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Operations Research Unit 5
Sikkim Manipal University Page No. 93
Unit 5 Transportation Problem
Structure:
5.1 Introduction
Learning objectives
5.2 Formulation of Transportation Problem (TP)
5.3 Transportation Algorithm (MODI Method)
5.4 The Initial Basic Feasible Solution
North west corner rule
Matrix minimum method
Vogel‟s approximation method
5.5 Moving Towards Optimality
Improving the solution
Modified distribution method / MODI method / U – V Method
Degeneracy in transportation problem
5.6 Summary
5.7 Terminal Questions
5.8 Answers to SAQs and TQs
Answers to Self Assessment Questions
Answers to Terminal Questions
5.9 References
5.1 Introduction
Welcome to the unit on transportation model in Operations Research
Management. Transportation model is an important class of linear
programs. For a given supply at each source and a given demand at each
destination, the model studies the minimisation of the cost of transporting a
commodity from a number of sources to several destinations.
The transportation problem involves m sources, each of which has available
ai (i = 1, 2… m) units of homogeneous product and n destinations, each of
which requires bj (j = 1, 2…., n) units of products. Here ai and bj are positive
integers. The cost cij of transporting one unit of the product from the ith
source to the jth destination is given for each i and j. The objective is to
develop an integral transportation schedule that meets all demands from the
inventory at a minimum total transportation cost.
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It is assumed that the total supply and the total demand are equal.
n
1j
m
1i
i bja (1)
The condition (1) is guaranteed by creating either a fictitious destination with
a demand equal to the surplus if total demand is less than the total supply or
a (dummy) source with a supply equal to the shortage if total demand
exceeds total supply. The cost of transportation from the fictitious
destination to all sources and from all destinations to the fictitious sources
are assumed to be zero so that total cost of transportation will remain the
same.
Learning objectives
By the end of this unit, you should be able to:
Formulate the transportation problem
Find the initial basic feasible solution
Compare the advantages of various methods of finding initial basic
feasible solution
Solve the degeneracy in the transportation problem
Apply the model to minimise the cost of transporting a commodity
5.2 Formulation of Transportation Problem
The standard mathematical model for the transportation problem is as
follows.
Let xij be number of units of the homogenous product to be transported from
source i to the destination j
Then objective is to
Minimise z = ij
m
1i
n
1j
ij xC
Subject to
n.,..........,2,1ij;bjx
m,......,2,1i,ax
m
1i
ij
n
1j
iij (2)
(2) (2)
With all xij 0 and integrals
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Theorem: A necessary and sufficient condition for the existence of a
feasible solution to the transportation problem (2) is:
n
1j
m
1i
i bja
Self Assessment Questions
Fill in the blanks
1. Transportation problems are a special type of ___________.
2. The number of rows and columns need not always be ___________.
3. Transportation problem develops a schedule at _______ and ________.
5.3 Transportation Algorithm (MODI Method)
The first approximation to (2) is integral. Therefore, you always need to find
a feasible solution. Rather than determining a first approximation by a direct
application of the simplex method, it is more efficient to work with the
transportation table given below. The transportation algorithm is the simplex
method specialised to the format of table involving the following steps:
i) Finding an integral basic feasible solution
ii) Testing the solution for optimality
iii) Improving the solution, when it is not optimal
iv) Repeating steps (ii) and (iii) until the optimal solution is obtained
The solution to TP is obtained in two stages.
In the first stage, you find the basic feasible solution using any of the
following methods a) North-west corner rule b) Matrix Minima Method or
least cost method c) Vogel‟s approximation method. In the second stage,
you test the basic feasible solution for its optimality either by MODI method
or by stepping stone method.
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Table 5.1: Transportation Table
D1 D2 Dn Supply ui
S1
x11
x12
x1n a1 u1
S2
x21
x22
x2n a2 u2
S3
x31
x32
x3n a3 u3
Sm
xm1
xm2
xmn am um
Demand b1 b2 bn ai = bi
vj v1 V2 vm
Self Assessment Questions
State Yes or No
4. In transportation problems, ai = bj is a sufficient and necessary
condition for getting a feasible solution.
