Download MOCK EXAM

I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
ECON 300
Advanced Macroeconomics
30 November 2011
Dr. Yetkiner
MOCK EXAM
Key to Midterm Exam
1. (20 Points) Calculate the GDP of Farmland, a fictitious economy whose numbers are
listed below. Do so using all three methods (value added approach, income approach, and
expenditure approach).
FarmLand, year 2000
Farmer Jones, (private firm)
Corn sold to Govt
Corn sold to Singapore
Corn sold to FoodCo, Inc
Paid workers
Tax on profit
Farmland Govt
Taxes
Purchase of Corn
Wages
Purchase of Corn Flakes
Value-added Approach:
Expenditure Approach:
Income Approach:
$25
$25
$20
$40
$15
$50
$25
$10
$15
FoodCo, Inc
Corn Flakes Sold to Consumers
Corn Flakes Sold to Japan
Corn Flakes Sold to Government
Corn bought from Farmer Jones
Corn Inventory
Beginning of Year
End of Year
Salt bought from Egypt
Paid workers
Tax on Profit
Households
Taxes on wage income
70 + [125-30-10] +10= 165
100 + -10 + 50 + (35-10) = 165
70 + 55 + 40 = 165
W Π
TA
Grading: 5 points each
1
$100
$10
$15
$20
$10
$0
$10
$20
$25
$10
5 points
5 points
5 points
I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
2. (10 Points-Partial Equilibrium) Suppose that utility function u of a representative
agent is u  c (3 / 4)  l (1/ 4) , where c is consumption of physical goods and l is consumption
of leisure. Suppose further that w  3 ,   24 , and h  24 . Find the optimal values of c
and l .
Initially, the Lagrangian is: L  C(3/4)l (1/ 4)  C  3l  72  24. The first order
conditions are:
L
(Equation 1)
 0  (0.75)C-0.25l 0.25    0
C
L
(Equation 2)
(3 points)
 0  (0.25)C0.75l 0.75    3  0
l
L
(Equation 3)
 0  C  3l  96  0

From the first two first-order conditions (i.e., from (1) and (2)), we obtain:
(0.75)C -0.25l 0.25


 C  9  l . Using this result in the third first-order condition:
0.75 0.75
3 
(0.25)C l
9  l  3l  96  l *  8  c*  72 .
2
I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
3. (25 Points- General Equilibrium) Suppose that utility function u of a representative
agent is u  c (1/ 4) l (3 / 4) , where c is consumption of physical goods and l is consumption
of leisure. Suppose that production technology is represented by y  (0.5) K 0.25  N 0.75
where K  16 is the physical capital stock and N is labor. We assume that h  24 ,
h  l  N and that there is no government in the economy (use w and  to denote the
real wage and profits, respectively). Find the optimal values of c , l , N , y , w ,  , and
u under the competitive equilibrium assumption.
This is a General Equilibrium Model. Let us start from the household’s problem. The
Lagrange is: L  C(1/4)l (3 / 4)  C  wl  wh   . The first order conditions are:
L
 0  (0.25)C -0.75l 0.75    0
C
L
 0  (0.75)C 0.25l 0.25  w  0
l
L
 0  C  wl  wh    0

(Equation 1)
(Equation 2)
(3 points)
(Equation 3)
From the first two first-order conditions (i.e., from (1) and (2)), we obtain:
(0.25)C-0.75l 0.75

wl

C 
0.25 0.25
(0.75)C l
w
3
Using this result in the third first-order condition:
wl
4wl
3
3
3
. In order to solve
 wl  wh   
 wh    l  h 
 l  18 
3
3
4
4w
4w
the model, we need labor supply, which may be obtained directly from N  l  24 :
3 
3

s
.
N s  24  l  N s  24  18 
 N  6
4 w
4w

We cannot solve the problem unless π is determined. For this, let us look at the
production side. From the firm’s profit maximization problem:
1
  (0.5) K
0.25
N
0.75
d
 0.75  0.25
 wN 
 (0.75) N 0.25  w  0  N d  
 .
dN
 w 
We need to also calculate profits:
w
1
  ( N 0.25  w) N    (
 w) N    (0.333) wN   w  N .
0.75
3
3
I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
Now we have enough information to solve the general equilibrium problem. First, use the
profit equation in the labor supply:
1
1
 w N
0.25
3
0
.
75


Ns  6 3
 N s  6  (0.25) N  6  (0.25)

4 w
 w 
Given that N s  N d at equilibrium,
N  6  (0.25) N  (1.25) N  6  N *  4.8
(2 points)
1
 0.75  0.25
Using this information at N d  
implies w*  0.506 .

 w 
(2 points)
The rest can be calculated by substitution:
 *  0.8096
(2 points)
(2 points)
(2 points)
(2 points)
(3 points)
l  19.2
C *  3.23
y *  3.23
*
u *  10.51
4
I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
4. (15 Points) Suppose that the government decides to increase taxes. Using the general
equilibrium model developed in chapter 5, determine the effects this has on aggregate
output, consumption, employment, and the real wage. Hint: Do not forget to draw a figure
and discuss in detail the impact of the exogenous shock.
p.161, figure 5.6
(Important Note: page numbers and/or figure numbers may change depending on the
edition of the textbook)
5. (15 Points) Suppose that Robinson Crusoe has a two-period life in a partial
equilibrium framework. Furthermore, we assume that Mr. Crusoe show “borrower”
behavior. Analyze the impact of an increase in the interest rate on current and future
consumption and saving.
p. 286 in chp.8, figure 8.14
(Important Note: page numbers and/or figure numbers may change depending on the
edition of the textbook)
6. (15 Points-Partial Equilibrium) Suppose that Daniel has income of
when
he is young and
when he is old. The real interest rate is r  0.1 . The overall
c1  1  1  c12  1
utility function of Daniel is U  1
, where   2 . Find the optimal


1
 1.21  1  
values of c1 , c 2 and s .
This
is a Partial Equilibrium Model. The household’s problem
c  1  1  c21  1 
c
990 
L

  c1  2  900 

 . The first order conditions are:
1
1.1
1.1 
 1.21   1

1
1
L
 0  c1-2    0
c1
(Equation 1)
5
is:
I. Hakan Yetkiner
http://www.hakanyetkiner.com/
Izmir University of Economics
Department of Economics
L
 1  -2 
0
0
c 2 
c2
1.1
 1.21 
c
L
 0  c1  2  1800  0

1.1
(Equation 2)
(3 points)
(Equation 3)
From the first two first-order conditions (i.e., from (1) and (2)), we obtain:
c2  0.953c1
c1*  964.44
0.953c1
 1800 
1.1
(2 points)
c2*  919.11
s*  64.44
(2 points)
(2 points)
Using this result in the third first-order condition: c1 
6