The document discusses transportation problems and assignment problems in operations research. It provides: 1) An overview of transportation problems, including the mathematical formulation to minimize transportation costs while meeting supply and demand constraints. 2) Methods for obtaining initial basic feasible solutions to transportation problems, such as the North-West Corner Rule and Vogel's Approximation Method. 3) Techniques for moving towards an optimal solution, including determining net evaluations and selecting entering variables. 4) The formulation and algorithm for solving assignment problems to minimize assignment costs while ensuring each job is assigned to exactly one machine.

1. Chapter 6.
1.0 TRANSPORTATION PROBLEM
The transportation problem is a special class of the linear programming problem. It
deals with the situation in which a commodity is transported from Sources to
Destinations. The objective is to determine the amount of commodity to be transported
from each source to each destination so that the total transportation cost is minimum.
EXAMPLE 1.1
A soft drink manufacturing firm has m plants located in m different cities. The total
production is absorbed by n retail shops in n different cities. We want to determine the
transportation schedule that minimizes the total cost of transporting soft drinks from
various plants to various retail shops. First we will formulate this as a linear
programming problem.
MATHEMATICAL FORMULATION
Let us consider the m-plant locations (origins) as O1 , O2 , …., Om and the n-retail shops
(destination) as D1 , D2 , ….., Dn respectively. Let ai ≥ 0, i= 1,2, ….m , be the amount
available at the ith plant Oi . Let the amount required at the jth shop Dj be bj ≥ 0, j=
1,2,….n.
Let the cost of transporting one unit of soft drink form ith origin to jth destination be Cij ,
i= 1,2, ….m, j=1,2,….n. If xij ≥ 0 be the amount of soft drink to be transported
from ith origin to jth destination , then the problem is to determine xij so as to
Minimize
m n
z = ∑∑ xij cij
i =1 j =1
Subject to the constraint
and xij ≥ 0 , for all i and j.
n
∑x
j =1
ij = ai , i = 1,2,...m
m
∑ xij = b j , j =1,2,...n.
i =1
This lPP is called a Transportation Problem.
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2. THEOREM 1.1
A necessary and sufficient condition for the existence of a feasible solution to the
transportation problem is that
m n
∑a = ∑b
i =1
i
j =1
j
Remark. The set of constraints
m n
∑ x = b and ∑ x = a
ij j ij i
i =1 j =1
Represents m+n equations in mn non-negative variables. Each variable xij appears in
exactly two constraints, one is associated with the origin and the other is associated with
the destination.
Note. If we are putting in the matrix from, the elements of A are either 0 or 1.
THE TRANSPORTATYION TABLE:
D1 D2 …… Dn supply
O1 c11 c12 ….. c1n a1
O2 c21 c22 ….. …. c2n a2
… …… ….. ….. ….. ….. :
Om cm1 cm2 …. … cmn am
Requirement b1 b2 … …. bn
Definition. (Loop). In a transportation table, an ordered set of four or more cells is said
to form a loop if :
(I) Any two adjacent cells in the ordered set lie in the same row or in the same
column.
(II) Any three or more adjacent cells in the ordered set do not lie in the same row or in
the same column.
RESULT:
A feasible solution to a transportation problem is basic if and only if the corresponding
cells in the transportation table do not contain a loop. To find an initial basic feasible
solution we apply:
(1) The North-West corner rule
(2) Vogel`s Approximation method.
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3. 1.1 THE NORTH-WEST CORNER RULE
Step (1). 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,
i.e; x11 = min( a1, b1 ) .
Step (2). If b1 > a1, the capacity of origin O 1 is exhausted but the requirement at D 1
is not satisfied. So move downs to the second row, and make the second allocation:
x21 = min ( a2 , b1 – x11 ) in the cell ( 2,1 ).
If a1 > b1 , allocate x12 = min ( a1 - x11 , b2 ) in the cell ( 1,2) .
Continue this until all the requirements and supplies are satisfied.
EXAMPL 1.1.1
Determine an initial basic feasible solution to the following transportation problem using
the North-West corner rule:
D1 D2 D3 D4 Availability
O1 6 4 1 5 14
O2 8 9 2 7 16
O3 4 3 6 2 5
Requirement 6 10 15 4
Solution to the above problem is: 6 8
6 4 1 5
2 14
8 9 2 7
1 4
4 3 6 2
Now all requirements have been satisfied and hence an initial basic feasible solution to
the transportation problem has been obtained. Since the allocated cells do not form a
loop, the feasible solution is non-degenerate. Total transportation cost with this allocation
is:
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4. Z = 6*6 + 4*8 + 2*9 +14*2 + 1*6 +4* = 128.
VOGEL’S APPROXIMATION METHOD (VAM ).
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 ith row and the minimum cost be Cij . Allocate a maximum feasible amount
xij = min ( ai , bj ) in the ( i, j )th cell, and cross off the ith row or jth column.
Step 3. Re compute 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.
Remark. VAM determines an initial basic feasible solution, which is very close to the
optimum solution.
PROBLEM 1.1.2
Obtain an initial basic feasible solution to the following transportation problem using
Vogels approximation method.
