transportation problem using russell's approximation method Unbalanced supply and demand example

3. Unbalanced supply and demand example

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Home > Operation Research calculators > Russell's approximation method example

6. russell's approximation method example ( Enter your problem )

( Enter your problem )

Algorithm and examples

Algorithm & Example-1
Example-2
Unbalanced supply and demand example

Other related methods

north-west corner method
least cost method
vogel's approximation method
Row minima method
Column minima method
Russell's approximation method
Heuristic method-1
Heuristic method-2
modi method (optimal solution)
stepping stone method (optimal solution)

2. Example-2
(Previous example)

7. Heuristic method-1
(Next method)

$D_(dummy)$

Unbalanced supply and demand
If the total supply is not equal to the total demand then the problem is called unbalanced transportation problem.

It's solution :
1. If the total supply is more than the total demand, then we add a new column, with transportation cost 0
2. If the total demand is more than the total supply, then we add a new row, with transportation cost 0

Example
Find Solution using Russell's Approximation method
D1 D2 D3 Supply
S1 4 8 8 76
S2 16 24 16 82
S3 8 16 24 77
Demand 72 102 41

Solution:
TOTAL number of supply constraints : 3
TOTAL number of demand constraints : 3
Problem Table is

	$D_1$	$D_2$	$D_3$	Supply
$S_1$	4	8	8	76
$S_2$	16	24	16	82
$S_3$	8	16	24	77

Demand	72	102	41

Here Total Demand = 215 is less than Total Supply = 235. So We add a dummy demand constraint with 0 unit cost and with allocation 20.
Now, The modified table is

Table-1: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8	8	0	76	$8$
$S_2$	16	24	16	0	82	$24$
$S_3$	8	16	24	0	77	$24$

Demand	72	102	41	20
$bar V_j$	$16$	$24$	$24$	$0$

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20}$

$2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24}$

$3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +24) = color{blue}{-24}$

$4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 0 - (8 +0) = color{blue}{-8}$

$5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24}$

$6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24}$

$7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +24) = color{blue}{-32}$

$8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 0 - (24 +0) = color{blue}{-24}$

$9. Delta_31 = c_31 - (bar U_3 + bar V_1) = 8 - (24 +16) = color{blue}{-32}$

$10. Delta_32 = c_32 - (bar U_3 + bar V_2) = 16 - (24 +24) = color{blue}{-32}$

$11. Delta_33 = c_33 - (bar U_3 + bar V_3) = 24 - (24 +24) = color{blue}{-24}$

$12. Delta_34 = c_34 - (bar U_3 + bar V_4) = 0 - (24 +0) = color{blue}{-24}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-20]	8 [-24]	8 [-24]	0 [-8]	76	$8$
$S_2$	16 [-24]	24 [-24]	16 [-32]	0 [-24]	82	$24$
$S_3$	8 [-32]	16 [-32]	24 [-24]	0 [-24]	77	$24$

Demand	72	102	41	20
$bar V_j$	$16$	$24$	$24$	$0$

The most negative

$Delta_(ij)$ is -32 in cell

$S_3 D_2$

The allocation to this cell is min(77,102) = 77.
This exhausts the capacity of

$S_3$ and leaves 102 - 77 = 25 units with

$D_2$

Table-1: This leads to the following table

Table-2: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8	8	0	76	$8$
$S_2$	16	24	16	0	82	$24$
$S_3$	8	16(77)	24	0	0	--

Demand	72	25	41	20
$bar V_j$	$16$	$24$	$16$	$0$

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20}$

$2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24}$

$3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +16) = color{blue}{-16}$

$4. Delta_14 = c_14 - (bar U_1 + bar V_4) = 0 - (8 +0) = color{blue}{-8}$

$5. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24}$

$6. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24}$

$7. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +16) = color{blue}{-24}$

$8. Delta_24 = c_24 - (bar U_2 + bar V_4) = 0 - (24 +0) = color{blue}{-24}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-20]	8 [-24]	8 [-16]	0 [-8]	76	$8$
$S_2$	16 [-24]	24 [-24]	16 [-24]	0 [-24]	82	$24$
$S_3$	8	16(77)	24	0	0	--

Demand	72	25	41	20
$bar V_j$	$16$	$24$	$16$	$0$

The most negative

$Delta_(ij)$ is -24 in cell

$S_2 D_(dummy)$

The allocation to this cell is min(82,20) = 20.
This satisfies the entire demand of

$D_(dummy)$ and leaves 82 - 20 = 62 units with

$S_2$

Table-2: This leads to the following table

Table-3: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8	8	0	76	$8$
$S_2$	16	24	16	0(20)	62	$24$
$S_3$	8	16(77)	24	0	0	--

