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1 LONDON SCHOOL OF ECONOMICS Department of Economics Leonardo Felli 32L.4.02; 7525 Solutions to Assignment 5 EC487 Advanced Microeconomics Part I 1. Sketch of the answers: (i) The map of isoquants for this technolog is: 1 2 (ii) An input requirement set for this technolog is clearl not conve as it 1

2 is apparent from the following graph. 1 1 V () 2 2 (iii) In this environment t cost minimization can onl lead to a corner solution. In other words for an given isoquant the cost minimization choice of inputs can onl be a point like (0, 1 ) or ( 2, 0) in the graph above. The result is that the cost function if (3 w 1 ) w 2 is: c(w 1, w 2, ) = w 2 2 = w 2 if instead (3 w 1 ) w 2 then the cost function is: c(w 1, w 2, ) = w 1 1 = w 1 3 2

3 in other words the cost function is: c(w 1, w 2, ) = min{3 w 1, w 2 } (iv) The cost function in (iii) is then represented in the following graph c(w, ) if 3 w 1 w 2 c(w, ) if 3 w 1 w 2 (v) Recall now that from the cost function we can onl recover the conve hull of each isoquant, hence since both versions of the cost function derived in (iv) are homogeneous of degree one in then we conclude that the conve hull of the isoquants are associated with a technolog that ehibits constant returns to scale. 2. The unique Walrasian equilibrium allocation in this econom is an allocation in which ever consumer s optimal consumption bundle is his/her endowment of the unique commodit:,i = ω i. This implies that in this econom there will be no trade. The Walrasian equilibrium price is indeterminate (notice that onl one commodit makes the definition of relative price vacuous). 3

4 3. Notice first that the allocation ( A, A, B, B ) = (2, 4, 3, 3) is non-wastefull: A + B = 2 and A + B = 3 and in the interior of the Edgeworth bo. Notice also that at the unique point in the Edgeworth bo that represents this allocation the slope of the indifference curves of A and B differ. The slope of the indifference curve of A is d A d A = A A = 2 while the slope of the indifference curve of B is d A d A = 2 3 In other words this is not a tangenc point. Notice last that the allocation ( A, A, B, B ) = (2, 4, 3, 3) is Pareto-dominated b the allocation (ˆ A, ŷ A, ˆ B, ŷ B ) = (3, 3, 2, 4) since: U A ( A, A ) = 8 < U A (ˆ A, ŷ A ) = 9, U B ( B, B ) = 15 < U B (ˆ B, ŷ B ) = 16 hence it cannot be Pareto efficient. 4. The set of Pareto efficient allocations is the solution to the following problem for an pair (λ A, λ B ) where λ i 0 for ever i {A, B}: ma λ A (3 A + A ) + λ B ( B + 3 B ) A, B, A, B s.t. A + B 2 A + B 3 (1) Alternativel the set of Pareto efficient allocations is the solution for all values of U of ma A, B, A, B s.t. 3 A + A B + 3 B U A + B 2 A + B 3 (2) 4

5 This is the set of allocations ( A, A, B, B ) such that: 0 A 2, A = 0, B = 2 A, B = 3 and A = 2, 0 A 3, B = 0, B = 3 A. 5. We shall start from the analsis of the no-grade regime. The student problem is: ma G() s.t The solution will depend on the shape of the function G( ). For eample: if G () > 0 for ever [0, 24] then = 24, if G () < 0 for ever [0, 24] then = 0, if G (0) > 0 and G (24) < 0 then G ( ) = 0 and G ( ) < 0. Assume from now on G () < 0 for all [0, 24]. Consider now the grade regime. The student problem is now: ( ) ma G() + F s.t We shall focus on the case in which this problem has an interior solution. 5

6 The first order conditions are: G ( ) + 1 ( ) F = 0 (3) and G ( ) + 1 ( ) F < 0. (4) 2 Since all students are identical, smmetr implies: = 1 n n i=1 i = hence (3) becomes: which given that b assumption G ( ) + 1 F (1) = 0 (5) 1 F (1) > 0 implies hence which if G < 0 implies: G ( ) < 0 G ( ) < G ( ) > that is over-investment in working hours. Moreover given that F (1) = 0 we get: G( ) + F (1) = G( ) < G( ) In fact, is the value of that maimizes G(). 6

7 We shall now derive the sociall optimal choice of hours. solution to the following problem: ma i n G( i ) + i=1 ( ) n i i i This is the (6) smmetr implies: i = ˆ. Therefore (6) can be re-written as: ma ˆ n G() + (1) = G(ˆ) (7) i=1 We obtain: ˆ =. A quota is eas to use to obtain the sociall optimal choice of hours just set an upper-bound on the amount of hours worked: h =. Consider now a ta/subsid scheme. Let t be the per unit of hour ta and g the lump-sum subsid. The student problem facing such a ta is: ( ) ma G() + F t + g s.t The first order conditions are: G () + 1 F (1) t = 0 hence t = F (1), or T = t = F (1) is the optimal ta. The optimal subsid is g = t. 7

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