Essential Microeconomics -1

Essential Microeconomics -1-

1.1 SUPPORTING PRICES

Production plan and production set 2

Efficient plan 6

Supporting hyperplane theorem 13

Supporting prices 22

Linear production set 20

Linear Programming problem 40

SUPPORTING PRICES

Key ideas: convex and non-convex production sets, price based incentives, Supporting Hyperplane Theorem

When do prices provide incentives to “support” desired outcomes?

Firm or plant: produces n different outputs qq= (1 ,..., qn )using m different inputs zz= (1 ,..., zm ).

mn+ A particular production plan is then the input-output vector (,zq )∈ +

Production set

A firm has available some collection of alternative feasible production plans (production vectors)

Let Y be the set of feasible production plans for this plant. This is the plant’s “production set.”

Production function and production set

Consider a firm producing q units of a single output.

Let fz() be the maximum output that a firm can produce using input vector z. Therefore output must satisfy the constraint q≤ fz(). The production set Y is the set of feasible plans so

Y={(,)|(,) zq zq ≥≤ 0, q f ()} z .

Example:

Y={(,)|(,) zq zq ≥≤ 0, q 2 z1/2 }, equivalently,

Y={( zq , ) | z ≥ 0, 4 z −≥ q2 0}

New notation: treat inputs as negative numbers 1 production vector

y=( y1 ,..., ynm+ ) =−− ( z 11 ,..., zqm , ,..., q n )

Any negative component of y is an input. Any positive component is an output. The production set Y is the set of feasible production vectors.

Fig.1.1-1a: Production set

New notation: treat inputs as negative numbers 1 production vector

y=( y1 ,..., ynm+ ) =−− ( z 11 ,..., zqm , ,..., q n )

Any negative component of y is an input. Any positive component is an output. The production set Y is the set of feasible production vectors.

Example: Fig.1.1-1a: Production set Y={( zq , ) | z ≥ 0, 4 z −≥ q2 0}.

Using the new notation, a production vector is (y12 , y )= ( − zq ,).

The production set is

2 Y ={(yy12 , ) | y 1 ≤−− 0, 4 y 1 y 2 ≥ 0}

This set is depicted in Fig 1.1-11b. Fig.1.1-1b: Production set ______

1 For a vector, our convention is that if yyjj≥ for all j we write yy≥ . If, in addition, the inequality is strict for some j we write yy> and if the inequality is strict for all j we write yy>> .

Efficient plan:

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. Formally the plan y is production efficient if there is no y ∈Y such that yy> .

Efficient plan:

Profit maximization

Let pp= (1 ,..., pmn+ ) be the vector of m input and n output prices.

mn+ total revenue TR = ∑ pyii im= +1

m m total cost TC = ∑ pzii=∑ pyii() − . i=1 i=1

mn+ m profit π =∑∑pyi i −=⋅ pz ii p y. im=+=11i

A production vector is profit maximizing if it solves

Max{| p⋅∈ y y Y } y

Supporting prices

When do prices provide an incentive for a firm to produce any efficient production plan? Mathematically, we seek prices that “support” an efficient production plan.

Example: a production plant that uses a single input to produce a single output. The input can be purchased at a price p1.

Suppose, furthermore, that the plant is part of a large firm.

0 Objective: Produce y2 units of output efficiently.

For delivery to another “down-stream” plant within the firm.

From the figure, zy11= − units of the input is efficient.

Fig.1.1-2a: Production set

Supporting prices

When do prices provide an incentive for a firm to produce any efficient production plan? Mathematically, we seek prices that “support” an efficient production plan.

Example: a production plant that uses a single input to produce a single output. The input can be purchased at a price p1.

Suppose, furthermore, that the plant is part of a large firm.

0 Objective: Produce y2 units of output efficiently.

For delivery to another “down-stream” plant within the firm.

From the figure, zy11= − units of the input is efficient.

