Part 3 Linear Programming

3.4 Transportation Problem

amount available at origin

amount required at destination

amount to be shipped from origin to

destination

cost of shipping one unit from origin to

i

j

ij

ij

a i

b j

x i

j

c i

destination j

The Transportation Model

1 1

1

1

1 1

min

. .

; 1,2, ,

; 1,2, ,

0; 1,2, , ; 1,2, ,

Note that .

Thus, there are -1 basic variables.

m n

ij iji j

n

ij ij

m

ij ji

ij

m n

i ji j

f c x

s t

x a i m

x b j n

x i m j n

a b

m n

x

Theorem

A transportation problem always has a solution, but there is exactly one redundant equality constraint. When any one of the equality constraints is dropped, the remaining system of n+m-1 equality constraints is linearly independent.

Constraint Structure

11 12 1 1

21 22 2 2

1 2

11 21 1 1

12 22 2 2

1 2

n

n

m m mn m

m

m

n n mn n

x x x a

x x x a

x x x a

x x x b

x x x b

x x x b

Problem Structure

11 12 1 21 2 1

1 2 1 2

1 1 1 1 1 1 1 1

;

T

T

T

n n n n n n

T

n n m mn

T

T

n

T T

m n

x x x x x x x

a a a b b b

aAx r

b

1

1

A

1

I I I

x

1

a b

Model Parameters

11 1

1

Since the problem structure is fixed, it is

only necessary to specify , and

n

m mn

c c

c c

a b

C

Transformation of Standard Form of Transportation Problem into the Primal Form

Given

min

. .

We can write it in the equivalent form

min

. .

T

T

s t

s t

c x

Ax b

x 0

c x

Ax b

which is in the primal form but with

coefficient matrix

Ax b

x 0

A

A

Asymmetric Form of Duality

let

Dual problem can be written as

max

. .

;

or ; ;

Let

max

. .

not restricted to be nonnegativ

T T T

T T

T T T

T

T T

s t

s t

z b Ay b A

w b A

y b z b w b

y A c y 0

z A w A c z 0 w 0

y z w

y b

y A c

y

e

Dual Transportation Problem

1 1

max

. .

1,2, , ; 1,2, ,

and unbounded

m n

i i j ji j

i j ij

i j

a u b v

s t

u v c

i m j n

u v

Interpretation of the Dual Transportation Problem

Let us imagine an entrepreneur who, feeling that he can ship more efficiently, come to the manufacturer with the offer to buy his product at origins and sell it at the destinations. The entrepreneur must pay -u1, -u2, …, -um for the product at the m origins and then receive v1, v2, …, vn at the n destinations. To be competitive with the usual transportation modes, his prices must satisfy ui+vj<=cij for all ij, since ui+vj represents the net amount the manufacturer must pay to sell a unit of product at origin i and but it back again at the destination j.

Example

12

x11

3

x12

8

x13

4

x14

7

x21

4

x22

6

x23

9

x24

8

x31

7

x32

3

x33

6

x34

D1 D2 D3 D4

O1

O2

O3

Amountrequired

Amount Available

a1=7

a2=10

a3=12

b1=4 b2=8 b3=11 b4=6

11 12 13 14 21 22 23 24 31 32 33 34

11 12 13 14

21 22 23 24

31 32 33 34

11 21 31

12 22 32

13 23 33

14 24 34

min 12 3 8 4 7 4 6 9 8 7 3 6

. .

7

10

12

4

8

11

6

0 1, 2,3 ij

x x x x x x x x 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

x x x

x i

1, 2,3, 4j

Solution Procedure

• Step 1: Set up the solution table.

• Step 2: “Northwest Corner Rule” – when a cell is selected for assignment, the maximum possible value must be assigned in order to have a basic feasible solution for the primal problem.

Northwest Corner Rule

12

4

3

3

8 4

7 4

5

6

5

9

8 7 3

6

6

6

7

10

12

4 8 11 6

Triangular Matrix

• Definition: A nonsingular square matrix M is said to be triangular if by a permutation of its rows and columns it can be put in the form of a lower triangular matrix.

• Clearly a nonsingular lower triangular matrix is triangular according to the above definition. A nonsingular upper triangular matrix is also triangular, since by reversing the order of its rows and columns it becomes lower triangular.

How to determine if a given matrix is triangular?

1. Find a row with exactly one nonzero entry.

2. Form a submatrix of the matrix used in Step 1 by crossing out the row found in Step 1 and the column corresponding to the nonzero entry in that row. Return to step 1 with this submatrix.

If this procedure can be continued until all rows have been eliminated, then the matrix is triangular.

The importance of triangularity is the associated method of back substitution in solving

and T Mx d M y c

Basis Triangularity

• Basis Triangularity Theorem: Every basis of the transportation problem is triangular.

