transportation in production and operation management

1. Transportation
Problem

2. Introduction
 The transportation algorithm is applied to
minimize the total cost of transporting a
homogeneous commodity (product) from
supply centers to Demand centers.
 It can also be applied to the maximization of
some total value or utility. For e.g., Financial
resources are distributed in such a way that
the profitable return is maximized.

3.  The Objective is to determine the number of
units of an item which is to be supplied from
the origin to a destination in order to satisfy
the required quantity of goods or services at
each destination centre, within the limited
quantity of goods or services available at
each supply centre, at a minimum
transportation cost &/or time.

4. Mathematical Model
Formulation
 Sources of supply are production facilities,
warehouses or supply points, characterized
by available capacities.
 The Destination are consumption facilities,
warehouse or demand points, characterized
by required level of demands.

5. The Transportation algorithm
STEP 1
Formulate the problem & arrange the data in matrix
form.
STEP 2
Obtain an initial basic feasible solution.
Different methods used :
a. North West Corner Method
b. Least Cost Method
c. Vogel’s Approximation (or penalty) Method.

6. Initial solution obtained must satisfy the following
conditions:
A. Rim Condition - SUPPLY= DEMAND
B. Allocations must be equal to m+n-1
If solution satisfies above condition is called Non-
degenerate basic feasible solution otherwise
degenerate solution.

7. STEP 3
Test the initial solution for optimality.
To test optimality of the solution obtained in step 2
Modified Distribution (MODI) method is used.
If the current solution is optimal, then stop, otherwise,
determine new improved solution.
STEP 4
Updating the solution
Repeat step 3 until an optimal solution is obtained.

8. Vogel’s approximation method
(VAM)
VA (penalty or regret) method is a heuristic method &
is preferred to the other two methods.
In this each allocation is made on the basis of the
opportunity (penalty or extra) cost that would have
been incurred if allocations in certain cells with
minimum unit transportation cost were missed.
In this method allocations are made so that the penalty
cost is minimized.
The advantage of this method is that it gives an initial
sol which is nearer to an optimal sol or is the optimal
sol itself.

9. Example
 A company has 3 production facilites S1,S2,&
S3 with production capacity of 7, 9, & 18 units
(in 100s) per week of the product,
respectively. These units are to be shipped to
4 warehouses D1,D2,D3 & D4 with
requirements 5,8,7,14units in (100s) per
week respectively.
The transportation costs (in rupees) per unit
between factories & warehouses are given in
table:

10. D1 D2 D3 D4 Capacity
S1 19 30 50 10 7
S2 70 30 40 60 9
S3 40 8 70 20 18
Demand 5 8 7 14 34

11. MODI Method
1. Check initial basic feasible solution using 1st
three methods & put allocations.
2. Use economic interpretation Ui & Vj
3. To start with any one of u’s or v’s is
assigned the value zero. Better to assign 0
where there are maximum no of allocations
in a row or column
4. For occupied cell : Cij = Ui + Vj
5. For unoccupied cell : Dij = Cij – (Ui + Vj)

12. 6. Examin sign of each Dij:
Dij >0 then current basic feasible sol is optimal
Dij = 0 then current basic feasible sol will reamin
unaffected but alternative sol exits.
If one or more Dij < 0 then an improved sol can be
obtained by entreing unocupied cell (I, j) in the
basis.
An unoccupied cell having the largest negative value
of Dij is chosen for entering into the solution mix.

13. 7. Construct closed path (or loop) for the unoccupied
largest negative oppurtunity cost
8. Assign + & - in loop.
9. Select smallest quality among loop (x= smallest
value)
10. Optain new improved solution by allocating units to
the unoccupied cell according to step 9 & calculate
new transportation cost.
11. Test the revised sol further for optimality. The
procedure terminates when all Dij > or = 0 for
unoccupied cells.

14. D1 D2 D3 D4 Capacity
S1 21 16 15 3 11
S2 17 18 14 23 13
S3 32 27 18 41 19
Demand 6 10 12 15

15. D1 D2 D3 D4 Capacity
S1 1 2 1 4 30
S2 3 3 2 1 50
S3 4 2 5 9 20
Demand 20 40 30 10