Skip to content

Business Mathematics and Statistics · Ch 10 — Operations Research (Transportation Problem, Assignment Problems, Decision Theory)

Least Cost (Matrix Minima) Method

3

Least Cost (Matrix Minima) Method

Least Cost (Matrix Minima) Method

The Least Cost Method (also called the Matrix Minima Method) improves on NWC by allocating to the cheapest cells first, so the starting solution is usually closer to optimal.

Rule: Find the cell with the lowest unit cost in the whole table (breaking a tie in favour of the cell that allows the larger allocation). Allocate the maximum possible there. Cross out the row or column whose supply/demand is now exhausted. Repeat on the remaining table until all supply and demand is used up.

Worked example — same cost matrix as the North-West Corner example

D1D_1D2D_2D3D_3Supply
O1O_1681030
O2O_27111150
O3O_3451220
Demand253540100

Step-by-step allocation:

  1. The lowest cost in the table is 4 at (O3,D1)(O_3,D_1): allocate min⁡(20,25)=20\min(20,25)=20. O3O_3 is exhausted; D1D_1 has 5 units left.
  2. Among the remaining cells, the lowest cost is 6 at (O1,D1)(O_1,D_1): allocate min⁡(30,5)=5\min(30,5)=5. D1D_1 is exhausted; O1O_1 has 25 units left.
  3. The next lowest remaining cost is 8 at (O1,D2)(O_1,D_2): allocate min⁡(25,35)=25\min(25,35)=25. O1O_1 is exhausted; D2D_2 has 10 units left.
  4. Only the O2O_2 row remains, against D2D_2 (10 left) and D3D_3 (40). The lower of its two costs is 11 at (O2,D2)(O_2,D_2) (tied with (O2,D3)(O_2,D_3), resolved here in favour of clearing the smaller remaining column first): allocate min⁡(50,10)=10\min(50,10)=10. D2D_2 is exhausted; O2O_2 has 40 units left.
  5. Final cell (O2,D3)(O_2,D_3): allocate min⁡(40,40)=40\min(40,40)=40.

Allocation table (LCM)

D1D_1D2D_2D3D_3Row total
O1O_1525—30
O2O_2—104050
O3O_320——20
Column total253540100
Definition 1Least Cost / Matrix Minima Method

An IBFS method that repeatedly allocates to the cheapest available cell in the table, normally giving a starting solution cheaper than the No …