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## Transportation

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### The objective to satisfy the destination requirements within the plants capacity ... is zero there is another alterative solution for the same transportation cost. ... – PowerPoint PPT presentation

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Title: Transportation

1
Transportation
• Transportation models deals with the
transportation of a product manufactured at
different plants or factories supply origins) to
a number of different warehouses (demand
destinations). The objective to satisfy the
destination requirements within the plants
capacity constraints at the minimum
transportation cost.

2
• A typical transportation problem contains

Inputs Sources with availability Destinations
with requirements Unit cost of transportation
from various sources to destinations Objective To
determine schedule of transportation to minimize
total transportation cost
3
How to solve?
• Define the objective function to be minimized
with the constraints imposed on the problem.
• Set up a transportation table with m rows
representing the sources and n columns
representing the destination
• Develop an initial feasible solution to the
problem by any of these methods a) The North west
corner rule b) Lowest cost entry method c)Vogels
approximation method

4
• 4. Examine whether the initial solution is
feasible or not.( the solution is said to be
feasible if the solution has allocations in (
mn-1) cells with independent positions.
• 5. Test wither the solution obtained in the above
step is optimum or not using a) Stepping stone
method b) Modified distribution (MODI) method.
• 6.If the solution is not optimum ,modify the
shipping schedule. Repeat the above until an
optimum solution is obtained.

5
Applications
• To minimize shipping costs from factories to
warehouses or from warehouses to retails outlets.
• To determine lowest cost location of a new
factor, warehouse or sales office
• To determine minimum cost production schedule
that satisfies firms demand and production
limitations.

6
North West corner method
• Select the northwest corner cell of the
transportation table and allocate as many units
as possible equal to the minimum between
availability supply and demand requirements
i.e.(min (s1,d1)
• Adjust the supply and demand numbers in the
respective rows and columns allocation
• A. If the supply for the first row is exhausted
,then move down to the first cell in the second
row and first column and go to step 2.
• If the demand for the first column is satisfied,
then move horizontally to the next cell in the
second column and first row and go to step 2

7
• 5. If for any cell, supply equals demand, then
the next allocation can be made in cell either in
the next row or column.
• 6. Continue the procedure until the total
available quantity is fully allocated to the
cells as required.
• Advantages it is simple and reliable. Easy to
compute ,understand and interpret.
• Disadvantages This method does not take into
considerations the shipping cost, consequently
the initial solution obtained b this method
require improvement.

8
• Problem 1 Obtain initial solution in the
following transportation problem by using
Northwest corner rule method

9
• Least cost method
• Select the cell with the lowest transportation
cost among all the rows or column of the
transportation table
• If the minimum cost is not unique, then select
arbitrarily any cell with this minimum cost.
(preferably where maximum allocation is possible)
• Repeat steps 1 and 2 for the reduced table until
the entire supply at different factories is
exhausted to satisfy the demand at different
warehouses.

10
• Problem 1 Obtain initial solution in the
following transportation problem by using Least
cost method

11
• Vegels Approximation Method (VAM)
• Compute a penalty for each row and column in the
transportation table. The penalty for a given
row and column is merely the difference between
the smallest cost and next smallest cost in that
particular row or column.
• Identify the row or column with the largest
penalty. In this identified row or column, choose
the cell which has the smallest cost and allocate
the maximum possible quantity to the lowest cost
cell in that row or column so as to exhaust
either the supply at a particular source or
satisfy demand at warehouse.( If a tie occurs in
the penalties, select that row/column which has
minimum cost. If there is a tie in the minimum
cost also, select the row/column which will have
maximum possible assignments)

12
• 3.Reduce the row supply or the column demanded
by the assigned to the cell
• 4.If the row supply is now zero, eliminate the
row, if the column demand is now zero, eliminate
the column, if both the row, supply and the
column demand are zero, eliminate both the row
and column.
• 5. Recompute the row and column difference for
the reduced transportation table, omitting rows
or columns crossed out in the preceding step.
• 6. Repeat the above procedure until the entire
supply at factories are exhausted to satisfy
demand at different warehouses.

13
• Problem 1 Obtain initial solution in the
following transportation problem by using VAM

14
• The Modified Distribution Method
• Determine an initial basic feasible solution
consisting of m n -1 allocations in independent
positions using any of the three methods
• Determine a set of number for each row and each
column. Calculate Ui ( i 1,2,..m)and Vj (j
1,2..n)for each column, and Cij (Ui Vj ) for
occupied cells.
• Compute the opportunity cost
• ?ij Cij - (Ui Vj ) for each
unoccupied cells.

15
• 4. Check the sign of each opportunity cost if
all the
• ?ij are positive or zero, the given solution is
optimum. If one of the values is zero there is
another alterative solution for the same
transportation cost. If any value is negative the
given solution is not optimum. Further
improvement is possible.
• 5. Select the unoccupied cell with the largest
negative opportunity cost as the cell to be
included in the next solution.
• 6. Draw a closed path or loop for the unoccupied
cell selected in step 5.It may be noted that
right angle turns in this path are permitted only
a occupied cells and at the original unoccupied
cell

16
• 7. Assign alternative plus and minus signs at the
unoccupied cells on the corner points of the
closed path with a plus sign at the cell being
evaluated.
• 8.Determine the maximum number of units that
should be shipped to this unoccupied cell. The
smallest one with a negative position on the
closed path indicates the number of units that
can be shipped to the entering cell. This
quantity is added to all the cells on the path
marked with plus sign and subtract from those
cells mark with minus sign. In this way the
unoccupied cell under consideration becomes an
occupied cell making one of the occupied cells as
unoccupied cell.
• 9.Repeat the whole procedure until an optimum
solution is attained i.e. ?ij is positive or
zero. Finally calculate new transportation cost.

17
• Problem 4 A distribution system has the
following constraints.
• Factory capacity (in
units)
• A 45
• B 15
• C 40
• Warehouse Demand (in units)
• I 25
• II 55
• III 20
• The transportation costs per unit( in rupees)
allocated with each route are as follows.

18
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19
• Find the optimum transportation schedule and the
minimum total cost of transportation.
• Problem 3
• A company is spending Rs.1,000 on transportation
of its units from these plants to four
distribution centres. The supply and demand of
units, with unity cost of transporataion are
given in the table.
• What can be the maximum saving by optimum
scheduling.

20
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21
• Special cases in Transportation
• Unbalanced transportation
• Maximisation
• Restricted routes

22
• A product is produced by 4 factories F1, F2,F3
and F4. Their unit production cost are
Rs.2,3,1,and 5 only. Production capacity of the
factories are 50,70,40 and 50 units respectively.
The product is supplied to 4 stores S1,S2,S3 and
S4., the requirements of which are 25,35,105 and
20 respectively. Unit cost of transportation are
given below

23
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24
• Find the optimal transportation plan such that
total production and transportation cost is
minimum.
• PROBLEM
• A particular product is manufactured in
factories A,B ,C and D it is sold at centres
1,2,and 3. the cost in rupees of product per unit
and capacity of each plant is given below

25
• The sales prices is Rs per unit and the demand
are as follows.
• Find the optimal solution