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Stepping Stone

The document outlines the steps of Vogel’s Approximation Method (VAM) for solving a transportation problem. It details the initial feasible solution, cost calculations, and the Stepping Stone Method for optimality checks, leading to an improved allocation with a new total cost of $560. The process involves calculating penalties, opportunity costs, and adjusting allocations to minimize transportation costs.

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Srihari Bhadabhagni
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74 views2 pages

Stepping Stone

The document outlines the steps of Vogel’s Approximation Method (VAM) for solving a transportation problem. It details the initial feasible solution, cost calculations, and the Stepping Stone Method for optimality checks, leading to an improved allocation with a new total cost of $560. The process involves calculating penalties, opportunity costs, and adjusting allocations to minimize transportation costs.

Uploaded by

Srihari Bhadabhagni
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Step 1: Vogel’s Approximation Method (VAM)

Given Cost Matrix

Factories → / Warehouses ↓ W1 W2 W3 Supply


F1 4 8 5 20
F2 6 10 7 30
F3 5 9 6 50
Demand 30 40 30

Step 1.1: Calculate Penalties (Row & Column Differences)

Initial Feasible Solution (VAM)

Factories → / Warehouses ↓ W1 W2 W3 Supply


F1 20 0 0 20 → 0
F2 0 30 0 30 → 0
F3 10 10 30 50 → 0
Demand 30 → 0 40 → 0 30 → 0

Initial Cost Calculation

(20×4)+(10×5)+(30×6)+(10×9)+(30×6)=80+50+180+90+180=580(20 \times 4) + (10 \times 5) +


(30 \times 6) + (10 \times 9) + (30 \times 6) = 80 + 50 + 180 + 90 + 180 =
580(20×4)+(10×5)+(30×6)+(10×9)+(30×6)=80+50+180+90+180=580

Initial Cost = $580

Step 2: Stepping Stone Method (Optimality Check)


We check opportunity costs for empty cells.

Check (F1, W2)

 Closed loop: (F1 → W1) → (F3 → W1) → (F3 → W2) → (F1 → W2)
 Opportunity Cost: 8−4+5−9=08 - 4 + 5 - 9 = 08−4+5−9=0

Check (F1, W3)

 Closed loop: (F1 → W1) → (F3 → W1) → (F3 → W3) → (F1 → W3)
 Opportunity Cost: 5−4+5−6=05 - 4 + 5 - 6 = 05−4+5−6=0

Check (F2, W1)


 Closed loop: (F2 → W2) → (F3 → W2) → (F3 → W1) → (F2 → W1)
 Opportunity Cost: 6−9+5−6=−46 - 9 + 5 - 6 = -46−9+5−6=−4

Negative cost (-4), so improvement is required!

Step 3: Improve the Solution

 Loop path:
(F2 → W1) (+X),
(F3 → W1) (-X),
(F3 → W2) (+X),
(F2 → W2) (-X)
 Minimum allocation: X = min(10, 10) = 10.

New Allocations:

Factories → / Warehouses ↓ W1 W2 W3 Supply

F1 20 0 0 20

F2 10 20 0 30

F3 0 20 30 50

Demand 30 40 30

Step 4: Recalculate Total Cost

(20×4)+(10×6)+(20×9)+(20×6)+(30×6)=80+60+180+120+180=560(20 \times 4) + (10 \times 6) + (20 \


times 9) + (20 \times 6) + (30 \times 6) = 80 + 60 + 180 + 120 + 180 =
560(20×4)+(10×6)+(20×9)+(20×6)+(30×6)=80+60+180+120+180=560

✅ New total cost = $560 ✅

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