Transportation
problem is a
special kind of LP problem in
which objective is to transport
various quantities of a single
homogenous commodity, to
different destinations in such a
way
that
the
total
transportation cost is minimized.

Sources
factories,
finished goods
warehouses ,
raw
materials
houses,
suppliers etc.

ware
Destinations
Markets
Finished goods ware
house
raw
materials
ware
houses,
factories,
Feasible Solution – Non negative values of xij,
where i=1,2,..,m and j=1,2,..,n which satisfy the
constraints of supply & demand is called feasible
to the solution.
 Basic Feasible Solution – A feasible solution to a
m origin and n destination problem is said to be
basic if no. of positive allocations = m+n-1.
 Optimal Feasible Solution – A feasible solution is
said to be optimal if it minimizes the total
transportation cost.
The optimal solution itself may or may not be a
basic solution.
After this no further decrease in transportation
cost is possible.


Balanced Transportation Problem – A
transportation problem in which the total
supply from all sources equals the total
demand in all destinations
m
n
 a  b
i 1

i
j 1
j
Unbalanced Transportation Problem – such
problems which are not balanced
m
n
 a  b
i 1

i
Matrix Terminology
j 1
j
 Availability
of the quantity.
 Transportation of items.
 Cost per unit.
 Independent cost.
 Objective.
Destination j

1
2
j
n
S 1 c11
O 2
U
R
C
E
c12
c1j
c1n
i ci1
ci2
i
m
Demand
a2
cij
cm1 cm2
b1
Supply
a1
b2
bj
cin
ai
cmn
am
bn
Where,
 m- number of sources
 n- number of destinations
 ai- supply at source I
 bj – demand at destination j
cij – cost of transportation per unit from
source i to destination j
Xij – number of units to be transported
from the source i to destination j
 Formulate
the problem and setup in
the matrix form.
 Obtain the Initial basic Feasible
Solution.
 Test the initial solution for optimality.
 Updating the solution.
Factories
(Sources)
Warehouses
(Destinations)
100 Units
D
A
300 Units
300 Units
E
B
200 Units
300 Units
F
C
200 Units
Capacities
Shipping Routes
Requirements
The first step is setting up the transportation
table
 Its purpose is to summarize all the relevant data
and keep track of algorithm computations

Transportation costs per desk for Executive Furniture
TO
FROM
A
B
C
D
5
4
3
E
8
4
3
F
9
7
5

