Chapter 11: Cost Minimisation and the Demand for Factors

Transcription

1 Chapter 11: Cost Minimisation and the Demand for Factors 11.1: Introduction We assume a very simple objective for firms namely, that they want to maximise profits 1. We will explore the implications of this in this chapter and the next two. We assume that the firm wants to choose its output and the quantity of the two inputs in such a way that profits are maximised. This problem can be considered in two parts: the choice of the optimal output and then, given that, the optimal choice of the two inputs to produce that output. We can split the problem up in this way since the two parts are independent in particular, whatever choice of output is made, the firm wants to produce that output in the most efficient way. To maximise its profits the firm must minimise its costs of producing whatever output it chooses to produce. This chapter considers the problem of choosing, for any given level of output, the cheapest way of producing that output. That is, we find in this chapter the cost-minimising input combination for any given level of output. This will enable us to identify, for any given level of output, the lowest cost of producing that output. This, for given output y, we will denote by C(y), and will be referred to as the cost function of the firm. We will study its properties in chapter 12. Then, in chapter 13, we will use this cost function to find the profit maximising output for the firms. 11.2: Isocost curves To help us find the cost-minimising input combination for producing any given level of output we introduce a useful expository and analytical device that of an isocost curve. This is a curve in the space we used in chapter 10 namely (q 1, q 2 ) space the space of input combinations. As you might be able to anticipate an isocost curve is a set of points in this space for which the cost of using that particular combination is constant. If we denote the price on input 1 by w 1 and that of input 2 by w 2, then the cost of purchasing the combination (q 1, q 2 ) is simply w 1 q 1 + w 2 q 2. An isocost curve is therefore defined by: w 1 q 1 + w 2 q 2 = constant (11.1) This is clearly a straight line in (q 1, q 2 ) space (it is linear in q 1 and q 2 ). If we rewrite the equation as q 2 = constant (w 1 /w 2 ) q 1 we see that it has slope equal to w 1 /w 2. So an isocost curve in (q 1, q 2 ) space is a straight line with slope equal to w 1 /w 2. Through each point in the space passes such an isocost curve and we can picture some of these in a figure: 1 For firms quoted on the stock exchange this is equivalent to maximising the value of the firm to the shareholders.

2 It is assumed in this figure that w 1 = w 2 = 1, for which w 1 /w 2 is equal to 1 so that all the isocost curves have slope 1. For each isocost curve we can calculate the cost. For the lowest curve going from (20, 0) to (0, 20) - the cost of each combination along the curve at prices w 1 = w 2 = 1 is obviously 20 (20 units of 1 and 0 of 2; 19 units of 1 and 1 of 2; 18 units of 1 and 2 of 2;, 1 unit of 1 and 19 of 2; 0 units of 1 and 20 of 2). For the second curve the cost everywhere along it is 40. For the third curve 60. And so on. Note, rather obviously, that the cost rises as we move outwards from the origin, and falls as we move inwards towards to the origin. If, instead of w 1 = w 2 = 1 the prices were w 1 = w 2 = 2, we would have exactly the same isocost curves but the costs associated with them would differ (40, 80, etc. instead of 20, 40, etc.). 11.3: The Cheapest Input Combination We can now proceed to find the cheapest way of producing any given level of output. Clearly the level of output that we want to produce has associated with it an isoquant. Suppose it is the one pictured in the next figure. Here I am assuming a symmetric Cobb-Douglas technology with a level of output equal to 40. So the equation of the curve below is q q = 40. (Note that the point (40, 40) is on this isoquant.) We ask the question: which is the cheapest point on this curve (on this isoquant)? Obviously the one on the lowest isocost curve. Let us put together the above two figures:

3 Clearly the point indicated with the asterisk is on the lowest possible isocost curve (consistent with producing the level of output indicated by the isoquant curve). We note in this example which is a nice symmetrical example - the optimal point is itself symmetrical, with 40 units of each input and a cost of 80. Note that there is no other point on the isoquant where the cost is lower. You may note the optimising condition 2. If we look at the asterisked point what do we see? That at that point the isocost line is tangential to the isoquant. It follows that the slope of the isocost line must be equal to the slope of the isoquant at the asterisked point. But we know these two slopes; the first is simply minus the ratio of the input prices (w 1 /w 2 ) and the second is minus the marginal rate of substitution (the mrs). So the condition for the optimal input combination is: w 1 /w 2 = mrs (11.2) 11.4: Input Demands as Functions of Input Prices and of Output We have now solved the problem in principle for any given input prices and for any given level of output we can now find the firm s cost-minimising combination of the inputs. More specifically, for any given input prices and for any given level of output we can now find the firm s demand for the two factors. In the example above, with prices 1 and 1 and a level of output 40, the firm demands 40 units of each input. Obviously these demands depend upon the technology as we shall see. We are also in a position to see how the input demands vary when (a) input prices vary and (b) when the desired output varies. Clearly once again the answers depends upon the technology. Let us stay with the symmetric Cobb-Douglas technology for the moment. Let us keep the desired level of output constant at 40 and let us keep the price of input 2 constant at w 2 = 1. Let us vary the price of input 1, w 1, starting at 1/4, and then going to 1/3, to 1/2, to 1, to 2, to 3 and finally going to 4. As you will be able to anticipate the slope of the isocost lines will vary starting at -1/4, and then going to -1/3, to -1/2, to -1, to -2, to -3 and finally going to -4. In the figure that follows you will see the first of these cases pictured and the slope of the isocost curves are all 1/4. 2 Obviously this condition assumes that the isoquant is smoothly convex and that the optimal point it not an extreme point. If these are not true then neither will be the optimising condition that follows.

4 In this case the optimal input combination is (80, 20) 80 units of input 1 and 20 of input 2. The firm takes advantage of the fact that in this scenario the price of input 1 is very low and buys lots of input 1 and very little of input 2. This turns out to be a cheaper way of producing the output 40. (Note that the combination (40, 40) at prices w 1 = 1/4 and w 2 = 1 would cost 50, while the combination (80, 20) costs just 40.) You can explore yourself what happens when the price of input 1 rises. The isocost curves get progressively steeper and the optimal point rotates around the isoquant. In particular, when w 1 = 1/3 the optimal input combination is (69.28, 23.09), when w 1 = 1/2 the optimal combination is (56.57, 28.28), when w 1 = 1 the optimal combination is (40, 40), when w 1 = 2 the optimal combination is (28.28, 56.57), when w 1 = 3 the optimal combination is (23.09, 69.28) and when w 1 = 4 the optimal combination is (20, 80). You may find it helpful to verify these in the figure above. If we now plot the optimal demands against the price of input 1 we get the input demand functions as functions of the input price. In this instance we get the following: Note that in this figure the price of input 1 is on the horizontal axis. The downward sloping curve (rather obviously) is the demand for input 1 and the upward sloping curve is the demand for good 2. As the price of input 1 rises the firm reduces the demand for input 1 and replaces it with an increased demand for good 2. The firm substitutes input 1 with input 2 as input 1 gets more expensive. We note that these input demand functions depend upon the technology. If, instead of having a symmetric Cobb-Douglas technology we had a non-symmetric technology (with weights 0.3 and 0.7) the problem to solve would be as follows (note the different position of the isoquant):

