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INITIAL BASIC
FEASIBLE SOLUTION
TRANSPORTATION PROBLEM


         Transport various
        quantities of a single
    homogeneous commodity
    to different destinations in
        such a way that total
       transportation cost is
             minimum.
TERMINOLOGY USED IN
             TRANSPORTATIONAL MODEL
Feasible solution: Non negative values of xij where i=1, 2……….m and j=1, 2,…n which
satisfy the constraints of supply and demand is called feasible solution.

Basic feasible solution: If the no of positive allocations are (m+n-1).

Optimal solution: A feasible solution is said to be optimal solution if it minimizes the
total transportation cost.

Balanced transportation problem: A transportation problem in which the total supply
from all sources is equal to the total demand in all the destinations.

Unbalanced transportation problem: Problems which are not balanced are called
unbalanced.

Matrix terminology: In the matrix, the squares are called cells and form columns
vertically and rows horizontally.

Degenerate basic feasible solution: If the no. of allocation in basic feasible solutions is
less than (m+n-1).
OPTIMAL SOLUTION OF TRANSPORTATION
              PROBLEM


                                     Obtain an
                                 optimal solution
                                    by making
        Initial Basic               successive
STEP1     Feasible      STEP2   improvements in
          Solution                 IBFS until no
                                further decrease
                                in transportation
                                  cost is possible
INITIAL BASIC FEASIBLE SOLUTION

     Methods Available

 NORTH WEST     LOWEST COST      VOGEL’S
   CORNER          ENTRY      APPROXIMATION
METHOD(NWCM)   METHOD(LCEM)    METHOD(VAM)
UNBALANCED TRANSPORTATION
         PROBLEM

             Demand>Supply

Add dummy column in matrix with zero cost


             Supply>Demand

 Add dummy row in matrix with zero cost
DEMAND>SUPPLY
PRODUCT   P    Q    SUPPLY
OFFICE
A         10   15   20

B         20   30   50

DEMAND    30   60             70
                         90

PRODUCT   P    Q    SUPPLY
OFFICE
A         10   15   20

B         20   30   50

DUMMY     0    0    20

DEMAND    30   60             90
                         90
SUPPLY>DEMAND
    PRODUCT        P         Q            SUPPLY
    OFFICE
    A              10        15           30

    B              20        30           60

    DEMAND         20        50                          90
                                               70


PRODUCT       P         Q         DUMMY        SUPPLY
OFFICE
A             10        15        0            20


B             20        30        0            50

DEMAND        20        50        20                          90
                                                    90
METHOD FOR OPTIMAL SOLUTION


             Stepping stone
                method


               Modified
              Distribution
             Method(MODI)
NORTH WEST CORNER METHOD




Most systematic and easiest
 method for obtaining initial
  feasible basic solution
STEPS TO SOLVE THE PROBLEM
Step1: construct an empty m*n matrix, completed with rows & columns.



Step2: indicate the rows and column totals at the end.


Step3: starting with (1,1)cell at the north west corner of the matrix, allocate maximum possible quantity
keeping in view that allocation can neither be more than the quantity required by the respective warehouses
nor more than quantity available at the each supply centre.


Step 4: adjust the supply and demand nos. in the rows and columns allocations.


Step5: 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 the step 4.


Step 6: if the demand for the first column is satisfied, then move to the next cell in the second column and first
row and go to step 4.


Step 7: if for any cell, supply equals demand then the next allocation can be made in cell either in the next
row or column.


Step 8: continue the procedure until the total available quantity is fully allocated to the cells as required.
MINIMIZATION USING NWCM
     WAREHOUSE W1   W2   W3   SUPPLY

PLANT
P1            7     6    9    20



P2            5     7    3    28



P3            4     5    8    17



DEMAND        21    25   19
Step1: Check whether problem is
                balanced
 WAREHOUSE W1    W2     W3      SUPPLY

PLANT
P1        7      6      9       20


P2        5      7      3       28


P3        4      5      8       17


DEMAND    21     25     19               65

                                65
Allocate P1W1 max quantity such that it
should not exceed supply/requirement
      WAREHOUSE W1         W2    W3         SUPPLY

