Friday 21 June 2019

Transportation Model


(i)              Degeneracy; Degeneracy occurs when the number of rows plus the number of columns less one is not equal to the number of occupied cells.
(ii)            A slack variable: This is a variable that is added to an inequality to make it equal e.g.  incase the supply and demand are not equal.
(iii)          Balanced transportation model: This is a model in which the total demand and total  supply are equal.

Initial distribution using LCM
 Source
D1
D2
D3
D4
D5
Supply
S1
10
12
2

25000
3

25000
4
50,000
S2
4

15000
10
9
2

5000
8
20,000
S3
9

10000
15

15000
3
12
1

10000
35,000
Demand
25,000
15,000
25,000
30,000
10,000
105,000 

Test for degeneracy
M= no of rows                  n = number of columns
M+n-1= number of occupied cells
3+5-1=7
No degenerate

Test for optimality
v + m=t
                                                               5                    12                      2                        3                    -3
 Source
D1
D2
D3
D4
D5
Supply
S1

               0
10
12
2

25000
3

25000
4
50,000
S2

              -1
4

15000
10
9
2

5000
8
20,000
S3

            4
9

10000
15

15000
3
12
1

10000
35,000
Demand
25,000
15,000
25,000
30,000
10,000
105,000 

Unoccupied cells          actual cost                -               v + m                                             difference
S1d1                                            10                                          0+5                                                       5
S1d2                                            12                                         0+12                                                     0
S1d5                                            4                                            0+-3                                                      7
S2d2                                            10                                        -1+12                                                     -1
S2d3                                            9                                            -1+2                                                      8
S2d5                                            8                                           -1+-3                                                     12
S3d3                                            3                                             4+2                                                       -4
S3d4                                            12                                          4+3                                                       5

The solution is not optimum as the negative numbers indicate that the unoccupied cells need to be allocated so as to reduce the cost further by the number I.e. s2d2 by 1 and s3d3 by 4
Initial basic feasible solution =(2 X 25000 + 3 X 25000 + 4 X 15000 + 2 X 5000 + 9 X 10000 + 15 X 15000 + 1 X 10000)
                             =KSH 520000

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