Chapter 3
Transportation Problem
A lot of applications can be modeled as transportation problems. And because the
transportation problem is just a special type of linear programming problem, so it can be
solved by applying the simplex method as described in chapter 1. However, these problems
often grow very large and therefore more efficient solution procedure is needed. The special
structure of the transportation problem has enabled management scientists to develop
special-purpose solution procedure that greatly simplify the computations. That is
transportation simplex method. For some problems, which are unrelated to transportation
problems from their physical meanings, can also be solved by transportation simplex
method because they have been fitted to this special model structure.
3.1
Transportation Problem and Its Mathematical
Model
Let us begin with a cotton distribution problem.
Example 3.1 Company APD has three cotton distribution stations (A1, A2, and A3 )
which are responsible for distributing cotton for three textile companies (B1, B2, and B3 ).
The respective supplies of the three stations are 50kt, 45kt, and 65kt and the respective
demands of the three textile companies are 20kt, 70kt, and 70kt. The cost per unit of cotton
distributed from each station to each textile company is shown in table 3-1. The problem
now is to determine which plan for distributing the cotton would minimize the total cost for
Company APD.
Table 3—1
Textile Company
Distribution Station
A1
A2
A3
B1
B2
B3
4
6
2
8
3
5
5
6
7
Let
xij = amount distributed form Ai to Bj.
Then we have the distribution variables table as table 3-2.
Table 3—2
Textile Company
Distribution Station
A1
A2
A3
Demand（kt）
B1
B2
B3
Supply（kt）
x11
x21
x31
20
x12
x22
x32
70
x13
x23
x33
70
50
45
65
160
To describe this problem, two kinds of restrictions must be considered. One is the total
61
amount of cotton distributed from a station to each textile company must be equal to the
supply of the station, another is the total amount of cotton distributed from each station to a
textile company must be equal to the demand of the textile company. Because the total
supply equals the total demand, so all the restrictions are equations, namely
x11+x12+x13 = 50
x21+x22+x23 = 45
x31+x32+x33 = 65
x11+x21+x31 = 20
x12+x22+x32 = 70
x13+x23+x33 = 70
Let f denotes total distributing coat. Thus the mathematical model for the cotton
distribution problem can be formulated as follows:
min f= 4x11+8x12+5x13+6x21+3x22+6x23+2x31+5x32+7x33
x11+x12+x13 = 50
x21+x22+x23 = 45
x31+x32+x33 = 65
x11+x21+x31 = 20
x12+x22+x32 = 70
x13+x23+x33 = 70
xij≥0，i=1，2，3；j=1，2，3
To describe the general model for the transportation problem, we need to use terms that
are considerably less specific than those for the cotton distribution problem.
Thus, in general, source Ai (i=1，…，m) has a supply of ai units to distribute to the
destinations, and destination Bj (j=1，…，n) has a demand for bj units to be received from
the resources. The cost per unit distributed from source Ai to destination Bj is cij. Assume
that Σ ai= Σ bj. These input data can be summarized conveniently in the cost and
requirements table shown in Table 3-3. The problem is to determine which plan for
distributing the commodity would minimize the total distribution cost.
Table 3—3 Cost and Requirements Table for Transportation Problem
Destination
B1
B2
Bn
Supply
…
Source
A1
c11
c12
…
C1n
a1
A2
c21
c22
…
c2n
a2
…
…
…
…
…
…
Am
cm1
cm2
…
cmn
am
demand
b1
b2
…
bn
Σai
Σbj
By letting f be total distribution cost and xij be the number of units to be distributed
from source Ai to destination Bj, the linear programming formulation of this problem
becomes
62
m
n
min f    c ij x ij
i 1 j 1

i  1,  , m
 x ij  a i
 j 1

m
j  1,  n
 x ij  b j
 i 1
 x ij  0,
i  1,  , m;
j  1,  , n



n
Note that the resulting table of constraints coefficients has the special structure shown
in following matrix. Any linear programming problem that fits this special formulation is of
the transportation problem type, regardless its physical context.
x11 x12  x1n x21 x22  x2n  xm1 xm2  xmn
1 1  1



