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

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written 8.2 years ago by

teamques10 ★ 69k

• modified 8.2 years ago

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

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