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© John Riley 3 October 2012
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
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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 1( ,..., )nq q q= using m different inputs 1( ,..., )mz z z= .
A particular production plan is then the input-output vector ( , )z q m n++∈
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.”
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Production function and production set
Consider a firm producing q units of a single output.
Let ( )f z be the maximum output that a firm can produce using input vector z.
Therefore output must satisfy the constraint ( )q f z≤ .
The production set Y is the set of feasible plans so
{( , ) | ( , ) 0, ( )}Y z q z q q f z= ≥ ≤ .
Example:
1/2{( , ) | ( , ) 0, 2 }Y z q z q q z= ≥ ≤ ,
equivalently,
2{( , ) | 0, 4 0}Y z q z z q= ≥ − ≥
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New notation: treat inputs as negative numbers
production vector1
1 1 1( ,..., ) ( ,..., , ,..., )n m m ny y y z z q q+= = − −
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
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New notation: treat inputs as negative numbers
production vector1
1 1 1( ,..., ) ( ,..., , ,..., )n m m ny y y z z q q+= = − −
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:
2{( , ) | 0, 4 0}Y z q z z q= ≥ − ≥ .
Using the new notation, a production vector is 1 2( , ) ( , )y y z q= − .
The production set is
21 2 1 1 2{( , ) | 0, 4 0}y y y y y= ≤ − − ≥Y
This set is depicted in Fig 1.1-11b.
__________________
1For a vector, our convention is that if j jy y≥ for all j we write y y≥ . If, in addition, the inequality is strict for some j we write y y>
and if the inequality is strict for all j we write y y>> .
Fig.1.1-1b: Production set
Fig.1.1-1a: Production set
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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 y y> .
*
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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 y y> .
Profit maximization
Let 1( ,..., )m np p p += be the vector of m input and n output prices.
total revenue TR = 1
m n
i ii m
p y+
= +∑
total cost TC = 1
m
i ii
p z=∑
1( )
m
i ii
p y=
= −∑ .
profit 1 1
m n m
i i i ii m i
p y p z p yπ+
= + =
= − = ⋅∑ ∑ .
A production vector is profit maximizing if it solves
{ | }y
Max p y y⋅ ∈Y
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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 1p .
Suppose, furthermore, that the plant is part of a large firm.
Objective: Produce 02y units of output efficiently.
For delivery to another “down-stream” plant within the firm.
From the figure, 01 1z y= − units of the input is efficient.
*
Fig.1.1-2a: Production set
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© John Riley 3 October 2012
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 1p .
Suppose, furthermore, that the plant is part of a large firm.
Objective: Produce 02y units of output efficiently.
For delivery to another “down-stream” plant within the firm.
From the figure, 01 1z y= − units of the input is efficient.
Transfer price
A price 2p , paid for each unit delivered to the downstream division.
The plant manager’s bonus will be based on the profit 1 1 2 2( )y p y p yπ = + .
Fig.1.1-2a: Production set
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Two contour sets (or “iso-profit” lines)
of 1 1 2 2( )y p y p yπ = + are depicted in Fig. 1.1-2a.
The steepness of any such line is the input-output
price ratio 1 2/p p .
As shown, the ratio is too low since profit is maximized
at a point to the North-West of 0y .
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
Fig.1.1-2a Transfer price too high
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Unfortunately, this approach does not always work.
Consider Fig. 1.1-3.
Suppose, once again that the output target is 02y units.
While 0y is locally profit maximizing, the profit
0 0 01 1 2 2p y p y p y⋅ = +
is negative. Profit is maximized by producing nothing.
*
Fig.1.1-3: No optimal transfer price
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Unfortunately, this approach does not always work.
Consider Fig. 1.1-3.
Suppose, once again that the output target is 02y units.
While 0y is locally profit maximizing, the profit
0 0 01 1 2 2p y p y p y⋅ = +
is negative. Profit is maximized by producing nothing.
Convex production sets
Y is convex if for any 0 1,y y ∈Y every convex combination 0 1(1 )y y yλ λ λ≡ − + ∈Y
As we shall see, prices can be used to support all efficient production plans if the production set is
convex.