 PLANT
 P1            7           6     9          20/0
                      20

 P2            5           7     3          28


 P3            4           5     8          17


 DEMAND        21/1        25    19         65



P1 can supply 20 units at max so W1 can be allocated 20
                   units only by P1
Check whether the requirements of W1
               fulfilled
     WAREHOUSE W1         W2   W3         SUPPLY

PLANT
P1            7           6    9          20/0
                     20

P2            5           7    3          28


P3            4           5    8          17


DEMAND        21/1        25   19         65




Now allocate the remaining requirement to next north
           west corner element i.e. P2W1
Allocate 1 to P2W1
       WAREHOUSE W1           W2   W3    SUPPLY

  PLANT
  P1            7             6    9     20/0
                         20

  P2            5             7    3     28/27
                         1

  P3            4             5    8     17


  DEMAND        21/1/0        25   19    65




In this way repeat it for all columns until demand of
              all warehouses is fulfilled
No. of allocations=No. of rows + No. of columns-1
         (For generating optimal solution)
      WAREHOUSE W1           W2          W3        SUPPLY

 PLANT
 P1            7             6           9         20/0
                        20

 P2            5             7           3         28/27/0
                        1           25        2

 P3            4             5           8         17/0
                                              17

 DEMAND        21/1/0        25/0        19/17/0   65




 Here no. of rows=3, no. of columns=3, total allocations=5
                         5=3+3-1
TOTAL TRANSPORTATION COST
PLANT        WAREHOUSE   COST

P1           W1          7*20=140

P2           W2          5*1=5

P2           W2          7*25=175

P2           W2          3*2=6

P3           W3          8*17=136

TOTAL COST               462
LEAST COST ENTRY METHOD


     This method takes into
  consideration the lowest cost
 and therefore takes less time to
       solve the problem
STEPS TO SOLVE THE PROBLEM
• Step1: select the cell with the lowest transportation cost
  among all the rows and columns of the transportation
  table. If the minimum cost is not unique then select
  arbitrarily any cell with the lowest cost.
• Step2: allocate as many units as possible to the cell
  determined in step 1 and eliminate that row in which
  either capacity or requirement is exhausted.
• Step3:adjust the capacity and the requirement for the
  next allocations.
• Step4: repeat the steps1to3 for the reduced table until
  the entire capacities are exhausted to fill the
  requirements at the different destinations.
MINIMIZATION USING LCEM
     WAREHOUSE W1       W2        W3         SUPPLY

PLANT
P1            7         6         9          20



P2            5         7         3          28



P3            4         5         8          17



DEMAND        21        25        19         65




           Step1 :This problem is balanced problem
Find out the cell having least cost
     WAREHOUSE W1      W2          W3          SUPPLY

PLANT
P1            7        6           9           20


P2            5        7           3           28/9
                                          19

P3            4        5           8           17


DEMAND        21       25          19/0        65




P2W3 is having least cost i.e. 3 so it is allocated 19 units
            as per the requirement of W3
In this way allocate next minimum costs
     until all requirements are fulfilled
      WAREHOUSE W1          W2           W3          SUPPLY

 PLANT
 P1            7            6            9           20/0
                                 20

 P2            5            7            3           28/9/5/0
                        4            5          19

 P3            4            5            8           17/0
                     17
 DEMAND        21/4/0       25/5/0       19/0        65



   Now P2W1 and P3W2 are having same cost, we can
    allocate any of them but in this case P2W1 will be
allocated because in P3W2 P3 cant supply anymore units
Final matrix with all allocations
     WAREHOUSE W1          W2           W3          SUPPLY

PLANT

P1            7            6            9           20/0
                                20

P2            5            7            3           28/9/5/0
                       4            5          19

P3            4            5            8           17/0
                    17

DEMAND        21/4/0       25/5/0       19/0        65
No. of allocations=No. of rows + No. of columns-1
         (For generating optimal solution)
      WAREHOUSE W1           W2           W3          SUPPLY

 PLANT
 P1             7            6            9           20/0
                                  20

 P2             5            7            3           28/9/5/0
                     4            5            19

 P3             4            5            8           17/0
                     17

 DEMAND         21/4/0       25/5/0       19/0        65



      Here no. of rows=3, no. of columns=3, total allocations=5
                              5=3+3-1
TOTAL TRANSPORTATION COST
PLANT        WAREHOUSE   COST

