Operation Research - Transportation and Assignment Problem by gsonkar

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									OPERATION RESEARCH   UNITED COLLEGE OF ENGG & RESEARCH




FOR MBA IST YEAR     1             FROM GAURAV SONKAR
OPERATION RESEARCH                                               UNITED COLLEGE OF ENGG & RESEARCH

    (Q1) Write short notes on “Transportation Problem”. Describe methods to obtain an initial
         feasible solution for a transportation problem.                       (03-04; 05-06)

    (A1) The transportation problem is a particular class of linear programming in which the objective is to
         transport a commodity from several supply origin to different demand destinations at a minimum
         total cost.
        A general transportation problem looks like:

              Destination               D1                D2                           Dn            Availability
                                                                      ……..
                  Origin
                                  C11               C12                          C1n
                      O1                x11               x12         ……..            x1n                  a1

                                  C21               C22                          C2n
                      O2              x21                 x22         ……..             x2n                 a2

                        .         .                 .                            .                          .
                        .         .                 .                 ……..       .                          .
                        .         .                 .                            .                          .
                                  Cm1               Cm2                          Cmn
                     Om                 xm1               xm2         ……..             xmn                 am


                Requirement              b1                b2         ……..              bn            Σai=Σbj

                               General Transportation Table
            Where    ai               amount of commodity available at origin to Oi

                     bj               amount of commodity required at destination Dj

                     Cij              cost of transporting one unit of commodity from origin Oi to destination Dj

                     xij              amount of commodity to be transported from origin     Oi to destination Dj
            Mathematical formulation of Transportation Problem:
                            Min z = ΣΣ cij xij
    subject to constraints
                   Σxij = ai    (i = 1, 2,……., m)                                2

                   Σcij = bj (j = 1, 2,……., n)                                   3
                   And xij ≥ 0

FOR MBA IST YEAR                                                  2                    FROM GAURAV SONKAR
OPERATION RESEARCH                                          UNITED COLLEGE OF ENGG & RESEARCH

         Constraints 2 and 3 are called availability constraints and requirement constraints
         respectively.

          Methods to obtain an Initial Feasible Solution




    NORTH WEST                           LEAST COST                                   VOGEL’S
      CORNER                                                                    APPROXIMATION
                                           METHOD
      METHOD                                                                          METHOD

    (I)NORTH WEST CORNER METHOD:
    This method does not take into account the cost of transportation on any route of transportation. This
    method consists of following steps:

     Step I:
     Start with north-west corner (uppermost left hand corner) of the transportation table allocate maximum
     possible amount to it, so that either the capacity of row is exhausted or the destination requirement of
     the first column is satisfied.
     Step II:
     (a) If the supply for the first row is exhausted. Then, move down vertically to the first cell in the second
         row.
     (b) If the demand for the first column is satisfied. Then, move right horizontally to the second cell in
         the first row.
     (c) If both supply in the row and demand in the column is satisfied. Then, move diagonally.
    Step III:
    Continuing in this way till an allocation is made in the south-east corner cell of the transportation table.
    (II) LEAST COST METHOD:
    This method takes into account the minimum unit cost of transportation for obtaining initial basic
    feasible solution. This method consists of following steps:
    Step I:
    Select the cell with lowest cost and allocate maximum possible amount to it; if there is a tie make an
    arbitrary selection.
    Step II:
    Delete the row or column or both which are satisfied by the allocation.
    Step III:
    Again search for the next lowest cost empty cell for which demand and supply are not exhausted and
    make the allocations. Repeat step I and II until all supply and demand conditions are satisfied.
FOR MBA IST YEAR                                             3                     FROM GAURAV SONKAR
OPERATION RESEARCH                                          UNITED COLLEGE OF ENGG & RESEARCH