5. Transportation problems can also be solved by simplex method.
6. Matrix-minima method gives optimum solution.
5.4 The Initial Basic Feasible Solution
Let us consider a TP involving m-origins and n-destinations. Since the sum
of origin capacities equals the sum of destination requirements, a feasible
solution always exists. Any feasible solution satisfying m + n – 1 of the
m + n constraints is a redundant one and hence it can be deleted. This also
means that a feasible solution to a TP can have only m + n – 1 positive
component; otherwise the solution will degenerate.
It is always possible to assign an initial feasible solution to a TP, satisfying
all the rim requirements. This can be achieved either by inspection or by
following some simple rules. You can begin by imagining that the
transportation table is blank that is initial xij = 0. The simplest procedures for
initial allocation are discussed in the following section.
C11
C21
C31
Cm
1
C12
C22
C32
Cm
2
C1n
C2n
C3n
Cm
n
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5.4.1 North West Corner rule
Step1: The first assignment is made in the cell occupying the upper left
hand (north-west) corner of the transportation table. The maximum feasible
amount is allocated here is
x11 = min (a1, b1)
Either the capacity of origin O1 is used up or the requirement at destination
D1 is satisfied or both. This value of x11 is entered in the upper left hand
corner (small square) of cell (1, 1) in the transportation table.
Step 2: If b1 > a1, the capacity of origin O is exhausted and the requirement
at destination D1 is still not satisfied. Then at least one variable in the first
column will have to take on a positive value. Move down vertically to the
second row and make the second allocation of magnitude:
x21 = min (a2, b1 – x21) in the cell (2, 1)
This either exhausts the capacity of origin O2 or satisfies the remaining
demand at destination D1.
If a1 > b1 ,the requirement at destination D1 is satisfied, but the capacity of
origin O1 is not completely exhausted. Move to the right in a horizontal
position to the second column to make the second allocation of magnitude:
x12 = min (a1 – x11, b2) in the cell (1, 2)
This either exhausts the remaining capacity of origin O1 or satisfies the
demand at destination D2.
If b1 = a1, the origin capacity of O1 is completely exhausted as well as the
requirement at destination is completely satisfied, then there is a tie at the
second allocation. An arbitrary tie breaking choice is made. Make the
second allocation of magnitude
x12 = min (a1 – a1, b2) = 0 in the cell (1, 2)
OR
x21 = min (a2, b1 – b2) = 0 in the cell (2, 1)
Step 3: Start from the new north-west corner of the transportation table
satisfying the destination requirements and exhausting the origin capacities
one at a time, moving down towards the lower right corner of the
transportation table until all the rim requirements are satisfied.
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Solved Problem 1
Determine an initial basic feasible solution to the following transportation
problem using the north west corner rule:
D1 D2 D3 D4
Availability
01 6 4 1 5 14
02 8 9 2 7 16
03 4 3 6 2 5
6 10 15 4 35
Requirements
Where Oi and Dj represent the ith origin and the jth destination respectively.
Solution: The transportation table of the given T.P. has 12 cells.
14
16
5
6 10 15 4
Following north west corner rule, the first allocation is made in the cell
(1,1), the magnitude being x11 = min (14, 6) = 6
The second allocation is made in the 6 10 15 4 cell (1, 2) and the
magnitude of allocation is given by x12 = min (14 – 6, 10) = 8
The third allocation is made in the cell (2, 2), the magnitude being
x22 = min (16, 10 – 8) = 2.
The magnitude of fourth allocation, in the cell (2, 3) is given by x23 = min
(16 – 2, 15) = 14.
The fifth allocation is made in the cell (3, 3), the magnitude being x33 = min
(5, 15 –14) =1.
The sixth allocation in the cell (3, 4) is given by x34 = min (5 –1, 4) = 4.
