Mathematical Foundations -1- Supporting hyperplanes. SUPPORTING HYPERPLANES Key Ideas: Bounding hyperplane for a convex set, supporting hyperplane

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1 Mathematical Foundations -1- Supporting hyperplanes SUPPORTING HYPERPLANES Key Ideas: Bounding hyperplane for a convex set, supporting hyperplane Supporting Prices 2 Production efficient plans and transfer prices 3 Bounding Hyperplane 7 Supporting Hyperplane 21 Supporting prices 23 Linear Model 25 John Riley September 18, 213

2 Mathematical Foundations -2- Supporting hyperplanes Supporting Prices A firm or plant can produce outputs q ( q1,..., q n ) using inputs z ( z1,..., z m ). Helpful to treat inputs as negative numbers and define a production vector y ( y1,..., ynm) ( z1,..., zm, q1,..., qn). TC = m i1 pz i i m pi( yi). TR = i1 mn im1 py i i Profit = mn p y p z p y m i i i i im1 i1 Production set Let Y be the set of feasible production plans for this plant. This is the plant s production set. John Riley September 18, 213

3 Mathematical Foundations -3- Supporting hyperplanes Production efficient plans and transfer prices A production plan y is wasteful if there is another plan in the production set for which outputs are larger and inputs are smaller. Non-wasteful plans are said to be production efficient. Fig.1-1-1: Transfer price too high * John Riley September 18, 213

4 Mathematical Foundations -4- Supporting hyperplanes Production efficient plans and transfer prices A production plan y is wasteful if there is another plan in the production set for which outputs are larger and inputs are smaller. Non-wasteful plans are said to be production efficient. Does profit maximizing at fixed prices provide appropriate incentives for all efficient production plans? Mathematically, we seek prices that support an efficient Fig.1-1-1: Transfer price too high production plan. To illustrate, consider a plant that uses a single input to produce a single output. The input can be purchased at a price p 1. The objective is to produce y 2 efficiently by announcing a transfer price p 2 John Riley September 18, 213

5 Mathematical Foundations -5- Supporting hyperplanes The plant manager s bonus will be based on the profit ( y) p y p y Two contour sets of ( y) or iso-profit lines are depicted in Fig The steepness of any such line is the input-output price ratio p1/ p 2. As shown, the ratio is too low since profit is maximized at a point to the North-West of y. Fig.1-1-2: Optimal transfer price However, with the transfer price lowered appropriately, as in Fig , the optimal production plan is achieved. The correct transfer price thus provides the manager with the appropriate incentive. John Riley September 18, 213

6 Mathematical Foundations -6- Supporting hyperplanes Class exercise: (a) For the production set (b) For the production set Y z q q z 1/2 {(, ) } solve for the supporting price vector (, ) r p. Y { y y, y y } solve for the supporting price vector ( p1, p 2) Exercise (a) Confirm that ( zq, ) (1,1,1) is on the boundary of the production set 1 11/ 11/ 11/ 1 2 Y {( z, q) ( z (1 ) z ) q } where, 1 (b) Solve for the supporting prices. (c) Draw the contour set for 1 11/ 11/ 11/ 1 2 f ( z) ( z (1 ) z ) through (1,1) if (i)) 2 (ii) 1/ 2. Does it touch the axes? If not, what happens as z1? If so what is the slope when z2? John Riley September 18, 213

7 Mathematical Foundations -7- Supporting hyperplanes Unfortunately, this approach does not always work. Consider Fig and suppose, once again, that the output target is y 2 units. In the first case the production set is convex. Fig.1-1-3: No optimal transfer price Proposition: Bounding Hyperplane Let n Y be a non-empty, convex set. Let y be a vector not in the closure of Y. Then there exists p such that, for all y iny, p y p y. Proposition C.5-2: Supporting Hyperplane Theorem Suppose p such n Y is convex and y does not belong to the interior ofy. Then there exists that for all y Y, p y p y John Riley September 18, 213

8 Mathematical Foundations -8- Supporting hyperplanes Special case: supporting hyperplane Suppose that y is a boundary point of Y and that Y can be written as follows: Y y f y f y { ( ) ( )} This is an upper contour set. Suppose that Y is convex. Y (That is, f is quasi-concave.) Proposition: Supporting hyperplane If Y y f y f y { ( ) ( )} is convex, then the tangent hyperplane f x { y ( y ) ( y y ) } is a supporting hyperplane. John Riley September 18, 213

