Question

Consider the transportation problem presented in the following table: Capacity 120 B 4 17 22 20 1 9 70 7 37 2 24 32 50 15 20 37 3 240 110 30 40 60 Demand Determine the optimal shipping schedule using: a) North-West Corner Method b) Least Cost Method c) Vogel Approximation Method

Answer 1

Solution:

a: North-West Corner Method.

Find Solution using North-West Corner method D2 D3 D4 Supply D1 |22 17 4 S1 20 120 37 97 S2 24 70 37 20 15 50 S3 32 Demand 60 40 30 110 Solution: TOTAL number of supply constraints : 3 TOTAL number of demand constraints : 4 Problem Table is D2 D3 D4 Supply 22 17 20 4 120 24 37 9 S2 7 70 32 37 20 15 S3 50 Demand 60 40 30 110 The rim values for S1 120 and D-60 are compared. 60 is assigned to S, D The smaller of the two i.e. min(120,60) This meets the complete demand of D and leaves 120 60 60 units with S Table-1 Supply 22 17 S1 20(60) 4 60 37 9 S2 24 7 70 37 20 15 S3 32 50 40 30 110 Demand 0 to


The rim values for s,-60 and D2=40 are compared The smaller of the two i.e. min(60,40) = 40 is assigned to si D2 This meets the complete demand of D2 and leaves 60 - 40 20 units with S Table-2 D3 D4Supply D1 D2 20(60) 22(40) S1 17 20 S2 24 37 7 70 S3 32 37 20 15 0 30 110 Demand 0 The rim values for S 20 and D,-30 are compared The smaller of the two i.e. min(20,30) = 20 is assigned to s D This exhausts the capacity of S and leaves 30 20 10 units with D3 Table-3 D2 D3 D& Supply 2240 20(60 47(20) 4 0 S2 24 37 9 7 70 S3 32 37 20 50 15 Demand 0 10 110 The rim values for S,-70 and D-10 are compared. The smaller of the two i.e. min(70,10) = 10 is assigned to S2 D3 This meets the complete demand of D3 and leaves 70- 10 = 60 units with s


Table-4 DI D2 D3 D4 Supply 20(60) 22(40) 47(20 4 0 24 9(10) 60 37 7 S3 32 15 50 37 20 0 Demand 0 0 110 The rim values for S, 60 and D 110 are compared. The smaller of the two i.e. min(60, 1 10) = 60 is assigned to S2 D4 This exhausts the capacity of S2 and leaves 110 60 50 units with D4 Table-5 D2 D3 D4 Supply 20(60) 22(40) 47(20) 4 0 S2 760) 24 37 0 37 32 20 15 50 Demand 0 0 0 50 The rim values for S 50 and D 50 are compared The smaller of the two i.e. min(50,50) = 50 is assigned to Sz D4 Table-6 D1 D2 D3 D4 Supply S1 20(60) 22(40) 47(20 4 C S2 760) 24 37 9f40) 0 45(50) 32 37 20 0 Demand 0 0 0 0 to


Initial feasible solution is D4 D1 D2 Supply 20 (60) 22 (40) 17 (20) 4 S1 120 S2 24 37 9 (10) 7 (60) 70 15 (50) 50 S3 32 20 37 Demand 60 40 110 The minimum total transportation cost 20 x 60 22 x 4017 x 209 x 10+7 x 6015 x 50 3680 Here, the number of allocated cells 6 is equal to m n 1 = 3 4 -1 6 This solution is non-degenerate 30

b: Least Cost Method

Find Solution using Least Cost method D1 D2 D3 D4 Supply 17 4 24 379 20 22 120 S1 S2 7 70 32 37 20 15 50 S3 Demand 60 4030 110 Solution: TOTAL number of supply constraints 3 TOTAL number of demand constraints : 4 Problem Table is D1 D2 D3 D4 Supply 22 4 20 17 120 24 37 9 S2 70 7 32 37 20 15 S3 50 Demand 60 40 30 110 The smallest transportation cost is 4 in cell SD4 The allocation to this cell is min(120,110) = 110. This satisfies the entire demand of D4 and leaves 120 110 10 units with S Table-1 D1 D2 Supply 17 4(110) 20 22 10 24 37 S2 9 70 32 37 S3 50 20 15 Demand 60 40 30 0 ST


The smallest transportation cost is 9 in cell S2D The allocation to this cell is min(70,30) 30. This satisfies the entire demand of D2 and leaves 70 - 30 40 units with s Table-2 D1 D2 D4 D3 Supply S1 10 20 17 4(110) 24 37 9(30) S2 40 7 S3 37 50 32 15 Demand 60 40 0 0 The smallest transportation cost is 20 in cell SD The allocation to this cell is min(10,60) = 10. This exhausts the capacity of S, and leaves 60 10 50 units with D Table-3 D2 D3 D4 Supply S1 4(110) 2010) 22 17 0 S2 24 37 9(30) 7 40 32 37 20 15 50 50 Demand 0 The smallest transportation cost is 24 in cell S2D1 The allocation to this cell is min(40,50) = 40. This exhausts the capacity of S, and leaves 50 40 10 units with D 40 20 22