Transportation table for Executive Furniture
TO
FROM
D FACTORY
E FACTORY
F FACTORY
WAREHOUSE
REQUIREMENTS
WAREHOUSE
AT
B
WAREHOUSE
AT A
300
WAREHOUSE
AT C
5
4
3
8
4
3
9
7
5
200
Cost of shipping 1 unit from C
F factory to B warehouse
warehouse
demand
200
Total supply
and
demand
D’s capacity
constraint
FACTORY
CAPACITY
100
300
300
700
Cell representing a
source-to-destination (E
to C) shipping
assignment that could
be made
Methods for
Transportation
Problems
Initial Basic
Feasible
Solution
North West
Corner
Method
(NWCM)
Lowest
Cost Entry
Method
(LCEM)
Optimality
Tests
Vogel’s
Approximati
on Method
(VAM)
MODI
Method
Stepping
Stone
Method
Construct an empty m х n matrix,
completed with rows and columns.
2) Indicate the row totals and column totals
at the end.
3) Starting with (1,1) cell at the North-West
Corner of the matrix, allocate maximum
possible quantity keeping in view that
allocation can neither be more than the
quantity required by the respective
warehouses nor more than the quantity
available at each supply centre.
1)
4)
5)
6)
7)
8)
Adjust the supply and demand numbers
in the respective rows and column
allocations.
If the supply for the first row is exhausted
then move down to the first cell in the
second row and first column and go to
step 4.
If the demand for first column is satisfied,
then move to the next cell in the second
column and first row and go to step 4.
If for any cell, supply equals demand
then the next allocation can be made
in cell either in the next row or column.
Continue the procedure until the total
available quantity is fully allocated to
the cells as required.
1. Beginning in the upper left hand corner, we
assign 100 units from D to A. This exhaust the
supply from D but leaves A 200 desks short. We
move to the second row in the same column.
TO
FROM
(D)
(A)
100
(B)
FACTORY
CAPACITY
(C)
$5
$4
$3
$8
$4
$3
$9
$7
$5
(E)
(F)
WAREHOUSE
REQUIREMENTS
300
200
200
100
300
300
700
2. Assign 200 units from E to A. This meets A’s
demand. E has 100 units remaining so we move
to the right to the next column of the second
row.
TO
FROM
(D)
(E)
(A)
100
200
(F)
WAREHOUSE
REQUIREMENTS
300
(B)
FACTORY
CAPACITY
(C)
$5
$4
$3
$8
$4
$3
$9
$7
$5
200
200
100
300
300
700
3. Assign 100 units from E to B. The E supply has now
been exhausted but B is still 100 units short. We
move down vertically to the next row in the B
column.
TO
FROM
(D)
(E)
(A)
100
200
$5
$8
100
$9
(F)
WAREHOUSE
REQUIREMENTS
(B)
300
200
FACTORY
CAPACITY
(C)
$4
$3
$4
$3
$7
$5
200
100
300
300
700
4. Assign 100 units from F to B. This fulfills B’s
demand and F still has 200 units available.
TO
FROM
(D)
(E)
(A)
100
200
$5
$8
$9
(F)
WAREHOUSE
REQUIREMENTS
(B)
300
100
100
200
FACTORY
CAPACITY
(C)
$4
$3
$4
$3
$7
$5
200
100
300
300
700
5. Assign 200 units from Fort Lauderdale to
Cleveland. This exhausts Fort Lauderdale’s
supply and Cleveland’s demand. The initial
shipment schedule is now complete.
TO
FROM
(D)
(E)
(A)
100
200
$5
$8
$9
(F)
WAREHOUSE
REQUIREMENTS
(B)
300
100
100
200
FACTORY
CAPACITY
(C)
$4
$3
$4
$3
$7
200
200
$5
100
300
300
700