Now all the rim requirements have been satisfied and hence an initial
feasible solution to the TP has been obtained. The solution is displayed as
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Table 5.2: Initial feasible solution to the TP
D1 D2 D3 D4
01
02
03
6
6
8
4
4
5 14
8
2
9
14
2
16
4
3
1
6
4
2 5
6 10 15 4
Clearly, this feasible solution is non-degenerate basic feasible solution as
the allocated cells do not form a loop. The transportation cost according to
the above loop is given by.
Z = x11 c11 + x12 c12 + x22 c22 + x23 c23 + x33 c33 + x34 c34
= 66 + 48 + 92 + 214 + 2461
= 128
5.4.2 Matrix minimum method
Step 1: Determine the smallest cost in the cost matrix of the transportation
table. Let it be cij . Allocate xij = min ( ai, bj) in the cell ( I, j )
Step 2: If xij = ai cross the ith row of the transportation table, decrease bj by
ai and proceed to step 3.
If xij = bj cross the ith column of the transportation table, decrease ai by bj and
proceed to step 3.
If xij = ai= bj cross either the ith row or the ith column, but not both.
Step 3: Repeat steps 1 and 2 to reduce transportation table until all the rim
requirements are satisfied. Whenever the minimum cost is not unique, make
an arbitrary choice among the minima.
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Solved Problem 2
Obtain an initial basic feasible solution to the following TP using the
matrix minima method.
Table 5.3: Initial table
D1 D2 D3 D4
01 1 2 3 4 6 Capacity
02 4 3 2 0 8
03 0 6 8 6 10
4 6 8 5 24
Demand
Where 0i and Di denote ith origin and jth destination respectively.
Solution: The transportation table of the given TP has 12 cells. Following
the matrix minima method,
The first allocation is made in the cells (3, 1), the magnitude being x21 = 4.
This satisfies the requirement at destination D1 and thus we cross the first
column from the table. The second allocation is made in the cell (2, 4),
the magnitude being x24 = min (6, 8) = 6. Cross the fourth column of the
table. This yields the table (i) shown below
There is again a tie for the third allocation. Arbitrarily choose the cell
(1, 2) and allocate x12 = min (6, 6) = 6. Cross the second column of the
first row. Next, choose to cross off the first row of the table. The next
allocation of magnitude x22 = 0 is made in the cell (3, 2), cross the
second column getting table(ii), as shown above
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Again you choose arbitrarily to make the next allocation in cell (2, 3) of
magnitude x23 = min (2, 8) = 2, cross the second row to get table (iii) as
show above. The last allocation of magnitude x23 = min (6, 6) = 6 is made
in the cell (3, 3).
Now that all the rim requirements have been satisfied, an initial feasible
solution has been determined. This solution is shown above in table (iv).
Since the cells do not form a loop, the solution is basic and degenerate.
The transportation cost according to the above route is given by
Z=6×2+2×2+6×0+4×0+0×2+6×2=28
5.4.3 Vogel’s approximation method
The Vogel‟s approximation method (VAM) takes into account the least cost
cij, but also the cost that just exceeds cij. The steps of the method are given
below.
Step 1: For each row of the transportation table, identify the smallest and
the next to smallest costs. Determine the difference between them for each
row. Display them alongside the transportation table by enclosing them in
parenthesis against the respective rows. Similarly, compute the differences
for each column.
Step 2: Identify the row or column with the largest difference among all the
rows and columns. If a tie occurs, use any arbitrary tie breaking choice. Let
the greatest difference correspond to the ith row and let Cij be the smallest
cost in the ith row. Allocate the maximum feasible amount xij = min (ai, bj) in
the (i, j)th cell and cross off the ith row or the jth column in the usual manner.
Step 3: Recompute the column and row differences for the reduced
transportation table and go to step 2. Repeat the procedure until all the rim
requirements are satisfied.
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Remarks:
1. A row or column “difference” indicates the minimum unit penalty incurred
by failing to make an allocation to the last smallest cell in that row or
column.
2. It is clear that VAM determines an initial basic feasible solution, which is
very close to the optimum solution. But the number of iterations required
to reach the optimal solution is small.