Fig.1.1-3: No optimal transfer price
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Proposition 1.1-1: Supporting Hyperplane Theorem
Suppose n⊂ Y is non-empty and convex and 0y lies on the boundary of Y . Then there exists
0p ≠ such that (i) for all y∈Y , 0p y p y⋅ ≤ ⋅ and (ii) for all int ,y∈ Y 0.p y p y⋅ < ⋅
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 1h∈ , that is { | ( ) 0}y h y= ≥Y .
Since 0y is on the boundary of 0( ) 0h y =Y, .
As long as the gradient vector is non-zero at 0y , the linear approximation of h at 0y is
0 0 0( ) ( ) ( ) ( )hh y h y y y yy∂
= + ⋅ −∂
.
Note that ( )h y and ( )h y have the same value and gradient at 0y
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In two dimensions, the contour set of the linear
approximation is the tangent plane as depicted in Fig. 1.1-4.
If the upper contour set 0{ | ( ) ( )}y h y h y≥Y = 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 0 0( ) ( ) ( ) ( ) 0hh y h y y y yy∂
≥ ⇒ ⋅ − ≥∂
.
*
Fig.1.1-4: Supporting hyperplane
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In two dimensions, the contour set of the linear
approximation is the tangent plane as depicted in Fig. 1.1-4.
If the upper contour set 0{ | ( ) ( )}y h y h y≥Y = 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 0 0( ) ( ) ( ) ( ) 0hh y h y y y yy∂
≥ ⇒ ⋅ − ≥∂
.
Formally, we have the following Lemma.
Lemma 1.1-2: If 0{ | ( ) ( )}y h y h y= ≥Y is convex then 0 0( ) ( ) 0h y y yy∂
∈ ⇒ ⋅ − ≥∂
y Y .
To prove Proposition 1.1-1, choose 0( )hp yy∂
= −∂
. Appealing to the lemma,
0( ) 0p y y∈ ⇒ − ⋅ − ≥y Y that is 0y p y p y∈ ⇒ ⋅ ≤ ⋅Y .
Fig.1.1-4: Supporting hyperplane
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Lemma 1.1-2: If 0{ | ( ) ( )}y h y h y= ≥Y is convex then 0 0( ) ( ) 0h y y yy∂
∈ ⇒ ⋅ − ≥∂
y Y .
Proof: Pick any y in Y . Since Y is convex, all convex combinations of 0y and y lie in Y . That is,
for all (0,1)λ∈ ,
0 0( ) ( ) 0 ( ) ( ) 0h y h y h y h yλ− ≥ ⇒ − ≥ where 0(1 )y y yλ λ λ= − + .
**
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Lemma 1.1-2: If 0{ | ( ) ( )}y h y h y= ≥Y is convex then 0 0( ) ( ) 0h y y yy∂
∈ ⇒ ⋅ − ≥∂
y Y .
Proof: Pick any y in Y . Since Y is convex, all convex combinations of 0y and y lie in Y . That is,
for all (0,1)λ∈ ,
0 0( ) ( ) 0 ( ) ( ) 0h y h y h y h yλ− ≥ ⇒ − ≥ where 0(1 )y y yλ λ λ= − + .
Define 0 0 0( ) ( ) ((1 ) ) ( ( ))g h y h y y h y y yλλ λ λ λ≡ = − + = + − .
Then
0 1 0 0( ) (0) ( ( )) ( ) 0g g h y y y h yλ λ
λ λ− + − −
= ≥ , for all (0,1)λ∈ .
*
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Lemma 1.1-2: If 0{ | ( ) ( )}y h y h y= ≥Y is convex then 0 0( ) ( ) 0h y y yy∂
∈ ⇒ ⋅ − ≥∂
y Y .
Proof: Pick any y in Y . Since Y is convex, all convex combinations of 0y and y lie in Y . That is,
for all (0,1)λ∈ ,
0 0( ) ( ) 0 ( ) ( ) 0h y h y h y h yλ− ≥ ⇒ − ≥ where 0(1 )y y yλ λ λ= − + .
Define 0 0 0( ) ( ) ((1 ) ) ( ( ))g h y h y y h y y yλλ λ λ λ≡ = − + = + − .
Then
0 1 0 0( ) (0) ( ( )) ( ) 0g g h y y y h yλ λ
λ λ− + − −
= ≥ , for all (0,1)λ∈ .
