Mathematical Foundations -- Constrained Optimization Constrained Optimization An intuitive approach First Order Conditions (FOC) 7 Constraint qualifications 9 Formal statement of the FOC for a maximum 6 Concave optimization problem 8 Sufficient conditions 9 Application: Joint Costs More on the Second Constraint Qualification 6 Exercises 9 Answer to Exercise 3 3 John Riley September 5,
Mathematical Foundations -- Constrained Optimization AN INTUITIVE APPROACH Max{ f ( x) b g( x), x } x We shall interpret this mathematical problem as the decision problem of a profit maximizing firm The revenue from the production plan x is f( x ) The firm can produce any vector of outputs x which satisfies the non-negativity constraint, x and a resource constraint gx ( ) b A simple example is a linear constraint Each unit of x requires a units of resource b Then n a x= ax b = John Riley September 5,
Mathematical Foundations -3- Constrained Optimization Suppose that x solves the optimization problem If the firm increases x, the marginal revenue is, f g ( x) However, the increase in x also utilizes additional resources The extra resource use is ( x ) x We now introduce a resource price λ and ask what price this must be if the profit maximizing decision is to continue to choose x g Note that given the resource price λ the marginal cost of x is λ ( x ) * John Riley September 5,
Mathematical Foundations -4- Constrained Optimization Suppose that x solves the optimization problem If the firm increases x, the marginal revenue is, f g ( x) However, the increase in x also utilizes additional resources The extra resource use is ( x ) x We now introduce a resource price λ and ask what price this must be if the profit maximizing decision is to continue to choose x g Note that given the resource price λ the marginal cost of x is λ ( x ) If x > marginal revenue must be equal to marginal cost so f g MR MC = ( x) λ ( x) = If x = the marginal profit must be negative so f g MR MC = ( x) λ ( x) If there is no market price we follow the same approach Instead of the market price we introduce a shadow price λ to reflect the opportunity cost of using the additional resources The FOC are then the conditions above with the as if market price if there is no actual market price John Riley September 5,
Mathematical Foundations -5- Constrained Optimization Summarizing our results f g ( x) - λ ( x), with equality if x > Equivalently: f g f g ( x) - λ ( x), and x ( ( x) - λ ( x)) = The second condition is called a complementary slackness condition as it can only hold if one but not f g both of the conditions (i) MR MC = ( x) - λ ( x) and (ii) x is not binding (ie slack ) John Riley September 5,
Mathematical Foundations -6- Constrained Optimization Since x must be feasible b gx ( ) Moreover, we have defined λ to be the opportunity cost of additional resource use Then if not all the resource is used, λ must be zero Summarizing, b gx ( ), with equality if λ > Equivalently: b gx ( ) and λ( b gx ( )) = Once again at most one of the inequalities b gx ( ) and λ can be slack So this is a second complementary slackness condition Generalization to multiple constraints Suppose that there are m constraints and we write these as follows h( x), i =,, m Arguing in the same way we introduce a shadow price λ i for the ith constraint i FOC: MR f m gi MC = i= with equality if x > =,, n John Riley September 5,
Mathematical Foundations -7- Constrained Optimization There is a convenient way to remember these conditions Write the i-th constraint in the form hi ( x), i =,, m In vector notation hx ( ) (Thus in our example we write the constraint as h( x) = b gx ( ) ) Then introduce a vector of Lagrange multipliers or shadow prices λ and define the Lagrangian L ( x, λ) = f( x) + λ hx ( ) The first order conditions are then all restrictions on the partial derivatives of L ( x, λ) f h (i) = + λ, i with equality if x >, =,, n L (ii) hi ( x), λ = i with equality if λ i >, i =,, m Exercise: Solve the following problem Max{ U ( x) = ln x+ ln( x + x3) px+ px + p3x3 6} (i) if p = (,, 6) (ii) p = (,,) (iii) p = (,,4) John Riley September 5,