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
21 25 17 17
1.2 MOVING TOWARDS OPTIMALITY
(1) DETERMINE THE NET-EVALUATIONS ( U-V METHOD)
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5. Since the net evaluation is zero for all basic cells, it follows that
zij - cij = ui +vj - cij , for all basic cells (i, j). So we can make use of this relation to find
the values of ui and vj . Using the relation ui +vj = cij , for all i and j which (i,j) is a
basic cell, we can obtain the values of ui `s and vj `s. After getting the values of ui `s and
vj `s, we can compute the net-evaluation for each non-basic cell and display them in
parenthesis in the respective cells.
(2) SELECTION OF THE ELECTING VARIABLES
Choose the variable xrs to enter the basis for which the net evaluation
zrs -crs = max { zij - cij > 0} .
After identifying the entering variable xrs , form a loop which starts at the non-basic cell
(r,s) connecting only basic cells . Such a closed path exists and is unique for any non-
degenerate solution. Allocate a quantity θ alternately to the cells of the loop starting +θ
to the entering cell. The value of θ is the minimum value of allocations in the cells
having -θ.
Now compute the net-evaluation for new transportation table and continue the above
process till all the net-evaluations are positive for non-basic cells.
1.3 DEGENCY IN TRANSPORTATION PROBLEM
Transportation with m-origins and n-destinations can have m+n-1 positive basic
variables, otherwise the basic solution degenerates. So whenever the number of basic
cells is less than m + n-1, the transportation problem is degenerate.
To resolve the degeneracy, the positive variables are augmented by as many zero-valued
variables as is necessary to complete m +n –1 basic variables.
UNBALANCED TRANSPORTATION PROBLEM
If
m n
,
∑ a ≠ ∑b
i =1
i
j =1
j
The transportation problem is known as an unbalanced transportation problem. There
are two cases
Case(1).
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6. m n
∑ a > ∑b
i =1
i
j =1
j
Introduce a dummy destination in the transportation table. The cost of transporting to
this destination is all set equal to zero. The requirement at this destination is assumed to
be equal to
m n
∑ a − ∑b
i =1
i
j =1
j.
Case (2) .
m n
∑ a < ∑b
i =1
i
j =1
j
Introduce a dummy origin in the transportation table, the costs associated with are set
equal to zero. The availability is
n m
∑b − ∑a
j =1
j
i =1
i
1.4 THE ASSIGNMENT PROBLEM
Suppose there are n-jobs for a factory and has n-machines to process the jobs. A job i
(i=1,2,…n ) when processed by machine j ( j=1,2,…n) is assumed to incur a cost c ij .The
assignment is to be made in such a way that each job can associate with one and only one
machine. Determine an assignment of jobs to machines so as to minimize the overall cost.
1.4.1 MATHEMATCAL FORMULATION
We can define xij = 0, if the ith job not assigned to jth machine.
= 1 , if the ith job is assigned to jth machine.
We can assign one job to each machine,
n n
∑x
i =1
ij = 1, and ∑ xij = 1
j =1
The total assignment cost is given by
n n
z = ∑∑ cij xij
j =1 i =1
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7. 1.4.2 THE ASSIGNMENT ALGORITHM
Step (1). Determine the effectiveness matrix. Subtract the minimum element of each row
of the given cost matrix from all of the elements of the row. Examine if there is at least
one zero in each row and in each column. If it is so, stop here, otherwise subtract the
minimum element of each column from all the elements of the column. The resulting
matrix is the starting effectiveness matrix.
Step (2). Assign the zeroes:
(a). Examine the rows of the current effective matrix successively until a row with
exactly one unmarked zero is found. Mark this zero, indicating that an assignment will be
made there. Mark all other zeroes lying in the column of above encircled zero. The cells
marked will not be considered for any future assignment. Continue in this manner until
all the rows have taken care of.
(b). Similarly for columns.
Step (3). Check for Optimality. Repeat step 2 successively till one of the following
occurs.
(a). There is no row and no column without assignment. In such a case, the current
assignment is optimal.
(b). There may be some row or column without an assignment. In this case the current
solution is not optimal. Proceed to next step.
Step (4). Draw minimum number of lines crossing all zeroes as follows. If the number of
lines is equal to the order of the matrix, then the current solution is optimal, otherwise it
is not optimal. Go to the next step>
Step (5). Examine the elements that do not have a line through them. Select the smallest
of these elements and subtract the same from all the elements that do not have a line
through them, and add this element to every element that lies in the intersection of the
two lines.
Step (6). Repeat this until an optimal assignment is reached.
PROBLEM 1.4.1
Consider the problem of assigning five jobs to five persons. The assignment costs are
given as follows:
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8. Jobs
Person 1 2 3 4 5
A 8 4 2 6 1
B 0 9 5 5 4
C 3 8 9 2 6
D 4 3 1 0 3
E 9 5 8 9 5
1.4.4.UNBALANCEED ASSIGNMENT PROBLEM
When the cost matrix of an assignment problem is not a square matrix, i.e; number
of sources is not equal to the number of destinations, the assignment problem is called an
unbalanced assignment problem. In such problems, dummy rows or columns are added
in the matrix so as to complete it to form a square matrix.
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