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Answers_HW_4 - Fall 2009 Truman Bewfey Economics a Answers...

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Unformatted text preview: Fall 2009 Truman Bewfey Economics a Answers to Homework #4 (Due Thursday, October 8) Problem: 2) Calculate a consumer‘s demand for goods 1 and 2 as a function of p1 and p2, the prices of goods 1 and 2, respectively, when the consumer’s utility function and initial endowment are as below. Draw the offer curve in each case. _. 5 2 _ a) u(x1, x2) _ x1x2, e— (10, 0), b) u(x1, x2) = min(5x1, 2x2), e= (10, 0), 'c) u(x,x)=5x +2x,e=(10, 0), 1 2 1 2 d) u(xj,x2) = ln(x) + ln(xz),e=(1,1). 1 Answer: a) Since the utility function is Cobb-Douglas, px =_§_w=_5_(10)p zflp. Therefore, if p1 is positive,. x=fl. ‘ 7 Similarly If p1 = 0, then demand is any point on the horizontai axis to the right of zero. If p2 :0, then demand is not defined. The offer curve consists of the heavy lines in the diagram below. 50/7 x b) Since the utility function is Leontief, we need only solve the two simultaneous linear equafions pX+pX =10p 11 22 1 and X =_5__X. 2 21 The solutions of these equations are 20 50 x zfiLandx =_,._.p_l_ 2 . 2p1+ 592 2p1+ 592 The offer curve is the heavy jagged line in the diagram below. emmmmmmmmwwmw 25 0 10 0) Suppose that both prices are positive. |f.p_2>3,thenx :10, x =0. 1 2 p15 If p2 = 3., demand is the set {(x1, x2) 6 i312|5x1 + 2x2 =10}. P1 5 10p1 5 1 2 p 1 2 E? i .r 5? % E 1‘: % If either price is zero, demand is not defined. The offer curve is the heavy tagged iine below. X 2 25 2:; 1; a i 3% ' t? 3?. a g a i 5% E :3 [2 § , i 3 it E Q i l: E 0 10 x 1 ”Wm . um , fifi’at‘vsWMfiWAMW-M d) if either price is zero, then demand is undefined. So suppose that _both prices are positive. Because the utiiity function is Cobb-Douglas, we know immediately that 'i' + X :LPiandX2=Ml We may normalize the price of commodity 1 to be 1. Substituting 1 for p1 in the equations for x1 and x2, we see that 1+ 1+ x 2 p2 andxzz p2. ‘ 2 2p2 Solving the first of these equations for p2, we find that p2=2x1—1. Subsituting this equation into that for x2, we see that X1 2 ———~—. 2x1—1 X: If we soive this equation for x1, we obtain the expected symmetric expression x2 x1=-——————. 2x2—1 Both of the last two equations express the offer curve. The curve goes through the point (1, 1) and has as asymptotes the vertical line x1 = 1 l2 and the horizontal line x2 = 1 2. We can obtain a more easily recognized formula for the offer curve by calcuiating that 1 x1 1 2x1—2x1+1 1 1 X——=——-—-—-————=———————-—-———-—::—-———-—————-——. 2 2 2x1—1 2 4x1—2 2 2x1—1 Since 1 1 x ——=———2x —1 , 1 2 2( 1 ) it follows that [X1 at —~;-i= 1—- That is, the offer curve is a hyperbola with origin at (1/2, 1/2). The curve tooks rcughiy as in the diagram below. aawmiwmwmgmmmwfiwmmumm mmwewmw «immatwwwmmm» mum-mm; Probtem: 4) In cases a and b below, find Tc(p) and 11(3)) for any vector p of appropriate dimension and such that p > 0, where Mp) = sup p.y and n( p) = {y 6 Y | p.y = 11(9)}. er w7 a)Y={(y,y.y) iy $0.3! Sofandosy S(—y)““(—y)“2}- 1 2 3 1 2 3 1 2 _ 2 2 b) Y— {(y1. y2, y3. yé) Iy1 20.31220. y3s0. ydsO. andy1 + yzs y3y4}. wmnmmwmmwwmmw “ MW Answer: a) If p3 = 0, then Mp) = 0 and mm = {(0, 0, 0.)}, since output has no value. if mums. p3> 0 and p1: 0 or p2 = 0,then1t(p)= ooandn(p) = o. Soassumep>> 0. The profit maximization probiem is _ 1r4_ 112 max [p3( y1) ( ya) + 131311 + Pat/21- s. so V10Y2 The first order conditions for this problem are )—3.’4 E a E; $2 g; 9f g i it 1 --—I93(—y1 (—312) “2 + p, = 0 4 and 1 1m —1.'2 _ —EpS(—y1) (—y2) + p2 — 0- These may be rewritten as 4p1 = p3(—y1)‘”4(-y2) ”2 and 1!4( )—11'2‘ 2p2 = psi—V1) —y2 Dividing the second equation into the first, we obtain 291 1?