Chapter 10: Firms and Technology 10.1: Introduction

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Chapter 10: Firms and Technology

10.1: Introduction

So far we have worked in a world without production – in a pure-exchange world where people have initial endowments and might want to exchange them. We must now add production to the story. Production is important in the real world. Production exists because people see that there are profits to be made by producing things that do not exist but which people want to buy.

Usually, but rather conventionally, we say that production is done by firms. Firms are there to produce. Indeed we can define a firm as an organisation that buys certain things, does something to them and then sells the resulting product. Rather trivially we can say that a firm buys inputs and converts the inputs into output through some production process and then sells the resulting output.

Usually a firm produces several outputs using several inputs, but we will be able to do the analysis that we need with a firm that produces one output with two inputs. All the results that we derive can be extended but this (one output, two input) world is sufficiently simple to make the analysis relatively easy.

The purpose of this chapter is to explore and define the relationships between the output and the two inputs. In the subsequent chapters we explore the implications.

10.2: Production Functions

In this chapter a firm is an organisation that buys two inputs and transforms them into an output. Let us denote the two inputs by input 1 and input 2, and let us denote the quantities of each by q

1 and q

2 .

Let us denote the quantity of output produced by y . Let us assume that both the inputs are genuine inputs in the sense that the higher the quantity of each used the higher is the output of the firm.

Central to the firm is the production process – which converts inputs into output. In general this can be described by a production function which we can write as y = f(q

1

,q

2

) (10.1) and where the function is increasing in both its arguments. We will find it useful to describe this function in two ways – as there are, in a sense, two dimensions to the description of the production process. The first is the relationship between the two inputs in the production process for a given scale of production; the second is the relationship between the scale of the inputs and the scale of the output.

To understand the relationship between the two inputs in the production process it is useful to introduce the concept of an isoquant . An isoquant is a curve in (q

1

, q

2

) space for which the level of output is constant. It is formally defined by f(q

1

,q

2

) = constant

We shall study this intensively later.

To understand the relationship between the scale of the inputs and the scale of the outputs, we ask what happens when we change the scale of production – moving from (q

1

, q

2

) to (sq

1

, sq

2

) – where s is some scaling factor, that is, we mutiply the quantity of both inputs by s . What happens?

Clearly output changes from f(q

1

,q

2

) to f(sq

1

, sq

2

) . The interesting question is whether the output is scaled up in the same proportion. In particular, is f(sq

1

, sq

2

) greater than, equal to or less than sf(q

1

,q

2

)

That is, is the output multiplied by the same amount as the two inputs?

Let us introduce some terminology: in the case when f(sq

1

, sq

2

) is equal to sf(q

1

,q

2

) , we say that the production process exhibits constant returns to scale – multiplying the inputs by s causes output to be multiplied by s. In contrast when f(sq

1

, sq

2

) is less than sf(q

1

,q

2

) , we say that the production process exhibits decreasing returns to scale – multiplying the inputs by s causes output to be multiplied by less than s. Finally when f(sq

1

, sq

2

) is more than sf(q

1

,q

2

) , we say that the production process exhibits increasing returns to scale – multiplying the inputs by s causes output to be multiplied by more than s.

It should be emphasised that these two dimensions (the shape of the isoquants and the returns to scale) exist independently of each other – we can have any shaped isoquant combined with different returns to scale, and we can have any kind of returns to scale combined with any shaped isoquant.

Let us therefore study them separately, beginning with different kinds of isoquant maps.

10.3: Isoquants

As we have already noted, an isoquant is the set of points in (q

1

, q

2

) space for which the level of output is constant. It is formally defined by f(q

1

,q

2

) = constant

The shape of an isoquant depends upon the relationship between the two inputs in the production process. Usually, but not always, an isoquant is convex reflecting the usual empirical fact that the two inputs are usually more efficiently used together. Moreover, there is an isoquant through every point in (q

1

, q

2

) space. So the usual case is a family of isoquants of the following form:

Isoquants may well remind you of indifference curves and indeed they have much in common. You might like to reflect on the fact that the consumer in chapters 3 to 7 of this course buys goods and

converts them into utility while the firm in this chapter buys inputs and converts them into output.

The analogy is almost complete – there is only one slight difference: output is measurable while utility is not. You might like to reflect on the implications.

One thing which we never explicitly proved in the material on indifference curves is that indifference curves cannot cross. The same is true for isoquants – perhaps it would be useful to prove here that they cannot cross. Consider the following:

In this figure there are two isoquants which cross at the point X. Is this possible? Consider first the isoquant which passes though the point B – according to it the output at point X is the same as at point B. Now consider the isoquant which passes through the point R – according to it the output at the point X is the same as at the point R. It follows that the output at B must be the same as the output at R – which is contradictory to the assumption that both inputs are useful to the firm in the sense that more of either leads to more output. (Note that at B there is more of both inputs than at

R.) We have a contradiction – which shows that isoquants can not cross.