9 Mathematical Foundations -9- Supporting hyperplanes Proof: For y y and 1 1 y Y define 1 g( ) f ( y( )) f ( y ( y y )). Since Y is convex, y( ) is in Y. Therefore g( ) g() and so g( ) g() dg It follows that (). d But as we have previously shown, dg d f x 1 () ( x ) ( x x ). Then for all 1 x S, f x 1 ( x ) ( x x ) Q.E.D.. John Riley September 18, 213

10 Mathematical Foundations -1- Supporting hyperplanes General proof of the bounding hyperplane theorem Let Y be the closure of Y. That is it is Y plus all its boundary points. The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p Bounding hyperplane through y lies strictly outside the set Y. John Riley September 18, 213

11 Mathematical Foundations -11- Supporting hyperplanes Proof: The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p through y lies strictly outside the set Y. Since y Y, the vector p y y. Bounding hyperplane Thus 2 y y ( y y ) ( y y ) p ( y y ). Hence p y p y John Riley September 18, 213

12 Mathematical Foundations -12- Supporting hyperplanes Proof: The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p through y lies strictly outside the set Y. Since y Y, the vector p y y. Bounding hyperplane Thus 2 y y ( y y ) ( y y ) p ( y y ). Hence p y p y For any y Y, consider the convex combination y y y y y y (1 ) ( ), 1. Since y and yy and Y is convex, y Y. Then y y y y. John Riley September 18, 213

13 Mathematical Foundations -13- Supporting hyperplanes Proof: The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p through y lies strictly outside the set Y. Since y Y, the vector p y y. Bounding hyperplane Thus 2 y y ( y y ) ( y y ) p ( y y ). Hence p y p y For any y Y, consider the convex combination y y y y y y (1 ) ( ), 1. Since y and yy and Y is convex, y Y. Then y y y y. that is ( y y ) ( y y ) ( y y ) ( y y ) John Riley September 18, 213

14 Mathematical Foundations -14- Supporting hyperplanes Proof: The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p through y lies strictly outside the set Y. Since y Y, the vector p y y. Bounding hyperplane Thus 2 y y ( y y ) ( y y ) p ( y y ). Hence p y p y For any y Y, consider the convex combination y y y y y y (1 ) ( ), 1. Since y and yy and Y is convex, y Y. Then y y y y. that is ( y y ) ( y y ) ( y y ) ( y y ) that is ( y y ( y y )) ( y y ( y y )) ( y y ) ( y y ) John Riley September 18, 213

15 Mathematical Foundations -15- Supporting hyperplanes Proof: The ball around the vector y just touches the set Y. Thus y is closer to y than any other point in Y. Then define p y y. The proof is completed by showing that the hyperplane orthogonal to p through y lies strictly outside the set Y. Since y Y, the vector p y y. Bounding hyperplane Thus 2 y y ( y y ) ( y y ) p ( y y ). Hence p y p y For any y Y, consider the convex combination y y y y y y (1 ) ( ), 1. Since y and yy and Y is convex, y Y. Then y y y y. that is ( y y ) ( y y ) ( y y ) ( y y ) that is that is ( y y ( y y )) ( y y ( y y )) ( y y ) ( y y ) ( ( )) ( ( )) p y y p y y p p John Riley September 18, 213

16 Mathematical Foundations -16- Supporting hyperplanes We have proved that ( ( )) ( ( )) p y y p y y p p Multiplying out p p p y y y y y y p p 2 2 ( ) ( ) ( ). John Riley September 18, 213

17 Mathematical Foundations -17- Supporting hyperplanes We have proved that ( ( )) ( ( )) p y y p y y p p Multiplying out p p p y y y y y y p p 2 2 ( ) ( ) ( ). Therefore 2 2 p ( y y ) ( y y ) ( y y ). John Riley September 18, 213

18 Mathematical Foundations -18- Supporting hyperplanes We have proved that ( ( )) ( ( )) p y y p y y p p Multiplying out p p p y y y y y y p p 2 2 ( ) ( ) ( ). Therefore 2 2 p ( y y ) ( y y ) ( y y ). Dividing by, 2 p ( y y ) ( y y ) ( y y ). John Riley September 18, 213