Mathematical Foundations -8- Constrained Optimization Restatement of FOC (i) L f h and x L = + λ = i, =,, n (ii) = hi( x) and λi = λ λ i i, i =,, m In vector notation (i) L f h and x L = + λ = (ii) = hx ( ) and λ = λ λ BUT It turns out that this intuitive argument is not quite correct It breaks down in some exceptional cases To avoid these we some very mild impose mild restrictions on the constraints These are called constraint qualifications John Riley September 5,
Mathematical Foundations -9- Constrained Optimization Constraint Qualifications Consider the maximization problem Max{ f ( x) x X} where X = { x x, h( x), i =,, m} x i The constraint qualifications holds at x X if hi (i) for each constraint that is binding at x the associated gradient vector ( x ) (ii) X, the set of non-negative vectors satisfying the linearized binding constraints has a non-empty interior L hi Note: The linear function hi ( x) = hi( x) + ( x) ( x x) has the same value and gradient as the function hi ( x ) at x Thus the linear approximation at x of the ith constraint hi ( x) is hi hi ( x) + ( x) ( x x) hi Since hi ( x ) = the linearized constraint is ( x ) ( x x ) John Riley September 5,
Mathematical Foundations -- Constrained Optimization Remark: Satisfying the second constraint qualification At the end of these slides is a proof of the following proposition Proposition: Suppose that we re-label the constraints so that it is the first I that are binding at x, that is h x = i = I Suppose also that there are J variables for which the non-negativity constraint i ( ),,, is binding Then consider the I linearized binding constraints and J binding non-negativity constraints This is a system of I + J linear constraints the second constraint qualification holds at A ( x x ) = If the rows of A are linearly independent then x As a practical matter, this result is almost never employed in economic analysis The reason is that economists typically make assumptions that imply that the feasible set X is convex and has a nonempty interior As we shall see, when these assumptions hold, checking the first constraint qualification is enough John Riley September 5,
Mathematical Foundations -- Constrained Optimization Example : Constraint qualification holds Max{ f ( x) = ln x x x, h( x) = x x } x As is readily confirmed, the maximizing value of x is x = (,) The feasible set and contour set for f through x = (,) are depicted The Lagrangian is L ( x, λ) = ln x+ ln x + λ( x x) The first order conditions are therefore Fig -: Constrained maximum (i) L x = λ, x with equality if x >, =, L (ii) x x, λ = with equality if λ > As is readily checked, the necessary conditions are all satisfied at ( x, λ ) = (,,) John Riley September 5,
Mathematical Foundations -- Constrained Optimization Example : Constraint qualification does not hold x 3 { ( ) ln ln, ( ) ( ) } Max f x = x + x x h x = x x Since the feasible set and maximand are exactly the same as in example, the solution is again x = (,) The Lagrangian is ( x, ) ln x ln x ( x x ) 3 L λ = + + λ Fig -: Constrained maximum Differentiating by x, L x = x x = x 3 λ( ) at x = (,) John Riley September 5,
Mathematical Foundations -3- Constrained Optimization Thus the first order condition does not hold at the maximum Our intuitive argument breaks down because h, the gradient of the constraint function, is zero at the maximum Thus the partial derivatives x of h no longer reflect the opportunity cost of the scarce resource More formally, at the maximum x, the linear approximation of the constraint is L hi hi ( x) = hi( x) + ( x) ( x x) Then, as long as the constraint is binding ( hi ( x ) = ) and the gradient vector is not zero, the linearized constraint is h i ( x ) ( x x ), i =,, m John Riley September 5,
Mathematical Foundations -4- Constrained Optimization There is a second (though unlikely) situation in which the linear approximations fail Consider the following problem Max{ f ( x) = x + x h( x) = ( x ) x, x } x 3 The feasible set is the shaded region in the figure Also depicted is the contour set for f through x = (,) From the figure it is clear that f takes on its maximum at x = (,) Fig -a: Original problem However, L x = = ( x) 3 λ( x ) Thus again the first order conditions do not hold at the maximum John Riley September 5,