The actual shape of isoquants is determined by the relationship of the two inputs in the production process. We can identify a number of different possibilities. The relevant one in any particular instance is an empirical issue – determined by the good being produced and the nature of the two inputs.

One simple case is when the two inputs are perfect 1:1 substitutes . This means that they are identical and can be used interchangeably in the production process

1 . The implied isoquants are in

this case given by:

1 In the UK in certain occupations the law says that this has to be the case when the two inputs are men and women.

Take any one of these – say the third, which goes from (60, 0) to (0, 60), This isoquant tells us that the same quantity of output can be produced with any combination of the two inputs along the line.

So 60 units of input 1 and 0 of input 2 produce a given output; the same output is produced with 59 units of input 1 and 1 of input 2; or with 58 units of input 1 and 2 units of input 2; or …; or with 1 unit of input 1 and 59 of input 2; or with 0 units of input 1 and 60 units of input 2. Every 1 unit less of input 1 needs to be substituted with 1 unit of input 2 to keep output constant; every 1 unit less of input 2 needs to be substituted with 1 unit of input 1 to keep output constant. Note that the slope of the isoquants is everywhere equal to –1. Note also that the magnitude of the slope indicates the rate at which input 1 needs to be substituted with input 2 to keep output constant. This is given the name

the marginal rate of substitution 2 . For perfect 1:1 substitutes this

mrs is everywhere equal to 1.

Of course, two inputs can be perfect substitutes – but not necessarily one-to-one. For perfect 1:2 substitutes (where everywhere 1 unit of input 1 is substitutable with 2 units of input 2, and 2 units of input 2 with 1 unit of input 1) we have the following isoquants:

Note that the mrs is everywhere equal to 2. The general perfect substitutes case is perfect 1:a substitutes in which the mrs is everywhere equal to a and where the isoquants all have slope –a .

The polar extreme to perfect substitutes is perfect complements. These could be 1-with-1 – that is where 1 unit of input 1 has always to be used with 1 unit if input 2

3 – for which the isoquant map is:

Of course we can have perfect 1-with-2 complements in which the isoquant map is

2 Or sometimes the marginal technical rate of substitution.

3 The classic example is men digging holes with spades. Every man needs a spade and every spade needs a man.

More generally we could have perfect 1-with-a complements for which 1 unit of input 1 should always be combined with a units of input 2, and for which the corner points lie along the line q

2

= aq

1

.

A case intermediate between these two extremes is the case of Cobb-Douglas technology. An isoquant here is given by q

1 a q

2

(1-a) = constant (10.2) the shape of which obviously depends on the parameter a . For example when a = 0.5 we have

Note the symmetry of the figure and also note how the mrs varies across the figure. A nonsymmetric case is when a = 0.3:

Obviously whether technology is perfect substitutes, perfect complements or Cobb-Douglas is an empirical issue. It has been found that some production functions can be well approximated by one

of these with an appropriate choice of the parameter a . Other production functions are more complicated and need more complicated formulae. This book can not present all the possibilities considered by economists but we should at least mention the Constant Elasticity of Substitution case, which appears to have good empirical credentials. An example is given here (the title gives the values of the parameters – which will mean a little more when we present a formula in the next section).

10.4: Production Functions and Returns to Scale

The isoquant map determines the relationship between the two inputs in the production process.

However the isoquant map does not determine the returns to scale of the two inputs. This is given by the production function. We have already noted that there are these two different dimensions in the production function. Let us be more specific.

Consider the case of perfect 1:1 substitutes. This is defined by the isoquant map in figure 10.4. An isoquant is defined by q

1

+ q

2

= constant (10.3)

We note that this is consistent with different returns to scale. Consider, for example, the production function y = f(q

1

, q

2

) = q

1

+ q

2

(10.4)

This has isoquants given by (10.3) and also displays constant returns to scale . Why? Well if we multiply both the inputs by s then the output gets multiplied by s.

What about the production function y = f(q

1

, q

2

) = (q

1

+ q

2

) 0.8

(10.5)

This has isoquants given by multiplied by s 0.8

(q

1

+ q

2

) 0.8

= constant , that is, by (q

1

+ q

2

)

– that is we have decreasing returns to scale .

= constant, that is, by

(10.3). What about its returns to scale? Well, if we multiply both the inputs by s then the output is

More generally consider the production function: y = f(q

1

, q

2

) = (q

1

+ q

2

) s (10.6)

This displays perfect 1:1 substitute technology and increasing, constant or decreasing returns to scale depending upon whether the parameter s is greater than, equal to or less than 1.

