part 3 linear programming 3.3 theoretical analysis
TRANSCRIPT
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Part 3 Linear Programming
3.3 Theoretical Analysis
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Matrix Form of the Linear Programming Problem
( )
where
|
Thus, the linear programming problem
can be written as
min
. .
0
0
T
Bm m n m m
D
T T TB D
T TB B D D
B D
B
D
f
s t
Ax b c x
xA B D x
x
c c c
c x c x
Bx Dx b
x
x
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LP Solution in Matrix Form
1
1 1
1 1
1 1
1
For a basic solution
| ; ;
For any solution
Let
T T T TB B B B
B D
T TB B D D
T TB D D D
T T TB D B D
T T TD D B
f
f
x x 0 x B b c x
x B b B Dx
c x c x
c B b B Dx c x
c B b c c B D x
r c c B D
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Tableau in Matrix Form
1 ( ) 1
1
1 1( ) 1
11
Initial Tableau
0 0
The canonical form
m n m m m m n m mT T T
n B D
m m m n m mT T
m D B
A b B D b
c c c
I B D B b
0 r c B b
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Criteria for Determining A Minimum Feasible Solution
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10 20 0
1 2 1
1, 1 1 10
2, 1 2 20
Suppose we have a basic feasible solution
0 0 0
together with a tableau having an identity matrix
appearing in the 1st m columns
1 0 0
0 1 0
0
T TB m
m m n
m n
m n
y y y
y y y
y y y
x 0
a a a a a b
, 1 0
0 1 10 2 20 0
0 1
The corresponding objective functionm m mn m
m m
y y y
f c y c y c y
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1 10 11
2 20 21
0 1
If arbitrary values are assigned to the
nonbasic variables, we can easily solve
for basic variables as
the basic variables from the objective func.
n
j jj m
n
j jj m
T
m
x y y x
x y y x
f
f c f
c x
1 1 2 2 2
1 1 2 2
where
and 1, 2, ,
m m m m m
n n n
j j j m mj
x c f x
c f x
f c y c y c y
j m m n
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Theorem (Improvement of Basic Feasible Solution)
• Given a non-degenerate basic feasible solution with corresponding objective function f0, suppose for some j there holds cj-fj<0. Then there is a feasible solution with objective value f<f0.
• If the column aj can be substituted for some vector in the original basis to yield a new basic feasible solution, this new solution will have f<f0.
• If aj cannot be substituted to yield a basic feasible solution, then the solution set K is unbounded and the objective function can be made arbitrarily small (negative) toward minus infinity.
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Optimality Condition
If for some basic feasible solution cj-fj or rj is larger than or equal to zero for all j, then the solution is optimal.
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Symmetric Form of Duality (1)
1 1
1 1
1
Primal
max
. .
Tn n
m n n m
n
f
s t
x c x
A x b
x 0
1 1
1 1
1
Dual
min
. .
Tm m
Tn m m n
m
g
s t
y b y
A y c
y 0
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Symmetric Form of Duality (2)
1. MAX in primal; MIN in dual.2. <= in constraints of primal; >= in constraints of dual.3. Number of constraints in primal = Number of variable in
dual4. Number of variables in primal = Number of constraints i
n dual5. Coefficients of x in objective function = RHS of constrai
nts in dual6. RHS of the constraints in primal = Coefficients of y in d
ual 7. f(xopt)=g(yopt)
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Symmetric Form of Duality (3)
1 1
1 1
1
Primal
min
. .
Tn n
m n n m
n
f
s t
x c x
A x b
x 0
1 1
1 1
1
Dual
max
. .
Tm m
Tn m m n
m
g
s t
y b y
A y c
y 0
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ExampleBatch
Reactor ABatch
Reactor BBatch
Reactor C
Raw materialsR1, R2, R3, R4
ProductsP1, P2, P3, P4
P1 P2 P3 P4 capacity time
A 1.5 1.0 2.4 1.0 2000
B 1.0 5.0 1.0 3.5 8000
C 1.5 3.0 3.5 1.0 5000profit /batch
$5.24 $7.30 $8.34 $4.18
time/batch
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Example: Primal Problem 1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
1 2 3 4
max 5.24 7.30 8.34 4.18
. .
1.5 2.4 2000
5 3.5 8000
3 3.5 5000
, , , 0
where , , and are the number of batches
of P
f x x x x
s t
x x x x
x x x x
x x x x
x x x x
x x x x
x
1, P2, P3 and P4 to be manufactured.
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Example: Dual Problem 1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
min 2000 8000 5000
. .
1.5 y 1.5 5.24
y 5 + 3.0 7.30
2.4y 3.5 8.34
3.5 4.18
, , 0
where , and are the changes in optiaml
tota
g y y y
s t
y y
y y
y y
y y y
y y y
y y y
x
l profit w.r.t. changes in capacity time
for batch reactors A, B and C.
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Property 1
For any feasible solution to the primal problem and any feasible solution to the dual problem, the value of the primal objective function being maximized is always equal to or less than the value of the dual objective function being minimized.
