Q.For the same transportation problem in Worked Examples 1 and 2, find the initial basic feasible solution using Vogel's Approximation Method (VAM), and prepare a table comparing the total cost from all three IBFS methods.

VAM allocates first in the row or column with the largest penalty (the cost of missing its own cheapest cell), rather than simply picking the cheapest cell in the whole table, so it typically lands closest to the optimal cost of the three IBFS methods.

Using the same cost matrix and supply/demand as Worked Examples 1–2:

Round 1 — penalties: Row $O_1$: $8-6=2$; Row $O_2$: $11-7=4$; Row $O_3$: $5-4=1$; Column $D_1$: $6-4=2$; Column $D_2$: $8-5=3$; Column $D_3$: $11-10=1$. Largest is 4 (row $O_2$); its lowest cost is $(O_2,D_1)=7$: allocate $\min(50,25)=25$. $D_1$ exhausted, $O_2$ has 25 left.

Round 2: Row $O_1$: $10-8=2$; Row $O_2$: $11-11=0$; Row $O_3$: $12-5=7$; Column $D_2$: $8-5=3$; Column $D_3$: $11-10=1$. Largest is 7 (row $O_3$); its lowest cost is $(O_3,D_2)=5$: allocate $\min(20,35)=20$. $O_3$ exhausted, $D_2$ has 15 left.

Round 3: Row $O_1$: $10-8=2$; Row $O_2$: $11-11=0$; Column $D_2$: $11-8=3$; Column $D_3$: $11-10=1$. Largest is 3 (column $D_2$); its lowest cost is $(O_1,D_2)=8$: allocate $\min(30,15)=15$. $D_2$ exhausted, $O_1$ has 15 left.

Round 4: only $D_3$ remains; $(O_1,D_3)=\min(15,40)=15$, then $(O_2,D_3)=\min(25,25)=25$.

Allocation table:

Feasible, 5 occupied cells $=m+n-1$.

Total cost:

Comparison table:

Recompute the row and column penalties afresh after every single allocation (not just once at the start) — a penalty calculated from an earlier round can become wrong once a row or column has been partly or fully used up.

Computing penalties only for rows and forgetting columns (or vice versa); and forgetting to recompute all penalties after each allocation, which can lead to allocating in a row/column that no longer has the largest penalty.