— pz V1 which is the same as 291v, P2 If we substitute this equation into the profit maximization problem, we obtain V2: 2p1 Pa max y!1 s 0 p3 (-31,)3” + 3p,y, This problem is of the form WWWwemwvmmmwml‘mwmwwmwmm max {ax""4 f bx], A) x20 where 2p1 Pa x=—yi,a=p3 ,andb=3p1. Solving problem A, we find that x = (SET, output = ax“ = 1_[£Ti, p3 4b p 4 b3 3 and the maximum value of the objective function is MEPET: 27 2:. 4 4b 256 b3 Substituting for a and b, we see that 4 4 3 4 y1=_. p3 ,yzz- pa ,y3= p3 ,andprofit=7£=_1_p3. 2 2 3 2 64 2 64p 1p2 32p1p2 16p1p2 p1 p2 In summary, 0, If p3 = 0, p4 Mp) = 3 ,ifp>>0,and 64p1p: oo,ifp3>0andp1=00rp2=0, and {(0, o, 0)}, if pa = 0 4 4 3 l0 P P . mp): " 322.- 32, 32 ,pr>>0,and 649192 329192 169.92 e,ifp3=0andp1=0 orp2=0. Answer: b) if p1 = p2 = 0, then Mp) = 0 and 11(9) = {(0, O, 0)}. lfp1> 0 or p2 > 0 and p3 = O orp4 = 0, then 7t(p) = wandmp) = o. Soassumethatp1> 0 or p2 > 0 andp3 > 0 and p4> 0. The profit maximization problem is max [p1y‘ + pave + pays + p4y4] 20, 20. so, so Y1 Y2 Y3 Y4 s.t.yf+y:s yy. 3 4 The Lagrangian of this problem is _ _ 2 2 _ < f — 913/1 + pzy2 + pays + 943/4 Alt/1 + 3/2 quayd ] From this, we see immediately that the first order conditions are p1—27ty1 =0, pz—Zly2=0, p —_1_ i =O,and 3 2 y 3 p —l 3.3. =0. 4 2 y 4 The first two of these equations imply that y=__2y. B) i, Ii 3 is i;- 1% l2, i if fig E g g 1% i? E g: it 3 y f The last two equations imply that p y =JY- p 4 3 4 If we substitute equations B and G into the profit maximization problem, it becomes 2 9 max {py+_£y+Py+Py] yszo'ysso 11 p 1 33 as 1 2 at y2 + gy?‘ s 1 2 1 P 1 which may be simplified to p2 + p2 max _1_____2_ + 2p y y120,y3s0 p1 ‘ 3 3 2 2 + s.t. p1 p2 2S—y Bi 2 1 3 to1 D4 2 112 114 $3 P y1= 2 1 2 i _y3 P + P P 2 4 If we substitute this equation into the objective function of the maximization problem, it becomes the unconstrained problem 9 1n'4 max 4p? + p: __3 I—ya + 2p3y3 . y 50 3 p4 This problem is of the form max [a5 — bx], x20 where 0) SWWWCV wwwwmwmlxwwmwm-wmwwmmxmwmawmwwwrmmwzm»» The soiution to this problem is “(iii and the maximum value of the objective function is Therefore :3 1 p2 + p y = _3y = -____1_____.. 4 3 1 {[2 312 I94 6 93 I04 Similarly pa + P p4 and 2 1 pt 4 p3p4 10 tn summary, O,pr1=p2= , p2+p2 g;(p)=_1_1 2,ifp>Oorp>0andp>0andp>0,and 8 W 1 2 3 4 34 no,Iipt>Oorp2>0andp3=00rp4=0, and {(0,0,O,0)},ifp1=;3220, 2 2 2 2 1_ D1 ,_1_ p2 I...-_1_p1+pzl_1_p1+pz = 4 4 16 3121!.2 16 112 312 I nip) W’p3p4 p394 I03 P4 p3 p4 ifp >Oorp >0,andp >0andp >0 1 2 3 4 a,ifp1>Oorp2>0,andp3=Oorp4:0. Problem: 5) Consider the Robinson Crusoe economy with utility function u(x1, x2) = |n(x1) + 2ln(x2), endowment vector e = (i, 0), and production function y2 = —2y1! where x1 and x2 are the consumptions of commodities 1 and 2, respectively, and (y1, y2) is the input-output vector of Robinson Crusoe’s firm. Find ali equilibria with the price of the first good equal to one. Answer: An easy way to solve this problem is to notice that the equilibrium allocation maximizes Robinson Crusoe’s utility and that by feasibility x =y =—2y1=2(1—x1). 2 2 Therefore the equilibrium allocation may be catcutated by solving the problem x20 1 max[in X1+ 2|n(2(1 —x1))]. 11 The first order conditions for this problem are L: 2 , x 1—x 1 1 sothat x=1/3. 1 It follows from feasibility that y =X —1=—2/3,andx =y =-—2y =4/3. 1 1 2 2 1 The equiiibrium price vector is the oniy one with p1 = 1 that gives maximum profits equal to zero at(y1, ya) = (~2/3, 4/3). That is, p = (1, 1/2). in summary, the equilibrium is ((x1, x2), (y1, ya), (pi, p2)) = ((1/3, 4/3), (—2/3, 4/3), (1, 1/2)). Problem: 8) Consider the following Edgeworth box economy. uA(x1, x2) = —13~|n( x1) + gm x2), 03A: (18, 0), uB(x1, x2) = ltn(x1) + —1—ln(x2), e8: (0, 20). 