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Pareto Efficiency
Draw Edge worth box & mark contract curve.
1. π’π΄ = π₯1π΄ π₯2π΄
π’π΅ = πππ. (π₯1π΅ π₯2π΅ )
Bundles
a. (100,100)
b. (180,100)
c. (100,50)
2. π’π΄ = π₯1π΄ + π₯2π΄
π’π΅ = π₯1π΅ + π₯2π
a. (50,50)
b. (50,100)
3. π’π΄ = π₯1π΄ + π₯2π΄
π’π΅ = 2π₯1 π΅ + π₯2π΅
a. (50,50)
b. (50,100)
4. π’π΄ = 2π₯1 π΅ + π₯2π΄
π’π΅ = π₯1π΅ + 2π₯2π΅
5. π’π΄ = πππ. (π₯1π΄ π₯2π΄ )
π’π΅ = πππ. (π₯1π΅ , π₯2π΅ )
a. (50,50)
b. (50,100)
6. π’π΄ = π₯2π΄
π’π΅ = π₯1π΅
7. π’π΄ = π₯1π΄
π’π΅ = π₯1π΅
8. π’π΄ = π₯1π΄ β π₯2π΄
π’π΅ = π₯2π΅ β π₯1π΅
9. π’π΄ = 2π₯1π΄ + π₯2π΄
π’π΅ = π₯1π΅ + π₯2π΅
10. π’π΄ = π₯1π΄ + π₯2π΄
π’π΅ = ππππππ’π (π₯1 π΅1 π₯2π΅ )
11. π’π΄ = π₯1π΄ + π₯2π΄
π’π΅ = π₯1π΅ + π₯2π΅
12. π’π΄ = ππππππ’π(π₯1π΄ π₯2π΄ )
π’π΅ = πππ₯πππ’π(π₯1π΅ π₯2π΅ )
13. π’π΄ = πππ₯πππ’ππ’(π₯1π΄ π₯2π΄ )
π’π΅ = πππ₯πππ’π(π₯1π΅ π₯2π΅ )
14. π’π΄ = π₯1π΄ π₯2π΄
π’π΅ = π₯1π΅ π₯2π΅
15. π’π΄ = πππ₯πππ’π(4π₯π΄ )
π’π΅ = 2π₯π΅ + π¦π΅
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1.
π’π΄ = π₯1π΄
π’π΅ = π₯1π΅ π₯2π΅
{5,20}
{5,5}
Find competitive equilibrium
2. Find competitive equilibrium for the following:1
1. π’π΄ = π₯1π΄ β 4 πππ(π₯)2
{10,10}
π΄ π΄
2. π’π΄ = π₯1 π₯2
{0,10}
π΄ π΄
3. π’π΄ = π₯1 π₯2
{60,10}
π΄ π΄
4. π’π΄ = π₯1 π₯2
{4,1}
π΄ π΄
5. π’π΄ = π₯1 π₯2
{3,2}
2
π΄
6. π’π΄ = π₯1 + βπ₯2π΄
{12,8}
π΄
7. π’π΄ = π₯1 2π₯2π΄
{0,12}
1
π΄
8. π’π΄ = π₯1 β (π₯2π΄ )2
4
{0,8}
π’π΅ = π₯1π΅ β (π₯2π΅ )2
{5,25}
π’π΅ = ππππππ’π(π₯1π΅ , π₯2π΅ )
{20,5}
π΅ π΅
π’π΅ = π₯1 π₯2
{20,30}
π’π΅ = ππππππ’π(π₯1π΅ , π₯2π΅ )
{0,7}
π΅ π΅
π’π΅ = π₯1 π₯2
{1,6}
4
π΅
π’π΅ = βπ₯1 + π₯2π΅
{8,4}
π’π΅ = ππππππ’π(π₯1π΅ , 2π₯2π΅ )
{12,0}
2
π΅
π΅
π’π΅ = π₯1 + βπ₯2
{16,8}
3. π’π΄ = ππππππ’π(π₯1π΄ π₯2π΄ )
{50,0}
π’π΅ = ππππππ’π(π₯1π΅ π₯2π΅ )
{0,100}
a. Find their contract curve.
b. If A consumes 25 units of both π₯1 &π₯2 and B consumes the rest, is it
equilibrium?
Sum 4
Unit β 1
1. Find amount if there is a competitive equilibrium or not & if there is then
find it.
a. π’π΄ = π₯1π΄ +π₯2π΄
π’π΅ = π₯1π΅ π₯2π΅
{5,0}
endowment {5,0}
π΄
π΄
π΅ π΅
b. π’π΄ = π₯1 +π₯2
π’π΅ = π₯1 π₯2
{16,0}
endowment {16,4}
π΄
π΄
c. π’π΄ = π1 because π₯1 + (1 β π1 ) because π₯2
π€1 , 0
π΅
π΅
π’π΅ = π2 because π₯1 + (1 β π2 ) because π₯2
0, π€2
d. π’π΅ = ππππππ’π (π₯1π΄ , π₯2π΄ )
{5,0}
π₯π΅
π’π΅ = ππππππ’π (π₯1π΅ , 42 )
{0,10}
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For the following find out pareto efficient allocations & competitive equilibrium.
a. π’π΄ = 2π₯ π΄ + π¦ π΄
π’π΅ = 3π₯ π΅ + π¦ π΅
{10,1}
{1,6}
π΄ π΄
π΅ π΅
b. π’π΄ = π₯ π¦
π’π΅ = π₯ π¦
{10,0}
{5,10}
π΄ π΄
π΅ π΅
c. π’π΄ = π₯ π¦
π’π΅ = π₯ π¦
{8,16}
{8,16}
π΄
π΄
d. π’π΄ = βπ₯ π¦
π’π΅ = 5π₯ π΅ π¦ π΅
{3,16}
{15,8}
π΄
π΄
e. π’π΄ = πππ₯πππ’π (π₯ π¦ )
π’π΅ = π₯π΅ π¦ π΅
{8,1}
{0,8}
f. π’π΄ = ππππππ’π (π₯π΄ 2π¦π΄ )
π’π΅ = π₯π΅ π¦π΅
{7,2}
{4,0}
2
π΄
π΄
g. π’π΄ = π₯1 + βπ₯2 (10,8)
π’π΅ = 4βπ₯1π΅ + π₯2π΅ (8,8)
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