General Equilibrium (Linear Models)

Lecture 10: Last updated 01/06/2010 Follow the lecture to revise the notes

Chapter 17 General Equilibrium (Linear Models) Assume: 2 goods, 2 factors in the economy. y1,y2: final goods, x1,x2: factors of production, p1,p2: prices of final goods (assumed to be fixed by the world market  small country (economy) assumption) xij = amount of factor i used in industry j. 1 2 max z  p1 f L1, K1  p2 f L2 , K2  s.t.

L1+ L 2 ｜ L 镲 K+ K K => endowments of L , K轾 GENERAL MODEL 1 2 �  ( ) 臌镲 L1, L 2 , K 1 , K 2 0 for simplification, we can assume that the functions f1 & f2 exhibit constant returns to scale.

1 1 f tL1,tK   tf L1, K1   ty1 for all t 1 let t  y1

1 L1 K1  f  ,  1  y1 y1  Input – Output table

Li aij  I – O coefficients (Technological Coefficients) y j

L L1 L2

K K1 K2

Y1 Y2 1 ==>f( aL1, a K 1 ) = 1 ==> unit isoquant so; the constant returns to scale enables us to completely describe the isoquants by the knowledge of one isoquant (unit – isoquant). Then the model can be written as

Max z  p1 y1  p2 y2 s.t. aL1 y1  aL2 y2  L aK1 y1  aK 2 y2  K

1 f( aL1, a K 1 ) = 1 2 f( aL2, a K 2 ) = 1

With fixed technological coefficients aij ,akj , the model becomes a “linear programming model”.

Max z  p1 y1  p2 y2 s.t. aL1 y1  aL2 y2  L & aK1 y1  aK 2 y2  K & y1, y2  0

(here; p1, p2 , L1, L2 , K1, K 2 and aij ==> constant

Fixed – Coefficients Production Function  L K  y  min j , j  j    aLj aKj 

If you want to increase to level of output, then you have to increase both of the onputs 4 y=2

Linear Activity Analysis Suppose the firm is faced with 3 production coefficient possibilities.

A: 1L,2K   aL1,aK1 

B:( 3,1) = ( bL1 , b K 1 )

C:( 2,3) = ( cL1 , c K 1 ) K Act. A Act. C C 3 E uses less labor and less capital although it is on activity C

A 2 E D Act. B

y = 1 1 B

L 1 2 3

Assumption: the firm can choose any one or more activities simultaneously. Suppose: 1 unit of output can be produced by ½ of A and ½ of B. ½ of A  ½ of L and 1 of K ½ of B  3/2 of L and ½ of K Total  2 of Labor and 1.5 of Capital  A new combination activity D. Using the connected line, we get another point E on the same activity line of C but using less of both inputs. C becomes irrelevant in a cost minimizing problem.

3 K K Tangency should be 1  2 at this point because A it should be >-1/2 L1 L2 2 2 1  1 3 y = 1 1 B

L 1 2 3

Production is done where slope of isoquant = slope of isocost MP P w MRTS  L L  MPK PK r 1 Slope AB   2 w 1  If  , tangency will occur at “A”  relatively more capital intensive activity. r 2 w 1  If  , tangency will occur at “B”  labor intensive activity will be used. r 2 w 1  If  , infinitely many number of solutions along AB r 2

Algebraic Formulation (Kuhn Tucker can be used) (Minimization Problem) Min z  wL  rK s.t. LA  LB  LC  L & K A  K B  KC  K K if A is used (1,2) then, L  A  y  L y 1 A 2 A A L if B is used (3,1) then, L  3K  y  B B B B 3 2 L if c is used (2,3) then, L  K  y  C C 3 C C 2

Problem becomes  1 3  Min z  wLA  LB  LC  r2LA  LB  LC   3 2  L L s.t L  B  C  y A 3 2 0

LA, LB , LC  0

 1 3   0 LB LC  L  wLA  LB  LC  r2LA  LB  LC    y  LA     3 2   3 2  Foc. L =w +2 r -l � 0, if = 0, LA 0 LA L r l =w + -�0, if = 0, LB 0 LB 3 3 L 3 l =w + r -�0, if = 0, LC 0 LC 2 2

L LB LC =y0 - LA - -�0, if = 0, l 0 Ll 3 2

  MC

If A  MC  w  2r If B  MC  3w  r If C  MC  2w  3r At the cost minimizing point, the process with minimum MC will be used.

