BUSINESS ECONOMICS, Production Functions

This section concentrates more on theory than on estimation, but we will get to the latter soon.

The advantages of using the Cobb-Douglas function for estimates of production are many -- and it is widely used in empirical work.  These advantages are listed below

        Cobb-Douglas # 1. According to whether the Sum of exponent terms in a Cobb-Douglas function is Greater Than, Less Than, or Equal To 1, this implies Increasing, Decreasing, or Constant returns to scale, respectively.

        Example:  Q = a*K^b*L^c = 2*K^(0.5)*L^(0.5)  

Since b + c  = 1, this implies constant returns to scale.  Say that K and L were both equal to 2.  Then, since 2^(0.5)*2^(0.5) = 2^(1.0) = 2, the value of Q from the production function must be 4.  Now let us double the inputs, so that K = L = 4.  Then we will find that the value of Q also doubles; it goes to 8.  

Now, we can verify that when b = 0.7 and c = 0.8, so that b + c = 1.5 > 1, that we will have increasing returns. This implies the following table:

Note in the above that when L and K double from 1 to 2, that Q more than doubles, from 2 to 5.657.  And when L and K again double, this time from 2 to 4, then Q again more than doubles, from 5.657 to 16.

Cobb-Douglas # 2.  The exponent for any input term in a Cobb-Douglas function represents the productive elasticity of that input.

Example: Q = a*K^b*L^c = 2*K^(0.5)*L^(1.5)  =>  log Q = (log 2) + 0.5*log(K) + 1.5*log(L)

(Note how taking logs allows the non-linear Cobb-Douglas form to be linearized, and thus be one that we can estimate by linear regression methods.)

Thus, the estimated coefficients for a linearized Cobb-Douglas function are also its estimated elasticities.

Here the exponent on L is c = 1.5.  This implies that, for any level of K, if L is increased by 1%, Q is increased by approximately 1.5%. Thus if L is increased by 10%, Q is increased by approximately 15%, ... etc.  An example is shown in the table below.  The first line shows that the elasticity prediction is for a 1.5%  It will be seen that the elasticity prediction is a little too pessimistic in this case.  The actual increase in Q, when L goes from 100 to 101, is an increase not by 300, as predicted by elasticity, but by 300.7.  But obviously the elasticity estimates are obviously close.  

Cobb-Douglas #3.  The fact that a production function shows constant or increasing returns to scale says nothing about whether or not there may be diminishing returns to a single input.

Returns to a single input will be increasing, decreasing, or constant according to whether the exponent on that input is greater than, less than, or equal to one.  Take, for example, the Cobb-Douglas function

Q = K^aL^b, where ^a and ^b are power terms; e.g., 

Q = K^(0.6)L^(0.4).

If the power terms add to 1 as above, there are constant returns to scale, i.e., doubling the inputs doubles the output. Try it: K=L=1 => Q=1, and K=L=2 => Q=2. But since the second partial derivative of each input is < 0, we have diminishing marginal products:

¶Q^2/(¶K)^2 = -0.6Q/K^2 < 0 , 

¶Q^2/(¶L)^2  = -0.4Q/L^2 < 0.

Or, just using Excel, we can determine that Q increases with each one unit increase in L, while K stays Fixed.  Q will indeed increase, but at a lower rate each time.  This declining change in Q is the diminishing marginal product.  One can construct the following table:

Cobb-Douglas #4.  The Derivative of a Cobb-Douglass function (i.e., its Marginal Product) is equal to the product of its exponent times its Average Product.

Example:  If Q = K^a*L^b  , then ¶Q/¶K =  a*Q/K

To show this, note that Q/K =  (K^a*L^b)/K  = K^(a-1)*L^b, and 

¶Q/¶K =  a*K^(a-1)*L^b

Cobb-Douglas #5.  (Really, for any function.)  A multiplicative function such as the Cobb-Douglas can be linearized so that it is possible to run linear regressions on it.