We can easily compute the cost of this shipping
assignment
ROUTE
UNITS
SHIPPED
PER UNIT
x COST ($)
TOTAL
COST ($)
FROM
TO
D
A
100
5
500
E
A
200
8
1,600
E
B
100
4
400
F
B
100
7
700
F
C
200
5
1,000
=
4,200
 This solution is feasible but we need to check to
see if it is optimal
1.
2.
3.
4.
Select the cell with the lowest transportation cost
among all the rows or columns of the
transportation table. If the minimum cost is nor
unique the select arbitrarily any cell with the
lowest cost.
Allocate as many units as possible to the cell
determined in step 1 and eliminate that row in
which either capacity or requirement is
exhausted.
Adjust the capacity and requirement for the next
allocations.
Repeat steps 1 to 3 for the reduced table until the
entire capacities are exhausted to fill the
requirement at different destinations.
2
3
5
6
5
2
1
3
5
10
3
12
8
8
4
4
6
6
15
6
5
3
2
5
5
3
1
2
2
8
12
X
6
4
8
3
4
6
15
2
3
5
6
5
2
1
3
5
X
2
8
3
10
8
X
4
4
6
6
15
2
3
5
6
X
5
2
1
3
5
X
2
8
3
5
8
X
4
4
6
6
15
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
4
6
5
X
X
4
6
10
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
6
4
X
X
X
6
6
2
3
5
6
X
5
2
1
3
5
X
2
8
3
8
5
4
4
X
X
6
6
X
X
X
Considered as the best method. The steps are
as follows:
1. For each row of the table identify the
lowest and the next lowest cost cell. Find
their differences and place it to the right of
that row. In case two cells contain the
same least cost then the difference shall
be zero.
2. Similarly find the difference of each
column and place it below each column.
These differences found in step 2 & 3 are
also called penalties.
3)
Looking at all the penalties, identify
highest of them and the row or column
relative to that penalty. Allocate the
maximum possible units to the least cost
cell in the selected row or column. Ties
should be broken in this order
Maximum difference least cost cell
Maximum difference
tie
Maximum Units allocation
Least Cost Cell.
tie
Arbitrary.
Adjust the supply & demand and cross
the satisfied row or column.
5) Recompute
the column and row
differences
ignoring
deleted
row/columns and go to step 3. repeat
the procedure until all the column and
row totals are satisfied.
4)
6
7
15
Demand
Column Penalty
Supply
Row Penalty
10
7-6=1
15
78-15=63
8
80
78
15
5
5
15-6=9
80-7=73
78-8=70
6
7
Supply
Row Penalty
5
8-6=2
15
78-15=63
8
5
15
Demand
Column Penalty
80
78
15
X
5
15-6=9
_
78-8=70
6
7
5
Column Penalty
Row Penalty
0
_
15
_
8
5
15
Demand
Supply
80
78
15
X
X
15-6=9
_
_
6
0
7
5
Supply
Row Penalty
X
_
15
_
8
5
15
80
78
Demand
15
X
X
Column Penalty
_
_
_
6
0
7
5
Supply
Row Penalty
X
_
X
_
8
5
15
80
78
15
Demand
X
X
X
Column Penalty
_
_
_
1)
2)
3)
4)
5)
Construct the transportation table with given
cost of transportation and rim conditions.
Determine initial basic feasible solution using
a suitable method.
For occupied cells, calculate Ri and Kj for rows
and columns respectively, where Cij = Ri + Kj
For occupied cells, the opportunity cost is
calculated using formula dij = Cij – (Ri + Kj)
Now, the opportunity cost of unoccupied cells
is determined, where
Opportunity cost = Actual cost – Implied Cost
dij = Cij – (Ri + Kj)
Examine unoccupied cells evaluation for dij
if dij > 0, cost will increase, i.e. optimal solution has
arrived.
b)
If dij = 0, cost will remain unchanged but there exists an
alternate solution.
c)
If dij < 0, then improved solution can be obtained by
moving to step 7.
7)
Select an unoccupied cell with largest negative
opportunity cost among all unoccupied cells.
8)
Construct a closed path for unoccupied cell determined in
step7 and assign (+) and (-) sign alternatively beginning
with plus sign for the selected unoccupied cell in any
direction.
9)
Assign as many units as possible to the unoccupied cell
satisfying rim conditions. The smallest allocation in a cell with
negative sign on the closed path indicated the number of
units that can be transported to the unoccupied cells. This
quantity is added to all the occupied cells on the path
marked with plus sign and substracted from those occupied
cells on the path marked with minus sign.
10) Go to step 4 and repeat procedure until all dij > 0,i.e., an
optimal solution is reached.
6)
a)


The stepping-stone method is an iterative
technique for moving from an initial feasible
solution to an optimal feasible solution
There are two distinct parts to the process
› Testing the current solution to determine if
improvement is possible
› Making changes to the current solution to obtain
an improved solution

This process continues until the optimal
solution is reached
There is one very important rule
 The number of occupied routes (or squares) must
always be equal to one less than the sum of the
number of rows plus the number of columns
 In the Executive Furniture problem this means the
initial solution must have 3 + 3 – 1 = 5 squares used

Occupied
shipping routes
(squares)
Number
= of rows
Number of
+ columns
–1
 When the number of occupied rows is less than
this, the solution is called degenerate
The stepping-stone method works by
testing each unused square in the
transportation table to see what would
happen to total shipping costs if one unit
of the product were tentatively shipped
on an unused route
 There are five steps in the process

1. Select an unused square to evaluate
2. Beginning at this square, trace a closed path back
to the original square via squares that are currently
being used with only horizontal or vertical moves
allowed
3. Beginning with a plus (+) sign at the unused square,
place alternate minus (–) signs and plus signs on
each corner square of the closed path just traced
4. Calculate an improvement index by adding
together the unit cost figures found in each square
containing a plus sign and then subtracting the unit
costs in each square containing a minus sign
5. Repeat steps 1 to 4 until an improvement index has
been calculated for all unused squares. If all indices
computed are greater than or equal to zero, an
optimal solution has been reached. If not, it is
possible to improve the current solution and
decrease total shipping costs.

