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Transcript of 5 Transportation Problem
Operations Research Unit 5
Sikkim Manipal University 77
Unit 5 Transportation Problem
Structure
5.1. Introduction
5.2. Formulation of T.P
5.3. Transportation Algorithm
5.4. The initial basic feasible solution
5.4.1 North West Corner Rule
5.4.2 Matrix Minimum Method
5.4.3 Vogel’s approximation method
5.5. Moving towards optimality
5.5.1 Improving the solution
5.5.2 Modified Distribution Method / MODI Method / U – V method.
5.5.3 Degeneracy in transportation problem
5.6. Summary
5.7. Terminal Questions
5.8. Answers to SAQs and TQs
5.1 Introduction Here we study an important class of linear programs called the transportation model. This model
studies the minimization of the cost of transporting a commodity from a number of sources to
several destinations. The supply at each source and the demand at each destination are known.
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 i th source to the j th 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.
It is assumed that the total supply and the total demand are equal.
i.e. ∑ ∑ = =
= n
1 j
m
1 i i bj a (1)
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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: After studying this unit, you should be able to understand the following
1. To formulate the transportation problem. 2. Find the initial basic feasible solution 3. Compare the advantages of various methods of finding initial basic feasible solution. 4. Handle and solve the degeneracy in the transportation problem. 5. Apply the model to minimize 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
Minimize z = ij
m
1 i
n
1 j ij x C ∑ ∑
= =
Subject to
= =
= =
∑
∑
=
=
n ., .......... , 2 , 1 ij ; bj x
m , ...... , 2 , 1 i , a x
m
1 i ij
n
1 j i ij (2)
with all xij ≥ 0 and integrals
Theorem: A necessary and sufficient condition for the existence of a feasible solution to the transportation problem (2) is that
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∑ ∑ = =
= n
1 j
m
1 i i bj a
Self Assessment Questions 1 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 __________ ________.
5.3 The Transportation Algorithm (MODI Method) The first approximation to (2) is always integral and therefore always 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 table given below called the transportation table. The transportation
algorithm is the simplex method specialized to the format of table it involves:
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 T.P is obtained in two stages. In the first stage we find Basic feasible solution by
any one of the following methods a) Northwest corner rale b) Matrix Minima Method or least
cost method c) Vogel’s approximation method. In the second stage we test the B.Fs for its
optimality either by MODI method or by stepping stone method.
D1 D2 Dn Supply ui
S1 x11 x12 x1n
a1 u1
S2 x21 x22 x2n a2 u2
S3 x31 x32 x3n a3 u3
C11
C21
C31
Cm1
C12
C22
C32
Cm2
C1n
C2n
C3n
Cmn
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Sm xm1 xm2 xmn
am um
Demand b1 b2 bn ∑ai = ∑bi
vj v1 V2 vm
Transportation Table
Self Assessment Questions 2
State Yes or No
1. In T.P ∑ai = ∑bj for getting a feasible solution.
2. T.P can also be solved by simplex method. 3. Matrixminima method gives optimum solution.
5.4 The Initial Basic Feasible Solution Let us consider a T.P involving morigins and ndestinations. 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 can be
deleted. This also means that a feasible solution to a T.P can have at the most only m + n – 1
strictly positive components, otherwise the solution will degenerate.
It is always possible to assign an initial feasible solution to a T.P. in such a manner that the rim
requirements are satisfied. This can be achieved either by inspection or by following some simple
rules. We begin by imagining that the transportation table is blank i.e. initially all xij = 0. The
simplest procedures for initial allocation discussed in the following section.
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 there, that is x11 = min
(a1,b1)
So that 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 but the requirement at destination D1 is
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still not satisfied , so that at least one more other 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 horizontally to the second column and 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. There is a tie for 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 destination requirements and exhausting the origin capacities one at a time, move down towards the lower
right corner of the transportation table until all the rim requirements are satisfied.
Example 1: Determine an initial basic feasible solution to the following transportation problem
using the north west corner rule:
D1 D2 D3 D4 01 6 4 1 5 14 02 8 9 2 7 16 Availability 03 4 3 6 2 5
6 10 15 4 35 Requirements
Where Oi and Dj represent the i th origin and the j th destination respectively.
