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

Vogel's Approximation Method (VAM)

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Vogel's Approximation Method (VAM)

Vogel's Approximation Method (VAM)

Vogel's Approximation Method is the most reliable of the three IBFS methods because it uses a penalty — the extra cost of not choosing the cheapest cell in a row or column — to decide where to allocate next, rather than looking only at the single cheapest cell in the whole table.

Rule (repeated until all rows/columns are exhausted):

  1. For every remaining row and every remaining column, compute the penalty = (second-lowest cost) − (lowest cost) in that row/column.
  2. Select the row or column with the largest penalty.
  3. In that row/column, allocate the maximum possible to its lowest-cost cell.
  4. Cross out the row or column that is now exhausted, and recompute penalties for what remains.

Worked example — same cost matrix as the earlier examples

D1D_1D2D_2D3D_3Supply
O1O_1681030
O2O_27111150
O3O_3451220
Demand253540100

Round 1 — penalties: Row O1O_1: 8−6=28-6=2. Row O2O_2: 11−7=411-7=4. Row O3O_3: 5−4=15-4=1. Column D1D_1: 6−4=26-4=2. Column D2D_2: 8−5=38-5=3. Column D3D_3: 11−10=111-10=1.

Largest penalty is 4 (row O2O_2). Lowest cost in that row is (O2,D1)=7(O_2,D_1)=7: allocate min⁡(50,25)=25\min(50,25)=25. D1D_1 is exhausted; O2O_2 has 25 units left.

Round 2 (remaining columns D2,D3D_2,D_3; supplies O1=30,O2=25,O3=20O_1=30,O_2=25,O_3=20) — penalties: Row O1O_1: 10−8=210-8=2. Row O2O_2: 11−11=011-11=0. Row O3O_3: 12−5=712-5=7. Column D2D_2: 8−5=38-5=3. Column D3D_3: 11−10=111-10=1.

Largest penalty is 7 (row O3O_3). Lowest cost in that row is (O3,D2)=5(O_3,D_2)=5: allocate min⁡(20,35)=20\min(20,35)=20. O3O_3 is exhausted; D2D_2 has 15 units left.

Round 3 (remaining D2=15,D3=40D_2=15,D_3=40; supplies O1=30,O2=25O_1=30,O_2=25) — penalties: Row O1O_1: 10−8=210-8=2. Row O2O_2: 11−11=011-11=0. Column D2D_2: 11−8=311-8=3. Column D3D_3: 11−10=111-10=1.

Largest penalty is 3 (column D2D_2). Lowest cost in that column is (O1,D2)=8(O_1,D_2)=8: allocate min⁡(30,15)=15\min(30,15)=15. D2D_2 is exhausted; O1O_1 has 15 units left.

Round 4 (only D3D_3 left; O1=15,O2=25O_1=15,O_2=25, D3=40D_3=40): allocate (O1,D3)=min⁡(15,40)=15(O_1,D_3)=\min(15,40)=15, then (O2,D3)=min⁡(25,25)=25(O_2,D_3)=\min(25,25)=25.

Allocation table (VAM)

D1D_1D2D_2D3D_3Row total
O1O_1—151530
O2O_225—2550
O3O_3—20—20
Column total253540100

Feasible, with 5 occupied cells (=m+n−1=m+n-1).

Total cost

ZVAM=25(7)+20(5)+15(8)+15(10)+25(11)Z_{VAM}=25(7)+20(5)+15(8)+15(10)+25(11)

=175+100+120+150+275=₹820=175+100+120+150+275=₹820

Comparing the three methods on the identical cost matrix …

Definition 1Penalty (in VAM)

For a row or column, the difference between its two lowest unit costs; VAM always allocates next in the row or column with the largest penalty, at that …