19 Mathematical Foundations -19- Supporting hyperplanes We have proved that ( ( )) ( ( )) p y y p y y p p Multiplying out p p p y y y y y y p p 2 2 ( ) ( ) ( ). Therefore 2 2 p ( y y ) ( y y ) ( y y ). Dividing by, 2 p ( y y ) ( y y ) ( y y ). Letting, we have at last 2 p ( y y ). Hence p y p y John Riley September 18, 213

20 Mathematical Foundations -2- Supporting hyperplanes We have proved that ( ( )) ( ( )) p y y p y y p p Multiplying out p p p y y y y y y p p 2 2 ( ) ( ) ( ). Therefore 2 2 p ( y y ) ( y y ) ( y y ). Dividing by, 2 p ( y y ) ( y y ) ( y y ). Letting, we have at last 2 p ( y y ). Hence p y p y But we have already shown that p y p y. Thus for all y Y, p y p y. Q.E.D. John Riley September 18, 213

21 Mathematical Foundations -21- Supporting hyperplanes We now show how this result can be extended to cases in which the vector Y. Proposition C.5-2: Supporting Hyperplane Theorem y is a boundary point of Suppose p such n Y is convex and y does not belong to the interior ofy. Then there exists that for all y Y, p y p y John Riley September 18, 213

22 Mathematical Foundations -22- Supporting hyperplanes t t Proof : Consider any sequence of points { y y Y } that approaches y. By the Bounding Hyperplane Theorem, there exists a sequence of vectors t t t t p such that for all t, and all y Y p y p y. Define p t p p t t. Then t t t p y p y and, for all t, each element of t p lies in the interval [ 1,1]. From the previous section we know that any bounded sequence of vectors in n has a convergent subsequence. Thus { p t } t 1... has a convergent subsequence, { p s } s Let p be the limit point of this subsequence. For all points in the convergent subsequence p t y t p t y Then, taking the limit, p y p y. Q.E.D. John Riley September 18, 213

23 Mathematical Foundations -23- Supporting hyperplanes Assumption: Free Disposal For any feasible production plan y Y and any, the production plan y is also feasible. Note that y the alternative plan y is a plan with a smaller output vector and larger input vector. Thus one way to achieve is to operate according to the plan y and throw away the extra output and unused inputs. Hence this assumption is immediately satisfied if the excess can be disposed of freely. Proposition 1.1-4: Supporting prices If y is a boundary point of a convex set Y and the free disposal assumption holds then there exists a price vector p such that p y p y for all y Y. Moreover, if Y, then p y. * John Riley September 18, 213

24 Mathematical Foundations -24- Supporting hyperplanes Assumption: Free Disposal For any feasible production plan y Y and any, the production plan y is also feasible. Note that y the alternative plan y is a plan with a smaller output vector and larger input vector. Thus one way to achieve is to operate according to the plan y and throw away the extra output and unused inputs. Hence this assumption is immediately satisfied if the excess can be disposed of freely. Proposition 1.1-4: Supporting prices If y is a boundary point of a convex set Y and the free disposal assumption holds then there exists a price vector p such that p y p y for all y Y. Moreover, if Y, then p y. Appealing to the supporting hyperplane theorem, there exists a vector p such that p y y ( ) for all y Y. By free disposal, y y Y for all vectors. Hence n ( ) ii i1 p y y p p. This holds for all. Setting for all j i, and 1it follows that p for each i 1,...,. n. j i i John Riley September 18, 213

25 Mathematical Foundations -25- Supporting hyperplanes Linear Model We now examine the special case of a linear technology. As will become clear, understanding this model is the key to deriving the necessary conditions for constrained optimization problems. A firm has n plants. It uses m inputs z ( z1,..., z m ) to produce a single output q. If plant j operates at activity level x j it can produce aoj xj units of output using axunits ij j of input i, i 1,.,.., m. Summing over the n plants, total output is n aojxj and the total input i requirement is j1 n ax ij j. j1 The production vector is then feasible if it is in the following set. Y n {( z, q) x, q a x, a x z, i 1,..., m} n oj j ij j i j1 j1 In matrix notation Y {( z, q) x, q a x, Ax z} Class Exercise: Show that the free disposal assumption holds. John Riley September 18, 213

