Specialization


Numerical example

Suppose education improves productivity (wages) at a decreasing rate. Specifically, let the education-wage relationship be the following:

educational expense     daily wage
__________________________________
0                       $100
$2,000                  $120
$4,000                  $130
Let's say a worker without any education is spending 150 days on activity A and 150 days on activity B. His total income is $30,000.

Suppose this worker is interested in improving his situation by investing in education. He invests $2,000 on activity A and $2,000 on activity B, and divides his time equally as before. Total earnings are now $36,000. Net income is $32,000, which is better than before. (It is not worth investing $4,000 in both activities, because net income would become $31,000.)

But he can do even better than that. Instead of investing in both activities, suppose he only acquires knowledge about activity A. His daily wage in activity A will be $120 while his wage in activity B will remain at $100. Obviously, this person will now spend all 300 days of work on activity A. His total earnings are still $36,000, but net income is now $34,000.

Once the person specializes, it is advantageous for him to invest even more. For example, investing $4,000 in activity A will generate a net income of $35,000.

The actual benefits may be even bigger than this. With a daily wage of $130 instead of $100, the person may want to work more days in a year. The costs of education do not change as he utilize it more intensively, but the returns do. For example, if he works 320 days, his net income would be $37,600.


A more general model

There are two type of activities. A person's productivity on each activity depends on his activity-specific knowledge/human capital. If one has k1 units of human capital on activity 1, and if one spends t units of time on this activity, his total earnings from this activity is w1k1t. So a person maximizes

w1k1t + w2k2(1-t) - C(k1, k2)
Note that the variable t does not enter into the cost function for knowledge acquisition. The cost of producing human capital does not depend on how intensively one is going to use it subsequently. Note also that we may assume Cii > 0, so that the marginal cost of acquiring human capital of the same type increases. This assumption will tend to induce diversification were it not for the utilization effect.

To solve this maximization problem, we need to compare "corner solutions" to "interior solutions." In an interior solution (i.e., t is strictly between 0 and 1), the person must be indifferent between spending extra time on either activity. So the marginal value of time on each activity must be the same. This implies

w1k1 = w2k2
We can therefore use this equation to eliminate either k1 or k2 and rewrite the value function as
V12(k) = w2 k - C( (w2/w1)k, k)
or
V12(k) = w1 k - C(k, k/(w2/w1))

Alternatively, in a "corner solution" in which the person specializes in activity 1 (i.e., t=1), obviously there is no need for k2. So the value function is

V1(k) = w1k - C(k,0)
Similarly, in a corner solution in which the person specializes in activity 2, the value function is
V2(k) = w2k - C(0,k)

Now it is clear that as long as C(k,0) < C(k, x) for x > 0, V1(k) > V12(k) for any k. Similarly, as long as C(0,k) < C(x,k) for x>0, V2(k) > V12(k) for any k.

Karl Marx wrote, "In a communist society, where nobody has one exclusive sphere of activity but each can become accomplished in any branch he wishes, society regulates the general production and thus makes it possible for me to do one thing today and another tomorrow, to hunt in the morning, fish in the afternoon, rear cattle in the evening, criticize after dinner, just as I have a mind, without ever becoming hunter, fisherman, shepherd, or critic." How romantic! And how inefficient!


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