Mathematical Foundations -5- Constrained Optimization This time the problem occurs because the feasible set, after taking a linear approximation of the constraint function, looks nothing like the original feasible set h At x, the gradient vector ( x ) = (, ) Thus the linear approximation of the constraint hx ( ) through x is Linearized feasible set Fig -b: Linearized problem h h h ( x) ( x x) = ( x)( x ) + ( x) x = x Since x must be non-negative, the only feasible value of x is x = In Figure -b the linearized feasible set is therefore the horizontal axis Then the solution to the linearized problem is not the solution to the original problem John Riley September 5,
Mathematical Foundations -6- Constrained Optimization As long as the constraint qualifications hold, the intuitively derived conditions are indeed necessary conditions This is summarized below Proposition: First Order Conditions for a Constrained Maximum Suppose x solves Max{ f ( x) x X} where X= { x x, hx ( ) } x If the constraint qualifications hold at x then there exists a vector of shadow prices λ such that and ( x, λ), =,, n with equality if x > ( x, λ), i =,, m with equality if λi > λ Kuhn-Tucker conditions This result is usually attributed to Kuhn and Tucker However the first formal statement of the conditions is by Karush (939) John Riley September 5,
Mathematical Foundations -7- Constrained Optimization Special case: Concave optimization problem For the optimization problem Max{ f ( x) x X} where X= { x x, hx ( ) } x suppose that f, h,, h m are all concave functions Then if we can find some ( x, λ ) satisfying the FOC we don t have to worry about the constraint qualifications Example: Max{ u( x) p x I, x n + } where u is a concave function x hx ( ) = I px n = px is a concave function and the sum of concave functions is concave therefore hx ( ) is concave John Riley September 5,
Mathematical Foundations -8- Constrained Optimization Proposition: If conditions (i) and (ii) hold at ( x, λ ) X= { x x, hx ( ) } (i) L f h and x L = + λ = then x solves Max{ f ( x) x X} where x (ii) = hx ( ) and λ = λ hx ( ) = λ λ, i =,, m Proof: The sum of concave functions is concave so L ( x, λ) = f( x) + λ hx ( ) is a concave function of x Hence L( x, λ) L ( x, λ) + ( x, λ) ( x x) As you can check from the FOC, the second term on the right hand side is negative Also, appealing to (ii) L ( x, λ ) = f( x) Therefore L ( x, λ) = f( x) + λ h( x) f( x) But λ hx ( ) for any feasible x Therefore f( x) f( x) John Riley September 5,
Mathematical Foundations -9- Constrained Optimization Proposition: Sufficient Conditions for a Maximum Suppose that f and h, i =,, m are quasi-concave and the i feasible set X= { x x, hx ( ) } has a non-empty interior f If the Kuhn-Tucker conditions hold at x, ( x) x hi and for each binding constraint, ( x ), then x solves Max{ f ( x) x, h ( x), i =,, m} x i Intuition: Under these assumptions the feasible set is convex and hence the linearized feasible set X contains the original feasible set X Then it is sufficient to show that X has a non-empty interior John Riley September 5,
Mathematical Foundations -- Constrained Optimization Application: Joint Costs An electricity company has an interest cost c o = per day for each unit of turbine capacity For simplicity we define a unit of capacity as megawatt It faces day-time and night-time demand price functions as given below The operating cost of running a each unit of turbine capacity is in the day time and at night p = q, p = q Formulating the problem Solving the problem Understanding the solution (developing the economic insights) John Riley September 5,
Mathematical Foundations -- Constrained Optimization Try a simple approach Each unit of turbine capacity costs + to run each day plus has an interest cost of so MC=4 With q units sold the sum of the demand prices is 3 q so revenue is TR = q(3 q) hence * MR = 3 4q Equate MR and MC q = 65 John Riley September 5,