P1           W2          6*20=120

P2           W1          5*4=20

P2           W2          7*5=35

P2           W3          3*19=57

P3           W1          4*17=68

TOTAL COST               300
MAXIMIZATION USING LCEM
INVESTMENT   P    Q    R    S    AVAILABLE


1            95   80   70   60   70

2            75   65   60   50   40

3            70   45   50   40   90

4            60   40   40   30   10

DEMAND       40   50   60   60   210




Convert maximization into minimization
Subtract all the elements from the largest
element thereby giving minimization case.
  INVESTMENT   P     Q    R       S       AVAILABLE



  1            95
                95   80   70      60      70

  2            75    65   60      50      40

  3            70    45   50      40      90

  4            60    40   40      30      10

  DEMAND       40    50   60      60      210



Here 95 is maximum element so subtract all elements
                      from 95
This is the matrix after subtracting all the
             elements from 95
 INVESTMENT   P    Q     R       S       AVAILABLE



 1            0    15    25      35      70

 2            20   30    35      45      40

 3            25   50    45      55      90

 4            35   55    55      65      10

 DEMAND       40   50    60      60      210



This is now minimization case which can be further
             solved as discussed before
TRANSPORTATION PROBLEM


  Categorized into two types:


 Minimization    Maximization
   problem        problem
MINIMIZATION PROBLEM


  In this transportation
 cost is given which is to
      be minimized.
MAXIMIZATION PROBLEM

        In this profit is given which is to be maximized.



To solve this problem we convert the problem into minimization.


Conversion is done by selecting the largest element from Profit
 Pay off matrix and then subtracting all elements from largest
                   element including itself.

  Reduced matrix obtain becomes minimization case and then
   same steps are taken to solve it as is done in minimization
                            problem
VOGEL’S APPROXIMATION METHOD
• BASIS OF ALLOCATION IS UNIT COST PENALTY
• THE SUBSEQUENT ALLOCATIONS IN CELLS ARE
  DONE KEEPING IN VIEW THE HIGHEST UNIT
  COST
• IBFS OBATINED BY THIS METHOD IS EITHER
  OPTIMAL OR VERY NEAR TO OPTIMAL
  SOLUTION
• SO AMOUNT OF TIME REQUIRED TO
  CALCULATE THE OPTIMUM SOLUTION IS
  REDUCED
STEPS TO SOLVE THE PROBLEM
Step1: for each row of the table identify the lowest and the next lowest cost
   cell. Find their least cost than the difference shall be zero.
Step 2: similarly find the difference of each column and place it below each
   column. These differences found in the steps 1 and 2 are also called
   penalties.
Step 3: looking at all the penalties. Identify the highest of them and the row
   or column relative to that penalty. Allocate the maximum possible units
   to the least cost cell in the selected row or column. Ties should be broken
   in this order
Maximum difference least cost cell.
Maximum difference            tie    least cost cell
Maximum unit’s allocations tie          arbitrary
Step 4: adjust the supply and demand and cross the satisfied row or column.
Step 5:Recompute the column and row differences ignoring deleted
   rows/columns and go to the step3. Repeat the procedure until the entire
   column and row totals are satisfied.
MINIMIZATION USING VAM
WAREHOUSE    W1      W2     W3      SUPPLY
PLANT

P1           7       6      9       20


P2           5       7      3       28


P3           4       5      8       17


DEMAND       21      25     19      65




     Step1 :This problem is balanced problem
Find out ‘Penalties’(P) for each row & column by
   subtracting the smallest element from next
      smallest element in the row/column
WAREHOUSE   W1      W2      W3          SUPPLY           P1
PLANT

P1          7       6       9           20               7-6=1

P2          5       7       3           28               5-3=2   In W3 3 is min cost
                                 19
P3          4       5       8           17               5-4=1

DEMAND      21      25      19          65

P1          5-4=1   6-5=1   9-3=6

                                      6 is max penalty

Find out the maximum penalty and in that row/column
         find minimum element and allocate it.
The row/column having zero demand/supply left
             will be cut by a line
     WAREHOUSE    W1      W2     W3          SUPPLY
     PLANT

