VI. INEQUALITY CONSTRAINED OPTIMIZATION Application of the Kuhn-Tucker conditions to inequality constrained optimization problems is another very, very important skill to your career as an economist. If your mathematical training comes from somehere other than an economics department, it s unlikely that you ve seen much emphasis put on this. Consider this problem. A person s utility function depends on consumption and leisure. Of his market time (the complement to leisure), h t, some share i t is devoted to human capital investment (training, schooling, and such). The lifetime utility maximization problem looks like: ( ) T max! t U c t,t " h t s.t.: t & c t t ( " i t )h t + rs t " " s t ( ' t + t + $ ( i t h t ) t ( ) 0 c t, 0 h t T, 0 i t This is a version of the Heckman model of human capital investment. The first to inequality constraints are typical, that consumption must be nonnegative (try eating negative a bagel!) and that market time is also nonnegative, but also less than the total amount of time in the year. Sometimes, lazy economists ill ignore corner solutions and just assume that the solution lies in the interior. After all, e never see people consuming absolutely zero in a year you d die and have utility of negative infinity if that happened. Though sometimes corner solutions might seem silly, in other cases they are not. The interpretation of the variable i t is that it is the share of your market time (nonleisure time) devoted to making yourself a more productive orker, and the constraint is that it must be beteen 0 and 00. The case here i t is very interesting, because it is interpreted as full-time schooling. You ve probably spent some time in this situation. Retirement is another interesting corner solution. People also study hat affects the decision hether to participate in the labor market, another corner solution. In fact, in this model, the most interesting things are the corner solutions. Generally, though, e orry about solutions that suggest optimal consumption is negative. Consider a person ith the quasi-linear utility function: U(, ) + ln Solving this the usual ay, e get the demand function: Fall 2007 math class notes, page 4
( )!, x(p,) (p,), (p,) " $ Here e have a potential problem. What if <? The first-order conditions tell our consumer that he should buy a negative amount of good one, but that s difficult. What makes more sense is to say these demands are optimal, provided that the person has enough ealth. If not, he just buys zero of good one and spends everything on good to: x ( p,! ) )"!, p + $ & ' if ( * +,( 0,) if < Demand functions should be ritten like this hen corner solutions are possible. We can also get unreasonable solutions hen e are dealing ith satiated preferences. Suppose that my demands for to goods (beer and ice cream) are given by the function: u(, ) 8! 2 + 8! 2 The price of each good is equal to one, and I have ealth 00. Solving the problem: & ' max8! 2 + 8! 2 s.t. + 00 gives demands of 50 and 50. Hoever, let s think a bit more carefully about this. My marginal utility of consuming the n-th unit of either good is 8! 2n. That means that consuming more than four units of either good gives me disutility. If I ant to survive the evening ithout puking, I should stop after four beers and four ice creams leaving some of my money unspent. There are to ays to do inequality-constrained optimization. One is the proper ay, and the other is the simpler ay that everyone does it. We ll do the proper version first, so let s set up the general frameork for an inequality-constrained optimization problem. We ant to maximize the objective f : X! R, ith the k inequality constraints g i ( x)! b i, and the requirement that each of the n elements x i of the vector x cannot fall belo a certain fixed value, c i : max x!x f (x) s.t.: g (x) " b $! & g m (x) " b m and s.t.: ' c $ " & x n ' c n Fall 2007 math class notes, page 42
