Question

A.

₹ 6060

B.

₹ 6660

C.

₹ 6000

D.

₹ 6600

Correct option is A

Introduction:

The correct answer is 1. ₹ 6060.

In Operations Research, the Vogel's Approximation Method (VAM) is a heuristic framework used to find an Initial Feasible Solution (IFS) to a transportation problem. It operates by minimizing the penalty cost resulting from a failure to allocate units to the lowest-cost cells. By systemically computing row and column penalties (the difference between the two lowest costs), identifying the maximum penalty, and making strategic allocations to the least expensive cell in that row or column, we achieve an optimized base distribution layout.  yields an optimal initial transportation cost of ₹ 6060.

Step-by-Step VAM Allocation Process:

Step 1: Calculate Initial Penalties

  • Row Penalties:

    • F1:1612=4F_1: 16 - 12 = 4​​

    • F2:148=6F_2: 14 - 8 = 6​​

    • F3:2416=8F_3: 24 - 16 = 8​​

  • Column Penalties:

    • W1:1614=2W_1: 16 - 14 = 2​​

    • W2:208=12W_2: 20 - 8 = 12​​

    • W3:1612=4W_3: 16 - 12 = 4​​

Step 2: First Allocation

  • The maximum penalty is 12 in column W2W_2​. The cell with the lowest cost in W2W_2​ is (F2,W2)(F_2, W_2)​ with a cost of 8.

  • Allocate min(Supply=160,Demand=130)=130\min(\text{Supply}=160, \text{Demand}=130) = 130​ units to (F2,W2)(F_2, W_2)​.

  • Demand for W2W_2​ is fully satisfied (130130=0130 - 130 = 0​). Remaining supply at F2=160130=30F_2 = 160 - 130 = 30​. Cross out column W2W_2​.

Step 3: Second Allocation (Recalculate Penalties for remaining $W_1, W_3$)

  • Row Penalties: F1=1612=4F_1 = 16 - 12 = 4​; F2=1814=4F_2 = 18 - 14 = 4​; F3=2616=10F_3 = 26 - 16 = 10​.

  • Column Penalties: W1=1614=2W_1 = 16 - 14 = 2​; W3=1612=4W_3 = 16 - 12 = 4​.

  • The maximum penalty is 10 in row F3F_3​. The lowest cost cell in F3F_3​ is (F3,W3)(F_3, W_3)​with a cost of 16.

  • Allocate min(Supply=100,Demand=150)=100\min(\text{Supply}=100, \text{Demand}=150) = 100​ units to (F3,W3)(F_3, W_3)​.

  • Supply at F3F_3​ is exhausted (100100=0100 - 100 = 0​). Remaining demand for W3=150100=50W_3 = 150 - 100 = 50​. Cross out row F3F_3​.

Step 4: Third Allocation (Recalculate Penalties for rows $F_1, F_2$ and columns $W_1, W_3$)

  • Row Penalties: F1=1612=4;F2=1814=4F_1 = 16 - 12 = 4; F_2 = 18 - 14 = 4​.

  • Column Penalties: W1=1614=2;W3=1812=6W_1 = 16 - 14 = 2; W_3 = 18 - 12 = 6​.

  • The maximum penalty is 6 in column $W_3$. The lowest cost cell in $W_3$ is $(F_1, W_3)$ with a cost of 12.

  • Allocate min(Supply=200,Demand=50)=50\min(\text{Supply}=200, \text{Demand}=50) = 50​ units to (F1,W3)(F_1, W_3)​.

  • Demand for W3W_3​ is fully satisfied (5050=050 - 50 = 0​). Remaining supply at F1=20050=150F_1 = 200 - 50 = 150​. Cross out column W3W_3​.

Step 5: Final Allocations (Only Column $W_1$ remains)

  • For cell (F1,W1)(F_1, W_1)​ with cost 16: Allocate remaining supply of F1=150F_1 = 150​ units. (Remaining demand of W1=180150=30W_1 = 180 - 150 = 30​).

  • For cell (F2,W1)(F_2, W_1)​ with cost 14: Allocate remaining supply of F2=30F_2 = 30​ units. (Demand of W1W_1​ becomes 3030=030 - 30 = 0​).

Step 6: Total Cost Calculation

Total Cost=(130×8)+(100×16)+(50×12)+(150×16)+(30×14)\text{Total Cost} = (130 \times 8) + (100 \times 16) + (50 \times 12) + (150 \times 16) + (30 \times 14)​​
Total Cost=1040+1600+600+2400+420=6060\text{Total Cost} = 1040 + 1600 + 600 + 2400 + 420 = ₹ 6060​​
Information Booster:
  • Vogel's Approximation Method (VAM): Recognized as the most efficient heuristic method for obtaining an initial feasible solution because it yields a starting cost very close to (and sometimes matching) the absolute optimal solution.

  • The Concept of Penalty: The penalty represents the economic opportunity loss or opportunity cost incurred if the allocator misses out on the best cell and is forced to use the next best available alternative path.

  • Feasibility Check Rule: For a transportation problem to have a non-degenerate basic feasible solution, the total number of occupied allocations must exactly equal m+n1m + n - 1​, where $m$ is the number of origins (rows) and $n$ is the number of destinations (columns). In this matrix, 3+31=53 + 3 - 1 = 5​ allocations, which perfectly satisfies the requirement.

Additional Knowledge:
  • Why Option 2 is incorrect: ₹ 6660 is an over-calculated value that typically arises if a student mistakenly allocates units across columns or rows using a generic structural matrix loop without properly dropping completed constraint headers.

  • Why Option 3 is incorrect: ₹ 6000 is an artificial mathematical rounding floor trap. VAM relies on precise multiplicative products of cell coefficients and structural units, which do not break down cleanly into an exact flat thousands marker here.

  • Why Option 4 is incorrect: ₹ 6600 represents a typical value obtained if an aspirant uses the simpler Least Cost Method (LCM) or North-West Corner Method (NWCM), which do not incorporate penalty opportunity evaluations and consequently generate less cost-efficient initial solutions than VAM.

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