MA3484 Methods of Mathematical
Economics
School of Mathematics, Trinity College
Hilary Term 2015
Lecture 27 (March 20, 2015)
David R. Wilkins

Farkas’ Lemma (continued)
Proposition
Proposition FK-F05 Let n be a positive integer, let I be a non-empty finite set, let ϕ :
R n →
R be a linear functional on
R and, for each i ∈ I , let η i
:
R n →
Suppose that ϕ ( v ) ≥ 0 for all v ∈
R be a linear functional on
R n
.
R n with the property that n
,
η i
( v ) ≥ 0 for all i ∈ I . Then there exist non-negative real numbers g i for all i ∈ I such that ϕ = P g i
η i
.
i ∈ I

Farkas’ Lemma (continued)
Proof
We may suppose that I = { 1 , 2 , . . . , m } for some positive integer m . For each i ∈ I there exist real numbers
A i , 1
, A i , 2
, . . . , A i , n such that
η i
( v
1
, v
2
, . . . , v n
) = n
X
A i , j v j j =1 for i = 1 , 2 , . . . , m and for all real numbers v
1
, v
2
, . . . , v n
. Let A be the m × n matrix whose coefficient in the i th row and j th column is the real number A i , j for i = 1 , 2 , . . . , m and j = 1 , 2 , . . . , n .
Then an n -dimensional vector v ∈
R n satisfies η i
( v ) ≥ 0 for all i ∈ I if and only if A v ≥ 0 .

Farkas’ Lemma (continued)
There exists an n -dimensional vector c ∈
R n for all v ∈
R n
. Then c
T v ≥ 0 for all v ∈
R n such that then follows from Corollary FK-F04 that there exists y ϕ
∈
( v ) =
R m c
T satisfying A v ≥ 0. It such v that y
T
A = c
T and y ≥ 0 . Let g i
= ( y ) i for i = 1 , 2 , . . . , m . Then g i
≥ 0 for i = 1 , 2 , . . . , m and P g i
η i
= ϕ , as required.
i ∈ I

Farkas’ Lemma (continued)
Remark
The result of Proposition FK-F05 can also be viewed as a consequence of Proposition FK-F01 applied to the convex cone in the dual space
R n ∗ of the real vector space
R n generated by the linear functionals η i for i ∈ I . Indeed let C be the subset of
R n ∗ defined such that
C =
(
X g i
η i i ∈ I
: g i
≥ 0 for all i ∈ I
)
.
It follows from Proposition FK-C02 that C is a closed convex cone in the dual space
R n ∗ of
R n
. If the linear functional ϕ did not belong to this cone then it would follow from Proposition FK-F01 that there would exist a linear functional V :
R n ∗ property that V ( η i
→
) ≥ 0 for all i ∈ I and V ( ϕ ) < 0.
R with the

Farkas’ Lemma (continued)
But given any linear functional on the dual space of a given finite-dimensional vector space, there exists some vector belonging to the given vector space such that the linear functional on the dual space evaluates elements of the dual space at that vector (see
Corollary LA-16 in the discussion of linear algebra). It follows that there would exist v ∈
R n such that V ( ψ ) = ψ ( v ) for all ψ ∈
R n ∗
.
But then η i
( v ) ≥ 0 for all i ∈ I and ϕ ( v ) < 0. This contradicts the requirement that ϕ ( v ) ≥ 0 for all v ∈
R n satisfying η i
( v ) ≥ 0 for all i ∈ I . To avoid this contradiction it must be the case that ϕ ∈ C , and therefore there must exist non-negative real numbers g i for all i ∈ I such that ϕ = P i ∈ I g i
η i
.

Farkas’ Lemma (continued)
Corollary
Corollary FK-F06 Let n be a positive integer, let I be a non-empty finite set, let ϕ :
R n and, for each i ∈ I , let η i
:
R n
→
→
R
R be a linear functional on be a linear functional on
R
R n
.
n
,
Suppose that there exists a subset I
0 all v ∈
R n with the property that η i
( v of I such that
) ≥ 0 for all i ϕ
∈
(
I v
0
) ≥ 0 for
. Then for all i ∈ I such that there exist non-negative real numbers g i ϕ = P g i
η i and g i
= 0 when i 6∈ I
0
.
i ∈ I
Proof
It follows directly from Proposition FK-F05 that there exist non-negative real numbers
Let g i
= 0 for all i ∈ I \ I
0 g i for all i ∈ I
0
. Then ϕ = P such that ϕ = P i ∈ I
0 g i
η i
, as required.
g i
η i
.
i ∈ I
0

Farkas’ Lemma (continued)
Definition
A subset X is said to be a convex polytope if there exist linear functionals η
1
, η
2
, . . . , η m on
R n such that and real numbers s
1
, s
2
, . . . , s m
X = { x ∈
R n
: η i
( x ) ≥ s i for i = 1 , 2 , . . . , m } .

Optimizing Linear Functions on Convex Polytope (continued)
Let ( η i
: i ∈ I ) be a finite collection of linear functionals on
R n indexed by a finite set I , let s i be a real number for all i ∈ I , and let
X =
\
{ x ∈
R
: η i
( x ) ≥ s i
} .
i ∈ I
Then X is a convex polytope in
R n
. A point x of
R n the convex polytope X if and only if η i
( x ) ≥ s i belongs to for all i ∈ I .