Typically, f ill be a utility function or a social elfare function; the function g ill be a budget constraint, or feasibility constraint; and c ill be the zero vector (for nonnegativity constraints). Once again, e define a Lagrangian: L(x,!,µ) " f (x) +!(b $ g(x)) + µ (x $ c) m n f (x) +! j ( b j $ g j (x)) + µ i x i $ c i j i ( ) The complete requirements for a maximum x * hen c 0 ) are: (called Kuhn-Tucker conditions.!l(x *,",µ)!x 0 2.!L(x *,",µ)!" 0 and! " 0, ith (!L(x *,",µ)!")" 0 3.!L(x *,",µ)!µ 0 and µ! 0, ith (!L(x *,",µ)!µ)µ 0 Which can also be ritten as:.!l!x i 0, for i,2,,n ( ) 0 2. b j! g j (x * ) " 0 and! j " 0, ith! j " b j g j (x * ) 3. x i *! c i " 0 and µ i! 0, ith µ i!(x i * " c i ) 0, for i,2,,n Though these look nasty, they have a simple interpretation. The first part of each of (2. ) and (3. ) says that each constraint must be satisfied e should certainly hope so, since that as the hole point. The second expression says that each constraint holds ith equality, or else the multiplier on the constraint is zero, hich e describe as a nonbinding constraint for instance, that either the person ends at a corner solution, or e might as ell not have orried about the constraint. This requirement that at least one of the terms equals zero is called complementary slackness. As an example of a nonbinding constraint, e could have a utility maximization problem ith the constraints the price of your consumption must be less than or equal to your ealth and the calorie values of the foods you eat must be greater than or equal to your required caloric intake. For most people the first of these constraints ould be binding, and so their budget balances exactly. On the other hand (at least in the U.S. at this time), the energy requirement ould probably be nonbinding, and they end up eating hatever they like. Okay, no it s time for an example. Hopefully, this problem ill be orked so ell (and not have any sign errors) that you can pattern your problems on this. The problem is to maximize a quasi-linear utility function ith respect to some variables x, subject to a budget constraint and some nonnegativity constraints: Fall 2007 math class notes, page 43
maxu(, ) + ln s.t.: +! and x i! 0. The Lagrangian function associated ith the solution to this problem is: L( x,!,µ ) " + ln +!( + ) + µ + µ 2 There are to x variables, hich give these partial derivatives: ()!L! " + µ $ 0 (2)!L! " + µ 2 $ 0 There are also three constraints, hich returns these conditions: (3)!L!" $ 0 and (!! )" 0 (4)!L!µ " 0 and: µ! 0 (5)!L!µ 2 " 0 and: µ 2! 0 I rerite the conditions () and (2) in this manner: (6) ( + µ )! (7) ( + µ 2 )! Well, no e can solve for in terms of and the multipliers and then derive to demand functions, inclusive of the multipliers: (8) ( ) ( )! ( ( + µ )! µ 2 )! ( + µ ) + µ 2 (9)! + µ! µ 2 (0) The next step is to hypothesize that each constraint is non-binding, and to see hat facts are consistent ith each hypothesis. If the hypothesis implies something illogical, e can determine that the constraint never binds. If the hypothesis implies something possible, then e have determined the conditions under hich this constraint doesn t bind. First, let s suppose that the budget constraint is non-binding, so! 0. Equation (6) ould become: () ( + µ )! 0 Fall 2007 math class notes, page 44
In order for the product to equal zero, either 0 or (+ µ ) 0. Provided that prices are finite, the first of these is impossible. The second ould require that µ 2!, but the Kuhn-Tucker conditions require that µ i! 0, so this is impossible as ell. Putting these together, e can determine that a non-binding budget constraint ould imply impossible things, so a non-binding budget constraint is impossible. Therefore,! > 0, and +. Next, let us ork ith the requirement that either µ 2 0 or 0. If 0, then equation (0) tells us that: (2) ( )! µ 2 0 + µ Multiplying both sides by (+ µ )! µ 2, e ould have that 0, hich is a very silly result this case is impossible. We can therefore conclude that > 0, so µ 2 0. This ill simplify the demand a bit. Finally, e ll examine the requirement that either µ 0 or 0. If 0 and µ > 0, then equation (9) tells us that: (3)! ( + µ )! 0 This means that: (4) ( + µ )! (5) (6) + µ! µ > 0 And so this implies that: (7) >! > There s no reason hy that couldn t occur. We have established that 0 is consistent ith >. What about the opposite case, here > 0 and µ 0? From equation (9), e have that this implies: (8)! > 0 That tells us to things: this case is consistent ith >, hich isn t impossible; and in this case, demand is. Fall 2007 math class notes, page 45