1 1  1







1 1  1

1

1
1




1
1
1









1
1
1





m






n



row
row
Any vector from the matrix above can be denoted by pij. Obviously, in pij only the i
element and the m+j element equal 1, all the other elements are equals zero) .
3.2
Transportation Simplex Method
3.2.1 Concepts of Transportation Problem Model
Because of the special structure, all such problems have the following properties.
Property 3.1 The transportation problem is certain to have optimal solution.
Obviously, the transportation problem has feasible solution and all the cost per unit cij
are greater than or equal to zero. So, the objective value corresponding to any feasible
solution must be greater than or equal to zero, namely the objective function has lower
boundary. So, the minimizing transportation problem is certain to have optimal solution.
Property 3.2 The rank of the matrix A that consists of constraints’ coefficients is
m+n－1, namely R（A）=m+n－1.
The property 3.1 tells us that we do not need to perform the test of no optimal solution
for transportation problems. The property 3.1 tells us that for any transportation problem
with m sources and n destinations, the number of its basic variables of any BF solution
equals m+n－1.
Definition 3.1 In distribution variables table, any set formed by following variables
xi1j1，xi1j2，xj2j3，…，xisjs，xisj1 , whose subscripts have a special relation, is called a closed
path (chain reaction、stepping-stone path). And any variable in the closed path is called
a corner. There are two closed paths shown in table 3-4.
63
Table 3—4
B1
B2
B3
B4
B5
B6
B7
A1
x11
x12
x13
x14
x15
x16
x17
A2
x21
x22
x23
x24
x25
x26
x27
A3
x31
x32
x33
X34
x35
x36
x37
Closed path 1：x11，x12，x22，x23，x33，x31，x11；
Closed path 2：x15，x16，x36，x37，x27，x25，x15；
The closed path has following features: ① There are two and only two corners in
each row or each column.② The line between two corners is vertical or horizontal.
Property 3.3 The sufficient and necessary condition of any m+n－1 variables xi1j1，
xi2j2，…，xisjs （s=m+n－1） being BF solution for the transportation problem is that any
subset of the variables cannot form a closed path.
This property tells us what kind of variable set can be a BF solution or whether or not a
feasible solution is a BF solution.
3.2.2 Transportation Simplex Method
Like any other linear programming problem, the optimal solution of transportation
problem exists in the BF solutions. So the procedure of transportation Simplex Method has
following three steps.
（1）Determine initial basic feasible solution
（2）Optimality testing
（3）Convert the BF solution
3.2.2.1 Determine Initial BF Solution
There are several methods for determining initial BF solution for transportation
problem, such as minimum-cost method、Vogel method and northwest corner rule. Here, we
just introduce the simplest method — minimum-cost method.
Now, we describe the steps of minimum-cost method through a simple example.
Example 3.2 A transportation problem has three sources (A1，A2，A3 ) and four
destinations (B1，B2，B3，B4). The supply of each source, the demand for each destination,
and the cost per unit distributed from each source to each destination are summarized in
table 3-5. Determine a plan for distributing the commodity would minimize the total
distribution cost.
Table 3—5
Destination
Source
A1
A2
A3
Demand
B1
B2
B3
B4
Supply
3
2
8
7
4
3
6
3
8
4
3
9
50
20
30
40
20
15
25
100
100
64
（1）Set up the transportation tableau as table 3-6；
（2）Identify the cell in the transportation tableau with the lowest cost, and allocate as
much number as possible to this cell (corresponding variable).
For example, in table 3—6, the cost per unit c21=2 is the lowest cost, so we write 20
units, the largest number we can write, in the cell corresponding to x21, which means
distributing 20 units of commodity from A2 to B1.
（3）Identify the cells in the row or the column corresponding to the cell in which we
have just written a number, and cross those cells whose distribution unit should be zero,
cross the cells just in row or in column, do not cross the cells in row and column
simultaneously.
In table 3-6, we have just written the number 20 in the cell corresponding to x21, that
means all the supply of source A2 has been distributed, so the distribution unit in the other
cells in the second row should be zero, then cross the corresponding cells.
（4）Continue with step (2) and step (3) for all unwritten and uncrossed cells.
（5）Write the appropriate number in the last unwritten and uncrossed row or column,
and just write number do not cross in the last unwritten and uncrossed row or column.
Table 3—6 Transportation Tableau
Destination
B1
B2
Source
A1
3
A2