     P1           7       6      9           20


     P2           5       7      3           28/9
                                        19

     P3           4       5      8           17

     DEMAND       21      25     19/0        65




W3 is having zero demand now so it will be cut and wont
                    be considered
Now find next penalty P2 in same way
WAREHOUSE         W1           W2      W3             SUPPLY   P2
PLANT

P1                7            6       9              20       7-6=1


P2                5    9       7       3              28/9/0   7-5=2   2 is max penalty

P3                4            5       8              17       5-4=1
                                             19
DEMAND            21/12        25      19/0

P2                5-4=1        6-5=1   ------------




         In W1 5 is min cost
In this way find out all the penalties until all
               allocations are not done
WAREHOUSE      W1            W2      W3             SUPPLY   P3
PLANT

P1             7             6       9              20       7-6=1


P2             5    9        7       3              28/9/0   ----------

P3             4    12       5       8              17/5     5-4=1
                                           19
DEMAND         21/12/0       25      19/0                                 In P3 4 is min cost

P3             7-4=3         6-5=1   ------------




          3 is max penalty
Final matrix with all allocations

WAREHOUSE    W1        W2           W3          SUPPLY

PLANT
P1           7         6            9           20/0
                            20

P2           5         7            3           28/9/0
                   9                       19

P3           4         5            8           17/5/0
                  12            5

DEMAND       21/12/0   25/5/0       19/0
No. of allocations=No. of rows + No. of columns-1
         (For generating optimal solution)
   WAREHOUSE   W1         W2           W3          SUPPLY

   PLANT
   P1          7          6            9           20/0
                               20

   P2          5          7            3           28/9/0
                     9                        19

   P3          4          5            8           17/5/0
                    12             5

   DEMAND      21/12/0    25/5/0       19/0



  Here no. of rows=3, no. of columns=3, total allocations=5
                          5=3+3-1
TOTAL TRANSPORTATION COST
PLANT        WAREHOUSE   COST

P1           W2          6*20=120

P2           W1          5*9=45

P2           W3          3*19=57

P3           W1          12*4=48

P3           W2          5*5=25

TOTAL COST               295
COMPARISON OF NWCM,LCEM,VAM
  WAREHOUSE   W1      W2     W3     SUPPLY
  PLANT
  P1          7       6      9      20

  P2          5       7      3      28

  P3          4       5      8      17

  DEMAND      21      25     19     65



           Minimum Transportation Cost
                  NWCM=462
                  LCEM=300
                   VAM=295
MAXIMIZATION CASE
            I    II    III    IV      CAPACITY

A           40   25    22     33      100

B           44   35    30     30      30

C           38   38    28     30      70

REQUIREMENT 40   20    60     30            200
                                      150




This is an unbalanced maximization problem for
      finding out maximum profit involved
To solve this first we will convert this in
minimization problem by subtracting the largest
           element from all elements
              I    II       III      IV        CAPACITY

A             4    19       22       11        100

B             0    9        14       14        30

C             6    6        16       14        70

REQUIREMENT   40   20       60       30               200
                                               150



44 is maximum element in the matrix so subtract all the
      elements from 44 and re-write in matrix form
Now we will balance this matrix
    Requirement<Capacity Dummy column will be
                       added
              I     II     III    IV      Dummy   CAPACITY

A             4     19     22     11      0       100


B             0     9      14     14      0       30


C             6     6      16     14      0       70


REQUIREMENT   40    20     60     30      50             200
                                                  200



     Now this can be solved by any of the IBFS methods
CHOCOLATE
   WINING
QUESTION FOR
  AUDIENCE
Obtain IBFS for the following
Maximization Problem using VAM:
         I    II   III    SUPPLY

A        6    4    4      100

B        5    6    7      25

C        3    4    6      75

DEMAND   60   80   85
SOLUTION:
         I           II        III        SUPPLY

A        6    60     4    40   4          100

B        5           6    15   7     10   25

C        3           4         6     75   75

DUMMY    0           0    25   0          25

DEMAND   60          80        85



    MAXIMUM PROFIT= Rs. 1130
Transportation Problem in Operational Research

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Transportation Problem in Operational Research

