Duality.ppt

Duality
for
linear programming

Illustration of the notion
• Consider an enterprise
producing r items:
fk = demand for the item k =1,…, r
using s components:
hl = availability of the component l = 1,…, s
• The enterprise can use any of the n process (activities):
xj = level for using the process j = 1,…, n
cj = the unit cost for using the process j = 1,…, n
• The process j
produces ekj units of the item k =1,…, r
uses glj units of the component l = 1,…, s
for each unit of its use

• Consider an enterprise
producing r items:
fk = demand for the item k =1,…, r
using s components:
hl = availability of the component l =
1,…, s
• The enterprise can use any of the n process
(activities):
xj = level for using the process j = 1,…, n
cj = the unit cost for using the process j =
1,…, n
• The process j
produces ekj units of the item k
=1,…, r
uses glj units of the component l =
1,…, s
each time it is used at level 1
• The enterprise problem: determine the
level of each process for satisfying the
demand without exceeding the
availabilities in order to minimize the total
production cost.
• Model
1
1
1
min
S. t. 1,2,..., (demands)
1,2,..., (availabilities)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
 
 




• A business man makes an offer to buy all the components and to sell the
items required by the enterprise to satisfy the demands.
• He must state proper unit prices (to be determined) to make the offer
interesting for the enterprise:
vk item unit price k = 1, 2, … , r
wl component unit price l = 1, 2, …, s.
vk
wl
1
1
1
min
S.t. 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
 
 




The business man must state
proper unit prices (to be
determined) to make the offer
interesting for the enterprise
To complete its analysis, the
enterprise must verify that for
each process j, the cost of making
business with him is smaller or
equal than using the process j. But
the cost of making business with
him is equal to the difference
between buyng the items required
and selling the components
unused in order to simulate using
one unit of process j (cj ).
1 1
buying the selling the
items components
r s
kj k lj l j
k l
e v g w c
 

  ≤
1
1
1
min
S. t 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
 
 



vk
wl

• The business man problem is to maximize his profit while maintaining
the prices competitive for the enterprise
1 1
buying the selling the
items components
r s
kj k lj l j
k l
e v g w c
 

  ≤
1 1
1 1
max
S. t 1,2,...,
0 1,2,...,
0 1,2,...,
r s
k k l l
k l
r s
kj k lj l j
k l
k
l
p f v h w
e v g w c j n
v k r
w l s
 
 
 
  
 
 
 
 

• The enterprise problem: multiply the availability constraints by -1
1
1
1
min
S. t. 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
 
 



1
1
1
min
S. t. 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
   
 



1
 

1 1
1 1
max
S. t. 1,2,...,
0 1,2,...,
0 1,2,...,
r s
k k l l
k l
r s
kj k lj l j
k l
k
l
p f v h w
e v g w c j n
v k r
w l s
 
 
 
  
 
 
 
 
Enterprise problem
Business man problem
sj
j
rj
j
g
g
e
e






1
1
kn
kj
k
k e
e
e
e 

2
1
1 2 ln
l l lj
g g g g
     






 G
E
1
1
1
min
S.t. 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
   
 



T T
E G

 
 
1 1
j rj j sj
kj lj
e e e g g g
     
r
s
n
n
r s

1
1
1
min
S. t. 1,2,..., (demands)
0 1,2,...,
n
j j
j
n
kj j k
j
n
lj j l
j
j
z c x
e x f k r
g x h l s
x j n




 
   
 



1 1
1 1
max
S. t. 1,2,...,
0 1,2,...,
0 1,2,...,
r s
k k l l
k l
r s
kj k lj l j
k l
k
l
p f v h w
e v g w c j n
v k r
w l s
 
 
 
  
 
 
 
 
Primal
Dual
T T
T T
max
S. t.
, 0
v
p f h
w
v
E G c
w
v w
x
 
 
 
   
 
 
 
 
   
 

T
min
S. t.
0
z c x
E f
x
G h
x
w


   

   
 
 
 
 
 
 

T
min
S. t.
0
c x
Ax b
x


T
T
max
S. t.
0
b y
A y c
y



Primal dual problems
Linear programming problem specified with equalities
Linear programming in standard form
T
min
S. t.
0
c x
Ax b
x


T
min
S. t.
0
c x
Ax b
x


Primal problem Dual problem
Primal problem Dual problem
y x
y x
T
T
max
S. t.
b y
A y c