    (III) VOGEL’S APPROXIMATION METHOD:
    This is the most preferred method as initial feasible solution obtained by this method is either optimal or
    very close to the optimal solution. Basic steps in VAM are as follows:
    Step I:
    Calculate the penalty cost i.e. the difference between the lowest and the second lowest cost values for each
    row and each column.
    Step II:
    Enter the penalty cost of each row to the right of corresponding row and the penalty cost of each column
    under the corresponding column.
    Step III:
    Select the row or column with the largest penalty and allocate maximum possible amount to the cell with
    the minimum cost in the selected row or column.
    If there are more than one largest penalty rows or columns we select any of them arbitrarily.
    Step IV:
    Cross out the particular row or column on which the requirement is satisfied.
    Step V:
    Recompute the penalty cost for each row and column for the reduced transportation table and go to the
    previous steps.
    Step VI:
    Repeat steps I to V until all the rim requirements are satisfied.
    (Q2) Explain the term ‘Degeneracy’ in the context of transportation problem.                     (03-04)

    (A2) Degeneracy:
          Degeneracy in an m x n transportation problem occurs whenever the number of positive
          allocations in a BFS is less than m+n-1.
          The degeneracy in the transportation problems may occur at the two stages:
          (a) At initial stages when the number of positive allocations in the initial BFS is less than m + n -1.
          (b) At any stage while moving towards optimal solution. This happens when two or more occupied
              cells with the same minimum allocation become unoccupied simultaneously.
    Necessity to remove degeneracy:
          (a) In such cases, the current solution can not be improved because it is not possible to draw a
              closed path (loop) for every occupied cell.
          (b) Also the values of dual variables ui and vj which are used to test the optimality can not be
              computed.
    Resolution of Degeneracy:
    To resolve degeneracy, we select as many unoccupied cells as is necessary to have exactly m + n – 1
    allocation.
    Now we allocate an extremely small positive amount to these unoccupied cells, denoted by ‘ε’ and is
    assumed to satisfy following conditions:
    (a)     0+ε=ε
    (b)     Xij ± ε = Xij
    The cells containing ε’s are kept in the transportation table until either degeneracy is removed or an
    optimal solution is attained, whichever occurs first. Then we set ε equals to zero.



FOR MBA IST YEAR                                             4                     FROM GAURAV SONKAR
OPERATION RESEARCH                                         UNITED COLLEGE OF ENGG & RESEARCH

    (Q3)What is meant by unbalanced transportation problem? Explain the method for solving such a
        problem with an example                                                      (03-04)


    (A3)
        A transportation problem is said to be unbalanced if total supply is not equal to total demand i.e.
               m      n
               ∑ ai ≠ ∑ bj
              i=1    j=1

    In this situation there are two cases arises.
    Case I: If the total supply is more than total demand i.e.
                                    ∑ ai > ∑ bj
    In this case, we introduce a dummy destination in the transportation table. The requirement at this
    destination is assumed to be equal to ∑ ai - ∑ bj.
    The unit transportation cost is taken to be zero.
    Case II: If the total supply is less than total demand i.e.
                 ∑ ai < ∑ bj
    In this case, we introduce a dummy origin in the transportation table. The availability at this origin is
    assumed to be equal to ∑bj - ∑ai.
    The unit transportation cost is taken to be zero.
     Thus an unbalanced transportation problem is converted into a balanced transportation problem which
    can be solved by either of North-West Corner Method, Least Cost Method or Vogel’s Approximation
    Method.
    EXAMPLE: Consider the following transportation problem:

            Plant                                 Market                                 Supply
                               A              B              C              D
              E                11             20             7              8               50
              F                21             16             10             12              40
              G                8              12             18             19              70
           Demand              30             25             35             40
    Solution: Here
              Total Supply (∑ai) = 160
       and Total Demand (∑bj) = 130
    Since, ∑ ai > ∑ bj , the given problem is unbalanced.
    To convert it into balanced transportation problem we introduce a dummy market E, with demand 160-
    130 = 30, such that the unit transportation cost from each plant to market is equal to zero.
    Thus, we obtain the following transportation problem.

