Primary education or IITs/IIMs

The economic theory mode of solving problems and investigating the implications for social welfare is to work out the planning solution first. That is to say, imagine that the purity of the exercise is not contaminated by the noise of the market and the problem of answering the question: where does the money come from? What if the reader was a benevolent despot and cared about the welfare of all her people? The economy has 100 teachers. It can assign them to primary schools, x, or higher educational institutions, y. x2 teachers are needed to run x schools and y2 faculty are needed at y institutes. The economys labour constraint is

x2 + y2 = 100

The planners preferences are represented by

U(x,y) = ax + by

where a, b, are given positive constants. In order to solve the problem we set up the Lagrangian

L(x,y,) = ax + by + (100 - x2 y2)

The first-order conditions are (Recall the rules In each case look for the variable that is bring perturbed on the right-hand side, keeping the rest in cold storage.)

L/x = a - 2x = 0

L/y = b - 2y = 0

L/ = 100 - x2 y2 = 0

Plugging in the first two equations into the third,

100 = ( a2 + b2)1/2/42

Simplifying,

= ( a2 + b2)1/2/20

and

x = 10a/( a2 + b2)1/2, y = 10b/( a2 + b2)1/2

We have supply functions for primary and higher education. It is convenient to regard a, b, as the weights that society or the planner attaches to primary education and higher education respectively. It can be seen that only the relative weights matter. If they are

increased in equal proportion x and y do not change. The supply of primary education can only increase with a planner with a higher a relative to b in her utility function.

Time and Money

Usually the constraint in optimization exercises is money. However, time is also of the essence. We do not have all the time in the world maximize our happiness. Some would blur the distinction and proclaim that time is money but we do not enter that area here. Consequently, along with the familiar budget constraint

p1x1 + p2x2 = m we introduce another constraint into our familiar exercise below, the time constraint.

t1x1 + t2x2 = T If the two xs denote apples and oranges, to return to our overworked fruit metaphor, there is a limited amount of time, T, say half an hour, that Avinash can spend with the fruit basket at the office buffet. t1 is the time taken to munch an apple, t2 the minutes taken to relish an orange. Recall the interpretation of the multiplier as being the shadow price or the marginal utility of the constraint. All along we have dealt with binding constraints but strictly speaking a budget constraint should read: the value of expenditures should not exceed the holding of money.

p1x1 + p2x2 m

Consider the case of the inequality being strict. That is, after the optimal choice of apples and oranges, there is cash left over. The budget constraint does not bite. The shadow price or the marginal utility of money is zero. On the other hand, if the inequality is an equality, the money in the pocket is a constraint. In that case, the shadow price or the marginal utility of money is positive. The same holds true for time. The associated Lagrangian multiplier would be the shadow price or the marginal utility of time. Assume that the preferences of our consumer are given by u(x1, x2) = x1x2. In that case the Lagrangian is L(x1, x2, , ) = x1x2 + (m - p1x1 - p2x2) + (T - t1x1 - t2x2) We consider only the first two equations of the first-order conditions (Note however that there are four equations to be solved).

L/x1 = x2 p1 t1 = 0

L/x2 = x1 p2 t2 = 0

Furthermore, we are provided the following data: p1 = 1, p2 = 2, t1 = t2 = 1, m = 12, T = 10. Then, our first-order conditions turn out to be

x2 = +

x1 = 2 + Now, we exclude the uninteresting case of both time and money not mattering. In that case, we are left with three possibilities.

1. = 0 and > 0 2. > 0 and > 0 3. > 0 and = 0

Let us take each in turn.

1. x1 = = x2. Since only time has a positive marginal utility, the associated constraint binds. Thus, 10 = x1 + x2. This means x1 = 5 = x2. However, this violates the money constraint! Check!

2. Here both time and money matter. Solving out the budget constraints,

12 x1 2x2 = 0

10 x1 x2 = 0

We get x1 = 8, x2 = 2. Our first-order conditions are, in this case,

2 = +

8 = 2 +

Solving these out simultaneously, the optimal values of the multipliers are = 6, = - 4. A negative shadow price is ruled out. This leaves the final scenario.

3. The first order conditions are

x2 =

x1 = 2

Since only the out-of-pocket costs kick in,

12 x1 2x2 = 0

The solution is x1 = 6, x2 = 3, = 3. It meets all the requirements.

Time is without a shadow marginal utility in this example. So, if the municipal authorities were to impose a cess on money holdings, using the proceeds to build a sophisticated road network that would save travel time, they would be ill-advised. The welfare of the citizens would fall.