 ? 4000) + (100 ? 0.2) - (300 ? 0.2) = 67 - 25 + 8 + 20 - 60 = 10 foreign goods per domestic good. The nominal exchange rate, e_nom, is e_nom = e ? P_FoJP = (10 foreign goods perdomesticgood)?( 115units of foreign currency per foreign good )/(230 units of domestic currency per domestic good), so e _nom = 5 units of foreign currency per unit of domestic currency.