Transportation Problem
The basic transportation problem was developed
in 1941 by F.I. Hitchaxic. However it could be
solved for optimally as an answer to complex
business problem only in 1951,when Geroge B.
Dantzig applied the concept of Linear Programming
in solving the Transportation models.
Transportation problems are primarily concerned with
the optimal (best possible) way in which a product
produced at different factories or plants (called
supply origins) can be transported to a number of
warehouses (called demand destinations).
The objective in a transportation problem is:-To fully
satisfy the destination requirements within the
operating production capacity constraints at the
minimum possible cost.
Whenever there is a physical movement of goods
from the point of manufacture to the final
consumers through a variety of channels of
distribution (wholesalers, retailers, distributors
etc.), there is a need to minimize the cost of
transportation so as to increase the profit on sales.
Transportation problems arise in all such cases. It
aims at providing assistance to the top
management in ascertaining how many units of a
particular product should be transported
from each supply origin to each demand
destinations to that the total prevailing
demand for the company’s product is
satisfied, while at the same time the total
transportation costs are minimized.
Formulation of Transportation Problem
as a Linear Programming Model
A Transportation problem is simply a special case of
LPP. Hence to convert into a LP model, some basic
steps are:1. P denote the plant (factory) where the goods are being
manufactured &
2. W denote the warehouse (godown) where the finished
products are stored by the company before shipping to
various destinations.
Formulation …..
Xij = no. of units of the product to be transported
from factory i to warehouse j or from plant Pi to
the warehouse Wj, and
Cij = transportation cost per unit of shipping from
plant Pi to the Warehouse Wj.
i.e.; The total supply available at the plants exactly
matches the total demand at the destinations.
Hence, there is neither excess supply nor excess
Such type of problems where supply and demand are
exactly equal are known as Balanced
Transportation Problem.
• Supply (from various sources) are written in the
rows, while a column is an expression for the
demand of different warehouses.
In general, if a transportation problem has m rows an
n columns, then the problem is solvable if there are
exactly (m + n –1) basic variables.
Unbalanced Transportation Problem :
A transportation problem is said to be unbalanced
if the supply and demand are not equal.
Two situations are possible:-
1. If Supply < demand, a dummy supply variable is
introduced in the equation to make it equal to
2. If demand < supply, a dummy demand variable is
introduced in the equation to make it equal to
Example :
Formulate the following transportation problem
as a LP model:
13 23
3 21
31 29
Formulating as a Linear Programming model, we have:
Minimise Z = 11X11 + 6X12 + 15X13 + 3X14+ 7X21 + 8X22 +4X23 + 13X24 +22X31 +
17X32 + 8X33 + 31X34
Subject to: Capacity constraints:
X11 + X12 + X13 + X14 =21
X21 + X22 + X23 + X24 = 23
X31 + X32 + X33 + X34 = 29
Notation used:
Xij = no. of units transported from source i to destination j
Requirement constraints:
X11 + X21 + X31 = 6
X12 + X22 + X32 = 20
X13 + X23 + X33 = 22
X14 + X24 + X34 = 25
Hence the transportation problem is successfully formulated as a LP model.
The various steps for solving a transportation
problem are as follows:
• From the given problem, express the objective
function which must be minimized along with the
relevant constraints.
• Set up a transportation table, where supply or
sources (factories, plants) are denoted as rows and
demand or destinations (warehouses, markets) are
denoted by columns.
• Find out an initial feasible solution by using one of
the three rules (discussed later in the chapter).
• A solution so obtained is aid to be feasible if and
only if, such a solution has allocations in the total
number of (m + n – 1) cells.
• By ascertaining the opportunity costs, the solution
obtained above can be checked for optimality.
Opportunity cost may be defined as the cost of
sacrifice. Here, it denoted the reduction in cost due
to the inclusion of a particular cell in the solution.
An optimum solution is obtained when there is a
positive opportunity cost for each of the empty
Where the solution so obtained is not found to be
optimal, that empty cell which results in the largest
saving is included and the above steps are again
Initial Feasible Solutions
Most commonly used methods are:
1. North-West corner Rule (NWC)
2. Least Cost method (LCM)
3. Vogel’s Approximation method (VAM)
Least Cost Method
As the name suggests, of all the routes select
the one where shipping cost is the least.
Now, consider the supply available at the
corresponding source and demand at the
corresponding destination and put the lower of
the two as the quantity to be transported
through that route. After that delete the source
or destination, whichever is satisfied.
Least Cost method…..
Consider the remaining routes and again choose
the one with the smallest cost and make
assignments. Continue in this manner until all
the units are assigned.
