2. TRANSPORTATION PROBLEM
This is a type of linear programming problem that may be solved using a
simplified version of the simplex technique called transportation method.
Because of its major application in solving problems involving several product
sources and several destinations of products, this type of problem is
frequently called the transportation problem.
3. MODELLINGTHETRANSPORTATION PROBLEM
The transportation problem is concerned with finding the minimum cost of
transporting a single commodity from a given number of sources (e.g.
factories) to a given number of destinations (e.g. warehouses) here we use a
specific transportation algorithm.
The data of the model include
1.The level of supply at each source and the amount of demand at each
destination.
2.The unit transportation cost of the commodity from each source to each
destination.
4. Since there is only one commodity, a destination can receive its demand from more
than one source.The objective is to determine how much should be shipped from
each source to each destination so as to minimize the total transportation cost.
5. BALANCEDV/S UNBALANCED
When the total supply is equal to the total demand
m ∑i=1 ai = n ∑j=1 bj
then the transportation model is said to be balanced.
A transportation model in which the total supply and total demand are unequal
m ∑i=1 ai = n ∑j=1 bj
then the transportation model is called unbalanced.
It is always possible to balance an unbalanced transportation problem.
If the total supply exceeds the total demand, then an additional dummy column is
added and if total demand exceeds the total supply, then an additional dummy row
is added.
7. TWO METHODS OF INITIAL FEASIBLE SOLUTION
NORTHWEST CORNER
METHOD
VOGEL’S
APPROXIMATION
METHOD
FOR OPTIMAL SOLUTION
MODIFIED
DISTRIBUTION (MODI)
METHOD
8. STEPS IN MODIFIED DISTRIBUTION
(MODI) 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 this value to find Pij = ui+vj-cij of all pij , if all are negative then it is the
optimal solution.
9. EXAMPLE
COMPANIES DESTINATIONS
JPTrading Company(Ahmedabad, Gujrat) Tender Depot(Gurgram)
Rathore & Co.(Delhi) Mulund Depot(Mumbai)
Shrivinayak industries suppliers Palampur Depot(Indore)
and manufacturing Ltd.(Udaipur) Malanpur Depot(MP)
The stock held at each supplier and the demand from each depot is known.The
cost of transportation of one lorry load of grit from each supplier to each depot is
also known.The information is given in table.
16. CALCULATION
Cij = ui+vj
Pij = ui+vj-cij
C11 = 0+30-180 = -50
C13 = 0+30-130 = -100
C14 = 0-40-180 = -330
C22 = 120+110-250 = -20
C24 = 120-40-280 = -200
C31= 160+30-240 = -50
All the pij (which is non basic cell values) are negative.Therefore it is an optimal solution.
So, Minimum transportation cost is same as that ofVogel’s approximation method i.e. Rs.
7120