Simplifying, using C> = An. and solving for C| gives

Substituting this expression into the solution for fji(t) and then using it in the solution for c(t) gives the optimal path of consumption chosen by the individual e~>" — I

Figure 25.5 Optima! consumption path. c(t), and bank account path. ,v(/). when x{T) = v0 and p < r

An interesting case arises when b = jr(). which requires the bank account at time T to equal its initial size. An obvious consumption path that satisfies this constraint is to just consume the interest. That is. set c(t) = rxQ. Then the bank account never grows or diminishes. However, it is apparent from our solution thai this situation will not be the utility-maximizing solution in general. To see this is so, sel b = Ay,. This gives c(t) = px» e '' " 'V'-"" e~f" — 1

In the special case in which the private rale of time preference is equal to the markel rate of interest, /; = r, this equation reduces to c(t) = rx< i

Figure 25.6 Optimal consumption paih, £•(/). and bank account path. x(t ), when v(T) = c,, and p > r

Thus consuming the interest maximizes utility only if p = /-, which occurs i the individual discounts future utility at a rate exactly equal to the market rate 0' return. This is shown as the line rxo in figures 25.5 and 25.6. On die oilier hand, i' the individual discounts future utility at a lower rate than the market rate, p < r. then the individual is better off foregoing consumption early in life (high savings) to lake advantage of the high return on saving and then consuming heavily later in life. In this case, consumption is a rising function of time. This consumption path and the associated path for the bank account, x(/), are shown in figure 25.5. Alternatively, if p > r, then consumption is a monotonicallv decreasing function of time. The individual consumes heavily early in life and saves later to bring the bank account back up to its required bequest level. This consumption path and the associated x(t) are shown in figure 25.6. ■

Inequality-Constrained Endpoint Problems: x(T) > b

When the terminal value of the state variable must satisfy an inequality constraint, we are free to choose i (T) optimally as long as our choice does not violate the constraint. The way we solve the optimal control problem then depends on whether the constraint binds or not. If our optimal (unconstrained) choice of .v(T) satisfies the constraint, then we effectively have a free-endpoint problem. The relevant boundary condition for .v(T') dien is >.('/') = 0. On the other hand, if our unconstrained choice of.* (7) is smaller than b. we have to settle for setting x(T) = b. and we effectively have a fixed-endpoint problem. The relevant boundary condition for x( T then, is.v(7") - b. As a result, the only thing that makes this a new kindof maximiza lion problem is that we have to decide whether ihe constraint on x(T) is binding.

In practice, the easiest and most reliable way to decide if the constraint on x(T) is binding is to first try to find the unconstrained, optimal value. jr*(7") and compare this value to /;. It is then easy to see whether x*(T) automatically satisfies the constraint.

Example 25.5 Solve the following problem:


Except for the constraint on jr( I), this is the same as the free-endpoint problem in example 25.1. After forming the Hamiltonian and maximizing with respect to the control variable y. we obtained the following system of differential equations:

The first boundary condition is x(0) = 2. We have to decide now whether the appropriate second boundary condition is A( I) = 0 (if the constraint on 1) is not binding) or .r( I) = b (if the constraint on x(l) is binding). To do this, we proceed as if there were no constraint on ,v( 11 (solve the free-endpoint problem). We will find ,v"(l). the optimal value, and then compare it to b to see if the constraint is binding or not. In example 25.1 we already solved the free-endpoint problem and found the optimal solution path for the state variable to be

«•0-2-4+2 Since 7 = 1. the optimal value for jc( I) is

Two cases can arise:

Case 1 b < 9/4, The constraint is not binding because v*( 1) > b. The problem is solved.

Case 2 b >9/4. The constraint is binding because x*{\) < h. Thus we cannot have our first choice for x( I). The next best we can do is to set x( l) = b, The differential equations now have to be re-solved using the correct boundary condi-tioas: a<0) - 2 and a(I) = b. This is the fixed-endpoint problem we solved in example 25.5. ■

Example 25.6 Solve the optimal consumption model with an inequality constraint on x(T):

max I e'pl\ncdl


We have already solved the model of optimal consumption with a fixed-endpoint constraint. We are now asked to reconsider this problem with an inequality con-straint on > (7~). The only thing that differs between this problem and its fixed-endpoint version is die boundary condition on x(T). As such, the solution up to the point of finding the boundary conditions is identical After using the initial condition x(0) = x0. we found the solution for a(f) to be x(t) = -C,-|(1 " V' +xnen (25.26)

0 0

Post a comment