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Math OPMT 5701 Assignment #2: IS-LM model: Answer Key Example: IS-LM Model: Closed Economy In this simple linear model of the economy we can think of it as being made up of two sectors: The real goods sector and the monetary sector. A) The goods market Y =C +I +G C = C0 + b(Y − T ) = C0 + b(1 − t)Y I = I0 − αr G = G0 T = tY The endogenous variables are Y,C,I and r. The exogenous variables are C0 , I0 and G0 . α, b and t are structured parameters. B) The money market Md = Ms M d = kY − βr M s = M0 Equilibrium Money Demand Money Supply where M0 is the exogenous stock of money and k and β are parameters. Putting them together gives us M0 = kY − βr Together, the two sectors give us the following system of equations: Y − C − I = G0 b(1 − t)Y − C = −C0 I + αr = I0 kY − βr = M0 Note that by further substitution the system could be further reduced to a 2 by 2 system of equations. For now, we will leave it as a 4 by 4 system. In standard matrix form 1 −1 −1 0 Y G0 b(1 − t) −1 0 0 C = −C0 0 0 1 α I I0 k 0 0 −β r M0 To find the determinant we can use Laplace expansion on one of the columns most zeros). Expanding the fourth column ¯ ¯ ¯ ¯ ¯ 1 −1 −1 1 −1 −1 ¯¯ ¯ ¯ |A| = (−α) ¯¯ b(1 − t) −1 0 ¯¯ − β ¯¯ b(1 − t) −1 0 ¯ ¯ 0 0 1 k 0 0 ¯ ¯ ¯ ¯ ¯ ¯ −1 ¯ −1 −1 ¯ −1 ¯¯ ¯ ¯ ¯ −β¯ |A| = (−α)(k) ¯ b(1 − t) 0 ¯ −1 0 ¯ |A| = αk − β [(−1) − (−1)b(1 − t)] |A| = αk − β(1 − b(1 − t)) 1 (preferably one with the ¯ ¯ ¯ ¯ ¯ ¯ Use Cramer’s rule to find equilibrium income, Y∗ . This is done by replacing the first column of the coefficient matrix, A, with the sector of exogenous variables and take the ratio of the determinant of the new matrix to the original determinant, or ¯ ¯ ¯ G0 −1 −1 0 ¯ ¯ ¯ ¯ −C0 −1 0 0 ¯¯ ¯ ¯ I0 0 1 α ¯¯ ¯ ¯ ¯ M 0 0 −β |A1 | 0 Y∗ = = |A| αk + β(1 − b(1 − t)) Using Laplace expansion on the second column of ¯ ¯ −C0 0 0 ¯ 1 α (−1)(−1)3 ¯¯ I0 ¯ M0 0 −β Y∗ = αk + β(1 − b(1 − t)) Y∗ = By further expansion ¯ ¯ −C0 (1) ¯¯ M0 Y∗ = ¯ ¯ −C0 ¯ ¯ I0 ¯ ¯ M0 the numerator produces ¯ ¯ ¯ ¯ G0 −1 0 ¯ ¯ ¯ (−1)(−1)4 ¯ I0 1 α ¯ ¯ ¯ ¯ M0 0 −β + αk + β(1 − b(1 − t)) ¯ ¯ 0 0 ¯¯ ¯¯ G0 −1 0 1 α ¯¯ − ¯¯ I0 1 α 0 −β ¯ ¯ M0 0 −β αk + β(1 − b(1 − t)) ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ½ ¯ ¯ ¯ 0 ¯¯ α ¯¯ 3 ¯ I0 4 ¯ G0 − (−1)(−1) + (−1)(−1) ¯ ¯ M0 ¯ ¯ M0 −β −β αk + β(1 − b(1 − t)) Y∗ = ¯ ¯ ¯ ¯ ¯ ¯ ¯¾ 0 ¯¯ −β ¯ C0 β − (I0 (−β) − αM0 ) − G0 (−β) αk + β(1 − b(1 − t)) Y∗ = β(C0 + I0 + G0 ) + αM0 αk + β(1 − b(1 − t)) Since the solution to Y∗ is linear with respect to the exogenous variables, we can rewrite Y∗ as µ µ ¶ ¶ α β ∗ Y = M0 + [C0 + I0 + G0 ] αk + β(1 − b(1 − t)) αk + β(1 − b(1 − t)) In this form, we can see that the Keynesian policy multipliers with respect to the money supply and government spending are α 4Y ∗ = 4M0 αk + β(1 − b(1 − t)) and 4Y ∗ β = 4G0 αk + β(1 − b(1 − t)) 2