Caselet
A company has four warehouses. Each warehouse supplies inventory to
five stores.
The company needs to develop an integral transportation schedule that
meets all demands from the inventory at a minimum total transportation
cost.
Assuming that the total supply and the total demand are equal, the
company can use the transportation model of a LPP to arrive at a basic
feasible solution.
Solved Problem 3
Obtain an initial basic feasible solution to the following TP using the
Vogel‟s approximation method:
Table 5.4: Transportation Table
Ware houses Stores
Availability I II III IV
A 5 1 3 3 34
B 3 3 5 4 15
C 6 4 4 3 12
D 4 -1 4 2 19
Requirement 21 25 17 17 80
Solution: The transportation table of the given TP has 16 cells. The
differences between the smallest and next to smallest costs in each row
and each column are computed and displayed inside the parenthesis
against the respective columns and rows. The largest of these differences
is (3) and is associated with the fourth row of the transportation table.
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The minimum cost in the fourth row is C42 = –1. Accordingly, you allocate
x42 = min (19, 25) = 19 in the cell (4, 2). This exhausts the availability at
warehouse D. Cross the fourth row. The row and column differences are
now computed for the resulting reduced transportation table (ii) as shown
below:
Table 5.5: Resulting reduced transportation table
5 1 3 3 34(2)
3 3 5 5 15(0)
6 4 4 3 12(1)
4 19-1 4 2 19(3)
21 (1)
25
(2)
17 (1)
17 (1)
(i)
Table 5.6: Resulting reduced transportation table
5 1 3 3 34 (2)
3 3 5 4 15 (0)
6 4 4 1 12 (1)
21 (2) 6 (2) 17 (1) 17 (0)
(ii)
The largest of this is (2) and is associated with the first row as well as the
first and second column. Arbitrarily, select the first row whose minimum
cost is C12 = 1. Thus the second allocation of magnitude x12 = min (34, 6)
= 6 is made in the cell (1, 2). Cross the second column from the table.
Continuing in this way, the subsequent reduced transportation tables and
the differences for the surviving rows and columns are shown below in
figure 5.1.
Figure 5.1: Surviving rows and columns
6
19
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Eventually, the basic feasible solution is obtained as shown in table 5.6
below:
Table 5.7
The transportation cost according to the above route is given by:
Z = 6 x 5 + 6 x 1 + 17 x 3 + 5 x 3 + 15 x 3 + 12 x 3 + 19 x (– 1) = 164
Self Assessment Questions
State True or False
7. In matrix-minima method, you start allocating from the left-top cell of
the table.
8. In Vogel‟s approximation method, you first construct penalty and then
start allocating.
9. North-west corner rule gives optimum solution.
10. Vogel‟s approximation method gives solution near to the optimum
solution.
5.5 Moving Towards Optimality
After evaluating an initial basic feasible solution to a transportation problem,
the next question is how to get the optimum solution. The basic techniques
are illustrated below.
1. Determine the net evaluations for the non–basic variables (empty cells)
2. Determine the entering variable
3. Determine the leaving variable
4. Compute a better basic feasible solution
5. Repeat steps (1) to (4) until an optimum solution has been obtained
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5.5.1 Improving the solution
Definition: A loop is the sequence of cells in the transportation table such
that:
i) Each pair of consecutive cells lie either in the same row or same
column
ii) No three consecutive cells lies in the same row or same column
iii) The first and the last cells of the sequence lies in the same row or
column
iv) No cell appears more than once in the sequence
Consider the non-basic variable corresponding to the most negative of the
quantities cij – ui – vj, calculated in the test for optimality; it is made the
incoming variable. Construct a loop consisting exclusively of this incoming
variable (cell) and current basic variables (cells). Then allocate the incoming
cell to as many units as possible after appropriate adjustments have been
made to the other cells in the loop. Avoid violating the supply and demand
constraints, allow all allocations to remain non-negative and reduce one of
the old basic variables to zero (where upon it ceases to be basic).
