Part 3 Linear Programming
3.4 Transportation Problem
ai  amount available at origin i
b j  amount required at destination j
xij  amount to be shipped from origin i to
destination j
cij  cost of shipping one unit from origin i to
destination j
The Transportation Model
m
n
min f  x    cij xij
i 1 j 1
s.t.
n
x
j 1
ij
m
x
i 1
ij
 ai ;
i  1, 2,
,m
 bj ;
j  1, 2,
,n
xij  0;
i  1, 2,
m
Note that
, m;
j  1, 2,
,n
n
a  b .
i 1
i
j 1
j
Thus, there are m  n -1 basic variables.
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
x11  x12

 x1n
x21
 x22

 x2 n
xm1  xm 2
 x m1
 xm 2
 x21
x11
 x22
x12
x1n
 a1
 a2
 x2 n

 xmn
 am
 b1
 b2
 xmn
 bn
Problem Structure
a
Ax  r   
b
 1T


A


 I nn
x   x11





1T 
I nn 
1T
I n n
x12
x1n
x21
x2 n

 1 1 1

n

1  1 1 1
1
a   a1 a2
am  ; b   b1 b2
T
T
xm1

1


T
bn 
T
xmn 
T
Model Parameters
Since the problem structure is fixed, it is
only necessary to specify a, b and
 c11

C

 cm1
c1n 


cmn 
Transformation of Standard Form of
Transportation Problem into the Primal Form
Given
min cT x
s.t.
Ax  b
x0
We can write it in the equivalent form
min cT x
s.t.
Ax  b
 Ax   b
x0
which is in the primal form but with
A
coefficient matrix  
A 
Asymmetric Form of Duality
z
let y   
w
b
b 
 b 
A
A 
A 
Dual problem can be written as
max y T b
T
T

z
b

w
b

s.t.
y T A  cT ; y  0
 or
zT A  w T A  cT ; z  0; w  0 
Let y  z  w
max y T b
s.t.
y T A  cT
y not restricted to be nonnegative
Dual Transportation Problem
n
 m

max   ai ui   b j v j 
j 1
 i 1

s.t.
ui  v j  cij
i  1, 2, , m; j  1, 2,
ui and v j unbounded
,n
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
D1
D2
12
O1
x11
3
x12
7
O2
O3
Amount
required
D3
x21
x13
x22
8
D4
8
4
4
a1=7
x14
6
x23
7
Amount
Available
9
x24
3
6
x31
x32
x33
x34
b1=4
b2=8
b3=11
b4=6
a2=10
a3=12
min 12 x11  3x12  8 x13  4 x14  7 x21  4 x22  6 x23  9 x24  8 x31  7 x32  3x33  6 x34 
s.t.
x11  x12
 x13
 x14

x21
 x22
 x23
 x24
 x22
x12
i  1, 2, 3
j  1, 2, 3, 4
 x33
 x34
 12
 4
 8
 11
 x34

 x33
 x24
x14
 x32
 x32
 x23
x13
xij  0
 10
x31
 x31
 x21
x11
7
6
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
8
4
7
3
7
4
5
6
10
5
8
7
3
6
4
9
8
6
6
11
6
12
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
Mx  d
and
M yc
T
Basis Triangularity
• Basis Triangularity Theorem: Every basis of
the transportation problem is triangular.
Given a basis B, the simplex multipliers y
can be founc by
y T B = cTB or BT y = c B
u
where y =  
v
If xij is basic, then the corresponding column in
A will be included in B. This column has exactly
two  1 entries:
(i) ith position of the top portion
(ii) jth position of the bttom portion.
 ui  v j  cij
Step 3: Find a basic feasible solution
of the dual problem – initial guess
This step is done by testing if the corresponding
simplex multipliers are feasible in the dual problem.
Notice that y T B  cTB . Thus,
to one of the constraints
u1  v1  12  Due
in the primal problem is
redundant!