Optimizing Linear Functions on Convex Polytope (continued)
Proposition
Proposition FK-07 Let n be a positive integer, let I be a non-empty finite set, and, for each i ∈ I , let η i
:
R n →
R be non-zero linear functional and let s i be a real number. Let X be the convex polytope defined such that
X =
\
{ x ∈
R
: η i
( x ) ≥ s i
} .
i ∈ I
(Thus a point x of
R n belongs to the convex polytope X if and only if η i
( x ) ≥ s i for all i ∈ I .) Let ϕ :
R linear functional on
R n
, and let x
∗ n →
R be a non-zero
∈ X . Then ϕ ( x
∗
) ≤ ϕ ( x ) for all i x ∈ X if and only if there exist non-negative real numbers g i
∈ I such that ϕ = P g i
η i and g i
= 0 whenever η i
( x
∗
) > s i
.
i ∈ I for all

Farkas’ Lemma (continued)
Proof
Let K = { i ∈ I : η i
( x
∗
) > s i
} . Suppose that there do not exist non-negative real numbers and g i g i for all i ∈ I such that ϕ = P g i
η i i ∈ I
= 0 when i ∈ K . Corollary FK-F06 then ensures that there must exist some v ∈ V such that η i
( v ) ≥ 0 for all i ∈ I \ K and ϕ ( v ) < 0. Then
η i
( x
∗
+ λ v ) = η i
( x
∗
) + λη i
( v ) ≥ s i for all i ∈ I \ K and for all λ ≥ 0. If i ∈ K then η i
( x
∗
) > s i
. The set K is finite. It follows that there must exist some real number λ
0 satisfying λ
0
> 0 such that η i
( x
∗
+ λ v ) ≥ s i for all i ∈ K and for all real numbers λ satisfying 0 ≤ λ < λ
0
.

Farkas’ Lemma (continued)
Combining the results in the cases when i ∈ I \ K and when i ∈ K , we find that η i therefore x
∗
+
(
λ x v
∗
+
∈
λ
X v ) ≥ s i for all i ∈ I for all real numbers and
λ
λ ∈ [0 , η
0 satisfying 0
], and
≤ λ ≤ λ
0
.
But ϕ ( x
∗
+ λ v ) = ϕ ( x
∗
) + λϕ ( v ) < ϕ ( x
∗
) whenever λ > 0. It follows that the linear functional ϕ cannot attain a minimum value in X at any point x
∗ for which either
K = I or for which K is a proper subset of I but there exist non-negative real numbers g i
P ϕ = for all g i
η i
. The result follows.
i ∈ I \ K such that i ∈ I \ K

Farkas’ Lemma (continued)
Strong Duality
Example
Consider again the following linear programming problem in general primal form:— find values of x
1
, x
2
, x
3 objective function and x
4 so as to minimize the c
1 x
1
+ c
2 x
2
+ c
3 x
3
+ c
4 x
4 subject to the following constraints:— a
1 , 1 x
1
+ a
1 , 2 x
2
+ a
1 , 3 x
3
+ a
1 , 4 x
4
= b
1
; a
2 , 1 x
1
+ a
2 , 2 x
2
+ a
2 , 3 x
3
+ a
2 , 4 x
4
= b
2
; a
3 , 1 x
1 x
1
+ a
3 , 2 x
≥ 0 and x
3
2
+ a
≥ 0 .
3 , 3 x
3
+ a
3 , 4 x
4
≥ b
3
;

Farkas’ Lemma (continued)
Now the constraint a
1 , 1 x
1
+ a
1 , 2 x
2
+ a
1 , 3 x
3
+ a
1 , 4 x
4
= b
1 can be expressed as a pair of inequality constraints as follows: a
1 , 1 x
1
+ a
1 , 2 x
2
+ a
1 , 3 x
3
+ a
1 , 4 x
4
− a
1 , 1 x
1
− a
1 , 2 x
2
− a
1 , 3 x
3
− a
1 , 4 x
4
≥ b
1
≥ − b
1
.
Similarly the equality constraint involving b
2 can be expressed as a pair or inequality constraints.

Farkas’ Lemma (continued)
Therefore the problem can be reformulated as follows:— find values of x
1
, x
2
, x
3 objective function and x
4 so as to minimize the c
1 x
1
+ c
2 x
2
+ c
3 x
3
+ c
4 x
4 subject to the following constraints:— a
1 , 1 x
1
+ a
1 , 2 x
2
+ a
1 , 3 x
3
+ a
1 , 4 x
4
≥ b
1
;
− a
1 , 1 x
1
− a
1 , 2 x
2
− a
1 , 3 x
3
− a
1 , 4 x
4
≥ − b
1
; a
2 , 1 x
1
+ a
2 , 2 x
2
+ a
2 , 3 x
3
+ a
2 , 4 x
4
≥ b
2
;
− a
2 , 1 x
1
− a
2 , 2 x
2
− a
2 , 3 x
3
− a
2 , 4 x
4
≥ − b
2
; a
3 , 1 x
1
+ a
3 , 2 x
2
+ a
3 , 3 x
3
+ a
3 , 4 x
4
≥ b
3
; x
1
≥ 0 ; x
3
≥ 0 .