You should notice what is happening: the technology determines the shape of the isoquants but the returns to scale determines the output associated with each isoquant. With the production function

(10.3) the levels of output associated with the 9 isoquants in figure 10.4 are 20, 40, 60, 80, 100, 120,

140, 160 and 180. You will see that every doubling in the scale of production doubles the implied output. However if we take the production function (10.5) – which we have already established has decreasing returns to scale - we have levels of production associated with the 9 isoquants in figure

10.4 equal to 10.99, 19.13, 26.46, 33.30, 39.81, 46.06, 52.11, 57.98 and 63.71 – doubling the scale of production less than doubles the level of output. Finally if we take the general form (10.6) and put the parameter s equal to 2 – which gives us perfect 1:1 substitutes with increasing returns to scale – we have levels of output associated with the 9 isoquants in figure 10.4 equal to 400, 1600,

3600, 6400, 10000, 14400, 19600, 25600 and 32400 – clear increasing returns to scale.

A quasi-general

4 case of perfect

1:a substitutes the production function is given by y = f(q

1

, q

2

) = (q

1

+ q

2

/a) s (10.7)

Here the parameter a determines the rate of substitution and the parameter s determines the returns to scale ( s is greater than, equal to or less than 1 according as the function displays increasing, constant or decreasing returns to scale). Note that the parameters a and s can obviously be chosen completely independently of each other.

For perfect 1-with-1 complements indifference curves are given by: min(q

1

, q

2

) = constant (10.8)

Obviously consistent with this is the production function y = f(q

1

, q

2

) = min(q

1

, q

2

) (10.9) which displays constant returns to scale. But so also is the more general form y = f(q

1

, q

2

) = [min(q

1

, q

2

)] s (10.10) which displays increasing, constant and decreasing returns to scale according as the parameter s is greater than, equal to or less than 1.

A quasi-general

5 case of perfect

1-with-a complements the production function is given by y = f(q

1

, q

2

) = [min(q

1

, q

2

/a)] s (10.11)

Here the parameter a determines the rate of complementarity and the parameter s determines the returns to scale ( s is greater than, equal to or less than 1 according as the function displays

4 Obviously an even more general case of perfect substitutes is y = f(q any increasing function.

1

, q

2

) = g(q

1

+ q

2

/a) where the function g(.) is

5 Obviously an even more general case of perfect complements is y = f(q g(.) is any monotonically increasing function.

1

, q

2

) = g(min[q

1

, q

2

/a]) where the function

increasing, constant or decreasing returns to scale). Note that the parameters a and s can obviously be chosen completely independently of each other.

The Cobb-Douglas case is interesting. We have already noted that an indifference curve is given by q

1 a q

2

(1-a) = constant

From this we get the general Cobb-Douglas case with a production function given by y = f(q

1

, q

2

) = A q

1 a q

2 b (10.12)

We note that the relative weights on input 1 and input 2 are a/(a+b) and b/(a+b) respectively, and what determines the returns to scale is the sum (a+b) . To be specific:

The general Cobb-Douglas production function (10.12) has increasing, constant or decreasing returns to scale depending on whether (a+b) is greater than, equal to or less than one.

You should check out this result.

The constant-elasticity of substitution case that we introduced above has the general production function: y = (c

1 q

1

ρ + c

2 q

2

ρ ) -s/ ρ (10.13)

We note that the parameter s determines the returns to scale (exactly as above) and the parameter ρ

is the constant elasticity of substitution 6 .

10.5: The Marginal Rate of Substitution and Marginal Products

For future reference we should conclude by defining the marginal product of each input. We can then relate the marginal rate of substitution to the two marginal products.

The marginal product of a particular input is simply the rate at which output increases when that input increases. If we were to draw a graph of output as a function of that particular input (with the quantity of the other output fixed) it would, in general, be an increasing function of that input and would generally be a concave function – as the input rises then so does the output but less than proportionately

7 . At any point on the graph the

marginal product of that input is simply the rate at which the output rises when the input rises – it is the slope of the curve. Mathematically

8 , the two

marginal products are defined by: the marginal product of input 1 = ∂ y/ ∂ q

1

= = ∂ f(q

1

, q

2

)/ ∂ q

1 the marginal product of input 2 = ∂ y/ ∂ q

2

= = ∂ f(q

1

, q

2

)/ ∂ q

2

(10.14)

6 You might like to know that the elasticity of substitution is d(ln(q

2

))/d(ln(q

1

)) which is equal to ( dq

2

/dq which is the marginal rate of substitution (the mrs ) divided by q

2

/q

1 that the mrs divided by q

2

/q

1 is constant – which implies that the mrs is proportional to q

1

)/(q

2

/q

1

)

. If the elasticity of substitution is constant it follows

7 Obviously this is not always true but it is generally true – it is very rare that we get increasing returns to a single input.