T
T
max
S. t.
0
b y
A y c
y



T
min
S. t.
0
c x
Ax b
x


T T
min 0
S. t.
0, 0
c x s
Ax Is b
x s

 
 
T
T
T
max
S. t.
0
b y
c
A y
I
   
   
  
 

T
T
max
S. t.
0
b y
A y c
Iy

 
T
T
max
S. t.
0
b y
A y c
y



Duality theorems
• It is easy to show that we can move from one pair of primal-dual problems
to the other.
• It is also easy to show that the dual of the dual problem is the primal
problem.
• Thus we are showing the duality theorems using the pair where the primal
primal is in the standard form:
T
min
S. t.
0
c x
Ax b
x


primal Dual
T
T
max
S. t.
b y
A y c


Duality theorems
• Weak duality theorem
If (i.e., x is feasible for the primal problem) and
if (i.e., y is feasible for the dual problem), then
Proof Indeed,
 
0
,
: 

 x
b
Ax
x
x
 
T
:
y y A y c
 
T T
b y c x

T T T T T
since and 0.
b y x A y x c A y c x
   

Duality theorems
• Corollary If and , and if
, then x* and y* are optimal solutions for the primal and dual
problems, respectively..
Proof It follows from the weak duality theorem that for any feasible
solution x of the primal problem
Consequently x* is an optimal solution of the primal problem.
We can show the optimality of y* for the dual problem using a similar
proof.
 
0
,
:
*


 x
b
Ax
x
x  
* T
:
y y A y c
 
T * T *
b y c x

T T * T *
.
c x b y c x
 

Duality theorems
• Strong duality theorem If one of the two primal or dual problem has a
finite value optimal solution, then the other problem has the same property,
and the optimal values of the two problems are equal. If one of the two
problems is unbounded, then the feasible domain of the other problem is
empty.
Proof The second part of the theorem follows directly from the weak
duality theorem. Indeed, suppose that the primal problem is unbounded
below, and thus cTx→ – ∞. For contradiction, suppose that the dual
problem is feasible. Then there would exist a solution ,
and from the weak duality theoren, it would follow that ; i.e., bTy
would be a lower bound for the value of the primal objective function cTx, a
contradiction.
 
T
:
y y A y c
 
T T
b y c x


Complementary slackness theory
• We now introduce new necessary and sufficient conditions for a pair of
feasible solutions of the primal and of the dual to be optimal for each of
these problems.
• Consider first the following pair of primal-dual problems.
T
min
S. t.
0
c x
Ax b
x


primal Dual
T
T
max
S. t.
b y
A y c
 x

• Complementary slackness theorem 1
Let x and y be feasible solution for the primal and the dual, respectively.
Then x and y are optimal solutions for these problems if and only if for all
j = 1,2,…,n
Poof First we prove the sufficiency of the conditions. Assume that
the conditions (i) et (ii) are satisfied for all j=1,2,…,n. Then
 
 
T
T
0
0
j j j
j j j
i x a y c
ii a y c x


  
  
T
[ ] 0 1, 2,...,
j j j
x a y c j n
    
T
1
Hence 0
n
j j j
j
x a y c


 
 
 

T
min
S. t.
0
c x
Ax b
x


T
T
max
S. t.
b y
A y c
 x
 


Consequently
and the corollary of the weak duality theorem implies that x et y are optimal
solutions for the primal and the dual problems, respectively.
T T T T T T T
1 1 1
But 0
n n n
j j j j j j j
j j j
x a y c x a y x c x A y c x b y c x
 
  
 
       
 
  
T
[ ] 0 1, 2,...,
j j j
x a y c j n
    
T
1
Hence 0
n
j j j
j
x a y c


 
 
 

T T
b y c x

 
T T T T
1 1 2 2
1
T
1
T
2
1 2
T
T T
, , ,
n
j j n n
j
n
n
x a y x a x a x a y
a
a
x x x y
a
x A y
   




 
   
 
 
 
 

 
 
 



Now we prove the necessity of the sonditions. Suppose that the
solutions x et y are optimal solutions for the primal and the dual problems,
respectively, and
Then referring to the first part of the theorem
T
T
Since 0 et 1,2,..., ,
it follows that 0 1,2,..., .
j j j
j j j
x a y c j n
x a y c j n


   
 
   
 
T T T T T T T
1 1 1
0
n n n
j j j j j j j
j j j
x a y c x a y x c x A y c x b y c x
 
  
 
       
 
  
 

T T
.
b y c x

The proof is completed.

