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Worked Examples · Example 1
Q.

A firm has three factories O1,O2,O3O_1, O_2, O_3 with per-unit transportation costs to three warehouses D1,D2,D3D_1, D_2, D_3 as given below. Supplies are 30, 50 and 20 units respectively, and demands at D1,D2,D3D_1, D_2, D_3 are 25, 35 and 40 units respectively.

D1D_1D2D_2D3D_3
O1O_16810
O2O_271111
O3O_34512

Find the initial basic feasible solution by the North-West Corner Method and compute the total transportation cost.

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Total supply =30+50+20=100=30+50+20=100 and total demand =25+35+40=100=25+35+40=100, so the problem is already balanced.

Starting at the north-west corner:

  1. (O1,D1)(O_1,D_1): allocate min⁡(30,25)=25\min(30,25)=25; D1D_1 done, O1O_1 has 5 left.
  2. (O1,D2)(O_1,D_2): allocate min⁡(5,35)=5\min(5,35)=5; O1O_1 done, D2D_2 has 30 left.
  3. (O2,D2)(O_2,D_2): allocate min⁡(50,30)=30\min(50,30)=30; D2D_2 done, O2O_2 has 20 left.
  4. (O2,D3)(O_2,D_3): allocate min⁡(20,40)=20\min(20,40)=20; O2O_2 done, D3D_3 has 20 left.
  5. (O3,D3)(O_3,D_3): allocate min⁡(20,20)=20\min(20,20)=20; both exhausted together.

Allocation table:

D1D_1D2D_2D3D_3Row total
O1O_1255—30
O2O_2—302050
O3O_3——2020
Column total253540100

Row totals (30, 50, 20) match the supplies exactly and column totals (25, 35, 40) match the demands exactly, so the allocation is feasible; 5 occupied cells =m+n−1=3+3−1=m+n-1=3+3-1, confirming a non-degenerate solution.

Total cost:

Z=25(6)+5(8)+30(11)+20(11)+20(12)=150+40+330+220+240=980Z=25(6)+5(8)+30(11)+20(11)+20(12)=150+40+330+220+240=980

✓Final answer

Total transportation cost by NWC = ₹980

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