5 and the solution in terms of the implied input demand functions (as functions of the price of input 1) would be: If you compare this figure with figure 11.5 you will see the differences. For reference purposes later it may be found useful to give the general formula for the optimal input demands for the general Cobb-Douglas case, where the production function is given by y = f(q 1, q 2 ) = A q 1 a q 2 b that is, by equation (10.12). The optimal input demands are found by minimising w 1 q 1 + w 2 q 2 with respect to y = A q 1 a q 2 b, that is, by minimising the cost of producing the level of output y. The proof can be found in the Mathematical Appendix. The resulting input demands are as follows. Be prepared for a bit of a shock these equations look rather nasty but we will explain what they say. q 1 = (y/a) 1/(a+b) (aw 2 /(bw 1 )) (b/(a+b)) and q 2 = (y/a) 1/(a+b) (bw 1 /(aw 2 )) (a/(a+b)) (11.3) While you are not expected to be able to derive these, you should be able to interpret them. More importantly you should be able to understand the economics behind the interpretation. There are three variables that influence the demand for the two inputs the desired level of output and the prices of the two inputs. Let us consider these in turn. 1) The effect of desired output on input demand: You will note that in both expressions in (11.3) input demands are proportional to y 1/(a+b) so as y increases then so do the input demands. Moreover what is crucial to the way that these demands increase with y is the sum of the parameters (a+b). But we already know what this sum indicates what kinds of returns to scale the technology exhibits. Recall from chapter 10 that the technology exhibits increasing, constant or decreasing returns to scale according as to whether the sum (a+b) is greater than, equal to or less than 1. Now the cxponent on y is 1/(a+b) this is less than, equal to or greater than 1 according as (a+b) is greater than, equal to or less than 1. So we have the result that the input demands are concave, linear or convex in the desired output according as the technology shows increasing, constant or

6 decreasing returns to scale. The constant returns to scale case is simple: in this to double the output we need to double the scale of the inputs so the input demands double: the relationship between output and input demand is linear. When instead we have increasing (decreasing) returns to scale we need to less (more) than double the input demands the relationship is concave (convex). 2) The effect of w 1 on input demand: If we look at the input demand functions (11.3) we see that in the demand for input 1, the term w 1 appears with exponent b/(a+b). This is negative which means that as w 1 increases then the demand for input 1 falls as the input becomes more expensive less of it is bought. Note also that the magnitude of the exponent is between 0 and 1, so the demand falls less and less as the price rises. If we look at the input demand functions (11.3) we see that in the demand for input 2, the term w 1 appears with exponent a/(a+b). This is positive which means that as w 1 increases then the demand for input 2 rises as one input becomes more expensive more of the other is bought. Note also that the magnitude of the exponent is between 0 and 1, so the demand rises less and less as the price rises. This is exactly what the figure above shows. 3) The effect of w 2 on input demand: If we look at the input demand functions (11.3) we see that in the demand for input 1, the term w 2 appears with exponent b/(a+b). This is positive which means that as w 2 increases then the demand for input 1 rises as one input becomes more expensive more of the other is bought. Note also that the magnitude of the exponent is between 0 and 1, so the demand rises less and less as the price rises. If we look at the input demand functions (11.3) we see that in the demand for input 2, the term w 2 appears with exponent -a/(a+b). This is negative which means that as w 2 increases then the demand for input 2 falls as the input becomes more expensive less of it is bought. Note also that the magnitude of the exponent is between 0 and 1, so the demand falls less and less as the price rises. These results are obviously just the converse of those discussed under 2) above. You might also like to think about the effect of a and b on the input demands. The technology obviously influences the input demands. If we take a different technology we get different demands. Consider, for example, perfect 1:2 substitutes. We begin with the usual isoquant/isocost analysis. Here we have a price of input 2 equal to 1 and a price of input 2 equal to 1/4. The desired output level is 40. I am assuming a production function of the form of (10.7) in chapter 10 with a = 2 and s = 1. Clearly the optimal point is where the firm uses just 40 units of input 1 and 0 units of input 2. Now consider what happens when the price of input 1 rises from 1/4 to 1/3, to 1/2, to 1, to 2, to 3 and finally to 4. When the price is less than 2 it is obviously best for the firm to continue to buy 40 units of input 1 and 0 of input 2. When the price gets to 2, the isocost and the isoquant coincide so any combination that produces the desired quantity of 40 is optimal. Above a price of 2 for input 1