Note that the limit of the left-hand side as 0λ → is the derivative of ( )g λ evaluated at 0λ = . Taking this derivative we obtain
0 0 0( ) ( ( )) ( )dg h y y y y yd y
λ λλ
∂= + − ⋅ −∂
.
Therefore 0 0( ) ( ) 0h y y yy∂
⋅ − ≥∂
.
Q.E.D.
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Example: Firm with two outputs
2 212 3 1 2 34{ | , 0, ( ) 0}y y y h y y y y= ≥ = − − − ≥Y .
The point 0 ( 25,8,3)y = − is on the boundary of this set and the gradient vector at this point is
0 0 012 32( ) ( 1, , 2 ) ( 1, 4, 6)h y y y
y∂
= − − − = − − −∂
.
**
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Example: Firm with two outputs
2 212 3 1 2 34{ | , 0, ( ) 0}y y y h y y y y= ≥ = − − − ≥Y .
The point 0 ( 25,8,3)y = − is on the boundary of this set and the gradient vector at this point is
0 0 012 32( ) ( 1, , 2 ) ( 1, 4, 6)h y y y
y∂
= − − − = − − −∂
.
Since the function ( )h y is the sum of three concave functions it is concave (and hence quasi-concave).
Define 0( ) (1,4,6)hp yy∂
= − =∂
. Then by the Lemma, the plane 0{ | }y p y p y⋅ = ⋅ is a supporting
plane.
*
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Example: Firm with two outputs
2 212 3 1 2 34{ | , 0, ( ) 0}y y y h y y y y= ≥ = − − − ≥Y .
The point 0 ( 25,8,3)y = − is on the boundary of this set and the gradient vector at this point is
0 0 012 32( ) ( 1, , 2 ) ( 1, 4, 6)h y y y
y∂
= − − − = − − −∂
.
Since the function ( )h y is the sum of three concave functions it is concave (and hence quasi-concave).
Define 0( ) (1,4,6)hp yy∂
= − =∂
. Then by the Lemma, the plane 0{ | }y p y p y⋅ = ⋅ is a supporting
plane.
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We can easily check this directly. To produce the output 2 3( , )y y , the minimum input requirement is
2 211 2 34y y y− = + . With the price vector (1,4,6)p = , the profit of the firm is
2 211 2 3 2 3 2 34( ) 4 6 4 6y y y y y y y yπ = + + = − − + + .
It is readily confirmed that profit is maximized at 2 3( , ) (8,3)y y = . Then the profit maximizing input
is 2 211 1 24 25y y y− = + = .
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From supporting hyperplanes to supporting prices
For 0p ≠ and 0y , a boundary point of Y , the hyperplane 0p y p y⋅ = ⋅ is a supporting hyperplane if
y∈ ⇒Y 0p y p y⋅ ≤ ⋅ .
To have a direct economic interpretation we must have 0p > .
Free Disposal Assumption
For any feasible production plan y∈Y and any 0δ > , 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− .
*
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From supporting hyperplanes to supporting prices
For 0p ≠ and 0y , a boundary point of Y , the hyperplane 0p y p y⋅ = ⋅ is a supporting hyperplane if
y∈ ⇒Y 0p y p y⋅ ≤ ⋅ .
To have a direct economic interpretation we must have 0p > .
Free Disposal Assumption
For any feasible production plan y∈Y and any 0δ > , 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 0y is a boundary point of a convex set Y and the free disposal assumption holds then there exists a
price vector 0p > such that 0p y p y⋅ ≤ ⋅ for all y∈Y . Moreover, if 0∈Y , then 0 0p y⋅ ≥ .
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Proof:
Appealing to the Supporting Hyperplane Theorem, there exists a vector 0p ≠ such that 0( ) 0p y y⋅ − ≥ for all y∈Y . By free disposal, 1 0y y δ= − ∈Y for all vectors 0δ > . Hence
0 1
1( ) 0
n
i ii
p y y p pδ δ=
⋅ − = ⋅ = ≥∑ .
This holds for all 0δ > . Setting 0jδ = for all j i≠ and 1iδ = , it follows that 0ip ≥ for each
1,...,.i n= .
If in addition 0∈Y , then, by the Supporting Hyperplane Theorem 0 0 0p y p⋅ ≥ ⋅ = .
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.
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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 1( ,..., )mz z to produce a single output q.
If plant j operates at activity level jx it can produce 0 j ja x units of output using ij ja x units of input
, 1,...,i i m= .
