0
1.5kviews
Determine the optimal distribution policy. Use VAM to find initial solution and MODI for finding optimal solution.

enter image description here

Mumbai University > Mechanical Engineering > Sem 7 > Operations Research

Marks: 10 Marks

Year: Dec 2015

1 Answer
0
3views

As demand is more than supply, we introduce a dummy row, whose supply is difference between total demand and total supply with cost coefficient of the cells as zero.

| a | W | X | Y | Z| | | |-|-|-|-|-|-|-|
|A| 17| 20| 14| 12| 60| [2]| |B| 15| 21| 25| 14| 75| [1]| |C| 15| 14| 15| 16| 105| [1]| |DUMMY| 0 (50)| 0| 0| 0| 50/0| [0]| | | 50/0| 65| 75| 100| | |
| | [15] ↑| [14]| [14]| [12]| | |

| | X| Y| Z| | | |-|-|-|-|-|-|
|A| 20| 14| 12| 60| [2]| |B| 21| 25| 14 (75)| 75/0| [7]←| |C| 14| 15| 16| 105| [1]| | | 65| 75| 100/25| | |
| | [6]| [1]| [2]|  | |

| | X| Y| Z| | | |-|-|-|-|-|-|
|A| 20| 14| 12| 60| [2]| |C| 14 (65) |15| 16| 105/40| [1]| | | 65/0| 75| 25| | |
| | [6] ↑| [1]| [4]| | |

| | Y| Z| | | |-|-|-|-|
|A| 14| 12 (25)| 60/35| [2]| |C| 15| 16| 40| [1]| | | 75| 25/0| | |
| | [1]| [4] ↑| | |

  Y  
A 14 (35) 35/0
C 15 (40) 40/0
  75/0  

The single final matrix showing allocations is shown below,

| | W| X| Y| Z| | | | | |
|-|-|-|-|
|A| 17| 20| 14 (35)| 12 (25)| 60| [2]| [2]| [2]| [2]| |B| 15| 21| 25| 14 (75)| 75/0| [1]| [7]←|  |  | |C| 15| 14 (65)| 15 (40)| 16| 105/40/0| [1]| [1]| [1]| [2]| |DUMMY| 0 (50)| 0| 0| 0| 50/0| [0]| | | |
| | 50/0| 65/0| 75/0| 100/25/0| | | | |
| | [15] ↑| [14]| [14]| [12] | | | | |
|  | | [6]| [1] |[2] | | | | |
| | | [6] ↑| [1]| [4] | | | | |
| | | | [1]| [4] ↑ | | | | |

Transportation cost :-

Z $= 14*35 + 12*25 + 14*75 + 14*65 + 15*40 + 0*50 \\ = 3350$

But, m + n – 1 = 7 and no. of allocations = 8, here degeneracy exists.

The minimum cost in the unallocated cells is 0. Selecting Dummy-Z cell arbitrarily for assigning ‘ɛ’ allocation to it. Also checking for optimality by UV rule/ MODI method.

  $v_1$ $v_2$ $v_3$ $v_4$
$u_1$ 17 20 14 (35) 12 (25)
$u_2$ 15 21 25 14 (75)
$u_3$ 15 14 (65) 15 (40) 16
$u_4$ 0 (50) 0 0 0 (ɛ)

$u_4 + v_1 = 0, u_3 + v_2 = 14, u_1 + v_3 = 14, u_1 + v_4 = 12, u_2 + v_4 = 14, u_3 + v_2 = 14, u_3 + v_3 = 15, u_4+v_4= 0 \\ Let u_1 = 0, \\ v_3 = 14, v_4 = 12, u_2 = 2, u_3 = 1, v_2 = 13, u_4 = -12, v_1 = 12.$

enter image description here

As we can see θ = 35, again checking for optimality with new allocations.

  $v_1$ = 13 $v_2$ = 12 $v_3$ = 13 $v_4$ = 12
$u_1$ = 0 17 (4) 20 (8) 14 (1) 12 60
$u_2$ = 2 15 35 21 (7) 25 (10) 14 40
$u_3$ = 2 15 (0) 14 65 15 40 16 (2)
$u_4$ = -13 0 15 0 (1) 0 35 0 (1)

No negative numbers are present, which shows we have obtained the optimal solution.

  W X Y Z
A 17 20 14 12 60
B 15 35 21 25 14 40
C 15 14 65 15 40 16
DUMMY 0 15 0 0 35 0

Minimum transportation cost $= 15*35 + 0*15 + 14*65 + 15*40 + 0*35 + 12*60 + 14*40 \\ = 3315.$

Please log in to add an answer.