IS-LM Relation - Economics

Topics: Macroeconomics, Liquidity preference, IS/LM model Pages: 5 (1281 words) Published: December 3, 2014
Assignment #1
Deriving the IS-LM Relation
Abstract
To find the IS-LM relation for an economy defined by six structural equations, algebra is used to derive the curves and the equilibrium conditions for these curves in relation to one another. The equations show and explain that if government spending (G) increases by EUR 150 billion, consumption (C) increases by EUR 50 billion, interest rates (i) increase by 0.05 (5%), and output (Y) increases by EUR 200 billion. This causes the IS curve to shift from IS to IS'. (EconLit E270, E620, E470)

Deriving the IS-LM Relation
The structure of the goods and financial markets of an economy given in Blanchard (2006, p. 111, problem 4) is represented by the following equations: C = 200 + 0.25YD (1)
I = 150 + 0.25 Y - 1000i (2)
G = 250(3)
T = 200 (4)
(M/P)d = 2Y – 8,000i(5)
M/P = 1,600(6)
Since no units of measurement are given in the problem; it will be assumed that Y, C, I, G, T, (M/P)d, and M/P are all measured in billions of US$. The interest rate i is expressed as a proportion and is measured as a decimal. Output (Y) is:

Y = Z = C(Y-T) + I(Y,i) + G + EX - IM(7)
Since there is no mention of exports (EX) or imports (IM), it is assumed that this is a closed economy and that EX = IM = 0. The subsequent equation is modified to: Y = Z = C(Y-T) + I(Y,i)+ G(8)

This equation (equation 8), is the formula for the IS curve. The IS curve represents all values of output (Y) and interest rate (i) where the goods market is in equilibrium. To derive the IS curve for a specific economy, substitute the functions given for C (equation 1), for I (equation 2), for G (equation 3), and for T (equation 4), to get an equation for output: Y = Z = 200 + 0.25YD + 150 + 0.25Y – 1000i + 250(9)

Since it is known that YD = disposable income, it is possible to substitute it for (Y – T): Y = Z = 200 + 0.25(Y – T) + 150 + 0.25Y – 1000i + 250 (10) As taxes (T) were given in equation 4, they must be plugged in to this equation:

Y = Z = 200 + 0.25(Y – 200) + 150 + 0.25Y – 1000i + 250(11) Now, solve for output (Y):
Y = 200 + 0.25Y – 50 + 150 + 0.25Y – 1000i + 250 , or
Y = 0.5Y + 550 – 1000i , or
0.5Y = 550 – 1000i , or
Y = 1100 – 2000i
Hence the equation for the IS curve (see figure 1) is:
Y = 1100 – 2000i (12)
To find the equilibrium between the goods and the financial market, the LM relation must also be derived. The LM relation shows all the values of output (Y) and interest rate (i) at which the financial market is in equilibrium. The following formula is that of the LM curve: M/P = Y L(i)(13)

At equilibrium, it is known that:
M/P = (M/P)d (14)
To derive the LM relation for this particular economy, substitute the equations for (M/P)d (equation 5) and M/P (equation 6): 1600 = 2Y – 8,000i(15)
Now, solve for interest rate (i):
8,000i = 2Y – 1600 , or
i = (Y – 800) / 4000
Hence, the equation for the LM curve (see figure 1) is:
i = (Y – 800) / 4000 (16)
To solve for equilibrium real output (Yeq) equation 16 must be plugged in to equation 12, so that output (Y) can be calculated:
Y = 1100 – 2000[(Y – 800) / 4000] , or
Y = 1100 – [(Y – 800) / 2] , or
Y = 1100 – 0.5Y + 400 , or
1.5Y = 1500 , or
Y = 1000
Hence, the equilibrium real output (Yeq) (see figure 1) is:
Y = 1000(17)
To solve for the equilibrium interest rate (ieq), the value from equation 17 must be plugged in to either equation 12 or equation 16. Since both equations describe an equilibrium, they would yield the same answer. Since equation 16 is already solved for the interest rate (i), it is easier to use equation 16. Here, it is plugged in to equation 16:

i = (1000...

References: Blanchard, O. (2006). Macroeconomics (4th ed.). Upper Saddle River, NJ: Pearson Prentice Hall.
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