  • 2. TRANSPORTATION PROBLEM Transport various quantities of a single homogeneous commodity to different destinations in such a way that total transportation cost is minimum.
  • 3. TERMINOLOGY USED IN TRANSPORTATIONAL MODEL Feasible solution: Non negative values of xij where i=1, 2……….m and j=1, 2,…n which satisfy the constraints of supply and demand is called feasible solution. Basic feasible solution: If the no of positive allocations are (m+n-1). Optimal solution: A feasible solution is said to be optimal solution if it minimizes the total transportation cost. Balanced transportation problem: A transportation problem in which the total supply from all sources is equal to the total demand in all the destinations. Unbalanced transportation problem: Problems which are not balanced are called unbalanced. Matrix terminology: In the matrix, the squares are called cells and form columns vertically and rows horizontally. Degenerate basic feasible solution: If the no. of allocation in basic feasible solutions is less than (m+n-1).
  • 4. OPTIMAL SOLUTION OF TRANSPORTATION PROBLEM Obtain an optimal solution by making Initial Basic successive STEP1 Feasible STEP2 improvements in Solution IBFS until no further decrease in transportation cost is possible
  • 5. INITIAL BASIC FEASIBLE SOLUTION Methods Available NORTH WEST LOWEST COST VOGEL’S CORNER ENTRY APPROXIMATION METHOD(NWCM) METHOD(LCEM) METHOD(VAM)
  • 6. UNBALANCED TRANSPORTATION PROBLEM Demand>Supply Add dummy column in matrix with zero cost Supply>Demand Add dummy row in matrix with zero cost
  • 7. DEMAND>SUPPLY PRODUCT P Q SUPPLY OFFICE A 10 15 20 B 20 30 50 DEMAND 30 60 70 90 PRODUCT P Q SUPPLY OFFICE A 10 15 20 B 20 30 50 DUMMY 0 0 20 DEMAND 30 60 90 90
  • 8. SUPPLY>DEMAND PRODUCT P Q SUPPLY OFFICE A 10 15 30 B 20 30 60 DEMAND 20 50 90 70 PRODUCT P Q DUMMY SUPPLY OFFICE A 10 15 0 20 B 20 30 0 50 DEMAND 20 50 20 90 90
  • 9. METHOD FOR OPTIMAL SOLUTION Stepping stone method Modified Distribution Method(MODI)
  • 10. NORTH WEST CORNER METHOD Most systematic and easiest method for obtaining initial feasible basic solution
  • 11. STEPS TO SOLVE THE PROBLEM Step1: construct an empty m*n matrix, completed with rows & columns. Step2: indicate the rows and column totals at the end. Step3: starting with (1,1)cell at the north west corner of the matrix, allocate maximum possible quantity keeping in view that allocation can neither be more than the quantity required by the respective warehouses nor more than quantity available at the each supply centre. Step 4: adjust the supply and demand nos. in the rows and columns allocations. Step5: 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 the step 4. Step 6: if the demand for the first column is satisfied, then move to the next cell in the second column and first row and go to step 4. Step 7: if for any cell, supply equals demand then the next allocation can be made in cell either in the next row or column. Step 8: continue the procedure until the total available quantity is fully allocated to the cells as required.
  • 12. MINIMIZATION USING NWCM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19
  • 13. Step1: Check whether problem is balanced WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 65
  • 14. Allocate P1W1 max quantity such that it should not exceed supply/requirement WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21/1 25 19 65 P1 can supply 20 units at max so W1 can be allocated 20 units only by P1
  • 15. Check whether the requirements of W1 fulfilled WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21/1 25 19 65 Now allocate the remaining requirement to next north west corner element i.e. P2W1
  • 16. Allocate 1 to P2W1 WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27 1 P3 4 5 8 17 DEMAND 21/1/0 25 19 65 In this way repeat it for all columns until demand of all warehouses is fulfilled