Fig.1.1-2a: Production set Transfer price

A price p2 , paid for each unit delivered to the downstream division.

The plant manager’s bonus will be based on the profit π ()y= py11 + py 2 2.

Two contour sets (or “iso-profit” lines)

of π ()y= py11 + py 2 2 are depicted in Fig. 1.1-2a.

The steepness of any such line is the input-output

price ratio pp12/ .

As shown, the ratio is too low since profit is maximized at a point to the North-West of y0 . Fig.1.1-2a Transfer price too high

However, with the transfer price lowered appropriately, as in Fig. 1.1-2b, the optimal production plan is achieved.

The correct transfer price thus provides the manager with

the appropriate incentive.

Fig.1.1-2b: Optimal transfer price

Unfortunately, this approach does not always work.

Consider Fig. 1.1-3.

0 Suppose, once again that the output target is y2 units.

While y0 is locally profit maximizing, the profit

000 p⋅= y py11 + py 2 2 is negative. Profit is maximized by producing nothing. Fig.1.1-3: No optimal transfer price

Unfortunately, this approach does not always work.

Consider Fig. 1.1-3.

0 Suppose, once again that the output target is y2 units.

0 While y is locally profit maximizing, the profit

000 p⋅= y py11 + py 2 2 is negative. Profit is maximized by producing nothing.

Fig.1.1-3: No optimal transfer price

Convex production sets

Y is convex if for any yy01, ∈Y every convex combination yλ ≡−(1λλ ) yy01 + ∈Y

As we shall see, prices can be used to support all efficient production plans if the production set is convex.

Proposition 1.1-1: Supporting Hyperplane Theorem

Suppose Y ⊂ n is non-empty and convex and y0 lies on the boundary of Y . Then there exists

0 0 p ≠ 0 such that (i) for all y ∈Y , py⋅≤⋅ pyand (ii) for all y ∈intY , py⋅<⋅ py.

For a general proof see EM Appendix C. Here we consider the special case in which the convex set Y is an upper contour set of the function h∈1 , that is Y ={yhy | ( ) ≥ 0}.

Since y0 is on the boundary of Y, hy()00 = .

As long as the gradient vector is non-zero at y0 , the linear approximation of h at y0 is

∂h hy()= hy ()0 + ()( y 00 ⋅− y y ). ∂y

Note that hy() and hy() have the same value and gradient at y0

In two dimensions, the contour set of the linear approximation is the tangent plane as depicted in Fig. 1.1-4.

≥ 0 If the upper contour set Y={yhy | ( ) hy ( )} is convex as depicted, all the points in Y lie in the upper contour set of h

(i.e. the lightly and heavily shaded areas.) In mathematical terms,

0∂h 00 hy()≥ hy () ⇒ ()( y ⋅− y y )0 ≥. Fig.1.1-4: Supporting hyperplane ∂y

In two dimensions, the contour set of the linear approximation is the tangent plane as depicted in Fig. 1.1-4.

0 If the upper contour set Y={yhy | ( )≥ hy ( )} is convex as depicted, all the points in Y lie in the upper contour set of h (i.e. the lightly and heavily shaded areas.) In mathematical terms,

0∂h 00 hy()≥ hy () ⇒ ()( y ⋅− y y )0 ≥. Fig.1.1-4: Supporting hyperplane ∂y

Formally, we have the following Lemma.

∂h Lemma 1.1-2: If Y ={yhy | ( ) ≥ hy (0 )} is convex then yY∈⇒(y00 )( ⋅ yy − ) ≥ 0. ∂y

∂h To prove Proposition 1.1-1, choose py= − ()0 . Appealing to the lemma, ∂y

yY∈ ⇒−pyy ⋅( −0 )0 ≥ that is y∈⇒⋅≤⋅Y py py0 .