2
A3
8
Demand
20
20
×
7
4
3
40
B3
×
5
6
×
3
20
×
10
8
20
B4
15
4
3
9
Supply
25
50
×
20
×
30
25
Now, we have a feasible solution for this transportation problem x11=20，x13=5，x14=25，
x21=20，x32=20，x33=10，the other xij=0. According Property 3.3, we know that this feasible
solution is a BF solution for the transportation problem. It can be proved that any feasible
obtained by minimum-cost method is a BF solution for the transportation problem. The
variables corresponding to those written cells are basic variables and the variables
corresponding to those crossed cells are nonbasic variables
3.2.2.2
Optimality Testing
Usually, There are two optimality testing methods: closed path method and index
method.
a．Closed Path Method
In the transportation tableau as table 3—6, we can draw such a closed path based on a
certain nonbasic variable that all the corners are basic variables except the nonbasic variable
we have chosen. It can be proved that such a closed path is the only path we can draw for a
certain nonbasic variable.
We can see that a new distribution plan obtained by adjusting the old plan in such
closed path is still a feasible solution for the transportation problem. In table 3-6, draw a
closed path based on nonbasic variable x12 shown in table 3-7.
65
Table 3—7
Destination
Source
A1
A2
A3
B1
3
2
20
20
B2
B3
B4
5
7
6
4
4
3
3
25
Supply
50
20
10
30
20
8
9
Demand
40
20
15
25
In the closed path in table 3-7, to remain the solution in the table still feasible, if x12
increases 1 unit of distribution, x13 must decease 1 unit of distribution, then x33 must
increase 1 unit of distribution, and then x32 must decease 1 unit of distribution.
Let us observe the change of objective value after such adjustment. When x12 increases
1 unit, then the cost (objective value) increases 7 units, x13 decreases 1 unit, then the cost
decreases 6 units, x33 increases 1 unit, then the cost increases 8 units, x32 decreases 1 unit,
then the cost decreases 7 units. So, the total change of objective value after adjustment is 7
－6+8－3=6, namely the objective value has increased 6 units. Therefore, such adjustment
is unprofitable to the transportation problem and choosing x12 as nonbasic variable is
correct.
The total change in objective value after adjusting 1 unit of distribution in such closed
path is actually the evaluation index of the corresponding nonbasic variable (e.g., the
evaluation index for x12 in table 3-7 is σ12=6). Since transportation problem is a
minimization problem, so the BF solution in transportation tableau is the optimal solution
only if all the evaluation indexes are greater than or equal to zero, otherwise, we must
convert the BF solution.
By using closed path method, the evaluation indexes of all nonbasic variables in table
3—6 are calculated in table 3-8.
Table3—8
Nonbasic variable
Closed path
Evaluation index
x12
X12→x13→x33→x32→x12
6
x22
x22→x21→x11→x13→x33→x32→x22
4
x23
x23→x21→x11→x13→x23
－2
x24
x24→x21→x11→x14→x24
0
x31
x31→x11→x13→x33→x31
3
x34
x34→x33→x13→x14→x34
3
8
3
Since the evaluation index for x23 is σ23=－2<0，so the BF solution in table 3—6 is not
the optimal solution of the transportation problem.
b．Index Method
The index method is usually used when the transportation problem is large and the
index method is a method for computer calculation.
Consider the following Transportation problem:
min f=CX
AX=b
X≥0
66
Assume that B is a feasible basis for the transportation problem, then the evaluation
indexes of variables should be calculated by following formulation
σij = cij－CBB-1pij
Since the dual problem for the transportation problem is
max z=Yb
YA≥C
Y (free)
Where Y=（u1，…，um，v1，…，vn）is the vector of dual variables and its elements correspond
to the m+n constraints.
According to dual theory
Y= CBB-1
So
σij = cij－Ypij
We know that in pij only the i element and the m+j element equal 1 but all the other
elements are equal to zero, namely, pij=ei+em+j
So
σij = cij－Ypij= cij－（u1，…，um，v1，…，vn）pij
=cij－（ui+vj）
And all the evaluation indexes of basic variables are equal to zero, so
cij－（ui+vj）= 0
Namely
ui+vj = cij
（i，j）∈I（The subscript set of basic variables）
Here, we call ui the row index of row i because it corresponds to the row i in the
transportation tableau and call vj the column index of column j because it corresponds to
the column j in the transportation tableau.
The steps of index method is as follows:
（1）List an equation according to formulation ui+vj = cij for each basic variable in the
transportation tableau, forming a system of equations.
（2）Choose any one of the indexes and determine its value( usually, choose u1=0) and
then solve the value of the other indexes from the system.