If there is a tie in the min. cost, so that 2 or more
routes have the same least cost of shipping,
then conceptually either of them may be
selected. However a better initial solution is
obtained if the route chosen is the one where
largest quantity can be transported.
Least Cost Method…
Thus if there are three cells for which the least
cost is equal, then consider all of these one by
one and determine the quantity which can be
dispatched, and choose the all with the largest
If there is still a tie, then either of them may be
Vogel’s Approximation Method (VAM)
Like the previous method, VAM also considers
the shipping cost, but in a relative sense, when
making allocations.
First consider the each row of the matrix
individually & find the difference between two
least cost cells in it. Then repeat this for each of
the column.
Identify the column or row with the max.
difference value.
Vogel’s Approximation Method (VAM)
Now, consider the cell with min. cost in that row
or column and assign the max. units possible.
In case of a tie in the largest cost difference,
although either of them may be chosen but it is
preferable to choose the cost difference
corresponding to which the largest no. of units
may be assigned or corresponding to which the
cell chosen has min. cost.
Delete the column or row which is satisfied.
Vogel’s Approximation Method (VAM)
Again find out the differences and proceed in the
same manner.
Continue until all units have been assigned.
Testing of Optimality
The Modified Distribution Method (MODI)
This method is an efficient method of testing the
optimality of a transportation solution.
Step1: Add to the transportation table a column
on the RHS titled ui and a row in the bottom of it
labelled vj.
• Assign any value arbitrarily to a row or column
variable ui or vj. Generally, u1=0 is assigned.
The Modified Distribution Method (MODI)
• Consider every occupied cell in the first row
individually and assign the column value vj
(when the occupied cell is in the jth column of
the row) which is such that the sum of the row
and the column values is equal to the unit cost
value in the occupied cell. With the help of
these values, consider other occupied cells one
by one and determine the appropriate values if
ui’s & vj’s, taking in each case ui+vj=cij
The Modified Distribution Method (MODI)
• After determining all the values of ui & vj,
calculate for each unoccupied cell Dij=ui+vj-cij.
The Dij’s represents the opportunity costs of
various cells.
• If all the empty cells have –ve opportunity cost,
the solution is optimal & unique.
• If some empty cells has a zero opportunity cost
but if none of the other empty cells have +ve
opportunity cost, then it implies that the given
The Modified Distribution Method (MODI)
solution is optimal but that is not unique there
exists other solutions that would be as good as
this solution.
• If the solution contains a +ve opportunity cost
for one or more of the empty cells, the solution
is not optimal. In such a case, the cell with the
largest opportunity cost value is selected, a
closed loop traced & transfers of units along
with the route are made.
The Modified Distribution Method (MODI)
Note: This method is workable only when solution is
non-degenerate, i.e. for a matrix of the order m*n,
there are exactly m+n-1 non zero (occupied) cells.
Special Topics of Transportation
• Unbalanced transportation Problem
• Prohibited Routes
• Unique vs multiple solutions
• Degeneracy
• Maximization problem
Unbalanced Transportation Problem
• For solving the transportation problem it is required
that the aggregate supply is equal to the aggregate
demand. Practically situation may arise when the
two are unequal.
• Two possibilities are: when the total supply exceeds
total demand and when total demand exceeds total
supply. Such problems are called unbalanced
transportation problems.
• Then before solving we must balance the demand &
Unbalanced Transportation Problem
• When supply exceeds demand, the excess supply is
assumed to go to inventory. A column of slack
variables is added to the transportation table which
represents dummy destination with a requirement
equal to the amount of excess supply and the
transportation cost equal to zero.
• When demand exceeds supply, balance is restored
by adding a dummy origin. The row representing it
is added with an assumed total availability equal to
the difference between total demand & supply and
Unbalanced Transportation Problem
with each cell having a zero unit cost.
• After balancing the transportation problem,
solution proceeds in exactly same manner as
discussed earlier.
Prohibited routes
• Sometimes in a given transportation problem some
routes may not available. This can happen due to
many reasons like the weather condition, strike etc.
• There is a restriction on the routes available for
transportation. To handle a situation of this type,
we assign a very large cost represented by M to
each of such routes which are not available.
• The effect of adding a large cost element would be
that such routes would automatically be eliminated
in the final solution.
Unique vs Multiple Solutions
• The optimal solution of a given problem may or may
not be optimal.
• If all Dij values are –ve, then the solution is unique.
• If some Dij=0, then multiple optimal solutions are
indicated so that there exist transportation patterns
other than the one obtained which can satisfies all
the rim requirements.
• To obtain an alternate solution, trace a closed loop
starting with cell Dij=0, and get the revised soluition