Solution: The transportation table of the given T.P. has 12 cells. 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
14
16
5
6 10 15 4
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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 T.P.
has been obtained. The solution is displayed as
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 nondegenerate basic feasible solution ; for 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 = 6 6 × + 4 8 × + 9 2× + 2 14 × + 2 4 6 1 × + ×
= 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 off the i th row of the transportation table and decrease bj by ai go to step 3.
if xij = bj cross off the i th column of the transportation table and decrease ai by bj go to step 3.
if xij = ai= bj cross off either the i th row or the i th column but not both.
Step 3: Repeat steps 1 and 2 for the resulting reduced 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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Example 2: Obtain an initial basic feasible solution to the following T.P. using the matrix minima
method.
D1 D2 D3 D4
01 1 2 3 4 6 02 4 3 2 0 8 Capacity 03 0 2 2 1 10
4 6 8 6 24 Demand
Where 0i and Di denote i th origin and j th destination respectively.
Solution: The transportation table of the given T.P. 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 off the first column from the table. The second allocation is made in the cell ( 2, 4 ) magnitude x24 = min( 6, 8 ) =6. Cross off the fourth column of the table. This yields the table (i)
1 2 3 4
4 3 2 8
0 2 2 2
(i) (ii)
There is again a tie for the third allocation. We choose arbitrarily the cell
(1, 2) and allocate x12 = min (6 , 6) = 6 there. Cross off either the second column of the first row
.We 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 off the second column getting table( ii)
(iii) (iv)
2
1 2 3 4
4 3 2 0
0 2 2 1
4
6
6 4 0 0
6
2
6
2
6
2
2
2 4
6
6
6
0 4
6
2
6
6
8
10 4 6 8 6 6 8
6
6
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We choose arbitrarily again to make the next allocation in cell (2,3) of magnitude x23 = min (2,8)
= 2 cross off the second row this gives table (iii). The last allocation of magnitude x23 = min ( 6 ,
6) = 6 is made in the cell (3,3).
Now all the rim requirements have been satisfied and hence an initial feasible solution has been
determined. This solution is shown in table ( iv)
Since the cells do not form a loop, the solution is basic one. Moreover the solution is degenerate
also. The transportation cost according to the above route is given by
Z = 2 6 × + 2 2× + 0 6 × + 0 4 × + 2 6 2 0 × + ×
= 28
5.4.3 Vogel’s Approximation Method The Vogels Approximation Method takes into account not only 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 nest 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. Similarity
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 i th
row and let Cij be the smallest cost in the i th row. Allocate the maximum feasible amount xij = min
(ai, bj) in the (i,j) th cell and cross off the i th row or the j th 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.
Remarks: 1. A row or column “difference” indicates the minimum unit penalty incurred by failing to make an
allocation to the smallest lost 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, that is, the number of iterations required to reach the optimal solution is
small in this case
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Example 3: Obtain an initial basic feasible solution to the following T.P using the Vogel’s approximation
method:
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 T.P 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.
The minimum cost in the fourth row is C42 = –1. Accordingly we allocate x42 = min (19, 25) = 19 in
the cell (4, 2). This exhausts the availability at ware house D. Cross off the fourth row. The row
and column differences are now computed for the resulting reduced transportation table (ii).
5 1 3 3
3 3 5 4
6 4 4 3
4 19
–1 4 2
the largest of this is (2) and is associated with the first row as well as the first and second column.