（3）Calculate the evaluation indexes of nonbasic variables according to formulation
σij =cij－（ui+vj）. If all the evaluation indexes of nonbasic variables are greater or equal to
zero, then the BF solution in the transportation tableau is the optimal solution of the
transportation problem.
Perform the optimality test for the BF solution in table 3—6 with index method.
Adding row and column index in table 3-6 leads to table 3-9.
Table 3—9
Destination
Row
B1
B2
B3
B4
Supply
Source
Index
A1
A2
A3
Demand
Column Index
3
2
20
20
8
7
6
4
3
20
3
40
v1
8
20
v2
5
3
10
15
v3
67
25
4
9
25
v4
50
u1
20
u2
30
u3
The system of index equations is
u1+v1=3
u1+v4=4
u3+v2=3
u1+v3=6
u2+v1=2
u3+v3=8
Choose u1=0，then solving the system above leads to the following value:
u1=0
v1=3
v2=1
u2=－1
v3=6
u3=2
v4=4
Calculate the evaluation index for each nonbasic variable
σ12 =c12－（u1+v2）=7－（0+1）= 4>0
σ22 =c22－（u2+v2）=4－（－1+1）= 2>0
σ23 =c23－（u2+v3）=3－（－1+6）= －2<0
σ24 =c24－（u2+v4）=3－（－1+4）= 0
σ31 =c31－（u3+v1）=8－（2+3）= 3>0
σ34 =c34－（u3+v4）=9－（2+4）= 3>0
Since the evaluation index for x23 is σ23=－2<0，so the BF solution in table 3—6 is not
the optimal solution of the transportation problem.
3.2.2.3
Convert the BF Solution
When there is negative evaluation index in the tableau, it means that we have not
obtained the optimal solution and we need to convert the BF solution. The steps of BF
solution converting is described as follows:
（1）Choose a variable xij whose evaluation index is σij <0（always the smallest one）
as entering variable.
（2）Draw a closed path based on xij. In the path, all the corners are basic variables
except the nonbasic variable xij. And give a number to each corner according to clockwise
or contra-clockwise ( begin with xij and give the number 1 to xij , then 2,3,4,……).
（3）Select the minimum distributing value among the corners whose number are even,
i.e., min｛
（xij）k|k is even｝, add the value to the allocation for each odd corner and subtract
the value from the allocation for each even corner, then a new BF solution can be obtained.
Convert the basic feasible solution in table 3—6 according to the steps above.
Since σ23 =－2<0，so x23 can be the entering basic variable. Based on x23 and
considering it as the first corner, draw the closed path shown in table 3-10.
Table 3—10
Destination
B1
Source
20
A1
3
20
A2
2
A3
Demand
8
B2
B4
5
7
6
4
3
3
40
B3
20
8
20
4
x23
10
15
68
25
Supply
50
20
3
30
9
25
In the closed path in table 3-10, the second corner (contra-clockwise) has the minimum
value 5 among the even corners. Adding 5 to the allocation for each odd corner and
subtracting 5 from the allocation for each even corner leads to a new BF solution of the
transportation problem shown in table 3-11.
Table 3—11
Destination
B1
Source
A1
A2
A3
Demand
3
2
25
15
8
B2
7
6
4
3
3
40
B3
20
8
20
B4
4
5
10
15
25
Supply
50
20
3
30
9
25
The new system of index equation is:
u1+v1=3
u2+v1=2
u3+v2=3
u1+v4=4
u2+v3=3
u3+v3=8
Choose u1=0，then solving the system above leads to the following value:
u1=0
v1=3
u2=－1
v2=－1
u3=4
v3=4
v4=4
Calculate the evaluation index for each nonbasic variable
σ12 =7－（0－1）= 8>0
σ24 =3－（－1+4）= 0
σ13 =6－（0+4）= 2>0
σ31 =8－（4+3）= 1>0
σ22 =4－（－1－1）= 6>0
σ34 =9－（4+4）= 1>0
Note that all the evaluation indexes are greater than or equal to zero, so the BF solution
in table 3-11 is the optimal solution, namely
x11=25，x14=25，x21=15，x23=5，x32=20，x33=10，other xij=0
The minimum value for objective function is
f*=3×25+4×25+2×15+3×5+3×20+8×10=360
Note from the procedure above that the solution procedure is actually the procedure of
solving linear programming problem, so it is called transportation simplex method.
Let us summarize the steps of transportation simplex method.
（1）Set up the transportation tableau.
（2）Determine an initial basic feasible solution using minimum-cost method in
transportation tableau.
（3）Perform optimality test with index method or closed path method. If σij≥0 for all
nonbasic variables, stop, you have reached the optimal solution. Otherwise, proceed to step
(4).
（4）Choose a variable whose evaluation index is less than zero（always the smallest
one）as entering variable, convert the BF solution using closed method, you can obtain a
new BF solution, and continue with step (3).
69
3.3
Total Supply Not Equal to Total Demand
Often the total supply is not equal to the total demand. In this case, we must do some
modification in the linear programming to make the problem have the same total supply and
the total demand before we solve it with transportation simplex method.
3.3.1 Total Supply Greater Than Total Demand
If total supply exceeds total demand(Σai >Σbj), the LP mathematical model would
be.
m
n
min f   cij xij
i 1 j 1