            Plant                                    Market                                    Supply
                             A            B           C              D             E
            X                11           20           7              8             0            50
            Y                21           16          10             12             0            40
            Z                8            12          18             19             0            70
          Demand             30           25          35             40            30           160


FOR MBA IST YEAR                                              5                   FROM GAURAV SONKAR
OPERATION RESEARCH                                                     UNITED COLLEGE OF ENGG & RESEARCH

    (Q4) Define the assignment problem.                 (02-03; 03-04; 05-06)
    (A4) The assignment problem is a particular case of transportation problem in which the objective is to
    assign a number of resources to an equal number of activities so as to minimize total cost or maximize
    total profit.
    Typical examples of decision making situations where assignment model can be used are:
                    workers to machines
                    sales personnel to different sales areas
                    products to factories
                    jobs to machines
                    classes to rooms
    A general assignment problem looks like:
                                                                Jobs                                  Supply
        Person                  1                           2          ………..                n
                 1                                    C12           …………..                 C1n        1
                               C11
                 2                                    C22          ……………                   C2n        1
                               C21
                 :                                                                                    :
                                .                       .                   .               .
                 :              .                       .                   .               .
                                                      Cn2          …………….                  Cnn        1
                 n             Cn1

     Demand                1               1
    This matrix is called cost matrix where Cij is the cost ………                  1
                                                            of assigning ith person to jth job
    The cost matrix is same as that of a transportation problem except that availability of the resource and
    the demand at each of the destination is unity.



    Let xij denotes the assignment of resource i to the activity j, such that

                                              1, if resource i is assigned to activity j
                              Xij =
                                              0, otherwise

    Then the mathematical model of assignment problem can be represented as:
                                n   n
               Minimize Z = ∑             ∑     Cij . xij
                                i=1 j=1

           Subject to constraints
                        n
                     ∑ xij = 1, for all i         (resource availability)
                  j=1



                     n
      and            ∑ xij = 1, for all j         (activity requirement)
                  i=1

     and                    xij = 0 or 1, for all i and j
FOR MBA IST YEAR                                                        6                        FROM GAURAV SONKAR
OPERATION RESEARCH                                          UNITED COLLEGE OF ENGG & RESEARCH

      Thus, it is clear that the assignment problem is a special case of transportation problem with the
      characteristics:
    (a) The cost matrix is a square matrix, and
    (b) The optimal solution for the problem would always be such that there would be only one assignment
        in a given row or column of the cost matrix.

    (Q5)Show       assignment      problem     is   a    special    case    of    transportation      problem.
                                                                                                   (07-08)
    (A5)            Same as above.

    (Q6) Explain Hungarian Method for solving an assignment problem.                (01-02)

    (A6)


           Step I:
               Row Operation:
                   Subtracting smallest element in each row from all other elements of that row.

                 Column Operation:
                  Subtracting smallest element in each column from all other elements of that column.

           Step II: MAKING ASSIGNMENTS:
           Search for an optimal assignment in the finally modified cost matrix as follows:
               Examine the first row. If there is only one zero in it, then enclose this zero in a box (     )
                  cross (X) all the zero of that column passing through the enclosed zero.
                  Similarly done with other rows.
               If any row has more than one zero, then do not touch that row & passes on the next row.
               Repeat the same procedure with columns.

           Step III:
           If each row and each column of the reduced matrix has exactly one enclosed zero, then the
           enclosed zero yield an optimal assignment.
           If not, then go to next step.

            Step IV:
           Draw minimum number of horizontal and/or vertical lines to cover all the zeros of the reduced
           matrix.
           Since the assignment is not optimal, the number of lines will be less than n (i.e. order of matrix).
           In order to move towards optimality, generates more zeros as follows:
                 i.    Find the smallest of the elements of the reduced matrix not covered by any of the lines.
                       Let this element be α.
                ii.    Subtract α from each of the element not covered by the lines and add α to the element at
                       the intersection of these lines.
                       Do not change the remaining elements.
           Step V:
           Go to step II and repeat the procedure till an optimal assignment is achieved.

FOR MBA IST YEAR                                             7                    FROM GAURAV SONKAR

								
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