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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
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