i  1,, m
 xij  ai
 j 1
 m
j  1, n
 xij  b j
 i 1
 xij  0, i  1,, m;
j  1,, n


n
In this case, a fictitious destination ( called the dummy destination) Bn+1 can be
introduced, the corresponding demand bn+1=Σai －Σbj. Let xi,n+1 be the number of units to
be distributed from source Ai to destination Bn+1, ci，n+1 equals zero or the cost per unit of
commodity stored by Ai. Because xi，n+1 is actually a slack variable that can be interpreted as
the undistributed supply in Ai. The new linear programming model after modification is
m
n 1
min f   cij xij
i 1 j 1

i  1,, m
 xij  ai
j 1


m
j  1, n  1
 xij  b j
 i 1
 xij  0,
i  1,, m;
j  1,, n  1



n 1
Thus the uneven transportation problem that has m sources and n destinations is
modified into an even transportation problem which has m sources and n+1 destinations and
can be solved with transportation simplex method. The column with zero cost per unit
should be considered after others cells when we determine the initial BF solution using
minimum-cost method.
3.3.2 Total Supply Less Than Total Demand
If total supply is less than total demand(Σai < Σb), the LP mathematical Model
would be.
70
m
n
min f   cij xij
i 1 j 1

i  1,, m
 xij  ai
j 1

 m
j  1, n
 xij  b j
 i 1
 xij  0, i  1,, m;
j  1,, n


n
In this case, a fictitious source ( called the dummy source) Am+1 can be introduced, the
corresponding supply is am+1=Σbj －Σai. Let xm+1,j be the number of units to be distributed
from source Am+1 to destination Bj, cm+1,j equals zero or the loss per unit of commodity
needed by Bj. Because xm+1,j is actually a shortage of commodity in destination Bj. The new
linear programming model after modification is
m 1 n
min f   cij xij
i 1 j 1

i  1,  , m  1
 xij  ai
 j 1
 m1
j  1,  n
 xij  b j
 i 1
 xij  0, i  1,  , m  1;
j  1,  , n