Mathematical Foundations -- Constrained Optimization Try a simple approach Each unit of turbine capacity costs + to run each day plus has an interest cost of so MC=4 With q units sold the sum of the demand prices is 3 q so revenue is TR = q(3 q) hence MR = 3 4q Equate MR and MC * q = 65 Lets take a look on the margin in the day and night time TR = q ( q ) then MR = q TR = q( q) then MR = q MR <!!!!!!!!!! Now what? John Riley September 5,
Mathematical Foundations -3- Constrained Optimization Key is to realize that there are really three variables Production in each of the two periods q and q and the plant capacity q Cq ( ) = q + q + q Rq ( ) = R( q) + R( q) = ( q) q+ ( q) q Constraints h( q) = q q h( q) = q q John Riley September 5,
Mathematical Foundations -4- Constrained Optimization Key is to realize that there are really three variables Production in each of the two periods q and q and the plant capacity q Cq ( ) = cq where c= ( c, c, c) = (,,) Rq ( ) = R( q) + R( q) = ( q) q+ ( q) q Constraints: h ( q) = q q h( q) = q q L = R ( q ) + R ( q ) c q cq c q + λ ( q q ) + λ ( q q ) Kuhn-Tucker conditions = MR ( q ) c λ, q with equality if q >, =, = c + λ+ λ, q with equality if q > = q qi, λ i with equality if λ i >, i=, John Riley September 5,
Mathematical Foundations -5- Constrained Optimization Kuhn-Tucker conditions = q λ, q = q λ, q with equality if q > with equality if q > = + λ+ λ, q with equality if q > = q q, λ = q q, λ with equality if λ > with equality if λ > Solve by trial and error (i) Suppose q > and both shadow prices are positive Then q = q = q (ii) Solve and you will find that one of the shadow prices is negative But this is impossible (iii) Inspired guess That shadow price must be zero Then the positive shadow price must be John Riley September 5,
Mathematical Foundations -6- Constrained Optimization Technical Remark: Satisfying the second constraint qualification Proposition: Suppose that we re-label the constraints so that it is the first I that are binding at x, that is h x = i = I Suppose also that there are J variables for which the non-negativity constraint i ( ),,, is binding Then consider the I linearized binding constraints and J binding non-negativity constraints This is a system of K = I + J linear constraints then the second constraint qualification holds at A ( x x ) = If the rows of A are linearly independent x Proof: Suppose that at x there are I binding constraints We re-label these so that h x = i = I and i ( ),,, i ( ), h x > i > I The linearized feasible set is therefore h {, i ( ) ( ),,, } X = x x x x x i = I Suppose that there are J variables for which x =, = I +,, I + J x = We re-label these variables so that Thus there are K = I + J binding constraints in the linearized feasible set Suppose we choose x = x > K (), John Riley September 5,
Mathematical Foundations -7- Constrained Optimization Then the K binding constraints can be re-written as a system of K linear equations in K unknowns h h h h ( x ) ( x ) ( x ) ( x ) I I+ K x x hi hi hi h I x I xi ( x ) ( x ) ( x ) ( x ) = I I I+ K x I+ xi+ x K xk () Choose b, k =,, K and consider the following linear equation system k h h h h ( x ) ( x ) ( x ) ( x ) b h h h h x x b = b x b K xk I I+ n x x I I I I I I I ( x ) ( x ) ( x ) ( x ) I I I+ I xi+ xi+ I + K (3) John Riley September 5,
Mathematical Foundations -8- Constrained Optimization If the columns (or rows) of the matrix are independent, then the matrix is invertible Then for each b there is a unique solution x ( b), k =,, K For k > K define Consider any b >> From (3) it follows that h x x b x b i I k x ( b) = x n i ( )( ( ) ) = i >, =,, } and ( ),,, = k k x b = b = I + K Remember that x b = x > k > K Also x () >, k =,, I k( ) k, k Then for b >> and sufficiently small x ( b) >, k =,, I Thus for b >> and sufficiently small h x x b x i I n i ( )( ( ) ) >, =,, and ( ) = k xb >> Thus X has a non-empty interior QED John Riley September 5,