5.5.2 Modified distribution method / Modi method / U–V method
Step 1: Under this method, you construct penalties for rows and columns
by subtracting the least value of row / column from the next least value.
Step 2: Then select the highest penalty constructed for both row and
column. Enter that row/column and select the minimum cost and allocate
min (ai, bj)
Step 3: Delete the row or column or both if the rim availability/ requirements
are met.
Step 4: You repeat steps 1 to 2 to till all allocations are over.
Step 5: For allocating all forms of equations ui + vj = cj, set one of the dual
variable ui / vj to zero and solve for others.
Step 6: Use this value to find ij = cij - ui - vj. If all ij 0, then it is the optimal
solution.
Step 7: If any ij 0, select the most negative cell and form loop. Starting
point of the loop is positive and alternative corners of the loop are negative
and positive. Examine the quantities allocated at negative places. Select the
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minimum, add it to the positive places and subtract from the negative
places.
Step 8: Form new table and repeat steps 5 to 7 till ij 0
Balanced TP
Solved Problem 4
Solve the following transportation problem with cost coefficients demands
and supplies as given in the following table:
Destinations
A B C Supply
Sources I
II
III
6 8 4 14
12
5
4 9 8
1 2 6
Demand 6 10 15
Solution: Since total demand = 31 = Total supply, the problem is
balanced. The initial basic feasible solution is obtained by Vogel‟s
approximation method. The following table gives the initial solution:
Supply
6
8
4
14 14
12
5
4
6
9
5
8
1
1 2
5
6
Demand 6 10 15
Destination
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D1 D2 D3
1 6 8 4 14
14
2 4 6
9 5
8 1
12
3 1 2 5
6 5
6 10 15
The optimum allocations are
x13 = 14, x21 = 6, x22 = 5, x23 = 1, x32 = 5
The minimum transportation cost is
144+64+59+18+52 = 143
For allocated cells
u1 + v3 = 4 u1 = -4
Set u2 = 0
u1 + v3 = 4 u1 = -4
u2 + v2 = 9 v1 = 4
u2 + v3 = 8 v2 = 9
u3 + v2 = 5 v3 = 8
Note: Select that variable ui / vi is repeated very often for easy calculation.
Here u2 is repeated often.
For unallocated cells
ij = cij – ui – vj
11 = 6 – (-4) – 4 = 6
12 = 8 – (-4) – 9 = 3
31 = 1 – (-4) – 4 = 1
33 = 6 – (-4) – 8 = 2
Since all ij 0, the optimum solution is
X13 = 14 x 4 = 56
X21 = 6 x 4 = 24
X22 = 5 x 9 = 45
X23 = 1 x 8 = 8
X32 = 5 x 2 = 10
Total cost = 143
Supply Sources
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Unbalanced T.P
A car company is faced with an allocation problem resulting from rental
agreement that allow cars to be returned to locations other than those
from where they were originally rented. At the present time there are two
cars with 15 and 13 simplex cars respectively and 4 locations requiring 9,
6, 7 and 9 cars respectively. The unit transportation costs (in dollars)
between the locations an given below:
Destinations
D1 D2 D3 D4
Sources S1 45 17 21 30
S2 14 18 19 31
Obtain a minimum cost schedule.
Solution: Since the supply and requirements are not equal it is called an
unbalanced TP. In general, if ai bj then it is called an unbalanced TP.
We introduce either a dummy row or a column with cost zero quantities
and bi aj respectively. Applying Vogel‟s approximation method we find
the basic feasible solution.