∂h Lemma 1.1-2: If Y ={yhy | ( ) ≥ hy (0 )} is convex then yY∈⇒(y00 )( ⋅ yy − ) ≥ 0. ∂y

Proof: Pick any y in Y . Since Y is convex, all convex combinations of y0 and y lie in Y . That is, for all λ ∈(0,1) ,

hy()−≥⇒−≥ hy ()000 hy (λ ) hy ()0 where yλ =−+(1λλ ) yy0 .

∂h Lemma 1.1-2: If Y ={yhy | ( ) ≥ hy (0 )} is convex then yY∈⇒(y00 )( ⋅ yy − ) ≥ 0. ∂y

Proof: Pick any y in Y . Since Y is convex, all convex combinations of y0 and y lie in Y . That is, for all λ ∈(0,1) ,

hy()−≥⇒−≥ hy ()000 hy (λ ) hy ()0 where yλ =−+(1λλ ) yy0 .

Define g()()((1)λ≡ hyλ =− h λλ y0 += y )( hy 00 +− λ ( y y )).

Then

g(λλ )− g (0) hy (0 + ( y 10 −− y )) hy ( 0 ) = ≥ 0, for all λ ∈(0,1) . λλ

∂h Lemma 1.1-2: If Y ={yhy | ( ) ≥ hy (0 )} is convex then yY∈⇒(y00 )( ⋅ yy − ) ≥ 0. ∂y

Proof: Pick any y in Y . Since Y is convex, all convex combinations of y0 and y lie in Y . That is, for all λ ∈(0,1) ,

hy()−≥⇒−≥ hy ()000 hy (λ ) hy ()0 where yλ =−+(1λλ ) yy0 .

Define g()()((1)λ≡ hyλ =− h λλ y0 += y )( hy 00 +− λ ( y y )).

Then

g(λλ )− g (0) hy (0 + ( y 10 −− y )) hy ( 0 ) = ≥ 0, for all λ ∈(0,1) . λλ

Note that the limit of the left-hand side as λ → 0is the derivative of g()λ evaluated at λ = 0. Taking this derivative we obtain dg∂ h ()(())()λλ=y0 + yy − 00 ⋅− yy. dyλ ∂ ∂h Therefore (y00 )(⋅− yy ) ≥ 0. ∂y Q.E.D.

Example: Firm with two outputs

1 22 Y ={y | y23 , y ≥ 0, hy ( ) =−− y14 y 2 − y 3 ≥0} .

The point y0 =( − 25,8,3) is on the boundary of this set and the gradient vector at this point is

∂h (y0 )=−− (1,1 yy 00 ,2 − ) =−−− (1,4,6). ∂y 2 23

Example: Firm with two outputs

1 22 Y ={y | y23 , y ≥ 0, hy ( ) =−− y14 y 2 − y 3 ≥0} .

The point y0 =( − 25,8,3) is on the boundary of this set and the gradient vector at this point is

∂h (y0 )=−− (1,1 yy 00 ,2 − ) =−−− (1,4,6). ∂y 2 23

Since the function hy() is the sum of three concave functions it is concave (and hence quasi-concave).

∂h Define py=−=(0 ) (1, 4, 6) . Then by the Lemma, the plane {|ypy⋅=⋅ py0 }is a supporting ∂y plane.

Example: Firm with two outputs

1 22 Y ={y | y23 , y ≥ 0, hy ( ) =−− y14 y 2 − y 3 ≥0} .

The point y0 =( − 25,8,3) is on the boundary of this set and the gradient vector at this point is

∂h (y0 )=−− (1,1 yy 00 ,2 − ) =−−− (1,4,6). ∂y 2 23

Since the function hy() is the sum of three concave functions it is concave (and hence quasi-concave).

∂h Define py=−=(0 ) (1, 4, 6) . Then by the Lemma, the plane {|ypy⋅=⋅ py0 }is a supporting ∂y plane.

We can easily check this directly. To produce the output (,)yy23, the minimum input requirement is

1 22 −=y14 yy 23 + . With the price vector p = (1, 4, 6) , the profit of the firm is

1 22 π ()46yyyy=++=−−++1234 yy 2346 yy 23.