Numerical Example

A firm produces 2 goods, y1 (food) and y2 (clothing). 3 inputs are land (H), Labor (L) and capital (K). inputs are combined in fixed proportions to produce 1 unit of output. 1 unit of food requires 3 ac/land, 2 w/h, 1 m/h 1 unit of clothing requires 2 ac/land, 2 w/h, 2 m/h I – O Matrix H 3 2

A= L 2 2 Also H  54, L  40, K  35& P1  $40, P2  $30 K 1 2 F C (food is more land intensive) Maximize Output

Max z  40y1  30y2

3y  2y  54 1  1 2   s. t. 2y1  2y2  40 2  y1, y2  0 “ A linear programming model”  y  2y  35 3 1 2    y 2

27 Land Cont. Slope = 3/2

20 Labor = Slope = 1 17.5

Capital Slope = 0.5 C We produce here

18 20 35 y 1

Slope of the obj function : 40/30 = 1.333 Comparing the slopes, we see that production will be at point C. it is the intersection of Land and Labor and capital is non – binding. Therefore, simultaneous solutions of (1) & (2) yields

* * y1 14 & y2  6 z*  40(14)  30(6)  740

3 2 314 26  54  y  2 2 1 2 14 2 6 40         y2  1 2 114 26  26

Derivation of Input Prices Suppose land is 55 (1 increased). The previous solution would yield;

3y1  2y2  55

2y1  2y2  40 * * y1 15, y2  5 z* =40( 15) + 30( 5) = 750 ( valueofoutput)  value of output increased by $10. VMPLand = 10 = Price of Land in competitive market. “Shadow Price” or “Imputed rent” Suppose change in Land: ΔH

3y1  2y2  54  H

2y1  2y2  40

* y 14  H  1 1 *  y2  6  H  z*  4014  H  306  H   740 10H z* 10  U H 1 similarly by increasing L by ΔL

3y1  2y2  54

2y1  2y2  40  L

y* 14  L  1  2 * 3   y2  6  L 2  z*  740  5L z* u = =5 ==> shadow price of labor. 2 L Finally doing the same for K, we will get z*  0  U L 3

The Rybczynski Theorem Effects of a change in factor endowment on the output.

* * y2  6  H y1 14  H 3 y* 14  L y*  6  L 1 2 2 3 dy*  dH  dL dy*  H  L 1 2 2 Theorem: if the endowment of some resources increases, the industry which uses that resources more intensively will increase its output, value of the other will decrease.

(recall in the example y1 (food) was made land intensive and y2 (clothing) labor intensive).

L j L y a j  j  Lj K j K j aKj y j

3 2 a a a a A 2 2 11 21 11 12        First industry is land intensive. a12 a22 a21 a22 1 2 Proof of the Theorem When land increases, output of land intensive industry will increase. a11 y1  a12 y2  H a21 y1  a22 y2  L Matrix Format

a11 a12  y1  H        a21 a22 y2   L  Cramer’s Rule

* a22 H  a12 L y1  a11a22  a12a21 for land intensive industry a a  a a  0 11 22 1221 denomin ator

y labor intensive 2

y land intensive 1

Rybczynski: (proof) y* a 1. 1  22  0 H 

* a11L  a21H 2. Similarly y2  a11a22  a12a21 For land intensive industry denominator > 0 y* a 2   21  0 H 

ECON 603 Chapter 17 continued Summary

1 2 Max p1 f H1, L1  p2 f H 2 , L2  s.t. H1  H 2  H

L1  L2  L

H1, H 2 , L1, L2  0 With technological coefficients

aH1 aH 2  A    aL1 aL2  Problem  Linear programming model

Max z  P1 y1  P2 y2

aH1 y1  aH 2 y2  H s.t. aL1 y1  aL2 y2  L

y1, y2  0 Rybz. Theorem

aH1 aH 2  y1  H        aL1 aL2 y2   L  by Cramer’s Rule

aL2 H  aH 2 L aH1L  aL1H y1  , y2  aH1aL2  aL1aH 2 aH1aL2  aL1aH 2 Suppose industry 1 is more land – intensive. aH1 aH 2 Then,  aL1 aL2

aH1aL2  aL1aH 2  0 Then, for a 1 unit increase in Land (H).

