1
Transportation Problem: A network Model
and a Linear Programming Formulation
Example:
1) Production capacities over the next 3-month planning
period for one particular type of generator are:
Origin
Plant
Production Capacity (units)
1
Cleveland, Ohio
5000
2
Bedford, Indiana
6000
3
York, Pennsylvania
2500
Total
13,500
2) The 3-month forecast of demand for the distribution
centers is:
Destination Distribution Center Demand Forecast (units)
1
Boston,
6000
2
Chicago,
4000
3
St. Louis,
2000
4
Lexington
1500
Total
13,500
3) Transportation cost per unit shipped via each route is:
Destination
Origin
Boston
Chicago
St. Louis Lexington
Cleveland
3
2
7
6
Bedford
7
5
2
3
York
2
5
4
5
Determine how much of the production should be shipped
from each plant to each distribution center to minimize the
transportation cost.
2
Plants
Distribution Centers
(Origin nodes)
(Destination nodes)
3
1
5000
Cleveland
1
Boston
6000
2
Chicago
4000
2
6 7
7
6000
2
Bedford
5
5
2
3
2
2500
3
York
5
3
St. Louis
5
4
4
Lexington
Supplies
2000
Distribution
routes (arcs)
1500
Demands
3
For a transportation problem having m origins and n destinations:
xij =number of units shipped from origin i to destination j
where i = 1, 2, …, m, and j = 1, 2, …, n
3 x11  2 x12  7 x13  6 x14
Min  7 x21  5 x22  2 x23  3 x24
 2 x31  5 x32  4 x33  5 x34
s.t. x11  x12  x13  x14  5000
x21  x22  x23  x24  6000
x31  x32  x33  x34  2500
x11  x21  x31  6000
x12  x22  x32  4000
x13  x23  x33  2000
x14  x24  x34  1500
xij  0 for i = 1, 2,3; j = 1, 2,3,4
The general linear programming model of the m origins, n destinations
transportation problem is:
m n
Min   cij xij
i 1 j 1
s.t
n
 xij  si i = 1, 2,…;m Supply
j 1
m
x
i 1
ij
 dj
xij  0
j = 1, 2,…,n
for all i and j
Demand
4
Summary of the transportation simplex method:
To apply the transportation simplex method, you must
have a transportation problem with total supply = total demand;
thus, for some problems you may need to add a dummy origin or
dummy destination to put the problem in this form.
The transportation simplex method applies a two-phase
solution procedure. In phase I, apply the minimum–cost method
to find an initial feasible solution. In phase II, begin with the
initial feasible solution and iterate until you reach an optimal
solution.
Start
Determine Initial
feasible solution
Is
Solution
Optimal
??
NO
Improve solution
by reallocating
YES
STOP
5
The steps of the transportation simplex method for a
minimization problem are summarized as follows.
Phase I: Minimum–Cost Method
Step 1
Identify the cell in the transportation tableau with the
lowest cost, and allocate as much flow as possible to
this cell. In case of a tie, choose the cell
corresponding to the arc over which the most units can
be shipped. If ties still exist, choose any of the tied
cells.
Step 2
Reduce the row supply and the column demand by the
amount of flow allocated to the cell identified in
Step1.
Step 3
If all row supplies and column demands have been
exhausted then STOP; the allocations made will
provide an initial feasible solution. Otherwise,
continue with Step 4.
Step 4
If all row supply is now zero, eliminate the row from
further consideration by drawing a line through it.
If the column demand is now zero, eliminate the
column by drawing a line through it.
Step 5
Continue with Step 1 for all unlined rows and
columns.
6
Phase II
Step 1
If the initial feasible solution is degenerate with less
than m+n-1 occupied cells, add an artificially cell or
cells so that m+n-1 occupied cells exist in locations
that enable use of the MODI method.
Step 2
Use the MODI method to compute row indexes, ui ,
and column indexes, v j .
Step 3
Compute the net evaluation index eij  cij  ui  v j
for each unoccupied cell.