7 it is best for the firm to buy only input 2 (80 units of it) and 0 of input 1. We thus get the following demand functions: The line which is horizontal at 40 until the price of 2 and then zero is the demand for input 1; the line which is zero until the price of 2 and then horizontal at 80 is the demand for input 2. Note that this case is perfect 1:2 substitutes which is why the price of 2 for input 1 is critical. (Remember that the price of input 2 is fixed at 1.) You might be able to verify the following demand functions for the general 1:a substitutes case, the production function for which is given by (10.7) is: if w 1 < aw 2 then q 1 = y 1/s and q 2 = 0 if w 1 > aw 2 then q 1 = 0 and q 2 = (ya) 1/s (11.4) You might like to reflect on the fact that with the Cobb-Douglas technology the substitution from input 1 to input 2 is continuous whereas with perfect substitutes it is all-or-nothing. The opposite extreme of no substitution is to be found in the perfect complements case. For example, with perfect 1-with-2 complements we have Notice what happens when the isocost curves change their slope the optimal combination is always at the asterisked point. So we get the following rather boring input demand functions as functions of the price of input 1 (the top one is the demand for input 2, the bottom the demand for good 1):

8 In the general perfect 1-with-a complements case, whose production function is given by the equation (10.11) y = [min(q 1, q 2 /a)] s it is clear that at the optimal point q 2 = aq 1 and at this point q s 1 = (q 2 /a) s = y from which it follows that the optimal input demands must be given by: Note the effect or returns to scale on the input demands. q 1 = y 1/s and q 2 = ay 1/s (11.5) Perhaps for completeness we should include one case of Constant Elasticity of Substitution technology if only to emphasise that the input demands depend upon the technology. If we consider the effect of changing the input price on the demands, we start with the isocost/isoquant analysis: from which we can derive in the usual way the input demands as functions of the price of input 1:

9 We can also find the effect of the desired output on input demands (at fixed prices): You will notice that these curves are linear. What can you infer as the returns to scale in the assumed technology? True that they are constant. More generally, it can be shown that the input demand functions for the general CES case (the production function for which is given in (10.13)) are given by the following expressions: q 1 s = y(c 1 /w 1 ) 1/(1+ρ) ((c 1 w 1 ρ ) 1/(1+ρ) +(c 2 w 2 ρ ) 1/(1+ρ )) q 2 s = y(c 2 /w 2 ) 1/(1+ρ) ((c 1 w 1 ρ ) 1/(1+ρ) +(c 2 w 2 ρ ) 1/(1+ρ )) (11.6) It would be a useful exercise for you to think about the implied relationship between the input demands and the input prices. As far as the effect of output on the input demands is concerned you should note that the relationship is increasing (obviously) and concave, linear or convex depending as to whether the technology exhibits increasing, constant or decreasing returns to scale. 11.5: Summary We started with defining an isocost curve. An isocost curve is the locus of points in quantity space where the cost of producing that input combination is constant. We then found, for technologies where the isoquants are smoothly convex that: The optimal input combination is (generally) where the MRS is equal to the ratio of the input prices. Obviously for non-smoothly convex isoquants this may not be true but we should always be able to find the point on the desired isoquant which is on the lowest possible isocost curve. In fact we did this for the cases of perfect substitutes and perfect complements. For perfect substitutes the cheapest input is used. For perfect complements both inputs are used. We found in general an important result concerning optimal input demands:

10 Input demands are a decreasing function of their own price and an increasing function of the price of the other input. We also found that: Input demands rise with output being concave, linear or convex if the technology displays everywhere increasing, constant or decreasing returns to scale. 11.6: Mathematical Appendix We present the derivation of the input demand functions for the Cobb-Douglas case. The problem is to find the combination (q 1, q 2 ) that minimises the cost w 1 q 1 + w 2 q 2 subject to the condition that the desired output is produced that is, subject to y = A q 1 a q 2 b. There are various ways to solve this but perhaps the simplest is to use the method of Lagrange and form the Lagrangian: L = w 1 q 1 + w 2 q 2 + λ(y - A q 1 a q 2 b ) We then minimise the Lagrangian with respect to q 1, q 2 and λ. The optimality conditions are: Solving these gives us the equations in the text. dl/dq 1 = w 1 - λaaq 1 a-1 q 2 b = 0 dl/dq 1 = w 2 - λabq 1 a q 2 b-1 = 0 dl/dλ = y - A q 1 a q 2 b = 0

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