Demand	72	25	41	0
$bar V_j$	$16$	$24$	$16$	--

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +16) = color{blue}{-20}$

$2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +24) = color{blue}{-24}$

$3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +16) = color{blue}{-16}$

$4. Delta_21 = c_21 - (bar U_2 + bar V_1) = 16 - (24 +16) = color{blue}{-24}$

$5. Delta_22 = c_22 - (bar U_2 + bar V_2) = 24 - (24 +24) = color{blue}{-24}$

$6. Delta_23 = c_23 - (bar U_2 + bar V_3) = 16 - (24 +16) = color{blue}{-24}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-20]	8 [-24]	8 [-16]	0	76	$8$
$S_2$	16 [-24]	24 [-24]	16 [-24]	0(20)	62	$24$
$S_3$	8	16(77)	24	0	0	--

Demand	72	25	41	0
$bar V_j$	$16$	$24$	$16$	--

The most negative

$Delta_(ij)$ is -24 in cell

$S_2 D_1$

The allocation to this cell is min(62,72) = 62.
This exhausts the capacity of

$S_2$ and leaves 72 - 62 = 10 units with

$D_1$

Table-3: This leads to the following table

Table-4: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8	8	0	76	$8$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	25	41	0
$bar V_j$	$4$	$8$	$8$	--

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +4) = color{blue}{-8}$

$2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +8) = color{blue}{-8}$

$3. Delta_13 = c_13 - (bar U_1 + bar V_3) = 8 - (8 +8) = color{blue}{-8}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-8]	8 [-8]	8 [-8]	0	76	$8$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	25	41	0
$bar V_j$	$4$	$8$	$8$	--

The most negative

$Delta_(ij)$ is -8 in cell

$S_1 D_3$

The allocation to this cell is min(76,41) = 41.
This satisfies the entire demand of

$D_3$ and leaves 76 - 41 = 35 units with

$S_1$

Table-4: This leads to the following table

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4	8	8 (41)	0	35
$S_2$	16 (62)	24	16	0 (20)	0
$S_3$	8	16 (77)	24	0	0

Demand	10	25	0	0

Table-5: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8	8(41)	0	35	$8$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	25	0	0
$bar V_j$	$4$	$8$	--	--

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (8 +4) = color{blue}{-8}$

$2. Delta_12 = c_12 - (bar U_1 + bar V_2) = 8 - (8 +8) = color{blue}{-8}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-8]	8 [-8]	8(41)	0	35	$8$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	25	0	0
$bar V_j$	$4$	$8$	--	--

The most negative

$Delta_(ij)$ is -8 in cell

$S_1 D_2$

The allocation to this cell is min(35,25) = 25.
This satisfies the entire demand of

$D_2$ and leaves 35 - 25 = 10 units with

$S_1$

Table-5: This leads to the following table

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4	8 (25)	8 (41)	0	10
$S_2$	16 (62)	24	16	0 (20)	0
$S_3$	8	16 (77)	24	0	0

Demand	10	0	0	0

Table-6: Calculate

$bar U_i$ and

$bar V_j$ (where

$bar U_i$ is the largest cost in row and

$bar V_j$ is the largest cost in column)

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4	8(25)	8(41)	0	10	$4$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	0	0	0
$bar V_j$	$4$	--	--	--

2. Compute reduced cost of each cell

$Delta_(ij)$ , where

$Delta_(ij) = c_(ij) - (bar U_i + bar V_j)$

$1. Delta_11 = c_11 - (bar U_1 + bar V_1) = 4 - (4 +4) = color{blue}{-4}$

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply	$bar U_i$
$S_1$	4 [-4]	8(25)	8(41)	0	10	$4$
$S_2$	16(62)	24	16	0(20)	0	--
$S_3$	8	16(77)	24	0	0	--

Demand	10	0	0	0
$bar V_j$	$4$	--	--	--

The most negative

$Delta_(ij)$ is -4 in cell

$S_1 D_1$

The allocation to this cell is min(10,10) = 10.
Table-6: This leads to the following table

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4 (10)	8 (25)	8 (41)	0	0
$S_2$	16 (62)	24	16	0 (20)	0
$S_3$	8	16 (77)	24	0	0

Demand	0	0	0	0

Initial feasible solution is

	$D_1$	$D_2$	$D_3$	$D_(dummy)$	Supply
$S_1$	4 (10)	8 (25)	8 (41)	0	76
$S_2$	16 (62)	24	16	0 (20)	82
$S_3$	8	16 (77)	24	0	77

Demand	72	102	41	20

The minimum total transportation cost

$= 4 xx 10 + 8 xx 25 + 8 xx 41 + 16 xx 62 + 0 xx 20 + 16 xx 77 = 2792$

Here, the number of allocated cells = 6 is equal to m + n - 1 = 3 + 4 - 1 = 6

$:.$ This solution is non-degenerate

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here

2. Example-2
(Previous example)

7. Heuristic method-1
(Next method)

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