26 Mathematical Foundations -26- Supporting hyperplanes The production set for the special case of two inputs and two plants is depicted in Fig Each crease in the boundary of the production set is a production plan in which only one plant is operated. For all the points on the plane between creases, both plants are in operation. Note that each point on the boundary lies on one or more planes. Thus there is a supporting There is a supporting plane for every such boundary point. Example: 1 a, 1, 1 o1 3 a11 a21 a 1, a 4, a We now show that this is true for all linear models. John Riley September 18, 213

27 Mathematical Foundations -27- Supporting hyperplanes Existence of supporting prices For any input vector, let, be the maximum possible output. Formally, q Max{ q a x A x z, x } (1.1-2) x o Thus ( zq, ) is a boundary point of the production set. Since the production set is convex and the free disposal assumption holds, there exists a positive supporting price vector ( r, p ) such that pq r z pq r z, for all ( z, q) Y (1.1-3) John Riley September 18, 213

28 Mathematical Foundations -28- Supporting hyperplanes Lemma: If the set Y has a non-empty interior, then the supporting output price, p, must be strictly positive. (Remark: If the set Y has a non-empty interior, there exists some xˆ such that ẑ A xˆ z.) Proof: Define qˆ a xˆ. Given the above assumption ( zq ˆ, ˆ) Y. o Therefore by the Supporting Hyperplane Theorem pq r z pqˆ r zˆ (1.1-4) We have already argued that p. To prove that it is strictly positive, we suppose that p and obtain a contradiction. First note that, if p it follows from (1.1-4) that r z r zˆ. Also, since ( rp, ), if p then r. But ẑ z, therefore r zˆ r z and so r zˆ r z. But this contradicts our previous conclusion. Thus p cannot be zero after all. QED Then, dividing by p and defining the supporting input price vector r/ p, condition (1.1-3) can be rewritten as follows. q z q z, for all ( z, q) Y (1.1-5) John Riley September 18, 213

29 Mathematical Foundations -29- Supporting hyperplanes Necessary conditions Appealing to the Supporting Hyperplane Theorem we have shown that there exists a positive vector ( rp, ) (,1) such that the boundary point is profit maximizing. We now seek to use this result to characterize the associated profit- maximizing activity vector x and price vector Proposition 1.1-4: Necessary conditions for a production plan to be on the boundary of the production set. Let be a point on the boundary of the linear production set. That is x arg Max{ q a x A x z, x } x o q ao x where Then, if the interior of the feasible set is non-empty, there exists a shadow price vector such that a A. (1.1-9) o Also x and satisfy the following complementary slackness conditions. (i) ( a A ) x (ii) ( z A x) o John Riley September 18, 213

30 Mathematical Foundations -3- Supporting hyperplanes Proof of (i): Since ( zq, ) is profit maximizing given price vector (,1) changing x j to xj xj lowers profit m R C a x a x ( a a ) x j j oj j i ij j oj i ij j i1 i1 Consider an increase in x j. Then R C MR MC a a m j j j j oj i ij xj xj i1 If x j, then x j can be positive or negative. Therefore m MR MC a a. j j oj i ij i1 m QED (i) John Riley September 18, 213

31 Mathematical Foundations -31- Supporting hyperplanes Proof of (ii): By construction q Max{ q a x A x z, x } and x arg Max{ q a x A x z, x } o o Define * z A x. Since the activity vector is feasible, From the Supporting Hyperplane Theorem * z x z A. q z q z * Rearranging, this inequality, * ( z z ). But z. * z and Combining these inequalities it follows that * ( z z ), that is ( z A x) QED (ii) John Riley September 18, 213

32 Mathematical Foundations -32- Supporting hyperplanes Summary We have shown that if x arg Max{ a x A x z, x } x o and if the interior of the feasible set is non-empty, then there exists a vector such that a A. (1.1-9) o where x and satisfy the following complementary slackness conditions. (i) ( a A ) x (ii) ( z A x) o Define L ( x, ) a x ( z Ax). o L ( x, ) zi aijx j i. (Feasibility) L ( x, ) a A x (1.1-9) j n L x ( x, ) ( a a ) x j oj i ij j x j j1 L i ( x, ) i ( zi aijx j ) i Thus for the linear model the Kuhn-Tucker conditions are indeed necessary conditions. (i) (ii) John Riley September 18, 213

33 Mathematical Foundations -33- Supporting hyperplanes Since the linear model is concave, the necessary conditions are also sufficient. That is, any ( x, ) satisfying the Kuhn-Tucker conditions is a solution to the linear maximization problem. John Riley September 18, 213