Knapsack problem of Greedy method [ understand with real life example].
Knapsack problem of Greedy method [ understand with real life example].
Contents
- 1 🎒 Knapsack Problem using Greedy Method — with Real-Life Example
- 2 ✅ What is the Knapsack Problem?
- 3 ⚙️ Greedy Method (Fractional Knapsack)
- 4 📌 Real-Life Example: Emergency Relief Backpack
- 5 🧠 Step-by-step Greedy Method:
- 6 🔄 If only 2 kg space left for Food Pack?
- 7 🧾 Why Greedy Works Here?
- 8 ❌ Where Greedy Fails?
- 9 💡 Summary:
- 10 Knapsack problem of Greedy method [ understand with real life example].
- 11 Greedy solution method for knapsack problems with R
🎒 Knapsack Problem using Greedy Method — with Real-Life Example
✅ What is the Knapsack Problem?
It’s a classic optimization problem where:
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You have a bag (knapsack) that can carry up to a certain weight/capacity.
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You have items, each with a weight and a value.
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The goal is to maximize total value without exceeding the capacity.
⚙️ Greedy Method (Fractional Knapsack)
In the Greedy approach, you:
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Choose items with the highest value per unit weight (value/weight ratio) first.
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You can take fractions of items (not whole only).
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It’s optimal for fractional knapsack, not for 0/1 version.
📌 Real-Life Example: Emergency Relief Backpack
Imagine you’re packing an emergency relief bag during a disaster. The bag can hold up to 15 kg. You have these items:
Item | Weight (kg) | Value (Importance) | Value/Weight |
---|---|---|---|
First Aid Kit | 3 | 60 | 20.0 |
Bottled Water | 5 | 100 | 20.0 |
Food Pack | 7 | 140 | 20.0 |
Flashlight | 2 | 30 | 15.0 |
🧠 Step-by-step Greedy Method:
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Sort items by value/weight ratio:
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All have 20.0, except flashlight (15.0)
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Start filling the bag with items of highest ratio:
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Take First Aid Kit (3 kg) → total weight = 3 kg, value = 60
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Take Bottled Water (5 kg) → total weight = 8 kg, value = 160
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Take Food Pack (7 kg) → total weight = 15 kg (max), value = 300
✅ Total weight used = 15 kg
✅ Total value gained = 300
(No room for Flashlight)
🔄 If only 2 kg space left for Food Pack?
Take 2/7 of it:
→ Value = (2/7) × 140 = 40
🧾 Why Greedy Works Here?
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Fractional choices are allowed.
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Maximizing value per kg works best when dividing is allowed.
❌ Where Greedy Fails?
In 0/1 Knapsack (items can’t be split), greedy may not give the best solution. There, Dynamic Programming is better.
💡 Summary:
📌 Concept | ✅ Greedy Approach Works |
---|---|
Fractional items | ✅ Yes |
0/1 items | ❌ Not always optimal |
Real-life fit | ✅ Packing, resource loading |
Would you like:
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A visual chart of this example?
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Practice problems with solutions?
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A version for 0/1 knapsack?