It is readily confirmed that profit is maximized at (yy23 , )= (8,3) . Then the profit maximizing input

1 22 is −=y14 yy 12 + =25.

From supporting hyperplanes to supporting prices

For p ≠ 0 and y0 , a boundary point of Y , the hyperplane py⋅=⋅ py0 is a supporting hyperplane if

y ∈⇒Y py⋅≤⋅ py0 .

To have a direct economic interpretation we must have p > 0.

Free Disposal Assumption

For any feasible production plan y ∈Y and any δ > 0, the production plan y −δ is also feasible.

Why “free disposal”? If y ∈Y The firm can purchase an additional input vector −δ and then simply dispose of it. Then the new production vector is y −δ .

From supporting hyperplanes to supporting prices

For p ≠ 0 and y0 , a boundary point of Y , the hyperplane py⋅=⋅ py0 is a supporting hyperplane if

y ∈⇒Y py⋅≤⋅ py0 .

To have a direct economic interpretation we must have p > 0.

Free Disposal Assumption

For any feasible production plan y ∈Y and any δ > 0, the production plan y −δ is also feasible.

Why “free disposal”? If y ∈Y the firm can purchase an additional input vector −δ and then simply dispose of it. Then the new production vector is y −δ .

Proposition 1.1-3: Supporting prices

If y0 is a boundary point of a convex set Y and the free disposal assumption holds then there exists a price vector p > 0 such that py⋅≤⋅ py0 for all y ∈Y . Moreover, if 0∈Y , then py⋅≥0 0.

Proof:

Appealing to the Supporting Hyperplane Theorem, there exists a vector p ≠ 0such that

py⋅(0 −≥ y )0 for all y ∈Y . By free disposal, yy10= −∈δ Y for all vectors δ > 0. Hence

n 01 py⋅() − y =⋅= pδδ∑ pii ≥0. i=1

This holds for all δ > 0. Setting δ j = 0 for all ji≠ and δi =1, it follows that pi ≥ 0 for each in=1,...,. .

If in addition 0∈Y , then, by the Supporting Hyperplane Theorem py⋅0 ≥ p ⋅=00.

Q.E.D.

Thus price-guided production decisions can always be used to achieve any efficient production plan if there is free disposal and set of feasible plans is convex.

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 (zz1 ,...,m ) to produce a single output q.

If plant j operates at activity level x j it can produce ax0 jj units of output using axij j units of input ii,= 1,..., m.

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 (zz1 ,...,m ) to produce a single output q.

If plant j operates at activity level x j it can produce ax0 jj units of output using axij j units of input ii,= 1,..., m.

Summing over the n plants, total output is

n n ∑ ax0 jj and the total input i requirement is ∑ axij j . j=1 j=1

The production vector y=( − zq ,) is then feasible if it is in the following set.

Y =−{(zq , ) | x ≥ 0, q ≤⋅ a0 x ,Α x ≤ z }. (1.1-1)

Class Exercise: Show that the free disposal assumption holds.

The production set for the special case of two inputs and two plants is depicted in Fig. 1.1-6. As we shall see below, 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 the two creases, both plants are in operation. Note that each point on the boundary lies on one or more planes. Thus there is a supporting plane for every such boundary point.

We now show that this is true for all linear models. Fig. 1.1-6: Production Set

Existence of supporting prices

For any input vector z , let q , be the maximum possible output1. Formally,

q= Max{ q =⋅ a0 x |A x ≤≥ z , x 0} (1.1-2) x

Thus (−zq ,) is a boundary point of the production set Y . Since the production set Y is convex and the free disposal assumption holds, there exists a positive supporting price vector (,rp )such that

pq−⋅ r z ≥ pq −⋅ r z for all (−∈zq ,) Y (1.1-3)

______

1Since all the constraints are weak inequality constraints X is closed. We assume that the feasible set X=≥≤{ xx | 0,A x z }is bounded. Then X is a compact set. Thus the maximum exists.

