Advanced Microeconomics FS 10
General Equilibrium
Solutions to Tutorial 2: General Equilibrium
Exercise 1 [Barter economy]
a) Consumer A:
The Lagrangian of consumer A is
1/2
L = x1/2
A1 xA2 − λ (pxA1 + xA2 − peA1 − eA2 ) .
First order conditions:
∂L
1/2
: 0.5x−1/2
A1 xA2 − λp = 0,
∂xA1
∂L
−1/2
: 0.5x1/2
A1 xA2 − λ = 0,
∂xA2
∂L
: pxA1 + xA2 − peA1 − eA2 = 0.
∂λ
Consumer B:
The Lagrangian of consumer B is
3/4
L = x1/4
B1 x B2 − λ (px B1 + x B2 − pe B1 − e B2 ) .
First order conditions:
∂L
3/4
: 0.25x−3/4
B1 x B2 − λp = 0,
∂xB1
∂L
−1/4
: 0.75x1/4
B1 x B2 − λ = 0,
∂xB2
∂L
: pxB1 + xB2 − peB1 − eB2 = 0.
∂λ
b) To get the price p we make a shortcut. We take the Marshallian demand functions from
Tutorial 1 exercise 3, substitute endowments for y and choose the right values for α.
1
1
Consumer A: xA1 = 2p
[2p + 2], xA2 = [2p + 2],
| {z }
2
y
Consumer B: : xB1 =
1
[2p
4p
3
+ 2], xB2 = [2p + 2].
4
At the equilibrium price demand equals endowment. We look at market 2 and solve for
p (because of Walras’s law, market 1 is also in equilibrium):
xA2 + xB2 = eA2 + eB2 ,
1
3
[2p + 2] + [2p + 2] = 4.
2
4
Solving for the price gives p = 3/5 in the Walrasian equilibrium.
Assistant Daniel Müller
1 of 7
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
c) Inserting the equilibrium price p = 3/5 into the Marshallian demand functions produces
the Walrasian equilibrium allocation
1 3
+ 2] = 83 , x∗A2 = [2 + 2] =
2 5
3 3
= 14 53 [2 35 + 2] = 43 , x∗B2 = [2 + 2] =
4 5
Consumer A: x∗A1 =
Consumer B: x∗B1
15
[2 35
23
8
.
5
24
.
10
d) For an internal Pareto Optima it has to be the case that:
A
B
MRS 1/2
= MRS 1/2
,
1/2
0.5x−1/2
A1 xA2
−1/2
0.5x1/2
A1 xA2
3/4
0.25x−3/4
B1 x B2
=
1
x−1
A1 xA2 =
−1/4
0.75x1/4
B1 x B2
,
1 −1 1
x x .
3 B1 B2
Given that the two feasibility conditions hold:
xA1{z
+ xB1
|
}
=
xA2{z
+ xB2
|
}
=
Total consumption of good 1
e|A1 +{z
eB1 = }4
,
e|A2 +{z
eB2 = }4
,
Total endowment of good 1
and
Total consumption of good 2
Total endowment of good 2.
we can insert these two equations into the MRS equation. Thereafter we get the following equation:
4 − xA1
xA1
=3
,
xA2
4 − xA2
and therefore
xA2 =
Assistant Daniel Müller
2xA1
.
6 − xA1
2 of 7
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
The function of the Pareto Optima is represented in figure 1:
˜ xB1 4
0
˜ xB2 -
- xA2 ™
3
e
2
Ω
1
Contract Curve
0
0
1
2
- xA1 ™
3
4
Figure 1: The Pareto Optima in an Edgeworth box
Assistant Daniel Müller
3 of 7
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
Exercise 2 [Kuhn-Tucker Theorem]
Consider a barter economy with 2 people and 2 goods: The utility functions of the two consumers are
1/2
uA (xA1 , xA2 ) = x1/2
A1 xA2
uB (xB1 , xB2 ) = 0.5xB1 + xB2 .
The original endowments are eA = (0, 3) and eB = (4, 1). Let good 2 be the numeraire good.