Example: Q = K^a*L^b  =>  log(Q) = log(K^a*L^b) = a*log(K) + b*log(L)

This has the further useful property that a = elasticity of K; i.e.  a = (dQ/dK)*(K/Q), and

                                                            and b = elasticity of L; i.e.  a = (dQ/dL)*(L/Q).

OTHER STUFF, not just on Cobb-Douglass

Note  the useful relationship between marginal and average products: where Average Product is at a maximum, the Marginal Product will be equal to this Average Product, with the Marginal curve crossing the Average curve from above. To see this, set the derivative of Average Product to 0, utilizing the rule for taking the derivative of a function with division terms:

¶ (Q(K,L)/L)/ ¶ L = (dQ/dL*L- Q)/L^2 = 0 => ¶ Q/¶ L*L = Q => ¶ Q/¶ L = Q/L

Or, in other words, MPL = APL.

The marginal conditions on an optimal mix of inputs are the same whether one is trying to

a) Maximize Output, subject to a Cost Constraint;

b) Minimize Cost, subject to an Output Constraint; or

c) Maximize Profit, with No Constraints.

Conditions a) and b) are exactly equivalent; each implies the other: a) <=> b). Condition c), on the other hand, implies the other two: c) => a) + b). That is, if c) is to hold true, then a) and b) must be met at every Q. As we'll see next chapter, this is why profit-seeking competition keeps the Marginal Cost curve as low as possible.

But a) + b) =/=> c), so a) and b) being true does NOT guarantee c). For c), one needs to take account of the effect a change in output may have on the price of that output, and hence on profits. But if the price of output is fixed, as in perfectly competitive conditions, (and in our linear programming problems) then this difficulty disappears, and we have perfect duality.

"INTEGRATING" PROBLEM:  The City Bus Company (CBC) has the following estimated Cobb-Douglas function, using monthly observations over the last two years:

            ln Q = 2.303 + 0.40 ln K  + 0.60 ln L  + 0.20 ln G

where Q is the number of bus miles driven, K (capital) is the number of busses operated, L (labor) is the number of bus drivers employed daily, and G is the gallons of gasoline used. Answer the following:

a)  Estimate Q if K = 200, L = 400, and G = 4,000.

b) Rewrite the estimated function in the form of a standard (non-linear) Cobb-Douglas Production function.  

c) Find the Marginal Product of Capital, Labor, and Gasoline at these input amounts as given.

MPK = .4*Q/K = 31.8617905823249

MPL = .6*Q/L = 23.8963429367437

MPG = .2*Q/G = 0.796544764558123

d) What is the output elasticity of K, L and G?  By how much does output increase if any input is increased by 10%?

Elasticities are simply the power coefficients, 0.4, 0.6, and 0.2 respectively.

Thus an increase of 10% in any input would mean output increases of

4%, 6% and 2% respectively.

e) What are the economies of scale?  What if all the inputs are increased simultaneously by 10%?

Since power terms add to 1.2, the output increase would be by 12%.

f) Say the CBC operates at the levels given, and that the rental price of a bus (r) is $40 a day, the wage of a driver (w) is $30 a day, and the price of gasoline (g) is $1 a gallon.  Is the firm using the optimal combination of inputs (to achieve a given output at minimum cost)?

The marginal conditions -- getting the same marginal output per marginal expenditure on each input ("getting the same bang per-last-buck-spent on each input" -- mean that all these terms must be equal:

MPK/r = 0.796544764558123

MPL/w = 0.796544764558123

MPG/g = 0.796544764558123 ..... and indeed they are.

g) If the average bus ride is 1 mile, what price should the company be charging for a ride (or per mile) in order to maximize profits?

Profit maximization will imply that Price*MPL = w, Price*MPK = r, and Price*MPG = g.   Taking the first partial differential with respect to ANY of the inputs K, L or G, we have that Price = w/MPL = r/MPK = g/MPG = 1.25542222, or $1.26 per mile.