*
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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 1( ,..., )mz z to produce a single output q.
If plant j operates at activity level jx it can produce 0 j ja x units of output using ij ja x units of input
, 1,...,i i m= .
Summing over the n plants, total output is
01
n
j jj
a x=∑ and the total input i requirement is
1
n
ij jj
a x=∑ .
The production vector ( , )y z q= − is then feasible if it is in the following set.
0{( , ) | 0, , }z q x q a x x z= − ≥ ≤ ⋅ ≤ΑY . (1.1-1)
Class Exercise: Show that the free disposal assumption holds.
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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
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Existence of supporting prices
For any input vector z , let q , be the maximum possible output1. Formally,
0{ | , 0}x
q Max q a x x z x= = ⋅ ≤ ≥A (1.1-2)
Thus ( , )z q− 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 ( , )r p such that
pq r z pq r z− ⋅ ≥ − ⋅ for all ( , )z q− ∈Y (1.1-3)
_______
1Since all the constraints are weak inequality constraints X is closed. We assume that the feasible set
{ | 0, }X x x x z= ≥ ≤A is bounded. Then X is a compact set. Thus the maximum exists.
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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 ˆ 0x >> such that ˆz x z≡ <<A
Proof: Define 0ˆ ˆq a x= ⋅ . Given the above assumption ˆˆ{ , )z q− ∈Y . Therefore by the Supporting
Hyperplane Theorem
ˆ ˆpq r z pq r z− ⋅ ≥ − ⋅ (1.1-4)
We have already argued that 0p ≥ . To prove that it is strictly positive, we suppose that 0p = and
obtain a contradiction. First note that, if 0p = it follows from (1.1-4) that ˆr z r z⋅ ≤ ⋅
Also, since ( , ) 0r p > , if 0p = then 0r > . Therefore, since z z<< , ˆr z r z⋅ < ⋅ .
But this contradicts our previous conclusion. Thus p cannot be zero after all. Then, dividing by p and
defining the supporting input price vector / 0r pλ = ≥ , condition (1.1-3) can be rewritten as follows.
q z q zλ λ− ⋅ ≥ − ⋅ for all ( , )z q− ∈Y (1.1-5)
Q.E.D.
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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 ˆ 0x >> such that ˆz x z≡ <<A
Proof: Define 0ˆ ˆq a x= ⋅ . Given the above assumption ˆˆ{ , )z q− ∈Y . Therefore by the Supporting
Hyperplane Theorem
ˆ ˆpq r z pq r z− ⋅ ≥ − ⋅ (1.1-6)
*
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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 ˆ 0x >> such that ˆz x z≡ <<A
Proof: Define 0ˆ ˆq a x= ⋅ . Given the above assumption ˆˆ{ , )z q− ∈Y . Therefore by the Supporting
Hyperplane Theorem
ˆ ˆpq r z pq r z− ⋅ ≥ − ⋅ (1.1-7)
We have already argued that 0p ≥ . To prove that it is strictly positive, we suppose that 0p = and
obtain a contradiction. First note that, if 0p = it follows from (1.1-4) that ˆr z r z⋅ ≤ ⋅
Also, since ( , ) 0r p > , if 0p = then 0r > . Therefore, since z z<< , ˆr z r z⋅ < ⋅ .
But this contradicts our previous conclusion. Thus p cannot be zero after all. Then, dividing by p and
defining the supporting input price vector / 0r pλ = ≥ , condition (1.1-3) can be rewritten as follows.
q z q zλ λ− ⋅ ≥ − ⋅ for all ( , )z q− ∈Y (1.1-8)
Q.E.D.
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Characterization of the activity vector
Appealing to the Supporting Hyperplane Theorem we have shown that there exists a positive
vector 1( , ) ( ,... ,1)mr p λ λ= such that the boundary point ( , )z q− 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 ( , )z q− be a point on the boundary of the linear production set. That is 0q a x= ⋅ where
0arg { | , 0}x
x Max a x x z x∈ ⋅ ≤ ≥A .
Then, if the interior of the feasible set is non-empty, there exists a supporting price vector 0λ ≥ such
that
0 0a λ′ ′− ≤A (1.1-9)
where x and λ satisfy the following “complementary slackness” conditions.