  • 17. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/27/0 1 25 2 P3 4 5 8 17/0 17 DEMAND 21/1/0 25/0 19/17/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  • 18. TOTAL TRANSPORTATION COST PLANT WAREHOUSE COST P1 W1 7*20=140 P2 W2 5*1=5 P2 W2 7*25=175 P2 W2 3*2=6 P3 W3 8*17=136 TOTAL COST 462
  • 19. LEAST COST ENTRY METHOD This method takes into consideration the lowest cost and therefore takes less time to solve the problem
  • 20. STEPS TO SOLVE THE PROBLEM • Step1: select the cell with the lowest transportation cost among all the rows and columns of the transportation table. If the minimum cost is not unique then select arbitrarily any cell with the lowest cost. • Step2: allocate as many units as possible to the cell determined in step 1 and eliminate that row in which either capacity or requirement is exhausted. • Step3:adjust the capacity and the requirement for the next allocations. • Step4: repeat the steps1to3 for the reduced table until the entire capacities are exhausted to fill the requirements at the different destinations.
  • 21. MINIMIZATION USING LCEM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Step1 :This problem is balanced problem
  • 22. Find out the cell having least cost WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28/9 19 P3 4 5 8 17 DEMAND 21 25 19/0 65 P2W3 is having least cost i.e. 3 so it is allocated 19 units as per the requirement of W3
  • 23. In this way allocate next minimum costs until all requirements are fulfilled WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Now P2W1 and P3W2 are having same cost, we can allocate any of them but in this case P2W1 will be allocated because in P3W2 P3 cant supply anymore units
  • 24. Final matrix with all allocations WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65
  • 25. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/5/0 4 5 19 P3 4 5 8 17/0 17 DEMAND 21/4/0 25/5/0 19/0 65 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  • 26. TOTAL TRANSPORTATION COST PLANT WAREHOUSE COST P1 W2 6*20=120 P2 W1 5*4=20 P2 W2 7*5=35 P2 W3 3*19=57 P3 W1 4*17=68 TOTAL COST 300
  • 27. MAXIMIZATION USING LCEM INVESTMENT P Q R S AVAILABLE 1 95 80 70 60 70 2 75 65 60 50 40 3 70 45 50 40 90 4 60 40 40 30 10 DEMAND 40 50 60 60 210 Convert maximization into minimization
  • 28. Subtract all the elements from the largest element thereby giving minimization case. INVESTMENT P Q R S AVAILABLE 1 95 95 80 70 60 70 2 75 65 60 50 40 3 70 45 50 40 90 4 60 40 40 30 10 DEMAND 40 50 60 60 210 Here 95 is maximum element so subtract all elements from 95
  • 29. This is the matrix after subtracting all the elements from 95 INVESTMENT P Q R S AVAILABLE 1 0 15 25 35 70 2 20 30 35 45 40 3 25 50 45 55 90 4 35 55 55 65 10 DEMAND 40 50 60 60 210 This is now minimization case which can be further solved as discussed before
  • 30. TRANSPORTATION PROBLEM Categorized into two types: Minimization Maximization problem problem
  • 31. MINIMIZATION PROBLEM In this transportation cost is given which is to be minimized.
  • 32. MAXIMIZATION PROBLEM In this profit is given which is to be maximized. To solve this problem we convert the problem into minimization. Conversion is done by selecting the largest element from Profit Pay off matrix and then subtracting all elements from largest element including itself. Reduced matrix obtain becomes minimization case and then same steps are taken to solve it as is done in minimization problem
  • 33. VOGEL’S APPROXIMATION METHOD • BASIS OF ALLOCATION IS UNIT COST PENALTY • THE SUBSEQUENT ALLOCATIONS IN CELLS ARE DONE KEEPING IN VIEW THE HIGHEST UNIT COST • IBFS OBATINED BY THIS METHOD IS EITHER OPTIMAL OR VERY NEAR TO OPTIMAL SOLUTION • SO AMOUNT OF TIME REQUIRED TO CALCULATE THE OPTIMUM SOLUTION IS REDUCED
