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.
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).
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
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