•Variable net exports
•Algebraic determination of
equilibrium GDP
•Problem
Now we make the more realistic assumption that
imports (M) depend on the domestic level of income
or GDP(Y).
The Marginal Propensity to Import (MPM) is the
fraction of the change in income that is spent on
imports. That is:
M
MPM 
DI
Note that:
0  MPM  1
Net exports and the aggregate expenditure line
Net exports
(trillions of dollars)
(a) Variable net export function
Net exports = X-M
X-M
0
6.0
7.0 8.0
9.0 10.0 11.0 12.0 13.0
Aggregate expenditure
(trillions of dollars)
(b) Aggregate expenditure lines
0
6.0
7.0 8.0
Real GDP
(trillions
of dollars)
Net exports are added
to C, I, and G to yield
AE.
C+I+G
The addition of net
C+I+G+(X-M) exports: rotates the
spending line about
the point where net
exports are zero (real
GDP is $10.0 trillion)
9.0 10.0 11.0 12.0 13.0 Real GDP
(trillions of dollars)
3
Let AE denote aggregate expenditure . Thus we can say:
AE = C + I + G + (X – M)
Let NT denote net taxes (taxes minus transfers).
Thus we can say:
DI = Y – NT
[1]
The consumption function can be written as:
C = a + b(Y – NT)
The above and be rewritten as
C = a –bNT + bY
[2]
Where a-bT is autonomous consumption and b is the marginal
propensity to consume.
Net exports are described by:
X – m(Y – NT)
[3]
Where m is the marginal propensity to import Now
substitute [2] and [3] into [1] to obtain:
AE = a – bT + bY + I + G + X – m(Y – NT)
In equilibrium AE = Y. Thus
Y = a – bT + bY + I + G + X – m(Y – NT)
We can rearrange [4] to obtain:
Y
1
(a  bNT  I  G  X  mNT )
1 b  m
[4]
Y
1
(a  bNT  I  G  X  mNT )
1 b  m
Multiplier
Autonomous expenditure
AE
AE
Slope = b - m
a- bNT + I + G + X +mNT
0
Y*
Y
Let :
C= 100 + 0.75(Y – NT)
I = 50
G = 30
X = 40
M = .15(Y – NT)
NT = 100
Let Y denote real GDP. Thus we can say:
Y = C + I + G + (X – M)
[1]
Let NT denote net taxes (taxes minus transfers).
Thus we can say:
DI = Y – NT.
Let NT = $1.0 trillion (net taxes are autonomous).
Our consumption function is given by:
C = 2.0 + 0.8(Y – NT) = 1.2 + 0.8Y
Autonomous C
[2]
Induced C
1
Y
(a  bNT  I  G  X  mNT )
1 b  m
1

{100  [(. 75)(100)]  50  30  40  [(. 15)(100)]}
1  .75  .15
 (2.5)(160)  400
AE
AE
Slope = .75 - .15 = .6
160
0
400
Y
Let ΔI = 5 . What is the resulting change in Y?
1
1
Y 
 I 
 5  12.5
1 b  m
1  .75  .15
AE
AE’
AE
165
160
0
400 412.5
Y