Lemma: If the following assumption is satisfied, then the supporting output price, p, must be strictly positive.

Assumption: The feasible set has a non-empty interior

There exists some xˆ >> 0 such that zˆ ≡A xzˆ <<

Proof: Define qaxˆˆ=0 ⋅ . Given the above assumption {−∈zqˆ ,)ˆ Y . Therefore by the Supporting Hyperplane Theorem

pq−⋅ r z ≥ pqˆ −⋅ r zˆ (1.1-4)

We have already argued that p ≥ 0. To prove that it is strictly positive, we suppose that p = 0 and obtain a contradiction. First note that, if p = 0 it follows from (1.1-4) that rz⋅≤⋅ rzˆ

Also, since (,rp )> 0, if p = 0 then r > 0. Therefore, since zzˆ << , rz⋅<⋅ˆ rz.

But this contradicts our previous conclusion. Thus p cannot be zero after all. Then, dividing by p and defining the supporting input price vector λ =rp/0 ≥ , condition (1.1-3) can be rewritten as follows.

q−⋅≥−⋅λλ zq z for all (−∈zq ,) Y (1.1-5)

Q.E.D.

Lemma: If the following assumption is satisfied, then the supporting output price, p, must be strictly positive.

Assumption: The feasible set has a non-empty interior

There exists some xˆ >> 0 such that zˆ ≡A xzˆ <<

Proof: Define qaxˆˆ=0 ⋅ . Given the above assumption {−∈zqˆ ,)ˆ Y . Therefore by the Supporting Hyperplane Theorem

pq−⋅ r z ≥ pqˆ −⋅ r zˆ (1.1-6)

Lemma: If the following assumption is satisfied, then the supporting output price, p, must be strictly positive.

Assumption: The feasible set has a non-empty interior

There exists some xˆ >> 0 such that zˆ ≡A xzˆ <<

Proof: Define qaxˆˆ=0 ⋅ . Given the above assumption {−∈zqˆ ,)ˆ Y . Therefore by the Supporting Hyperplane Theorem

pq−⋅ r z ≥ pqˆ −⋅ r zˆ (1.1-7)

We have already argued that p ≥ 0. To prove that it is strictly positive, we suppose that p = 0 and obtain a contradiction. First note that, if p = 0 it follows from (1.1-4) that rz⋅≤⋅ rzˆ

Also, since (,rp )> 0, if p = 0 then r > 0. Therefore, since zzˆ << , rz⋅<⋅ˆ rz.

q−⋅≥−⋅λλ zq z for all (−∈zq ,) Y (1.1-8)

Q.E.D.

Characterization of the activity vector

Appealing to the Supporting Hyperplane Theorem we have shown that there exists a positive vector (rp , )= (λλ1 ,...m ,1)such that the boundary point (−zq ,) is profit maximizing. We now seek to use this result to characterize the associated profit- maximizing activity vector.

Proposition 1.1-4: Necessary conditions for a production plan to be on the boundary of the production set.

Let (−zq ,) be a point on the boundary of the linear production set. That is q= ax0 ⋅ where

x∈arg Max { a0 ⋅ x |A x ≤≥ z , x 0}. x Then, if the interior of the feasible set is non-empty, there exists a supporting price vector λ ≥ 0 such that

′ a0 −≤λ′A 0 (1.1-9) where x and λ satisfy the following “complementary slackness” conditions.

′ (i) (ax0 −=λ′A )0 and (ii) λ′(zx−=A )0.