y1 L2   0 output of y1 which is land intensive good increases. H A y a Also, 2   L1  0  y  H A 2

y labor intensive 2

y land intensive 1 in addition for y1  k1H  k2 L where

aL2 k1  aH1aL2  aL1aH 2 a k   H 2 then, 2 A

y1 y 1 H y1 H k1H  .  k1  y1 y1 H y1 k1H  k2 L

 k1H  k1H  k2 L So, when some factor endowment changes, the output that is intensive in that factor, will change in greater absolute proportion than the parameter. The Stolper Samuelson Theorem Effect of an increase in the price of output on the factor prices. Theorem. If the price of (say) labor intensive good increases, the price of labor will increase and in greater proportion. The price of the other factor falls, (but not necessarily in greater proportion).

Proof: Consider the unit factor cost of y1 and y2. remember we found ui as a shadow prices of factors. a11 u 1+ a 21 u 2 + a 31 u 3 =3(10) + 2(5) + 1(0) = 40 = P 1 a12 u 1+ a 22 u 2 + a 32 u 3 =2(10) + 2(5) + 1(0) = 30 = P 2

u3=0 ==> a 11 u 1 + a 21 u 2 = p 1

==>a12 u 1 + a 22 u 2 = p 2

轾a11 a 21 轾 u 1 轾 p 1 ==>犏犏 = 犏臌a12 a 22 臌 u 2 臌 p 2 Here the coefficient matrix is the transpose of technical coefficients matrix of the primal problem. So the algebra of the relations between the factor and output prices is virtually identical to the relations between physical output and resource endowment. 3 2 Solving the system with A  we get 2 2

p1 a 21

p2 a 222 p 1- 2 p 2 u1= = = p 1 - p 2 , a11 a 22- a 12 a 21 2

a11 p 1

a22 p 23 p 2- 2 p 1 3 u2= = = p 2 - p 1 a11 a 22- a 12 a 21 2 2

抖u1 a 11 u 2 a 12 P1 �, > = 0 and - < 0 抖P1 A P 1 A U 2

a11 u 1+ a 21 u 2 = p 1

a12 u 1+ a 22 u 2 = p 2

The Dual Problem Original problem  revenue was maximized constraints were statements of limited resource endowments. (Primal problem) shadow factor prices derived. z* u = i H z* L Recall    (envelope theorem) H H Lagrance multipliers in primal problem are shadow factor prices.

L= p1 y 1 + p 2 y 2 + u 1( H - a 11 y 1 - a 12 y 2) + u 2( L - a 21 y 1 - a 22 y 2) + u 3( K - a 31 y 1 - a 32 y 2 ) L =p1 - a 11 u 1 - a 21 u 2 - a 31 u 3�0, = 0, y 1 0 y1 L =p2 - a 21 u 1 - a 22 u 2 - a 32 u 3�0, = 0, y 2 0 y2 These are the non-positive profit conditions. The remaining foc. are the inequality constraints: L =H - a11 y 1 - a 21 y 2�0, >0, u 1 0 u1 L =L - a21 y 1 - a 22 y 2�0, = 0, u 2 0 u2 L =K - a31 y 1 - a 32 y 2�0, = 0, u 3 0 u3 For a competitive economy we expect revenue maximization to imply (and to be implied by) “minimization of total value of recourses subject to the constraint that profits are non-positive”. So, Rearranging Lagrange function we get;

L= Hu1 + Lu 2 + Ku 3 + y 1( p 1 - a 11 u 1 - a 21 u 2 - a 31 u 3) + y 2( p 2 - a 12 u 1 - a 22 u 2 - a 32 u 3 )

A minimization problem w.r.t. u1, u 2 , u 3 and objective function w= Hu1 + Lu 2 + Ku 3 , the total value of recourses.