For the Executive Furniture Corporation data
Steps 1 and 2. Beginning with Des Moines–Boston
route we trace a closed path using only currently
occupied squares, alternately placing plus and
minus signs in the corners of the path
 In a closed path, only squares currently used for
shipping can be used in turning corners
 Only one closed route is possible for each
square we wish to test
Step 3. We want to test the cost-effectiveness of
the Des Moines–Boston shipping route so we
pretend we are shipping one desk from Des Moines
to Boston and put a plus in that box
 But if we ship one more unit out of Des Moines
we will be sending out 101 units
 Since the Des Moines factory capacity is only
100, we must ship fewer desks from Des Moines
to Albuquerque so we place a minus sign in that
box
 But that leaves Albuquerque one unit short so we
must increase the shipment from Evansville to
Albuquerque by one unit and so on until we
complete the entire closed path

Evaluating the unused
Des Moines–Boston
shipping route
Warehouse A
Factory
D
Factory
E
TO
FROM
ALBUQUERQUE
DES MOINES
100
EVANSVILLE
200
WAREHOUSE
REQUIREMENTS
$5
$8
$9
FORT LAUDERDALE
300
BOSTON
100
100
200
CLEVELAND
$4
$3
$4
$3
$7
200
200
$5
Warehouse B
$5
100
–
+
+
$8
200
FACTORY
CAPACITY
100
300
300
700
$4
–
100
$4

Evaluating the unused
Des Moines–Boston
shipping route
Warehouse A
Factory
D
Factory
E
TO
FROM
ALBUQUERQUE
DES MOINES
100
EVANSVILLE
200
WAREHOUSE
REQUIREMENTS
$5
$8
$9
FORT LAUDERDALE
300
BOSTON
100
100
200
CLEVELAND
$4
$3
$4
$3
$7
200
200
$5
Warehouse B
$5
99
1
100
–
201
+
$8
200
FACTORY
CAPACITY
100
300
300
700
$4
99
+
–
100
$4

Evaluating the unused
Des Moines–Boston
shipping route
Warehouse A
Factory
D
Factory
E
TO
FROM
ALBUQUERQUE
DES MOINES
100
EVANSVILLE
200
WAREHOUSE
REQUIREMENTS
$5
$8
$9
FORT LAUDERDALE
300
BOSTON
CLEVELAND
$4
100
100
200
$3
$4
$7
$3
200
200
$5
Warehouse B
$5
99
1
100
–
201
+
$4
$8
200
99
+
–
$4
100
FACTORY
Result
CAPACITY
of Proposed
Shift in Allocation
100
= 1 x $4
– 1 x $5
300 + 1 x $8
– 1 x $4 = +$3
300
700
Step 4. We can now compute an improvement
index (Iij) for the Des Moines–Boston route
 We add the costs in the squares with plus signs
and subtract the costs in the squares with minus
signs
Des Moines–
Boston index = IDB = +$4 – $5 + $5 – $4 = + $3
 This means for every desk shipped via the Des
Moines–Boston route, total transportation cost will
increase by $3 over their current level
Step 5. We can now examine the Des Moines–
Cleveland unused route which is slightly more
difficult to draw
 Again we can only turn corners at squares that
represent existing routes
 We must pass through the Evansville–Cleveland
square but we can not turn there or put a + or –
sign
 The closed path we will use is
+ DC – DA + EA – EB + FB – FC

Evaluating the Des Moines–Cleveland shipping
route
TO
FROM
ALBUQUERQUE
DES MOINES
–
100
EVANSVILLE
+
200
WAREHOUSE
REQUIREMENTS
$5
$8
$9
FORT LAUDERDALE
300
BOSTON
$4
–
100
100
+
200
CLEVELAND
Start
+
$4
$7
$3
$3
–
200
200
$5
FACTORY
CAPACITY
100
300
300
700
Des Moines–Cleveland
= IDC = + $3 – $5 + $8 – $4 + $7 – $5 = + $4
improvement index
 Opening the Des Moines–Cleveland route will
not lower our total shipping costs
 Evaluating the other two routes we find
Evansville= IEC = + $3 – $4 + $7 – $5 = + $1
Cleveland
index
 The closed path is
+ EC – EB + FB – FC
Fort Lauderdale–
= IFA = + $9 – $7 + $4 – $8 = – $2
Albuquerque
index
 The closed path is
+ FA – FB + EB – EA
 So opening the Fort Lauderdale-Albuquerque
route will lower our total transportation costs
In the Executive Furniture problem there is only one
unused route with a negative index (Fort
Lauderdale-Albuquerque)
 If there was more than one route with a negative
index, we would choose the one with the largest
improvement
 We now want to ship the maximum allowable
number of units on the new route
 The quantity to ship is found by referring to the
closed path of plus and minus signs for the new
route and selecting the smallest number found in
those squares containing minus signs

