Question

(HO model, with two goods and with two factors of production) Suppose that at current factor prices a ton of steel is produced using 20 hours of labor and 1 acre of land, and a ton of rice is produced using only 5 hours of labor and 1 acre of land. Suppose that the economy’s total resources are 600 hours of labor and 60 acres of land. Initially, PS = $600/ton & Pr =$300/ton. Draw the production possibility frontier with steel on the vertical axis and rice on the horizontal axis. ( 6 points) Write down the unit cost of producing one ton of steel and one ton of rice as a function of the hourly wage rate (w) and the rental rate (r). In a competitive market, these costs will be equal to the prices of steel and rice. Solve for the factor prices w and r. ( 7 points) Now suppose that the labor supply increases to 900 hours. State and prove the Rybczynski theorem. ( 6 points) Suppose Pr increases to $450/ton (from the initial situation).State and prove the Stolper-Samuelson theorem. (6 points)

Answer 1

Factor Goods	Land (Acre)	Labor (hours)
1 Ton of Steel	1	20
1 Ton of Rice	1	5

Again, let Quantity of steel produced be Q _S and Quantity of rice produced be Q _R . Therefore equations of resource constraint are -

• Q _S x 1 + Q _R x 1 =< 60 ...........for land
• Q _S x 20 +Q _R x 5 =< 600 ......for labor

Now when Q _R= 0 then Q _S=60 and when Q _S =0 then Q _R = 60 for land.

And Q _R= 0 then Q _S=30 and when Q _S =0 then Q _R = 120 for labor.

a.) The P P F in this H- O model is shown below,

Note : Figure 1 shows P P F when land and labor cannot be substituted for each other and it depicts maximum production that can be obtained given resource constraints.

Figure 2 shows P P F when land and labor can be substituted for each other and the curve becomes curved.Here we use prices of Steel and Rice.Given P _S =600 per ton and P _R = 300.

b.) The value of total Production in the economy is given by -

Maximize V = 600 x Q _S + 300 x Q _R

subject to -

• Q _S x 1 + Q _R x 1 =< 60 ...........for land
• Q _S x 20 +Q _R x 5 =< 600 ...
• and Q _S >=0 ,  Q R >=0

On solving for maximization we get

Q _S = 20 units and Q _R = 40 units.This implies

Total steel production consumes 20 acre of land and 400 labor hours.

Similarly, total rice production consumes 40 acre of land and 200 labor hours.

To find w and r ,we have

40 r + 200 w = 12000

20r+400w =12000

hence w= 20; r=200

Here Total cost of steel production = 12000

therefore unit cost = 12000/20=600 per unit.

Here Total cost of rice production = 12000

therefore unit cost = 12000/40=300 per unit.

--------------------------------------------------------------------------

I did not have sufficient time to answer last two questions.Please attach them again separately.