L =H - a11 y 1 - a 21 y 2�0, if = 0, u 1 0 u1 L =L - a21 y 1 - a 22 y 2�0, if = 0, u 2 0 u2 L =K - a31 y 1 - a 32 y 2�0, if = 0, u 3 0 and, u3 L =p1 - a 11 u 1 - a 21 u 2 - a 31 u 3�0, if = 0, y 1 0 y1 L =p2 - a 21 u 1 - a 22 u 2 - a 32 u 3�0, if = 0, y 2 0 y2

Primal Problem Dual Problem

Max z  P1 y1  P2 y2 min w  HU1  LU 2  KU 3 s.t. a11 y1  a12 y2  H s.t a11U1  a21U 2  a31U 3  P1 a21 y1  a22 y2  L a12U1  a22U 2  a32U 3  P2 a31 y1  a32 y2  K

y1, y2  0 U1,U 2 ,U 3  0 obviously, when these two problems are solved, the values of objective functions are identical. Using the numerical example.

* * * z  P1 y1  P2 y2  40(14)  30(6)  740

* * * * w  HU1  LU 2  KU 3  54(10)  40(5)  35(0)  740 actually, by Euler’s Theorem z* z* z* z*  H  L  K  U *H U *L U *K H L K 1 2 3 The Fundamental Theorem of Linear Programming General Linear Programming Problem Primal Dual Max z  Px Min w  bU s.t. Ax  b s.t. ATU  P x  0 U  0 Theorem: suppose there exists an x0  0 such that Ax0  b (x0 is the feasible solution) and U 0  0 such that ATU 0  P (U0 is the feasible solution of minimum problem). Then both problems possess an optimal solution (i.e a finate solution) and these two solution values are identical. Meaning, suppose x*  0 is the vector that maximizes z  Px , such that Ax  b . The max value of Px is z  Px* Similarly, let U *  0 be the vector for which w  bU is minimum and w*  bU * . Then z*  w* .

Simplex Algorithm for Solving Linear Programming Models

Max z  40y1  30y2 s.t. 3y1  2y2  54

2y1  2y2  40

y1  2y2  35

y1, y2  0

let y3 , y4 , y5 be the slack variables, then problem becomes:

Max z  40y1  30y2 s.t. 3y1  2y2  y3  54 (1)

2y1  2y2  y4  40 (2)

y1  2y2  y5  35 Choose y1, y2 , y3 as basis. Solve y1, y2 , y3 in terms of y4 , y5

(2) & (3)  y1  y4  y5  5  y1  5  y4  y5

(3)  5  y4  y5  2y2  y5  35  y2  15  0.5y4  y5

(1)  y3  54  3(5  y4  y5 )  2(15  0.5y4  y5 )  9  2y4  y5

set y4  y5  0  we get y1  5, y2 15, y3  9 . This corresponds to point B in the previous diagram.

This is a solution but is it optimal? Does it maximize the objective function?

* z  40(5  y4  y5 )  30(15  0.5y4  y5 )  650  25y4 10y5 z*  10  0 (if y5 increases, max value increases so that is not optimal because by increasing y5 we can y5 increase z*).

Instead of choosing y5=0, we must increase it but how much? PAY ATTENTION

For y4=0.

y1  5  y5  0  y5 15

y2 15  y5  0  y5  9

y3  9  y5  0  y5  9 for y5  9 let y3 be out of the system.

New basis; y1, y2 , y5 and write in terms of y3 & y4

(6) y5  9  y3  2y4 and put into (4)

y1  5  y4  9  y3  2y4  14  y3  y4 8

(5) y2 15  0.5y4  9  y3  2y4   6 1.5y4  y3 (9) for y3  y4  0

y1 14, y2  6, y5  9 is it optimal? z*  40(14)  30(6)  740

* z  40(14  y3  y4 )  30(6  y3 1.5y4 )  740 10y3  5y4 it is definitely optimum. Deriving Shadow Prices

Let factor endowments be H, L, K where y3 and y4 are non basis variables.

3y1  2y2  H

2y1  2y2  L

y1  H  L 3 y  L  H 2 2 3 z*  40y  30y  40(H  L)  30( L  H ) 10H  5L 1 2 2 z* U *  10 1 H z* U *   5 2 L z* U *   0 3 K