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Proof
1 1
1 1 1 1
1 1 1 1
1 1 1 1
1 1 1 1
or
m n n m
T Tm m n n m m
T T Tn m m n m m n n
T Tm m n n n n
T Tm m n n
A x b
y A x y b
A y c y A c
y A x c x
y b c x
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Property 2
ˆIf is a feasible solution to the primal problem
ˆand is a feasible solution to the dual problem
such that
ˆ ˆ
ˆthen is an optimal solution to the primal problem
and
T T
x
y
c x b y
x
ˆ is an optimal solution to the dual problem.y
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Proof
According to the previous property, no feasible
ˆcan make larger than . Since achieves this value,
it is optimal (maximum).
Similarly, no feasible can bring by below the number
, and
T T
T
x
c x b y x
y
c x ˆ any that achieves this minimum must be optimal.y
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Duality Theorem
If either the primal or dual problem has a finite optimal solution, so does the other, and the corresponding values of objective functions are equal. If either problem has an unbounded objective, the other problem has no feasible solution.
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Additional Insights
If there is an optimal solution to the primal problem,
we can prove by construction that there is an optimal
solution to the dual problem. In fact, if is the basis
matrix for the primal problem corr
B
1
esponding to an
optimal solution and if contains the prices in the basis,
then the optimal solution to the dual is
B
T TB
c
y c B
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Symmetric Form of Duality (3)
1 1
1 1
1
Primal
min
. .
Tn n
m n n m
n
f
s t
x c x
A x b
x 0
1 1
1 1
1
Dual
max
. .
Tm m
Tn m m n
m
g
s t
y b y
A y c
y 0
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LP Solution in Matrix Form
1
1 1
1 1
1 1
1
For a basic solution
| ; ;
For any solution
Let
T T T TB B B B
B D
T TB B D D
T TB D D D
T T TB D B D
T T TD D B
f
f
x x 0 x B b c x
x B b B Dx
c x c x
c B b B Dx c x
c B b c c B D x
r c c B D
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Relations associated with the Optimal Feasible Solution of the
Primal problem
1
1
1
1
let
This vector is a basic feasible solution
of dual problem. Since
In addition,
T T TD D B
T TB
T T T T
T T T T TB B B D
T T TB B Bg
r c c B D 0
y c B
y A y B D y B y D
c c B D c c c
y y b c B b c x
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Example
1 2 3
1 2 3
1 2 3
1 2
1 2
1 2
1 2
max 4 3
. .
2 2 4
2 2 6
min 4 6
. .
2 1
2 2 4
2 3
x x x
s t
x x x
x x x
y y
s t
y y
y y
y y
PRIMAL
DUAL
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1 2 3 4 5
4
5
1 2 3 4 5
2
5
1 2 3 4 5
2
3
2 2 1 1 0 4
1 2 2 0 1 6
1 4 3 0 0 0
1 1 1/ 2 1/ 2 0 2
1 0 1 1 1 2
3 0 1 2 0 8
3/ 2 1 0 1 1/ 2 1
1 0 1 1 1 2
2 0 0 1 1 10
x x x x x
x
x
x x x x x
x
x
x x x x x
x
x
1B
1TB
c B
1
2
3
1
2
0
1
2
1
1
x
x
x
y
y
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Tableau in Matrix Form
1 ( ) 1
1
1 1( ) 1
11
Initial Tableau
0 0
The canonical form
m n m m m m n m mT T T
n B D
m m m n m mT T
m D B
A b B D b
c c c
I B D B b
0 r c B b
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1 2
1
1
1
1
1 11 2
1 11
1 11
11
T T TD D B
T T T TD D B B
T T T TD B B
T T TD B B
T T TD B
r c c B D
c c c B D c B D
c 0 c B D c B I
c c B D c B
c c B D y
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Example: The Primal Diet Problem
How can we determine the most economical diet that satisfies the basic minimum nutritional requirements for good health? We assume that there are available at the market n different foods that the ith food sells at a price ci per unit. In addition, there are m basic nutritional ingredients and, to achieve a balanced diet, each individual must receive at least bj unit of the jth nutrient per day. Finally, we assume that each unit of food i contains aji units of the jth nutrient.
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Primal Formulation
1
1
min
. .
1,2, ,
or
nT
i ii
n
ji i ji
c x
s t
a x b
j m
c x
Ax b
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The Dual Diet Problem
Imagine a pharmaceutical company that produces in pill form each of the nutrients considered important by the dietician. The pharmaceutical company tries to convince the dietician to buy pills, and thereby supplies the nutrients directly rather than through purchase of various food. The problem faced by the drug company is that of determining positive unit prices y1, y2, …, ym for the nutrients so as to maximize the revenue while at the same time being competitive with real food. To be competitive with the real food, the cost a unit of food made synthetically from pure nutrients bought from the druggist must be no greater than ci, the market price of the food, i.e. y1 a1i + y2 a2i + … + ym ami <= ci.
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Dual Formulation
1
1
max
. .
1,2, ,
or or
mT T
j jj
m
ji j ij
T T T
b y
s t
a y c
i n
y b b y
y A c A y c
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Shadow Prices
How does the minimum cost change if we change the right hand side b?
If the changes are small, then the corner which was optimal remains optimal. The choice of basic variables does not change. At the end of simplex method, the corresponding m columns of A make up the basis matrix B.
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1minimum cost ( *)
Thus, a small shift of size changes
the minimum cost by ( *) . The
solution * to the dual problem gives
the rate of change of minimum cost of
the primal problem wrt ch
T TB
T
c B b y b
b
y b
y
anges in .b