Farkas’ Lemma (continued)
Let ϕ ( x
1
, x
2
, x
3
, x
4
) = c
1 x
1
+ c
2 x
2
+ c
3 x
3
+ c
4 x
4
, and let
η
1
+
( x
1
, x
2
, x
3
, x
4
) = a
1 , 1 x
1
+ a
1 , 2 x
2
+ a
1 , 3 x
3
+ a
1 , 4 x
4
,
η
1
−
( x
1
, x
2
, x
3
, x
4
) = − η
1
( x
1
, x
2
, x
3
, x
4
) ,
η
+
2
( x
1
, x
2
, x
3
, x
4
) = a
2 , 1 x
1
+ a
2 , 2 x
2
+ a
η
2
−
( x
1
, x
2
, x
3
, x
4
) = − η
3
( x
1
, x
2
, x
3
, x
4
) ,
2 , 3 x
3
+ a
2 , 4 x
4
,
η
3
( x
1
, x
2
, x
3
, x
4
) = a
3 , 1 x
1
+ a
3 , 2 x
2
+ a
3 , 3 x
3
+ a
3 , 4 x
4
,
ζ
1
( x
1
, x
2
, x
3
, x
4
) = x
1
,
ζ
3
( x
1
, x
2
, x
3
, x
4
) = x
3
,

Farkas’ Lemma (continued)
Then ( x
1
, x
2
, x
3
, x
4
) is a feasible solution to the primal problem if and only if this element of
R
4 belongs to the convex polytope X , where X is the subset of
R
4 consisting of all points x of
R
4 that satisfy the following constraints:—
η
1
+
η
1
−
( x ) ≥ b
1
;
( x ) ≥ − b
1
;
η
+
2
( x ) ≥ b
2
;
η
2
−
( x ) ≥ − b
2
;
η
3
( x ) ≥ b
3
;
ζ
1
( x ) ≥ 0;
ζ
3
( x ) ≥ 0.

Farkas’ Lemma (continued)
An inequality constraint is said to be binding for a particular feasible solution x if equality holds in that constraint at the feasible solution. Thus the constraints on the values of η
+
1
, η
1
−
, η
+
2 and η are always binding at points of the convex polytope X , but the
−
2 constraints determined by η
3
, ζ
1 and ζ
3 need not be binding.
Suppose that the linear functional ϕ attains its minimum value at a point p
1
−
, p x
∗ of X , where x
∗
= ( x
∗
1
, x
∗
2
, x
∗
3
, x
∗
4
). It then follows from
Proposition FK-07 that there exist non-negative real numbers p
1
+
2
, p
−
2
, p
3
, q
1 and q
3 such that
+
, p
+
1
η
1
+
+ p
−
1
η
1
−
+ p
+
2
η
2
+
+ p
2
−
η
2
−
+ p
3
η
3
+ q
1
ζ
1
+ q
3
ζ
3
= ϕ.
Moreover p
3
= 0 if η
3
( x if ζ
3
( x
∗
) > 0.
∗
) > b
3
, q
1
= 0 if ζ
1
( x
∗
) > 0, and q
3
= 0

Farkas’ Lemma (continued)
Now η
−
1
= − η
1
+ and η
−
2
= − η
2
+
. It follows that p
1
η
+
1
+ p
2
η
+
2
+ p
3
η
3
+ q
1
ζ
1
+ q
3
ζ
3
= ϕ, where p
1
4
P a
3 , j x j
∗ i =1
= p
+
1
− p
1
−
> b
3
, q
1 and
= 0 if p x
2
∗
1
= p
+
2
− p
−
2
. Moreover p
3
> 0, and q
3
= 0 if x
∗
3
> 0.
= 0 if

Farkas’ Lemma (continued)
It follows that p
1 a
1 , 1
+ p
2 a
2 , 1
+ p
3 a
3 , 1
≤ c
1
, p
1 a
1 , 2
+ p
2 a
2 , 2
+ p
3 a
3 , 2
= c
2
, p
1 a
1 , 3
+ p
2 a
2 , 3
+ p
3 a
3 , 3
≤ c
3
, p
1 a
1 , 4
+ p
2 a
2 , 4
+ p
3 a
3 , 4 p
3
= c
4
,
≥ 0 .
Moreover p
3
= 0 if
4
P a
3 , j x j
∗
> b
3
,
3
P p i a i , 1
= c
1 if x
∗
1
> 0, and i =1 i =1
3
P p i a i , 3
= c
3 if x
∗
3
> 0. It follows that ( p
1
, p
2
, p
3
) is a feasible i =1 solution of the dual problem to the feasible primal problem.

Farkas’ Lemma (continued)
Moreover the complementary slackness conditions determined by the primal problem are satisfied. It therefore follows from the
Weak Duality Theorem (Theorem DT-05) that ( p
1
, p
2
, p
3
) is an optimal solution to the dual problem.