Proof of (i):

Since (−zq ,) is profit maximizing given price vector (λ ,1) increasing x j by ∆ lowers profit

mm MRj− MC j = a00 j ∆−∑∑λλi a ij ∆=( aj −i a ij )0 ∆≤ . ii=11=

m Therefore aa0 j −≤∑ λi ij 0, jn=1,..., (*) i=1

In matrix notation a0 −≤λ′A 0

Proof of (i):

Since (−zq ,) is profit maximizing given price vector (λ ,1) , increasing x j by ∆ lowers profit.

mm MRj− MC j = a00 j ∆−∑∑λλi a ij ∆=( aj −i a ij )0 ∆≤ . ii=11=

m Therefore aa0 j −≤∑ λi ij 0, jn=1,..., (*) i=1

In matrix notation a0 −≤λ′A 0

Suppose x j > 0. Then lowering x j by ∆ also lowers profit for all ∆ sufficiently small.

mm MRj− MC j = a00 j () −∆ −∑∑λλia ij ()( −∆ = aj −i a ij )()0 −∆ ≤ ii=11=

m Therefore aa0 j −≥∑ λi ij 0. i=1

m Appealing to (*) it follows that aa0 j −=∑ λi ij 0. Q.E.D. i=1

Proof of (ii):

By construction

q= Max{ q =⋅≤ a0 x |}A x z x≥0 and

x∈arg Max { q =⋅≤ a0 x |A x z } x≥0

Define zx* = A . Since the activity vector x is feasible, z− Ax =−≥ z z* 0.

From the Supporting Hyperplane Theorem

qzqz−λλ′′* ≤− .

* * * Rearranging, this inequality, λ′(zz−≤ )0. But zz−≥0and λ ≥ 0 so λ′(zz−≥ )0. Combining these inequalities it follows that λλ′′(z− Ax ) = ( z −= z* )0. QED

Example: Two plants and 2 inputs

Consider the following 2 plant example. If plant 1 operates at the unit activity level it produces

1 a01 = 3 units of output and has an input requirement vector of (1,1) . If plant 2 operates at the unit

1 activity level it produces a02 = 2 a unit of output and has an input requirement of (4,1) . Then given the vector x of activity levels, total input requirements are

14x 1 Ax =  11x2

Maximum output with activity vector x is

11 qax=0 ⋅=32 x 12 + x.

The feasible outputs are depicted opposite. If only one plant is used, the production vector is on one of the creases in the boundary of the production set. Fig. 1.1-6: Production Set

Suppose that the vector of available inputs is z = (11, 5) . What is the maximum output of the firm? Note that to produce an additional unit of output requires increasing the activity level of plant 1 by three so the input requirement vector for each unit of output is

14  3   3 

zˆ1 =   =  . 11  0   3 

Similarly, for plant 2 the unit input requirement vector is

14  0  8 zˆ2 =  = . 11  2  2 Fig. 1.1-7: Isoquants These two input vectors are depicted in Fig. 1.1-7.

Using convex combinations of these two input vectors also yields 1 unit of output. Thus the line joining

zˆ1 and zˆ2 is a line of equal quantity or “isoquant.”

Since the constraints are all linear, the production set must therefore be as depicted in Fig. 1.1-6. The input output vector (,)zq lies on the boundary of the production set.

Since the set is convex there are supporting prices. That is, for some output price p ≥ 0and input price vector r ≥ 0, and any feasible vector (,zq ) pq−⋅≤ r z pq −⋅ r z . Fig. 1.1-7: Isoquants

Class Exercise: Solve for the input-output price vector (r ,1) that supports the production plan y=−=−( zq , ) ( 11, 2) .

HINT: Write down MR() x j - MC() x j for each plant.

Linear Programming Problem

Max{ a0 ⋅ x |A x ≤≥ z , x 0} x

Necessary conditions for a maximum

If x∈arg Max { a0 ⋅ x |A x ≤≥ z , x 0}and the interior of the feasible set is non-empty, there exists a x shadow price vector λ ≥ 0 such that

′ a0 −≤λ′A 0 where x and λ satisfy the following “complementary slackness” conditions.

′ (i) (ax0 −=λ′A )0 and (ii) λ′(zx−=A )0.

Proof: Identical to the proof of Proposition 1.1-4.