Destinations
S1 D1 D2 D3 D4 P1 P2 P3
P4
Sources
S2
6
3
6
15/9/6
4 4 4 11
S3
9
4
13/4
4 4 4 12
1st
Cancel
3
3
0 - - -
Demand 9/0 6 7/3 9/6
P1 14 17 19 30
P2 31 1 2 1
P3 – 1 2 1
P4 – – 2 1
45
14
4
0
17
18
0
21
19
0
30
31
0
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X34 = Min [3, 9] = 3 x 10 = 0
X21 = Min [13, 9] = 9 x 14 = 126
X12 = Min [15, 6] = 6 x 17 = 102
X23 = Min [4, 7] = 4 x 9 = 76
X14 = Min [6, 6] = 6 x 30 =
180
Total Cost 547
Testing for optimality
u1 + v2 = 17 Set u1 = 0
u1 + v3 = 21 u1 = 2
u1 + v4 = 30 u3 = 30
u2 + v1 = 14 v1 = 16
u2 + v3 = 19 v2 = 17
u3 + v4 = 0 v3 = 21
v4 = 30
For unallocated cells
ij = cij - ui - vj
11 = 45 – 0 – 16 = 29
22 = 8 + 2 – 17 = 3
24 = 31 +2 – 30 = 3
31 = 0 + 30 – 26 = 4
32 = 0 + 30 – 17 = 13
33 = 0 + 30 – 21 = 9
For non-allocated cells, determine cij – uI – vi. Since all then quantities are
non-negative, the current solution is optimal. The minimum transportation
cost is equal to
617+321+630+914+419+3.0 = 470.
This is achieved by transporting x12 = 6 cars from source 1 to destination
2, x13 = 3, x14 = 6 cars from sources 1 to destinations 3 and 4
respectively; x21 = 9 and x33 = 4 cars from sources 2 to destinations 1
and 3 respectively.
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5.5.3 Degeneracy in transportation problem
It is shown that a basic solution to an m-origin, n destination; transportation
problem can have at the most m+n-1 positive basic variables (non-zero),
otherwise the basic solution degenerates. It follows that whenever the
number of basic cells is less than m + n – 1, the transportation problem is a
degenerate one. The degeneracy can develop in two ways:
Case 1: The degeneracy develops while determining an initial assignment
via any one of the initial assignment methods discussed earlier.
To resolve degeneracy, you must augment the positive variables by as
many zero-valued variables as is necessary to complete the required
m + n – 1 basic variable. These zero-valued variables are selected in such a
manner that the resulting m + n – 1 variable constitutes a basic solution.
The selected zero valued variables are designated by allocating an
extremely small positive value ε to each one of them. The cells containing
these extremely small allocations are then treated like any other basic cells.
The ε‟s are kept in the transportation table until temporary degeneracy is
removed or until the optimum solution is attained, whichever occurs first. At
that point, we set each ε = 0.
Case 2: The degeneracy develops at the iteration stage. This happens
when the selection of the entering variable results in the simultaneous drive
to zero of two or more current (pre-iteration) basic variables.
To resolve degeneracy, the positive variables are augmented by as many
zero-valued variables as it is necessary to complete m+n-1 basic variables.
These zero-valued variables are selected from among those current basic
variables, which are simultaneously driven to zero. The rest of the
procedure is exactly the same as discussed above in case 1.
Note: The extremely small value ε is infinitely small and it never affects the
value it is added to or subtracted from. Introduce „‟ in unallocated minimum
cost cell to avoid forming a loop.
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Solved Problem 5
Obtain an optimum basic feasible solution to the following degenerate T.P.
Table 5.7: Initial table
From
7 3 4 2 Available 2 1 3 3
3 4 6 5
4 1 5 10
Demand
Solution: Following the North West Corner rule, an initial assignment is
made as shown in the table 5.7. Since the basic cells do not form a loop,
the solution is basic. However, since the number of basic cells is 4, which
is less than 5. (= m + n - 1) the basic solution degenerates.
Table 5.8: Initial assignment
7
3 4 2
2 1 3 3
3 4 6 5
In order to complete the basis and remove degeneracy, you require only
one more positive basic variable. Select the variable x23 and allocate a
negligibly small positive quantity in the cells (2, 3) as shown in the table
5.8.
Table 5.9: Augmented solution
7
3 4 2
2 1 3 3 + = 3
3 4 6 5
4 1 5 + = 5
2
2
1
5
1 2
1
5
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Notice that even after the inclusion of cell (2, 3) in the basis; the basic
cells do not form a loop, that is, the augmented solution remains basic.