We select arbitrarily 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 off 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 in fig (iii)
5 6
1
3 3
3 3 5 4
6 4 4 3
19
34 (2)
15(0)
12(1)
19(3)
34 (2)
15(0)
12(1)
21 25 17 17 (1) (2) (1) (1)
(i)
21 6 17 17 (2) (2) (1) (0)
(ii)
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28 (0)
15 (1)
12(1)
21 (2)
17 (1)
17 (0)
Fig (iii) Eventually, the basic feasible solution is obtained as shown in table (iv) below:
Table (iv)
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 3
State True or False
1. In Matrixminima method we start allocating from lefttop cell of the table. 2. In vogel’s approximation method we first construct penalty and then start allocating. 3. Northwest corner Rule gives optimum solution. 4. Vogel 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:
5 3 3
3 5 4
6 4 3
15
5 1 3
6 6 5
3
17
3 3 4
15
5
6 4 3
12
4
4 –1 2
18
4
5 3 3 28(0)
6 4 12
3 12(1)
6 17 17 (1) (1) (0)
6
5
17
3
5
3 6 17 5
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1. Determining the net evaluations for the non – basic variables (empty cells)
2. Determining the entering variable
3. Determining the leaving variable
4. Computing a better basic feasible solution
5. Repeating steps (1) to (4) until an optimum solution has been obtained.
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 nonbasic 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 to the incoming cell as many units as possible such that, after appropriate
adjustments have been made to the other cells in the loop, the supply and demand
constraints are not violated, all allocations remain nonnegative and one of the old basic
variables has been reduced to zero. (where upon it ceases to be basic)
Modified Distribution Method / Modi Method / U – V Method.
Step 1: Under this method we construct penalties for rows and columns by subtracting the least
value of row / column from the next least value.
Step 2: We 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 is met.
Step 4: We repeat steps 1 to 2 to till all allocations are over.
Step 5: For allocation all form equation ui + vj = cj set one of the dual variable ui / vj to zero and solve for others.
Step 6: Use these value to find ∆ij = cij ui vj of all ∆ij ≥, then it is the optimal solution.
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Step 7: If any ∆ij ≤ 0, select the most negative cell and form loop. Starting point of the loop is +ve
and alternatively the other corners of the loop are –ve and +ve. Examine the quantities allocated
at –ve places. Select the minimum. Add it at +ve places and subtract from –ve place.
Step 8: Form new table and repeat steps 5 to 7 till ∆ij ≥ 0
Balanced T.P
Example 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
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 given 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
Sources
The optimum allocations are x13 = 14, x21 = 6, x22 = 5, x23 = 1, x32 = 5. The minimum
transportation cost is 14×4+6×4+5×9+1×8+5×2 = 143.
For allocated cells.
u1 + v3 = 4 u1 = 4
PSet u2 = 0
Pu1 + v3 = 4 u1 = 4
P u2 + v2 = 9 v1 = 4
P u2 + v3 = 8 v2 = 9
u3 + v2 = 5 v3 = 8
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 u
X13 = 14 x 4 = 56
X21 = 6 x 4 = 24
X22 = 5 x 9 = 45
1 6 8 4
14
14
2 4
6
9
5
8
1
12
3 1 2
5
6 5
6 10 15
Supply
Note:
Select that variable ui / vi is repeated very
often for easy calculation. Here u2 is
repeated often
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X23 = 1 x 8 = 8
X32 = 5 x 2 = 10
Total cost = 143
Unbalanced T.P
Example 5
A car company is faced with an allocation problem resulting from rental agreement that allow cars to be returned to locations other than those which 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 T.P. in general if
Σai ≠ Σbj then it is called an unbalanced T.P. we introduced either a dummy row or a dummy
column with cost zero an quantity Σbi ≠ Σaj respectively. Applying VAM – Method we find
B.F.S as Destinations
D1 D2 D3 D4 P1 P2 P3 P4
45 17 6
21 3
30 6
15 /9 /6 4 4 4 11
←
14 9
18 19 4
31 13 / 4 4 4 1 12 ←
0 0 0 0 3 3
0
9 /0 6 7/3 9/6
P1 14 17 19 30↑
S1
Sources S2
S3 1 st Cancel
Demand
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P2 31↑ 1 2 1
P3 1 2 1
P4 2 1
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
Pu1 + v2 = 17 PSet u1 = 0
Pu1 + v3 = 21 u1 = 2
P u1 + v4 = 30 u3 = 30
P u2 + v1 = 14 v1 = 16
Pu2 + v3 = 19 v2 = 17
Pu3 + 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 nonallocated cells determine cij – uI – vi. Since all then quantities are nonnegative, the
current solution is optimal. The minimum transportation cost is =
6×17+3×21+6×30+9×14+4×19+3.0 = 470.