u1  v2  3
 7 variables
u2  v2  4 
 6 equations
u2  v3  6 
 set u1  0
u3  v3  3 

u3  v4  6 
u2  1
u3  2
v1  12

v2  3
v3  5
v4  8
Step 3
v1=12
u1 = 0
u2 = 1
u3 = -2
12
4
v2=3
v3=5
v4=8
12 3
3
3 5
7 4
VIOLATION 5
4 6
5
6 9
7 3
OK
6
3 6
6
OK
13
10
8 1
VIOLATION
4
8 8
8
4
VIOLATION
9
10
OK
11
7
6
12
6
Cycle of Change
v1
v2
v3
v4
u1
-1 c11 +1 c12
x11
x12
0
u2
c23
+1 c21 -1 c22
0
x22
x23
0
c31
u3
0
a1
c24
a2
c33
x33
b2
c14
0
c32
0
b1
c13
b3
c34
x34
b4
a3
Selection of the New Basic
Variable
The change in the objective function of primal problem is
f  c21  c11  c12  c22
 c21   u1  v1    u1  v2    u2  v2 
 c21  u2  v1
Thus,
if c21  u2  v1 a. Constraint of the dual problem is satisfied.
b. Objective function of the primal problem
cannot be reduced.
if c21  u2  v1 a. Constraint of the dual problem is violated.
b. Objective function of the primal problem
can be reduced.
Step 4: Find a basic feasible solution of
the dual problem – Loop identification
Loop 1: 21  11  12  22  21
u2  v1  c21  13  7  6  -f  4  6  24
Loop 2: 14  34  33  23  22  12  14
u1  v4  c14  8  4  4  -f  3  4  12
Loop 3: 31  11  12  22  23  33  31
u3  v1  c31  10  8  2  -f  4  2  8
Step 4:
Move 4 unit around loop 1
v1=6
v2=3
v3=5
v4=8
u1 = 0
6
0
12 3
7
3 5
0
8 8
0
4
u2 = 1
7
4
7 4
1
4 6
5
6 9
0
9
u3 = -2
4
0
8 1
0
7 3
6
3 6
6
6
4
8
11
7
10
12
6
Repeat Step 3
u1  v2  3 
 v2  3


u2  v1  7
u2  1


u2  v2  4 
 v1  6
 u1  0  
u2  v3  6 
v

5
3

u3  2
u3  v3  3 


u3  v4  6 
 v4  8
Violation: Cell 14
Repeat Step 4:
Move 5 unit around the loop
v1=6
v2=3
v3=1
v4=4
12 3
2
3 5
0
8 8
5
4
u1 = 0
6
0
7 4
6
4 6
0
6 9
0
9
u2 = 1
7
4
u3 = 2
4
0
8 1
0
7 3
11
3 6
1
6
4
8
11
NO VIOLATION!!!
7
10
6
12
Solution
x11  x13  x23  x24  x31  x32  0
x12  2; x14  5; x21  4; x22  6; x33  11; x34  1
f *  3  2  4  5  7  4  4  6  3  11  6  1
 117
g *  7  0  10  1  12  2  4  6  8  3  11  1  6  4
=117
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
csik  cold stream i in interval k ;
hs jl  hot stream j in interval l ;
aik  the heat needed by csik ;
b jl  the heat available from hs jl ;
qikjl  the heat transferred from hs jl to csik .
Transportation Formulation
C
L
H
L
min  cikjl qikjl
qikjl
i 1 k 1 j 1 l 1
s.t.
H
L
 q
j 1 l 1
C
ikjl
L
 q
i=1 k 1
ikjl
i  1, 2, , C
 aik ;
k  1, 2, , L
 b jl ;
j  1, 2, , H
l  1, 2, , L
qikjl  0 for all i, j, k and l
Cost Coefficients
cikjl  0 for i and j are both process streams and match
is allowed (i.e. k  l );
 0 for i and j are both utility streams (i  C , j  H );
 1 only i or j is a unitlity stream;
 M otherwise, where M is very large (think infinity) number.
Additional Constraints
H 1 L
aCL   b jl
j 1 l 1
C 1 L
bHL   aik
i 1 k 1
C 1 L
H 1 L
aCL   aik  bHL   b jl
i 1 k 1
j 1 l 1
Solution