Step 4
If eij  0 for all unoccupied cells, STOP; you have
reached the optimal solution. Otherwise, proceed to
Step 5.
Step 5
Identify the unoccupied cells with the smallest (most
negative) net evaluation index and select it as the
incoming cell.
Step 6
Find the stepping-stone path associated with the
incoming cell. Label each cell on the stepping-stone
path whose flow will increase with a plus sign and
each cell whose flow will decrease with a minus sign.
Step 7
Choose as the outgoing cell the minus-sign cell on the
stepping-stone path with the smallest flow. If there is
a tie, choose any one of the tied cells. The tied cells
that are not chosen will be artificially occupied with a
flow of zero at the next iteration.
Step 8
Allocate to the incoming cell the amount of flow
currently given to the outgoing cell; make the
appropriate adjustments to all cells on the steppingstone path, and continue with step 2.
7
Transportation tableau
Origin
Cleveland
Boston
3
Bedford
York
Demand
Destination
Chicago St. Louis Lexington
2
7
6
7
5
2
3
2
5
4
5
6000
4000
2000
1500
Supply
5,000
6,000
2,500
13,500
Final Tableau showing the initial feasible solution using the
Minimum-Cost Method
Origin
Boston
3
Cleveland
1000
7
Bedford
2500
2
York
2500
Demand
6000
Total Cost =
Destination
Chicago St. Louis Lexington
2
7
6
4000
5
2
3
2000
1500
5
4
5
4000
2000
1500
Supply
5,000
6,000
2,500
13,500
1000  $3  4000  $2
 2500  $7  2000  $2
=$42000
 1500  $3  2500  $2
8
Net Evaluation Indexes for the initial feasible solution using
the MODI method
Destination
Origin
0
4
-1
Demand
3
2
3
1000
-2
2
7
5
-1
2500
2
2500
6000
7
9
4000
Supply
-1
7
2
2000
6
5,000
3
6,000
1500
5
4
5
4
7
7
4000
2000
1500
2,500
13,500
Stepping-Stone Path
Origin
Cleveland
Bedford
York
Demand
Boston
+
3
1000
-
Destination
Chicago St. Louis Lexington
2
7
6
4000
7
+
5
2500
2
2000
2
5
3
1500
4
5
2500
6000
4000
2000
1500
Supply
5,000
6,000
2,500
13,500
9
New Solution after one iteration
Origin
Destination
Chicago St. Louis Lexington
2
7
6
Boston
3
Cleveland
3500
5,000
1500
7
Bedford
5
2
2500
2000
2
York
Supply
3
6,000
1500
5
4
5
2,500
2500
Demand
6000
4000
2000
1500
13,500
MODI Evaluation of each cell in solution
Origin
0
3
-1
Destination
2
-1
2
7
3
3
3500
1
5
2500
2
2500
6
2
2000
5
4
6
8
1500
7
0
3
1500
4
6
3500  $3  1500  $2
 2500  $5  2000  $2
Total Cost =
 1500  $3  2500  $2
5
6
=$39500
10
1) Total supply not equal to total demand.
Origin
Cleveland
Boston
3
Bedford
York
Dummy
Demand
Destination
Chicago St. Louis Lexington
2
7
6
7
5
2
3
2
5
4
5
0
0
0
0
6000
4000
2000
1500
Supply
4,000
6,000
2,500
1,000
13,500
2) Maximization objective function
Pick the incoming cell for which eij is largest. If eij  0 for all
unoccupied cells, we STOP; the maximization solution has been
reached.
3) Unacceptable routes.
a) FOR MINIMIZATION problem, assign the unacceptable
route an extremely high cost M.
b) FOR MAXIMIZATION problem, assign the
unacceptable route a profit per unit of -M.
Origin
Cleveland
Boston
3
7
5
2
3
M
5
4
5
Bedford
York
Demand
Destination
Chicago St. Louis Lexington
2
7
6
6000
4000
2000
1500
Supply
5,000
6,000
2,500
13,500