MSIS 685: Linear Programming
Lecture 7
10/22/98
Scribe: Konstantinos Moyssiadis
In this lecture will have some examples from the economic and financial world and we'll
try to give an interpretation of duality through them.
The arbitrage theorem in finance:
Let's see a random experiment with 1,2,…,m different outcomes and n wagers 1,2,…,n
The return matrix is: R  R nm
1 2 ……j…..m experiment
1
wager 2
:
i
:
n
rij
rij is the return (profit or loss) if the wager is
the ith one and the actual outcome is the jth one.
Let X=(x1, x2,…,xn) denote the wager vector.
Suppose that the j outcome occurred. The return in this case is:
ΣxiRij=xTrj
m
If the probability for the jth outcome is Pj (
 P  1)
j
then the expected return is:
j 1
Ep(X)=ΣPj(xTrj)
The theorem says: Either,
Pr
j ij
j
0
(1) There exists a probability vector P , Σpi=1, pi > 0 such that:
for all i.
(Rp=0)
Or
(2) There is a betting scheme X=(x1,x2,…,xn) such that: Σxirij=xTrij>0 for every outcome
j.
(Equivalent to Rx>0)
Example 1
Let's say that the outcomes are: 1,2,…,m and the wagers are: {1},{2},…,{n}
Let also RT denote the matrix:
 r1 - 1 - 1
 1 r2 - 1
RT  R  


 1
The question is: Is there a vector like:
P1 
P2 
 
 
 
Pm 
with ΣPi=1 which satisfies the equation:
 r1 - 1 - 1
 1 r2 - 1
RT  R  


 1
-1 
- 1


rm 
P1 
P2 
 
 
 
Pm 
0 
0 
=  
 
 
0 
The ith equation is:
-P1-P2-…+riPi-Pi+1-…-Pm=0
or
riPi=P1+P2+…+Pi-1+Pi+1+…Pm
-1 
- 1


rm 
or
riPi=1-Pi
or
Pi=1/(1+ri)
( 0 < Pi < 1 )
 1 r  1
1
with ΣPi = 1 or
i
Suppose you have a stock with present value $100 and the possible outcomes after a
certain period are:
Outcome 1: double  value = $200
Outcome 2: half  value = $50
Let C denote the price of the option.
Including C in the outcomes, we have:
Outcomes
200
50
Buy option
wagers
buy stock
50  C
 100

-C 
- 50
Does there a exist (P, 1-P) with P>0, and 1-P>0
such that:
50  C
 100

- C   P  0 

- 50 1  P  0
from which we get C=50/3
Let's try now to form the dual problem in the case of the diet problem
Food 1 …………………….. Food n
Cost per unit:
C1
Nutrients
Nut1
Minimum
Requirements
of nutrients
………………………………..
Cost n
……………………. Nutm
b1
…………………………………
bm
Let aij denote the amount of nutrient i in 1 unit of food j.
For the nutrient aI the amount offered by the food is:
ai1x1 + …. + ainxn > bi
xi > 0
(I=1,…,m)
the objective function is :
min c1x1 + …. + cnxn
for the dual problem we have:
subject to:
max b1y1 + …. + bmym
a1jy1 + … + amjym < cj
j=1, … ,n
yi > 0
where:
yi is the cost / unit of nutrient
The dual problem in the case of the transportation problem.
Supply
S1
:
:
:
:
:
Factories
Warehouses
1
1
Cij
Demand
d1
:
:
:
n
m
Sn
dm
Let Cij be the cost of sending an item from factory I to warehouse j.
The demand constraints are:
x
x
ij
 dj
j=1,…,m
ij
 Sj
i=1,…,n
i
And the supply constraints:
j
min ijcijxij
The objective function is:
The dual problem for this case is:
m
Max
n
 d u  s v
j
j
i 1
St.
i
i
i 1
uj +vi < cij
Where uj is the corresponding variable for the variable dj and vi is the corresponding
variable for the Si variable.