n
Thus the uneven transportation problem that has m sources and n destinations is
modified into an even transportation problem which has m+1 sources and n destinations and
can be solved with transportation simplex method. The row with zero cost per unit should
be considered after others cells when we determine the initial BF solution using
minimum-cost method.
3.4
Applications of Transportation Problem
3.4.1 General Uneven Transportation Problem
Example 3.3 Table 3—12 shows a transportation problem which has 3 sources and 4
destinations. All the data are summarized in table 3-12. Determine a plan for distributing the
commodity to minimize the total distribution cost.
Table 3—12
Destination
B1
B2
B3
B4
Supply (t)
Supply
A1
3
2
4
5
200
A2
7
5
2
1
100
A3
9
6
3
5
150
450
Demand (t)
50
100
150
50
350
71
Solution
Note that the total demand is 450 t and the total supply is 350 t, i.e., the total supply is
less than total demand. Introduce a dummy destination B5, its demand is 450－350=100(t),
the cost per unit distributed from each source to B5 should be zero(ci5=0, i=1，2，3). Then we
obtain an even transportation problem shown in table 3-13.
Table 3—13
Destination
B1
B2
B3
B4
B5
Supply (t)
A1
A2
A3
3
7
9
2
5
6
4
2
3
5
1
5
0
0
0
Demand (t)
50
100
150
50
100
200
100
150
450
450
Supply
Solving the problem with transportation simplex method leads to the following optimal
solution: x11=50，x12=100，x15=50，x23=50，x24=50，x33=100，x35=50。Where x15=50 and
x35=50 represent the respective storage of source A1 and A3. The total cost of this problem is
800.
Example 3.4 A transportation problem has 3 sources and 3 destinations. The total
supply is less than the sum of maximum demand of each destination but greater than the
sum of minimum demand of each destination. The minimum demand of each destination
must be met. If the demand between the maximum and the minimum cannot be met, it will
cause some loss. Now, the demand for B1 must be met, the loss per unit of shortage in B2
and B3 are 3 Yuan and 2 Yuan respectively, other data are summarized in table 3-14.
Determine the optimal distribution plan.
Table 3—14
Destination
B1
B2
B3
Supply
A1
A2
A3
5
6
3
1
4
2
7
6
5
200
800
150
Minimum Demand
Maximum Demand
600
600
120
200
300
430
Supply
1150
Solution
The demand for B1 must be met. The maximum demand and minimum demand for B2
is not equal, so we can consider B2 as two destinations B21 and B22; the demand for B21 is
the minimum demand of B2; the demand for B22 is given by the minimum demand
subtracted from the maximum demand, namely 200－120 = 80; this demand can be unmet
if necessary. For B3, it can be conducted as the same way as B2. Thus the problem now is a
3-source 5-destination transportation problem; the total supply is 1150; the total demand is
1230; so a dummy source A4 needs be introduced and its supply is 1230－1150 = 80. In
order to meet the minimum demand of each destination, let the cost per unit of commodity
72
distributed from A4 to B1, B21 and B31 equal M (a huge positive number). The conduction
above leads to the following transportation tableau shown in table 3-15.
Source
Table 3—15 Transportation Tableau
Destination
B1
B21
B22
B31
A1
A2
A3
A4
5
6
3
M
1
4
2
M
1
4
2
3
7
6
5
M
B32
Supply
7
6
5
2
200
800
150
80
1230
Demand
600
120
80
300
130
1230
Solve the problem using transportation simplex method, we obtain the optimal
distribution plan as follows.
Table 3—16 The Optimal Distribution Plan
Destination
Source
A1
A2
A3
A4
Demand
B1
B21
B22
120
80
450
150
B31
300
B32
50
80
600
120
80
300
130
Supply
200
800
150
80
1230
1230
Note that all the demand of B1 and B2 have been met; 350 out of the demand of B3
have been met. The total cost (including the loss of shortage of B3) is 8100 Yuan.
3.4.2 Production Scheduling
Example 3.5 A high-tech company produces a kind of electronic product. The
contracted number of product for next four seasons, the facilities that will be available for
producing the product, and the cost per product in each season are summarized in table 3-17.
Because of the variation in production costs, it may be worthwhile to produce some of the
product a season more before they are scheduled for delivering. The drawback is that such
products must be stored until the scheduled season for delivering. And the storage cost per
season for each product is 0.1 thousand Yuan. The production manager wants a schedule
developed for the number of the products to be produced in each of the 4 seasons so that the
total of the production and storage costs will be minimized.