The net evaluations can now be computed and the solution is tested for
optimality.
Starting table:
Since all the net evaluations for the non-basic variables are not non-
positive, the initial solution is not optimum. The non basic cell (1, 3) must
enter the basis. The exit criterion removes the basic cell (2, 3) from the
basis max .
Table 5.10: Starting table
–
7
(3)
3
(4)
4
5
+
2
1 -
3 0
(2)
3
(0)
4
6 8
-
7
(3)
3 +
4
0
2
1
(– 4)
3
-
(6)
(4)
4 –
6
2
First iteration: Introduce the cell (1, 3) into the basis and drop the cell (2,
3) from it. Determine the current net evaluations. Since all of them are
not non-positive the current solution can be improved.
Second iteration: Introduce the cell (3, 1) and drop the cell (1, 1) from the
basis. Since some of the current net evaluations are still positive, the
current solution can further be improved.
2 1
2 2
2
1
6
Vi 2 1 3 Vj 7 6 4
Starting table First iterated table
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Table 5.11: Second iterated table
Ui
(-6)
7
(-3)
3
4 –1
–
2
1
(2)
3
0
+
3
(–2)
4 –
6 1
vj 2 1 5
Table 5.12: Optimum table
ui
(– 6)
7
(–1)
3
4 4
(–2)
2
1
3 3
3
(0)
4
6 6
Vj – 3 – 2 0
Third iteration: Introduce the cell (2, 3) and drop the cell (2, 1) from the
basis. Since all the current net evaluations are non positive, the current
solution is an optimum one. The transportation cost according to the
above route is given by
6134321142z = 33
Self Assessment Questions
Fill in the blanks
11. All the values of Cij - ui - vj should be __________ or _________ for the
solution to be optimum.
12. In unbalanced transportation problem ai is _________ _______ to bj.
13. If the number of allocation is less than _________ then it is said to be a
degenerate transportation problem.
1
1
2
2
4
2 1
3
2
2
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Sikkim Manipal University Page No. 114
5.6 Summary
The transportation problem is a special type of linear programming problem
in which the objective is to transport a homogeneous product manufactured
at several plants (origins) to a number of different destinations at a minimum
total cost. In this unit, you have learnt several different techniques for
computing an initial basic feasible solution to a transportation problem, such
as north-west corner rule, matrix minimum method and Vogel‟s
approximation method. Further, you studied the degeneracy in
transportation problem with examples on obtaining an optimum basic
feasible solution.
5.7 Terminal Questions
1. Solve the following transportation problem.
2. A company has three cement factories located in cities 1,2,3 which
supply cement to four projects located in towns 1,2,3,4. Each plant can
supply daily 6,1,10 truck loads of cement respectively and the daily
cement requirements of the projects are respectively 7,5,3,2 truck loads.
The transportation cost per truck load of cement (in hundreds of rupees)
from each plant to each project site is as follows.
Factories
Determine the optimal distribution for the company so as to minimise the
total transportation cost.
1 2 3 4
1 2 3 11 7
2 1 0 6 1
3 5 8 15 9
Operations Research Unit 5
Sikkim Manipal University Page No. 115
3. Solve the following transportation problem.
9 12 9 6 9 10 5
7 3 7 7 5 5 6
6 5 9 11 3 11 2
6 8 11 2 2 10 9
4 4 6 2 4 2 22
5.8 Answers to SAQs and TQs
Answers to Self Assessment Questions
1. LPP
2. Equal
3. Minimum cost
4. Yes
5. Yes
6. No
7. False
8. True
9. False
10. True
11. zero
12. Not equal to
13. m + n – 1
Answers to Terminal Questions
1. The optimal transportation cost is Rs. 796
2 Optimal transportation cost is Rs. 10, 000
3 The minimum transportation cost is Rs. 112 as 0
5.9 References
No external sources have been referred for this unit.