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Which achieved by transporting x12 = 6 cars from sources 1 to dentition 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.
5.5.3 Degeneracy In Transportation Problem It was shown that a basic solution to an morigin, n destination Transportation problem can have
at most m+n1 positive basic variables (nonzero) and so that 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, we augment the positive variables by as many zerovalued variables as
is necessary to complete the required m + n – 1 basic variables. These zerovalued variables are
selected in such a manner that the resulting m + n – 1 variables constitute 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 some 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 zerovalued variables
as is necessary to complete m+n1 basic variables. These zerovalued 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 in case 1 above.
Note: The extremely small value ε is infinitesimally small and it never affects the value it is added to or subtracted from.
We introduce ‘∈’ in unallocated minimum cost cell such that we cannot form a loop
Example: Obtain an optimum basic feasible solution to the following degenerate T.P.
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7 3 4 2
From 2 1 3 3 Available
3 4 6 5
4 1 5 10
Demand
Solution: Following the North West Corner rule, an initial assignment is made as shown in table
1. 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.
7 3 4 2
2 1 3 3
3 4 6 5
Table 1 (Initial assignment)
In order to complete the basis and thereby remove degeneracy, we require only one more
positive basic variable. We select the variable x23 and allocate a negligibly small positive quantity
∈ in the cells (2, 3) as shown in table 2.
7 3 4 2
2 1 3 3 +∈ = 3
3 4 6 5
4 1 5 + ∈ = 5
Table 2
2
2
1
5
1
2 1
5
∈
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(Augmented solution)
Notice that even after the inclusion of cell (2, 3) in the basis the basic cells do not form a loop. i.e.
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 nonbasic 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 ε θ = .
uI
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 nonpositive 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.
Ui
(6)
7
(3)
3 4 –1
–θ 2 1
θ (2)
3
0
–θ
7
(3)
3
θ
(4)
4
5
+θ
2 1
θ
3 0
(2)
3
(0)
4 6 8
θ
7
(3)
3
+θ
4 0
2 1
(– 4)
3 ε
θ
(6)
3
(4)
4
– θ
6 2
2
2 1 ε
2
1
5
2
6
5
ui
Vi 2 1 3 Vj 7 6 4 Starting table First iterated table
2 1
2
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+θ
3
(–2)
4
–θ
6 1
vj 2 1 5
(Second Iterated Table)
ui (– 6)
7
(–1)
3 4 4
(–2)
2 1 3 3
3
(0)
4 6 6
Vj – 3 – 2 0
(Optimum table)
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
6 1 3 4 3 2 1 1 4 2 z × + × + × + × + × =
= 33 Self Assessment Questions 4
Fill in the blanks
1. All the values of ∆Cij ui vj should be __________ or _________ for the solution to be
optimum.
2. In unbalanced T.P ∑ai is __________ __________ to ∑bj.
3. If the number of allocation is less than _________ then it is said to be a degenerate T.P
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 chapter, you learned 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 approximation method. Further we
1
1
2
2
4
3 2
Operations Research Unit 5
Sikkim Manipal University 96
studied the degeneracy in transportation problem with examples on obtaining an optimum basic
feasible solution.
Terminal Questions 1. Solve the following transportation problem
Destination
A B C D 21 16 25 13 11
17 18 14 23 13
32 27 18 41 19
6 10 12 15 43
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 are as follows.
Factories
Determine the optimal distribution for the company so as to minimize the total transportation
cost.
1 2 3 4
1 2 3 11 7
2 1 0 6 1
3 5 8 15 9
Availability
Source I
II
III
Requirement
Operations Research Unit 5
Sikkim Manipal University 97
3. Solve the following transportation problem
To
Answer To Self Assessment Questions Self Assessment Questions 1
1. L.P.P 2. equal 3. minimum cost
Self Assessment Questions 2 a) Yes b) Yes c) No
Self Assessment Questions 3
a) False b) True c) False d) True
Self Assessment Questions 4
1. ≥ zero
2. Not equal to 3. m + n – 1
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 → ε
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