(i) 0 ( ) 0a xλ′ ′− =A and (ii) ( ) 0z xλ′ − =A .
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Proof of (i):
Since ( , )z q− is profit maximizing given price vector ( ,1)λ increasing jx by ∆ lowers profit
0 01 1
( ) 0m m
j j j i ij j i iji i
MR MC a a a aλ λ= =
− = ∆ − ∆ = − ∆ ≤∑ ∑ .
Therefore 01
0m
j i iji
a aλ=
− ≤∑ , 1,...,j n= (*)
In matrix notation 0 0a λ′− ≤A
*
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Proof of (i):
Since ( , )z q− is profit maximizing given price vector ( ,1)λ , increasing jx by ∆ lowers profit.
0 01 1
( ) 0m m
j j j i ij j i iji i
MR MC a a a aλ λ= =
− = ∆ − ∆ = − ∆ ≤∑ ∑ .
Therefore 01
0m
j i iji
a aλ=
− ≤∑ , 1,...,j n= (*)
In matrix notation 0 0a λ′− ≤A
Suppose 0jx > . Then lowering jx by ∆ also lowers profit for all ∆ sufficiently small.
0 01 1
( ) ( ) ( )( ) 0m m
j j j i ij j i iji i
MR MC a a a aλ λ= =
− = −∆ − −∆ = − −∆ ≤∑ ∑
Therefore 01
0m
j i iji
a aλ=
− ≥∑ .
Appealing to (*) it follows that 01
0m
j i iji
a aλ=
− =∑ . Q.E.D.
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Proof of (ii):
By construction
00{ | }
xq Max q a x x z
≥= = ⋅ ≤A
and
00arg { | }
xx Max q a x x z
≥∈ = ⋅ ≤A
Define *z x= A . Since the activity vector x is feasible, * 0z Ax z z− = − ≥ .
From the Supporting Hyperplane Theorem
*q z q zλ λ′ ′− ≤ − .
Rearranging, this inequality, *( ) 0z zλ′ − ≤ . But * 0z z− ≥ and 0λ ≥ so *( ) 0z zλ′ − ≥ . Combining
these inequalities it follows that
*( ) ( ) 0z Ax z zλ λ′ ′− = − = .
QED
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Example: Two plants and 2 inputs
Consider the following 2 plant example. If plant 1 operates at the unit activity level it produces 1
01 3a = units of output and has an input requirement vector of (1,1). If plant 2 operates at the unit
activity level it produces 102 2a = a unit of output and has an input requirement of (4,1) . Then given
the vector x of activity levels, total input requirements are
1
2
1 41 1
xx
x
=
A
Maximum output with activity vector x is
1 10 1 23 2q a x x 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
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Suppose that the vector of available inputs is (11,5)z = .
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
1
1 4 3 3ˆ
1 1 0 3z = =
.
Similarly, for plant 2 the unit input requirement vector is
2
1 4 0 8ˆ
1 1 2 2z = =
.
These two input vectors are depicted in Fig. 1.1-7.
Fig. 1.1-7: Isoquants
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Using convex combinations of these two input vectors
also yields 1 unit of output. Thus the line joining
1z and 2z 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 ( , )z q lies on the boundary of the production set.
Since the set is convex there are supporting prices. That is,
for some output price 0p ≥ and input price vector 0r ≥ ,
and any feasible vector ( , )z q pq r z pq r z− ⋅ ≤ − ⋅ .
Class Exercise: Solve for the input-output price vector ( ,1)r that supports the production plan
( , ) ( 11,2)y z q= − = − .
HINT: Write down ( )jMR x - ( )jMC x for each plant.
Fig. 1.1-7: Isoquants
Essential Microeconomics -40-
© John Riley 3 October 2012
Linear Programming Problem
0{ | , 0}x
Max a x x z x⋅ ≤ ≥A
Necessary conditions for a maximum
If 0arg { | , 0}x
x Max a x x z x∈ ⋅ ≤ ≥A and the interior of the feasible set is non-empty, there exists a
shadow price vector 0λ ≥ such that
0 0a λ′ ′− ≤A
where x and λ satisfy the following “complementary slackness” conditions.
(i) 0 ( ) 0a xλ′ ′− =A and (ii) ( ) 0z xλ′ − =A .
Proof: Identical to the proof of Proposition 1.1-4.