The price for good 1 is p.
a) Consumer B:
The Kuhn-Tucker Lagrangian of consumer B is
f(xB1 , xB2 , λ) = 0.5xB1 + xB2 − λ[pxB1 + xB2 − peB1 − eB2 ],
L
and the first order conditions are:
∂L
:
1/2 − λp ≤ 0, [1]
xB1 (1/2 − λp) = 0, [2]
∂xB1
∂L
:
1 − λ ≤ 0, [3]
xB2 (1 − λ) = 0, [4]
∂xB2
∂L
: pxB1 + xB2 − peB1 − eB2 ≤ 0, [5] λ (pxB1 + xB2 − peB1 − eB2 ) = 0. [6]
∂λ
b) If the Walrasian solution is an internal solution then equations [1] and [3] are exactly
equal to zero. Dividing equation [1] by equation [3] gives the equilibrium price p = 0.5.
c) If the equilibrium price is 0.5 then agent B is indifferent between all possible consumption sets that lie on the budget constraint. Therefore, we have to start with agent A and
agent B gets the rest.
To receive the Walrasian equilibrium allocation, we take the Marshallian demand function (xAi (p, e) = pαi [p1 eA1 + p2 eA2 ] for i = 1, 2) for Cobb-Douglas utility functions of
|
{z
}
y
consumer A in exercise 1). Then, we take the endowments from exercise 2) and insert
the equilibrium price p = 0.5.
Agent A: x∗A1 =
12
(3)
21
1
3
= 3, x∗A2 = (3) = .
2
2
Agent B gets the remains (Remember that aggregate demand equals aggregate endowment: x∗Ai + x∗Bi = eAi + eBi )
Consumer B: x∗B1 = 4 − 3 = 1, x∗B2 = 4 −
Assistant Daniel Müller
4 of 7
3 5
= .
2 2
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
Now assume the following endowments: eA = (1, 4) and eB = (3, 0).
d) Once again, we take the Marshallian demand function of consumer A and adapt the
endowment eA = (1, 4). With the equilibrium price p = 0.5 we get
"
#
h
i
1 1
9
12 1
Agent A: xA1 (p = 0.5) = 2 1 2 1 + 4 = 4.5, xA2 (p = 0.5) =
1+4 = .
2 2
4
The demand of consumer A is greater then the aggregated endowment of the two agents.
Therefore the price p = 0.5 can not be the equilibrium price. The solution with these
endowments will be a corner solution.
e) In the corner solution agent B will consume nothing from good 1, since the price is
greater than 0.5. The equilibrium price can be derived by setting the Marshallian demand
function of agent A for good 1 equal to 4 and then solve for the equilibrium price.
Agent A: x∗A1 =
1
2p
4
p1 + 4 = 4, ⇒ p∗ = .
7
The equilibrium allocation is
1 4
16
Agent A: x∗A1 = 4, x∗A2 = ( 1 + 4) = .
2 7
7
Agent B gets the remains
Consumer B: x∗B1 = 0, x∗B2 = 4 −
Assistant Daniel Müller
5 of 7
16 12
= .
7
7
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
f) The function of the Pareto Optima is represented in figure 2:
The Edgeworth box
˜ xB1 4
e1
Ω2
˜ xB2 -
- xA2 ™
3
0
e2
2
Ω1
1
Contract Curve
0
0
1
2
- xA1 ™
3
4
Figure 2: The Pareto Optima in an Edgeworth box
Assistant Daniel Müller
6 of 7
Part 1 (Lengwiler)
Advanced Microeconomics FS 10
General Equilibrium
Exercise 3 [Walras’ law]
a) Aggregate surplus demand for good 1:
z1 (p) = xA1 + xB1 − eA1 − eB1 ,
α (p1 eA1 + p2 eA2 ) β (p1 eB1 + p2 eB2 )
=
+
− eA1 − eB1 ,
p1
p1
p2
p2
= (α − 1) eA1 + αeA2 + (β − 1) eB1 + βeB2 .
p1
p1
Aggregate surplus demand for good 2:
z2 (p) = xA2 + xB2 − eA2 − eB2 ,
(1 − α) (p1 eA1 + p2 eA2 ) (1 − β) (p1 eB1 + p2 eB2 )
=
+
− eA2 − eB2 ,
p2
p2
p1
p1
(1 − α) eA1 − αeA2 +
(1 − β) eB1 − βeB2 .
=
p2
p2
b) Walras’ law states p1 z1 (p) + p2 z2 (p) = 0 for all p1 , p2 .
p1 z1 (p) + p2 z2 (p) = p1 (α − 1) eA1 + p2 αeA2 + p1 (β − 1) eB1 + p2 βeB2 ,
+p1 (1 − α) eA1 − p2 αeA2 + p1 (1 − β) eB1 − p2 βeB2 ,
= 0.
We showed that Walras’ law is true.
Assistant Daniel Müller
7 of 7
Part 1 (Lengwiler)