  • 34. STEPS TO SOLVE THE PROBLEM Step1: for each row of the table identify the lowest and the next lowest cost cell. Find their least cost than the difference shall be zero. Step 2: similarly find the difference of each column and place it below each column. These differences found in the steps 1 and 2 are also called penalties. Step 3: looking at all the penalties. Identify the highest of them and the row or column relative to that penalty. Allocate the maximum possible units to the least cost cell in the selected row or column. Ties should be broken in this order Maximum difference least cost cell. Maximum difference tie least cost cell Maximum unit’s allocations tie arbitrary Step 4: adjust the supply and demand and cross the satisfied row or column. Step 5:Recompute the column and row differences ignoring deleted rows/columns and go to the step3. Repeat the procedure until the entire column and row totals are satisfied.
  • 35. MINIMIZATION USING VAM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Step1 :This problem is balanced problem
  • 36. Find out ‘Penalties’(P) for each row & column by subtracting the smallest element from next smallest element in the row/column WAREHOUSE W1 W2 W3 SUPPLY P1 PLANT P1 7 6 9 20 7-6=1 P2 5 7 3 28 5-3=2 In W3 3 is min cost 19 P3 4 5 8 17 5-4=1 DEMAND 21 25 19 65 P1 5-4=1 6-5=1 9-3=6 6 is max penalty Find out the maximum penalty and in that row/column find minimum element and allocate it.
  • 37. The row/column having zero demand/supply left will be cut by a line WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28/9 19 P3 4 5 8 17 DEMAND 21 25 19/0 65 W3 is having zero demand now so it will be cut and wont be considered
  • 38. Now find next penalty P2 in same way WAREHOUSE W1 W2 W3 SUPPLY P2 PLANT P1 7 6 9 20 7-6=1 P2 5 9 7 3 28/9/0 7-5=2 2 is max penalty P3 4 5 8 17 5-4=1 19 DEMAND 21/12 25 19/0 P2 5-4=1 6-5=1 ------------ In W1 5 is min cost
  • 39. In this way find out all the penalties until all allocations are not done WAREHOUSE W1 W2 W3 SUPPLY P3 PLANT P1 7 6 9 20 7-6=1 P2 5 9 7 3 28/9/0 ---------- P3 4 12 5 8 17/5 5-4=1 19 DEMAND 21/12/0 25 19/0 In P3 4 is min cost P3 7-4=3 6-5=1 ------------ 3 is max penalty
  • 40. Final matrix with all allocations WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/0 9 19 P3 4 5 8 17/5/0 12 5 DEMAND 21/12/0 25/5/0 19/0
  • 41. No. of allocations=No. of rows + No. of columns-1 (For generating optimal solution) WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20/0 20 P2 5 7 3 28/9/0 9 19 P3 4 5 8 17/5/0 12 5 DEMAND 21/12/0 25/5/0 19/0 Here no. of rows=3, no. of columns=3, total allocations=5 5=3+3-1
  • 42. TOTAL TRANSPORTATION COST PLANT WAREHOUSE COST P1 W2 6*20=120 P2 W1 5*9=45 P2 W3 3*19=57 P3 W1 12*4=48 P3 W2 5*5=25 TOTAL COST 295
  • 43. COMPARISON OF NWCM,LCEM,VAM WAREHOUSE W1 W2 W3 SUPPLY PLANT P1 7 6 9 20 P2 5 7 3 28 P3 4 5 8 17 DEMAND 21 25 19 65 Minimum Transportation Cost NWCM=462 LCEM=300 VAM=295
  • 44. MAXIMIZATION CASE I II III IV CAPACITY A 40 25 22 33 100 B 44 35 30 30 30 C 38 38 28 30 70 REQUIREMENT 40 20 60 30 200 150 This is an unbalanced maximization problem for finding out maximum profit involved
  • 45. To solve this first we will convert this in minimization problem by subtracting the largest element from all elements I II III IV CAPACITY A 4 19 22 11 100 B 0 9 14 14 30 C 6 6 16 14 70 REQUIREMENT 40 20 60 30 200 150 44 is maximum element in the matrix so subtract all the elements from 44 and re-write in matrix form
  • 46. Now we will balance this matrix Requirement<Capacity Dummy column will be added I II III IV Dummy CAPACITY A 4 19 22 11 0 100 B 0 9 14 14 0 30 C 6 6 16 14 0 70 REQUIREMENT 40 20 60 30 50 200 200 Now this can be solved by any of the IBFS methods
  • 47. CHOCOLATE WINING QUESTION FOR AUDIENCE
  • 48. Obtain IBFS for the following Maximization Problem using VAM: I II III SUPPLY A 6 4 4 100 B 5 6 7 25 C 3 4 6 75 DEMAND 60 80 85
  • 49. SOLUTION: I II III SUPPLY A 6 60 4 40 4 100 B 5 6 15 7 10 25 C 3 4 6 75 75 DUMMY 0 0 25 0 25 DEMAND 60 80 85 MAXIMUM PROFIT= Rs. 1130