• Now consider the other pair of primal-dual problems
• Complementary slackness theorem 2
Let x and y be feasible solution for the primal and dual problems,
recpectively. Then x and y are opyimal solutions of these problems
if and only if
for all j = 1,2,…,n for all i=1,2,…,m
T
min
S. t.
0
c x
Ax b
x


 
 
T
T
0
0
j j j
j j j
i x a y c
ii a y c x


  
  
 
  i
i
i
i
i
i
b
x
a
y
iv
y
b
x
a
iii








0
0
T
max
S. t.
0
T
b y
A y c
y


y x

Proof This theorem is in fact a corollary of the complementary slackness
theorem 1. Transform the primal problem into the standard form using the
slack variables si , i=1,2,…,m:
The dual of the primal problem in standard form
T
min
Sujet à
, 0
c x
Ax Is b
x s
 

T T
T T
max max
S. t S. t.
0 0
b y b y
A y c A y c
I y I y
  
  
T
min
S. t.
0
c x
Ax b
x



Use the result in the preceding theorem to this pair of primal-dual problems
For j=1,2,…,n
and for i=1,2,…,m
T
min
S. t.
, 0
c x
Ax Is b
x s
 

 
 
T
T
0
0
j j j
j j j
i x a y c
ii a y c x


  
  
 
  0
0
0
0








i
i
i
i
s
y
iv
y
s
iii
T
T
max
S. t.
0
b y
A y c
I y

 
x
s
y

For j=1,2,…,n
and for i=1,2,…,m
and then the conditions become
 
 
T
T
0
0
j j j
j j j
i x a y c
ii a y c x


  
  
 
  0
0
0
0








i
i
i
i
s
y
iv
y
s
iii
But i i i
s a x b

 
 
  i
i
i
i
i
i
b
x
a
y
iv
y
b
x
a
iii








0
0
T
min
S. t.
, 0
c x
Ax Is b
x s
 


Dual simplex algorithm
• The dual simplex method is an iterative procedure to solve a linear
programming problem in standard form.
1 1 2 2
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
min
S. t.
...
...
. . . .
. . . .
...
0 1,2,...,
n n
n n
n n
m m mn n m
j
z c x c x c x
a x a x a x b
a x a x a x b
a x a x a x b
x j n
  
   
   
   
 

• At each iteration, a basic infeasible solution of problem is available, except
at the last iteration, for which the relative costs of all variables are non
negatives.
• Exemple
min 3/ 2 1/ 2 27
S. t. 1/ 4 1/ 4 6/ 4
1/ 4 3/ 4 15/ 2
1/12 5/12 13/ 2
, , , , 0
z u h
x u h
u p h
y u h
x y u p h
  
   
   
  

basicvar. r.h.s.

Analyse one iteration of the dual simplex algorithm, and suppose that the
current solution is as follows:
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.

Leaving criterion
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
If 0 1,2, , , then the solution is feasible
and optimal. The algorithm stops.
i
b i m
  

Leaving criterion
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
 
1,2, ,
1
Otherwise, let min . If 0 1,2, , , then
the feasible domain is empty. Indeed, since
for all 0, 0 and
it is not possible to find 0 such that
r i rj
i m
n
rj j r
j
rj
b b a j n
x a x b
x a


   
   


1
= .
n
j r
j
x b



Leaving criterion
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
 
1,2, ,
Otherwise, let let min 0. is the leaving variable.
The pivot will be completed with an element in the row.
r
r i j
i m
th
b b x
r

 

Leaving criterion
1 1s s
r rs s
m ms s
b a x
b a x
b a x



m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
We select the entering variable in such a way that
1) the value of the leaving variable increases when
the value of increases
ii) the relative costs of all the variables remains non
nega
r
s
j
s
x
x
x
tive when the pivot on is completed to modify
the tableau.
rs
a
0
rs
a



Leaving criterion
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
0
rs
a  
nega
r
s
j
s
x
x
x
tive when the pivot on is completed to modify
the tableau.
rs
a
1 2 1
0 1 0 0
rn
r r r
rs rs rs rs rs
a
a a b
a a a a a
0,
1,2, ,
rj
j s
rs
a
c c
a
j n
 