2 2 Find alt competitive equilibria with prices that sum to one. Answer: Because the utility functions are iinear and each consumer is endowed with only one commodity, we can calculate immediately that xA1 = 1 8/3 = 6 and hence feasibility implies that xB1 = 12. Similarly xB2 = 20/2 : 10, and hence xA2 = 10. We may calculate the price p1 from the budget equation of consumer A ms, + xmlt -i0,) = (13,, 1 —p,).eA- On substituting numbers for the allocation variabies, we obtain the equation 6p +10(1——p1)=18pi. 1 The solution of this equation is p1 = 5/11, so that p2 = 6/11. In summary, the equilibrium is (xA,xB,p)=((6,10),(12,10),(5/11,6/11)). 12 “momma/team mummmwwwvmamwwwwwm Problem: 9) Consider an Edgeworth box economy with utility functions uA(x1, x2) = min(3x1, x2) and uB(x1, x2) : min(xt, 3x2). a) Find the unique equilibrium when eA = (4, 0) and eB = (0, 4) and the price of good1 equal to 1. Compute the utility of person A at the equilibrium. b) Find the unique equilibrium when eA = (6, 0) and e8 2 (0, 4) and the price of good1 equai to 1. 0) Compute the utility of person A at the equilibrium of part (b) and compare it with his utility in the equilibrium of part (a). Is there anything paradoxical about your finding? Can you explain intuitively why person A's utility ieve! changes in the way that it does from part (a) to part (b)? Answer: a) This probtem may be solved algebraically as foltows. Because of the nature of the utility functions, x =3x andx =3x. A2 A1 81 82 Feasibility implies that x +x =4andx +x :4. A! B1 A2 B2 The soiution to these linear equations is A A which is 1 + 3p2 = 4, so that p =1. In summary, the equilibrium is (xAprip) :((1s3)!(3v1)!(1l1)) 13 The utility of person A in this equilibrium is 3. Answer: b) The relevant linear equations now become x =3x ,x =3x ,x +x =4,andx +x =6. A2 A181 32 A1 81 A2 B2 Proceeding as in part a, we find that the equilibrium is (xA, xB, p) = ((3/4, 9/4), (21/4,7/4),(1,7/3)). 0 1 :2. 3 4 5 6 A : Answer: 0) The utility of person A in the equilibrium of part b is 9/4, which is less than 3, the utility of person A in the equilibrium of part a. This is somewhat surprising, as person A’s endowment is larger in part b than in part a. The explanation can be understood visually by considering the superposition of two Edgeworth boxes in the diagram above. The origins of person B in the models of parts a and b are labeied as 0 Ba and 03b, respectiveiy. The endowment points of the two models are indicated by e ande , and the equiiibrium allocations are indicated b a '14 by E and E . The dashed tines are indifference curves. The heavy slanted fines are the locus of b . a corners of the indifference curves of persons A and B and are the effective part of the offer curves of the two people. In the model of part b, the right-hand edge of the box is disptaced to the right, and this movement brings the intersection of the offer curves downward along the offer curve of person A and therefore makes him or her worse off. This movement atso increases the price of good 2, which is the commodity that person A must purchase. It is this increase in price that makes person A worse off. The increase in the supply of the commodity that person A provides turns the terms of trade against him or her. The change in terms of trade may be seen from the change in the slope of the budget lines in the two equitibria. 15 memflmk‘r’rTVWflé’bt,WKWWM”MWW‘W ...
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