Download Math OPMT 5701 Assignment #2: IS-LM model: Answer Key

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