To obtain a new solution, that number is added to
all squares on the closed path with plus signs and
subtracted from all squares the closed path with
minus signs
 All other squares are unchanged
 In this case, the maximum number that can be
shipped is 100 desks as this is the smallest value in a
box with a negative sign (FB route)
 We add 100 units to the FA and EB routes and
subtract 100 from FB and EA routes
 This leaves balanced rows and columns and an
improved solution


Stepping-stone path used to evaluate route FA
TO
A
FROM
D
100
E
200
F
WAREHOUSE
REQUIREMENTS
B
$5
$8
–
+
100
–
100
$9
+
300
200
FACTORY
CAPACITY
C
$4
$3
$4
$3
$7
200
200
$5
100
300
300
700

Second solution to the Executive Furniture problem
TO
FROM
A
D
100
E
100
F
100
WAREHOUSE
REQUIREMENTS
300
B
$5
$8
200
$9
$4
$3
$4
$3
$7
200
FACTORY
CAPACITY
C
200
200
$5
100
300
300
700
 Total shipping costs have been reduced by (100
units) x ($2 saved per unit) and now equals
$4,000
This second solution may or may not be optimal
 To determine whether further improvement is
possible, we return to the first five steps to test
each square that is now unused
 The four new improvement indices are

D to B = IDB = + $4 – $5 + $8 – $4 = + $3
(closed path: + DB – DA + EA – EB)
D to C = IDC = + $3 – $5 + $9 – $5 = + $2
(closed path: + DC – DA + FA – FC)
E to C = IEC = + $3 – $8 + $9 – $5 = – $1
(closed path: + EC – EA + FA – FC)
F to B = IFB = + $7 – $4 + $8 – $9 = + $2
(closed path: + FB – EB + EA – FA)
 Path to evaluate for the EC route
TO
A
FROM
D
100
E
100
F
WAREHOUSE
REQUIREMENTS

B
$5
$8
–
100
+
300
$4
200
$9
$4
$3
Start
$3
+
$7
200
FACTORY
CAPACITY
C
–
200
200
$5
100
300
300
700
An improvement can be made by shipping the
maximum allowable number of units from E to C
 Total cost of third solution
ROUTE
DESKS
SHIPPED
PER UNIT
x COST ($)
TOTAL
COST ($)
FROM
TO
D
A
100
5
500
E
B
200
4
800
E
C
100
3
300
F
A
200
9
1,800
F
C
100
5
500
=
3,900
 Third and optimal solution
TO
FROM
D
A
100
WAREHOUSE
REQUIREMENTS
$5
$8
E
F
B
200
300
$4
200
$9
$4
$7
200
FACTORY
CAPACITY
C
$3
100
100
200
$3
$5
100
300
300
700

This solution is optimal as the improvement
indices that can be computed are all greater
than or equal to zero
D to B = IDB = + $4 – $5 + $9 – $5 + $3 – $4 = + $2
(closed path: + DB – DA + FA – FC + EC – EB)
D to C = IDC = + $3 – $5 + $9 – $5 = + $2
(closed path: + DC – DA + FA – FC)
E to A = IEA = + $8 – $9 + $5 – $3 = + $1
(closed path: + EA – FA + FC – EC)
F to B = IFB = + $7 – $5 + $3 – $4 = + $1
(closed path: + FB – FC + EC – EB)
1. Set up a balanced transportation table
2. Develop initial solution using either the northwest
corner method or Vogel’s approximation method
3. Calculate an improvement index for each empty cell
using either the stepping-stone method or the MODI
method. If improvement indices are all nonnegative,
stop as the optimal solution has been found. If any
index is negative, continue to step 4.
4. Select the cell with the improvement index indicating
the greatest decrease in cost. Fill this cell using the
stepping-stone path and go to step 3.