Mathematical Foundations -9- Constrained Optimization Exercises Utility maximization ux ( ) = x x + 4x x The price vector is p = (,) Solve for the utility maximizing consumption vector if (i) I = 6 (ii) I = Utility maximization ux ( ) = ln x+ 4x x The price vector is p = (,) (a) Solve for the utility maximizing consumption vector if I = 7 (b) For what income levels (if any) is consumption of commodity equal to zero? 3 Cost minimization The maximum output that a firm can produce with input vector z = ( z, z) is mapping is called a production function) q = ( z + z ) (This / (a) Confirm that the constraint qualifications hold for any feasible input vector x > (b) If the input price vector is r, solve for the cost minimizing inputs and hence the minimum cost under the assumption that q> ( r / r ) (c) Show that marginal cost is an increasing function for q r r > ( / ) John Riley September 5,
Mathematical Foundations -3- Constrained Optimization (d) Show also that an increase in one of the two input prices has no effect on marginal cost Do you find this puzzling? (e) Solve also for minimized cost if q r r < ( / ) (f) Is marginal cost MC( q ) continuous on +? 4 Peak and off-peak pricing An electric power company divided the day into three equal periods The demand prices in the three periods are as follows p = a q, p = q, p = q 3 The unit cost of producing electricity in each period is If the capacity of the firm is q (so that the maximum output in each period is q) the interest cost per day is 4q That is, the unit cost of capacity is 4 (a) If a = 4 solve for the profit maximizing outputs and prices in each period (b) Solve again if a = 6 (c) Returning to the first case, show that the total (ie gross) benefit of the production plan ( q, q, q 3) is B= 4q q + q q + q q 4 3 3 (d) Show that social surplus (SS = B-C) is maximized with a plan twice as large as the profit maximizing plan John Riley September 5,
Mathematical Foundations -3- Constrained Optimization Answer to Exercise 3 The production function q= F( z) is the maximum output that can be produced using input vector z Thus the set Z = { z F( z) q } is the set of feasible input vectors As depicted, cost is minimized at z >> If you check the slopes at A and B you will see why one of these points will be cost minimizing for some range of input price ratios while the other is never cost minimizing Note that if q = the cost minimizing input vector is z = Henceforth we consider q > Then for feasibility, z > Define hz ( ) = ( z+ z ) q Then / h = / / / (( z + z ),( z + z ) z ) z John Riley September 5,
Mathematical Foundations -3- Constrained Optimization Constraint Qualifications Note that h ( z ) > z if z > Then the first CQ is satisfied Note next that z + z is concave since it is the sum of two concave functions Since / y is strictly increasing on + it follows that hz ( ) = ( z+ z ) qis strictly quasi-concave Then we olny need to / check that the feasible set Z = { z F( z) q } has a non-empty interior Note that if / z = ( q, q) then hz ( ) = ( z+ z ) q= 8q> Thus the interior of Z is indeed non-empty / Minimization problem Min{ r z + r z ( z + z ) q } z / Note that if the constraint is not binding, then cost can be reduced by reducing z Thus at z the constraint must be binding Maximization problem Max{ r z r z ( z + z ) q } z / John Riley September 5,
Mathematical Foundations -33- Constrained Optimization The Lagrangian for this optimization problem is L = rz rz + λ(( z + z ) q) / Kuhn-Tucker conditions L = / r + λ( z + z ) z with equality if z > () z = r + λ( z + z ) z / / with equality if z > () L = / ( z + z ) q λ with equality if λ > (3) Since q >, z > If z > () is an equality and so λ > If z > () is an equality and so again λ > John Riley September 5,
Mathematical Foundations -34- Constrained Optimization We can rewrite () and () as follows: λ ( z + z ) r with equality if z > () / λ ( z + z ) rz with equality if z > () / / Case (i) Try z >> Then both () and () are equalities and so r = rz / Then z r = and so / r z r r = ( ) From (3) ( z + z ) = q Therefore / z r r / = q ( ) Note that since we assumed that z >, q r ( ) >, that is r / q r r > 4( ) Total cost is r C( q) = r z = rq r / John Riley September 5,
Mathematical Foundations -35- Constrained Optimization Case (ii) Try z = Then from (3) 4z Then = q C( q) rq 4 = rz = for q r r 4( ) Note: For a complete solution you should check that equality () and inequality () are both satisfied Marginal cost MC MC dc r r = =, q 4( ) dq 4 r dc r r = =, q > 4( ) dq q r / If you check you will find that MC is continuous John Riley September 5,