Table-4 D3 DA D2 Supply 22 2010) 17 4(110) 37 24(40) (oe)6 S3 32 37 20 15 0 Demand 10 40 0 The smallest transportation cost is 32 in cell S3D1 The allocation to this cell is min(50,10) = 10. This satisfies the entire demand of D, and leaves 50 10 40 units with S Table-5 DI D4 D2 D3 Supply 20(10) 22 17 4(110) 0 9(30) 24(40) 37 0 32(10) 20 37 15 40 Demand 0 40 0 The smallest transportation cost is 37 in cell SD The allocation to this cell is min(40,40) 40. Table-6 D2 Supply S1 22 17 20(10) 4(140) 0 S2 24(40) 37 37(40) 3240) 20 15 0 Demand 0 0 0 0 50


Initial feasible solution is D1 D2 D3 Supply 4 (110) 120 20 (10) 22 17 24 (40) 37 9 (30) 7 70 32 (10) 37 (40) 20 15 50 Demand 60 40 30 110 20 x 104x 110 +24 x 409 x 30+32 x 1037 x 40 The minimum total transportation cost 3670 Here, the number of allocated cells 6 is equal to m n-1 3 4- 1 = 6 This solution is non-degenerate

c: Vogel Approximation Method

Find Solution using Voggel's Approximation method D1 D2 D3 D4 Supply |22 17 4 S1 20 120 37 97 S2 24 70 37 20 15 32 S3 50 Demand 60 40 30 110 Solution: TOTAL number of supply constraints : 3 TOTAL number of demand constraints : 4 Problem Table is D2 D3 D4 D1 Supply 22 17 4 20 120 37 9 S2 7 70 37 20 15 S3 32 50 40 30 110 Demand 60 Table-1 Supply Row Penalty DI D2 D3 D4 13 17-4 20 22 17 4 120 S2 24 37 2 9-7 9 7 70 5 20 15 S3 32 37 15 50 Demand 60 40 30 110 Column 4 24-20 15 37- 22 8 = 17 -93 7 - 4 Penalty The maximum penalty, 15, occurs in column D, The minimum c in this column is c12 = 22. The maximum allocation in this cell is min(120,40) 40 20 24


It satisfy demand of D2 and adjust the supply of S, from 120 to 80 (120 40 80) Table-2 Supply Row Penalty D4 D2 D3 13 17-4 S1 20 22(40) 17 4 80 2 9-7 S2 24 37 7 70 20 15 5 32 37 20 15 50 Demand 60 0 30 110 Column 8 17-9 3 7 - 4 4 24 -20 Penalty The maximum penalty, 13, occurs in row S The minimum c In this row is c14 = 4. The maximum allocation in this cell is min(80,110) = 80. It satisfy supply of si and adjust the demand of D4 from 110 to 30 (110 - 80 30) Table-3 D2 D4 Supply Row Penalty 22(40 20 47 4(80) 0 2 9-7 S2 24 37 7 70 20 15 5 32 37 20 15 50 Demand 60 0 30 30 Column 8 32-24 11 20-9 8 15 7 Penalty The maximum penalty, 11, occurs in column D3. The minimum c in this column is c23 = 9. !


The maximum allocation in this cell is min(70,30) = 30. It satisfy demand of D3 and adjust the supply of S2 from 70 to 40 (70 - 30 = 40). Table-4 Supply Row Penalty D2 D& D3 22(40) 17 4(80) 20 0 17 24-7 S2 24 9(30) 40 37 7 20 50 17 32- 15 37 15 0 Demand 60 30 Column 8 32-24 8 15 7 Penalty The maximum penalty, 17, occurs in row S The minimum C in this row is c24 = 7. The maximum allocation in this cell is min(40,30) = 30 It satisfy demand of D4 and adjust the supply of s, from 40 to 10 (40 30 = 10) Table-5 Supply Row Penalty D1 D2 D4 D3 22(40) 20 47 4(80) 0 24 9(30) 7(30) 24 37 10 S3 20 50 32 32 37 15 0 Demand 60 0 Column8= 32 - 24 Penalty The maximum penalty, 32, occurs in row S3. = 32 The minimum c, in this row is it The maximum allocation in this cell is min(50,60) = 50. 32


The maximum allocation in this cell is min(50,60) = 50. It satisfy supply of S3 and adjust the demand of D from 60 to 10 (60 - 50 10) Table-6 Supply Row Penalty D2 D4 D'3 20 22(40) 480) 47 0 24 7(30) 24 37 9(30) 10 20 32150 37 15 0 Demand 10 0 0 Column 24 Penalty The maximum penalty, 24, occurs in row S, The minimum ci in this row is c21 = 24 The maximum allocation in this cell is min(10,10) = 10 It satisfy supply of s2 and demand of D. Initial feasible solution is D2 D3 D4 Supply Row Penalty -- | S1 22(40) 4(80) 120 13 | 13 17 9(30) 7(30) 24(10) 2| 2| 2| 17| 24 | 24| 37 70 32(50) 37 15 51 5 517|32| -- | 40 Demand 60 110 4 15 8 3 4 8 3 Column 11 8 8 Penalty 8 24 50 ! 20 30 20


The minimum total transportation cost = 22 x 40 +4 x 80 24 x 10 9 x 30 3520 7 x 30 +32 x 50= Here, the number of allocated cells 6 is equal to m n 1 3 4-1 6 This solution is non-degenerate