Putting this altogether, e can conclude that there are to possible cases. If <, then (9)! ( + µ )! 0 (0) µ! Which means that ( ( ))! (0) + µ + ( (! ))! p ( )! ( )! You spend nothing on good one, and you spend all your ealth on good to. The other possible case is that >, ith µ µ 2 0. This gives demands of: (9)! ( + µ )!! (0) + µ ( ( ))! ( ) ( )! And that s it. We might rite these demand functions as: " $ p!, if > $ 0 otherise p! p 2 if > " $ otherise There is a lot of intuition involved in dealing ith corner solutions. Sometimes e can rule certain kinds of binding constraints. Here are some general guidelines. Rule : Almost never is the number of binding constraints greater than the number of choice variables. Rule 2: The budget constraint or feasibility constraint is alays binding hen a utility or elfare function is strictly increasing in any variable:!i : "x i u x i > 0 $ > 0 Rule 3: Any variable hose marginal utility goes to infinity as that variable goes to zero ill not have a binding nonnegativity constraint: lim "u "x i $ µ i 0 x i!0 Here are some more practice problems. Exercise: max x u, Exercise: max x u c ( ) x! " x 2, s.t.: +!, x! 0 ( )! ", subject to: p! x ",! 0 Fall 2007 math class notes, page 46
Exercise: max x, u, Exercise: max q,x ( ) +, subject to: +!, x!! 0! ( pq! x), subject to: q! x + 4 " 2, q! 0, x! 0 Exercise: min x, ( + 2 ), subject to: +! q,! 0,! 0 ( q fixed). There is another ay to handle corner solutions, omitting the second multiplier µ. I ill use! L to denote this less-than-rigorous Lagrangian ithout µ. Instead, they allo the first order condition for!! L!x to have an inequality this is the ay that most people do it (and likely you ll end up doing, too). Except in eird cases ith, you ll get the same anser either ay, provided you re careful. The problem is the same: max f (x) s.t.: g(x)! b and to x " c Set up the Lagrangian: ( ) m!l(x,!) " f (x) +!(b $ g(x) f (x) +! i b i $ g i (x) Take the derivative of this ith respect to x i. For each x i, either the derivative of the Lagrangian ith respect to that variable equals zero (the usual first-order condition) or that variable is at a corner:! f (x * ) k!g " $ j (x * )( ' j j &!x i!x * i ) + 0 and:! f (x * ) k!g " $ j (x * )( ' j j &!x i!x * i ) xi* " c i i ( ) 0 The rules for derivatives ith respect to! j are the same as before. Pretend that the usual condition holds ith equality and solve. Look at the implied value for x i. Does it go outside the permissible range? If so, set it equal to the corner value. Unless you have some strange things going on ith a bunch of corners, that s all you have to orry about. The end. As an example, once more the problem is to maximize a quasi-linear utility function ith respect to some variables x, subject to a budget constraint and some nonnegativity constraints: maxu(, ) max + ln s.t.: x! p + " and to: x! 0. The Lagrangian function associated ith the solution to this problem is:!l ( x,! ) " + ln +!( + ) There are to x variables and one constraint, hich give these partial derivatives: Fall 2007 math class notes, page 47
( ) 0 ( )!! L! " $ 0 and:! " ( ) 0 (2 )! L! " $ 0 and:! " (3 )! L!" $ 0 and: (!! )" 0 Solve these as if the inequality holds ith equality. We get the demand for good 2: (9)! Substituting into the budget constraint, the demand for good is then: (20) x`!! ( ) "! This is fine (that is, certainly nonnegative) only if!. In any other case, e ll have to set 0. So e mention this possibility hen riting don the demand function: " $ p (2)!, if > $ 0 otherise This also suggests that the predicted optimal amounts of (that is, the amount that came from solving as if the equality held) ill also be okay, but only if!. In the other case, e kno that 0, so from the budget constraint e can infer that! 0 ( ) ". So the demand function is: p! p (22) 2 if > " $ otherise These solutions both coincide ith the solutions from the other method, hich is a good thing. A fe more practice problems: Exercise: maxu c,! Exercise: maxu,, x 3 Exercise: maxu c ( ) c +! ln (!), subject to: p! c " h! and h +!! T. ( ) x! " x 2 " x $ 3, subject to: + + p 3 x 3! ( ) c! "c 2, subject to: p! c " Fall 2007 math class notes, page 48