Given a basis , the simplex multipliers

can be founc by

or

where

If is basic, then the corresponding column in

will be included in . This column ha

T T TB B

ijx

B y

y B = c B y = c

uy =

v

A B s exactly

two 1 entries:

(i) th position of the top portion

(ii) th position of the bttom portion.

i j ij

i

j

u v c

Step 3: Find a basic feasible solution of the dual problem – initial guess

1 1

1 2

2 2

2 3

3 3

3 4

This step is done by testing if the corresponding

simplex multipliers are feasible in the dual problem.

Notice that . Thus,

12

37 variables

46 equat

6

3

6

T TB

u v

u v

u v

u v

u v

u v

y B c

2

3

1

21

3

4

1

2

12ions

3set 0

5

8

u

u

v

vu

v

v

Due to one of the constraintsin the primal problem is redundant!

Step 3

12 12

4

3 3

3

5 8

OK

8 4VIOLATION

13 7VIOLATION

4 4

5

6 6

5

9 9

OK

10 8 VIOLATION

1 7

OK

3 3

6

6 6

6

7

10

12

4 8 11 6

v1=12 v2=3 v3=5 v4=8

u1 = 0

u2 = 1

u3 = -2

Cycle of Change

-1 c11

x11

+1 c12

x12

c13

0

c14

0

+1 c21

0

-1 c22

x22

c23

x23

c24

0

c31

0

c32

0

c33

x33

c34

x34

v1 v2 v3 v4

u1

u2

u3

a1

a2

a3

b1 b2 b3 b4

Selection of the New Basic Variable

21 11 12 22

21 1 1 1 2 2 2

21 2 1

21 2 1

The change in the objective function of primal problem is

Thus,

if a. Constraint of the dual problem is satisfied.

b. Objective functi

f c c c c

c u v u v u v

c u v

c u v

21 2 1

on of the primal problem

cannot be reduced.

if a. Constraint of the dual problem is violated.

b. Objective function of the primal problem

can be reduced.

c u v

Step 4: Find a basic feasible solution of the dual problem – Loop identification

2 1 21

1 4 14

3 1 31

Loop 1: 21 11 12 22 21

13 7 6 - 4 6 24

Loop 2: 14 34 33 23 22 12 14

8 4 4 - 3 4 12

Loop 3: 31 11 12 22 23 33 31

10 8 2 - 4 2 8

u v c f

u v c f

u v c f

Step 4: Move 4 unit around loop 1

6 12

0

3 3

7

5 8

0

8 4

0

7 7

4

4 4

1

6 6

5

9 9

0

4 8

0

1 7

0

3 3

6

6 6

6

7

10

12

4 8 11 6

v1=6 v2=3 v3=5 v4=8

u1 = 0

u2 = 1

u3 = -2

Repeat Step 3

1 2 2

2 1 2

2 2 11

2 3 3

3 3 3

3 4 4

3 3

7 1

4 60

6 5

3 2

6 8

u v v

u v u

u v vu

u v v

u v u

u v v

Violation: Cell 14

Repeat Step 4: Move 5 unit around the loop

6 12

0

3 3

2

5 8

0

8 4

5

7 7

4

4 4

6

6 6

0

9 9

0

4 8

0

1 7

0

3 3

11

6 6

1

7

10

12

4 8 11 6

u1 = 0

u2 = 1

u3 = 2

v1=6 v2=3 v3=1 v4=4

NO VIOLATION!!!

Solution

11 13 23 24 31 32

12 14 21 22 33 34

*

*

0

2; 5; 4; 6; 11; 1

3 2 4 5 7 4 4 6 3 11 6 1

117

7 0 10 1 12 2 4 6 8 3 11 1 6 4

=117

x x x x x x

x x x x x x

f

g

Application – Minimum Utility Consumption Rates and Pinch

Points

Cerda, J., and Westerberg, A. W., “Synthesizing Heat Exchanger Networks Having Restricted Stream/Stream Matches Using Transportation Formulation,” Chemical Engineering Science, 38, 10, pp. 1723 – 1740 (1983).

Example - Given Data

Temperature Partition

Definitions

cold stream in interval ;

hot stream in interval ;

the heat needed by ;

the heat available from ;

the heat transferred from to .

ik

jl

ik ik

jl jl

ikjl jl ik

cs i k

hs j l

a cs

b hs

q hs cs

Transportation Formulation

1 1 1 1

1 1

i=1 1

min

. .

1,2, ,;

1,2, ,

1,2, ,;

1,2, ,

0 for all , , and

ikjl

C L H L

ikjl ikjlq

i k j l

H L

ikjl ikj l

C L

ikjl jlk

ikjl

c q

s t

i Cq a

k L

j Hq b

l L

q i j k l

Cost Coefficients

0 for and are both process streams and match

is allowed (i.e. );

0 for and are both utility streams ( , );

1 only or is a unitlity strea

ikjlc i j

k l

i j i C j H

i j

m;

otherwise, where M is very large (think infinity) number.M

Additional Constraints

1

1 1

1

1 1

1 1

1 1 1 1

H L

CL jlj l

C L

HL iki k

C L H L

CL ik HL jli k j l

a b

b a

a a b b

Solution