Part 3 Linear Programming

3.3 Theoretical Analysis

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

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

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

Criteria for Determining A Minimum Feasible Solution

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

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

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.

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.

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

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)

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

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

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.

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.

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.

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

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

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

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.

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

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

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

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

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

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

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

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

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.

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

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.

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

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.

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