8 If y is a function of just one variable, we denote the derivative of y with respect to x by dy/dx , and if y is a function of several variables x

1

, x

2

, … then we denote the derivates of y with respect to x

1

, x

2

, … by ∂

2

/q

1

.

y/ ∂ x

1

, ∂ y/ ∂ x

2

,…

We can now relate the marginal rate of substitution to these marginal products. The proof is given in the Mathematical Appendix and yields the following: dq

2

/dq

1

= - [ ∂ f(q

1

, q

2

)/ ∂ q

1

]/ [ ∂ f(q

1

, q

2

)/ ∂ q

2

] which says that the slope of an isoquant ( dq

2

/dq

1

) is equal to minus the ratio of the marginal product of input 1 to the marginal product of input 2. So we get the important result that: the marginal rate of substitution = the ratio of the marginal product of input 1 to the marginal product of input 2

You might like to think about this. Perhaps it is simplest to take a particular example. Suppose we are at a point where the marginal product of input 1 is 5 and the marginal product of input 2 is 2 .

This means that an increase of 1 unit in input 1 increases output by 5 units and an increase of 1 unit in input 2 leads to an increase in output of 2 units. Now ask yourself – if we decrease the amount of good 1 by 1 unit how much do we have to increase the employment of good 2 to keep output unchanged? Well, decreasing input 1 by 1 unit will lower output 5 units (as the marginal product of input 1 is 5). To compensate we need to increase the employment of input 2 by 2.5 units (as each increase of 1 unit of good 2 increases output by 2). So to compensate for a fall in input 1 by 1 unit we need to increase input 2 by 2.5 units. Hence the mrs is 2.5.

10.6: Concave and Convex Technology

Throughout this chapter we have rather implicitly assumed that the technology is convex in the sense that the isoquants are convex. This is probably empirically more common than the contrary case – when the isoquants are concave. This latter is possible but unlikely. Here is an example:

You should think what this kind of technology implies. (Do remember that we are still assuming that both of the inputs are proper inputs in that more of either implies more output.) The technology implies that the two inputs are more efficiently employed separately rather than together. To see this, suppose that we have 100 units of input 1 and 100 units of input 2 and we are thinking of having two firms using these inputs. What is best: to have one firm use 100 of input 1 and 0 of input

2 and the other to use 0 of input 1 and 100 of input 2 – or to have both firms use 50 of input 1 and

50 of input 2? In which circumstance would total output be higher? Clearly in the first. We get the conclusion we stated above – the inputs are better used separately rather than together. You might

like to ask yourself whether this is likely to be empirically valid 9 . Let me note that in practice we

9 It creates no real problem to our subsequent analysis – in fact it almost trivialises it to such an extent that there is really no problem left to consider.

usually observe firms in an industry all employing the same inputs, rather than some firms using one input and other firms using others.

10.7: Summary

As you have been reading this chapter you probably had a feeling of deja vu . This is hardly surprising as much of what we have done is familiar from the material of chapter 5 – on the preferences of the individual. There are strong parallels – to which we have already alluded. The consumer buys goods and produces utility; the firm buys inputs and produces output. The mapping from goods to utility for the consumer is represented through indifference curves; the mapping from inputs to output for a firm is represented through isoquants. The main difference is that utility is not measurable whereas output is.

In this chapter we started with production functions which show the relationship between inputs and output. We then introduce the concept of an isoquant in (q

1

, q

2

) space.

Isoquants show the locus of input combinations for which output is constant.

We showed that isoquants cannot cross and introduced the marginal rate of substitution which measures the slope of an isoquant.

We then considered some special cases.

The isoquants for perfect substitutes are parallel straight lines.

The isoquants for perfect complements are in the form of an L.

We argued that convex technology is much more common in practice than concave, and we gave two examples of convex technology which appear to have empirical validity: the Cobb-Douglas and the CES technologies.

Finally we defined returns to scale.

The technology has increasing, constant or decreasing returns to scale depending upon whether, when we increase the scale of inputs by s, the output goes up by more, the same or less than this scale factor s.

We saw that the returns to scale and the type of technology (the shape of the isoquants) are completely independent attributes of the production function.

10.8: Mathematical Appendix

We find the expression for the marginal rate of substitution, that is, the slope of an isoquant.

Recall that an isoquant is given by f(q

1

, q

2

) = constant

If we take the total derivative of this expression we get

[ ∂ f(q

1

, q

2

)/ ∂ q

1

]dq

1

+ [ ∂ f(q

1

, q

2

)/ ∂ q

2

]dq

2

= 0

from which it follows that dq

2

/dq

1

= - [ ∂ f(q

1

, q

2

)/ ∂ q

1

]/ [ ∂ f(q

1

, q

2

)/ ∂ q

2

] which is the result stated in the text.

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