 


Leaving criterion
0
rs
a  
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
negat
s
r
s
x
x
x
ive when the pivot on is completed to modify
the tableau.
rs
a
1 2 1
0 1 0 0
rn
r r r
rs rs rs rs rs
a
a a b
a a a a a
0,
1,2, ,
rj
j s
rs
a
c c
a
j n
 
 

If 0, then the value of can only increase
since 0 and 0.
rj j
s rs
a c
c a

 

Leaving criterion
0
rs
a  
m
i
c
n
j
c
i
j
j
,...,
2
,
1
0
,...,
2
,
1
0






basic var. r.h.s.
negat
s
r
s
x
x
x
ive when the pivot on is completed to modify
the tableau.
rs
a
1 2 1
0 1 0 0
rn
r r r
rs rs rs rs rs
a
a a b
a a a a a
0,
1,2, ,
rj
j s
rs
a
c c
a
j n
 
 

For all such that 0, we have to inforce the non negativity
of the relative cost by selecting properly the pivot element .
rj
rs
j a
a


Entering criterion
For all such that 0, we have to inforce the non negativity
of the relative cost; i.e.,
rj
j a 
0, such that 0
rj
j s rj
rs
a
c c j a
a
   
0, such that 0
j s
rj
rj rs
c c
j a
a a
   
1,2, ,
1,2, ,
Then the index of the entering variable is such that
max : 0 or min : 0
j j
s s
rj rj
j n
j n
rs rj rs rj
s
c c
c c
a a
a a a a


   
   
   
   
 
   
   
, such that 0
j s
rj
rj rs
c c
j a
a a
  

Pivot
• To obtain the simplex tableau associated with the new basis where the
entering variable xs remplaces the leaving variable xr we complete the pivot
on the element 0.
rs
a 

Exemple
• x is the leaving variable, and consequantly, the pivot is completed in the
first row of the tableau
• h is the entering variable, and consequently, the pivot is completed on the
element -1/4
• After pivoting, the tableau becomes
This feasible solution
is optimal
basic var.
basic var.
r.h.s.
r.h.s.

Convergence when the problem is non degenerate
• Non degeneracy assumption:
the relative costs of the non basic variables are positive at each iteration
• Theorem: Consider a linear programming problem in standard form.
If the matrix A is of full rank, and if the non degeneracy assumption is
verified, then the dual simplex algorithm terminates in a finite number of
iterations.
T
min
Subject to
0
, ,
matrix
n m
z c x
Ax b
x
c x R b R
A m n



 


• Proof:
Since the rank of matrix A is equal to m, then each basic feasible solution
includes m basic variables strictly positive (non degeneracy assumption).
But there is a finite number of ways to select columns among the
columns of to specify an sub matrix of :
!
! ( )!
m
n A m m A
n
n
m m n m

 

 

 
But the non feasible basis of are a subset of these. Then
!
! ( )!
is an upper bound on the number of non feasible basis of .
A
n
n
m m n m
A
 

 

 

• The influence of pivoting on the objective function during an iteration of
the simplex
→ s
c

rs
r
a
b
Substact from
since 0, 0, and 0 under the non degeneracy ass.
r
s
rs
r rs s
b
z z z c z
a
b a c
       
  
z

basic var. r.h.s.
Deviding row
by rs
r
a

Then and the value of the objective function increases stricly at
each iteration.
Consequently, the same basic non feasible solution cannot repeat during the
completion of the dual simplex algorithm.
Since the number of basic non feasible solution is bounded, it follows that
the dual simplex algorithm must be completed in a finite number of
iterations.
,
z z

since 0, 0, and 0 under the non degeneracy ass.
r
s
rs
r rs s
b
z z z c z
a
b a c
       
  

Comparing
(primal) simplexe alg. and dual simplexe alg.
Simplex alg.
Search in the feasible domain
Search for an entering variable to
reduce the value of the objective function
Search for a leaving variable preserving
the feasibility of the new solution
Stop when an optimal solution is found
or when the problem is not bounded
below
Dual simplex alg.
Search out of the feasible domain
Search for a leaving variable to eliminate
a negative basic variable
Search for an entering variable preserving
the non negativity of the relative costs
Stop when the solution becomes feasible
or when the problem is not feasible

Duality.ppt

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Duality.ppt