Table 3—17
Scheduled Number
Maximum
Cost per Product
Season
（set）
Production（set） （Thousand Yuan）
1
230
270
3.2
2
265
260
3.33
3
255
280
3.31
4
245
270
3.42
73
Solution
Let xij be the number of products produced in season i for delivering in season j. After
added the storage costs, the corresponding coefficients in objective function cij are shown in
table 3-18.
i
Table 3—18 Coefficients cij in objective function corresponding to xij
j
1
2
3
1
2
3
4
3.2
3.3
3.33
3.4
3.43
3.31
4
3.5
3.53
3.41
3.42
The linear programming model can be formulated as follows:
min f = 3.2x11+3.3x12+3.4x13+3.5x14+3.33x22+3.43x23
+3.53x24+3.31x33+3.41x34+3.42x44
x11+x12+x13+x14≤270
x22+x23+x24≤260
x33+x34≤280
x44≤270
x11
=230
x12+x22
=265
x13+x23+x33
=255
x14+x24+x34+x44 =245
xij≥0，i=1，…，4；j=1，…，4；i≥j
The former 4 constraints are facilities restrictions of 4 seasons; the later 4 constraints
are contract restrictions of 4 seasons. Since it is impossible to produce the product in one
season for delivery in earlier season, xij must be zero if i > j . Therefore, there is no real cost
that can be associated with such xij. In order to have a well-defined transportation problem,
it is necessary to assign some value for the unidentified costs. Fortunately, we can use the
Big M method to assign this value. Thus we assign a very large number to the unidentified
cost cells in transportation tableau to force the corresponding value of xij to be zero in the
final solution. After introducing a dummy destination, we obtain an even transportation
problem that is shown in table 3-19.
Table 3—19
Destination
1
2
3
4
5
Supply
Source
1
3.2
3.3
3.4
3.5
0
270
2
M
3.33
3.43
3.53
0
280
3
M
M
3.31
3.41
0
260
4
M
M
M
3.42
0
270
1080
Demand
230
265
255
245
85
1080
74
Solve the problem using transportation simplex method, we obtain the optimal
production schedule as follows.
Table 3—20
Destination
1
2
1
2
3
4
230
40
225
Demand
230
Source
3
4
5
Supply
270
280
260
270
55
255
265
255
5
240
30
245
85
1080
1080
Note that season 1 produces 270 sets, 40 of them will be delivered in season 2; season
2 produces 225 sets; season 3 produces 260 sets, 5 of them will be delivered in season 4;
and season 4 produces 240 sets. The total production costs is 3299.15 thousands Yuan
3.4.3 Network Representation of Transportation Problem
Example 3.6 A company has two factories that are respectively in Dalian and
Guanzhou. The two factories are responsible for producing a kind of high-tech products.
The company also has two sales stations in Shanghai and Tianjing that are responsible for
delivering the products from the two factories to 4 cities —— Nanjing, Jinan, Nanchang,
and Qingdao. The products can be directly delivered from Dalian to Qingdao because of
the short distance between the two cities. The capacity of each factory, the demand of each
city, and the cost per unit of product delivered between factories and sales stations, sales
stations and cities, and factories to cities are shown in figure 4-1 (cost is expressed in 100
Yuan). Now, the problem is to determine a transportation plan for minimizing the total
transportation cost.
5
Nanjing
600
1
2
Guang zhou
2
6
3
6
3
3
Shanghai
4
400
2
Dalian
4
6
5
4
3
Tianjing
1
4
200
6
Jinan
150
7
Nanchang 350
8
Qingdao
↑
Supply
300
↑
Demand
Figure 3—1 Network Re
75
Solution
Give the number to each city , i=1，2，…，8 represent Guangzhou, Dalian, Shanghai,
Tianjing, Nanjing, Jinan, Nanchang, and Qingdao.
Let xij be the number of products delivered from city i to city j , then the objective
function can be described as follows:
min f = 2x13+3x14+3x23+x24+4x28+2x35+6x36+3x37
+6x38+4x45+4x46+6x47+5x48
The constraints for the two supply cities are:
x13+x14≤600
x23+x24+x28≤400
The constraints for the two transshipment cities are:
x13+x23 x35 x36 x37 x38 = 0
x14+x24 x45 x46 x47 x48 = 0
The constraints for the four destination cities are:
x35+x45 = 200
x36+x46 = 150
x37+x47 = 350
x38+x48 +x28 = 300
So, the linear programming mathematical model for this transshipment problem is as
following:
min f = 2x13+3x14+3x23+x24+4x28+2x35+6x36+3x37
+6x38+4x45+4x46+6x47+5x48
x13+x14≤600
x23+x24+x28≤400
x13+x23 x35 x36 x37 x38 = 0
x14+x24 x45 x46 x47 x48 = 0
x35+x45 = 200
x36+x46 = 150
x37+x47 = 350
x38+x48 +x28 = 300
xij≥0，for i、j
Obviously, we can solve the problem with simplex method. But it is easier and
intuitional to solve the problem using transportation simplex method if we transform the
model into transportation problem. The transformation can be realized by following steps:
each transshipment station can be considered as a destination for the sources and its demand
is the sum of demands of all the sources that can reach the station; each transshipment
station can also be considered as a source for the destinations and its supply is the sum of
supplies of all the destinations that the station can reach; Thus, the problem becomes a
4-souece 6-destination transportation problem. The cost per unit of distributing from
transshipment station to itself equals zero; the cost per unit of distributing between two
notes that cannot reach directly is M; other cost per unit of distributing are shown in the
figure. The transportation tableau is as following:
76
Table 3—21
Destination
Source
1(Guangzhou)
2(Dalian)
3(Shanghai)
4(Tianjing)
3
(Shanghai)
4
(Tianjing)
5
(Nanjing)
5
(Jinan)
7
(Nanchang)
8
(Qindao)
2
3
0
M
3
1
M
0
M
M
2
4
M
M
6
4
M
M
3
6
M
4
6
5
Demand
1000
1000
200
150
350
300
Supply
600
400
1000
1000
3000
3000
Solve the problem with transportation simplex method and the optimal distributing
plan is shown in table 3-22.
Table 3—22
Destination
Source
1(Guangzhou)
2(Dalian)
3(Shanghai)
4(Tianjing)
3
(Shanghai)
4
(Tianjing)
550
50
100
Demand
1000
450
5
(Nanjing)
5
(Jinan)
8
(Qindao)
300
200
350
850
1000
7
(Nanchang)
150
200
150
350
300
Supply
600
400
1000
1000
3000
3000
Namely, Guanzhou to Shanghai : 550 sets; Guangzhou to Tianjing : 50 sets; Dalian to
tianjing: 100 sets; Dalian to Qingdao : 300 sets; Shanghai to Najing :200 sets, to Nanchang :
350 sets; Tianjing to Jinan 150 sets. The minimum total cost is 4600 yuan.
77
Problem 3
1．Can the distributing plan in table 3-23 and table 3-24 be BF solution transportation
simplex method? Why?
Table 3—23
Destination
B1
B2
B3
B4
Supply
Source
A1
50
150
200
A2
0
100
0
100
A3
50
50
Demand
50
100
150
50
Table 3—24
Destination
Source
A1
A2
A3
Demand
B1
B2
B3
60
B4
140
100
10
70
10
150
100
50
50
2．Solve following problem using transportation simplex method
Table 3—25
Destination
B1
B2
B3
B4
Cost per unit
Supply
200
100
70
Supply
Source
A1
A2
A3
3
12
6
9
20
11
8
7
13
6
10
14
Demand
300
500
900
600
600
1000
700
3．Table 3—26 shows a transportation problem which has 2 sources and 3 destinations.
Assume that the destinations are allowed to be short of commodity and the sources are
allowed to store commodity. All the data are summarized in table 3-12. Determine a plan for
distributing the commodity to minimize the total distribution cost. Set up the mathematical
model for this problem and solve it with transportation simplex method.
表 3—26
Destination
Cost per
Cost per unit
B1
B2
B3
Supply
storage of
Source
commodity
A1
4
6
8
200
5
A2
6
2
4
200
4
Demand
50
100
100
Cost per shortage
3
8
5
of commodity
78
4．A department store wants to purchase four types of clothes : type A——1500
suits；type B——2000 suits; type C——3000 suits; type D——3500 suits. Three cities can
supply these types of clothes: city I——2500 suits; city II——2500 suits; city III——5000
suits. The profits per suit purchased from different cities are different because of different
service quality and transportation costs. The profits per suit are shown in table 3-27.
Table 3—27
A
B
C
D
I
10
5
6
7
II
8
2
7
6
III
9
3
4
8
Identify a purchasing plan for the department store to maximize the total profit.
5．Three cities need coal every year: city I——3,200,000 t; city II——2,500,000 t; city
III——3,500,000 t. Two coal mines can supply this type of coal: mine A——4,000,000 t;
mine B——4,500,000 t. The cost per unit of coal distributed from each mine to each city is
shown in table 3-28.
Table 3—28
I
15
12
A
B
II
18
25
III
22
16
Note that the total demand is greater than total supply. Now the decision is: the supply
to city I is no less than 2,800,000 t; all the demand of city II should be met; the supply to
city III is no less than 2,700,000 t. And the cost per unit of shortage of city I is 100,000
yuan; the cost per unit of shortage of city III is 120,000 yuan; Identify a distributing plan for
the coal supply department to minimize the total cost.
6．A factory need to complete the order for some kind of products during the future
three stages. Now the production capacity in each stage, the order for each stage, and the
cost per unit of product are summarized in table 3-29. And the cost per unit of storage for
this kind of product is 1.
Table 3—29
Stage
Product Capacity
Order
Cost per unit of product
1
30
20
8
2
20
15
12
3
25
20
11
Set up the mathematical model for this problem, transform it into transportation problem
and solve it with transportation simplex method.
79