Fractional Knapsack problem is defined as, “Given a set of items having some weight and value/profit associated with it. The knapsack problem is to find the set of items such that the total weight is less than or equal to a given limit (size of knapsack) and the total value/profit earned is as large as possible.”

Knapsack problem has two variants.

**Binary or 0/1 knapsack :**Item cannot be broken down into parts.**Fractional knapsack :**Item can be divided into parts.

Given a set of items, each having some weight and value/profit associated with it. The goal is to find the set of items such that the total weight is less than or equal to a given limit (size of knapsack) and the total value/profit earned is as large as possible.

- The knapsack is an optimization problem and it is useful in solving resource allocation problem.
- Let X = <x
_{1}, x_{2}, x_{3}, . . . . . , x_{n}> is the set of n items. W = <w_{1}, w_{2}, w_{3}, . . . , w_{n}> and V = <v_{1}, v_{2}, v_{3}, . . . , v_{n}> are the set of weight and value associated with each items in x, respectively. Knapsack capacity is M. - Select items one by one from the set of items x and fill the knapsack such that it would maximize the value. Knapsack problem has two variants. 0/1 knapsack does not allow breaking of items. Either add entire item in a knapsack or reject it. It is also known as a binary knapsack. Fractional knapsack allows the breaking of items. So profit will also be considered accordingly.

- Knapsack problem can be formulated as follow :

Maximize \sum_{i=1}^{n} v_i x_i subjected to \sum_{i=1}^{n} v_i x_i \le M

x_i \in (0, 1) for binary knapsack

x_i \in [0, 1] for fractional knapsack

## Algorithm

```
Algorithm GREEDY_FRACTIONAL_KNAPSACK(X, V, W, M)
// Description : Solve the knapsack problem using greedy approach
// Input:
X: An array of n items
V: An array of profit associated with each item
W: An array of weight associated with each item
M: Capacity of knapsack
// Output :
SW: Weight of selected items
SP: Profit of selected items
// Items are presorted in decreasing order of pi = vi / wi ratio
S ← Φ // Set of selected items, initially empty
SW ← 0 // weight of selected items
SP ← 0 // profit of selected items
i ← 1
```**while** i ≤ n **do**
**if** (SW + w[i]) ≤ M **then**
S ← S ∪ X[i]
SW ← SW + W[i]
SP ← SP + V[i]
**else**
frac ← (M - SW) / W[i]
S ← S ∪ X[i] * frac // Add fraction of item X[i]
SP ← SP + V[i] * frac // Add fraction of profit
SW ← SW + W[i] * frac // Add fraction of weight
**end**
i ← i + 1
**end**

## Complexity Analysis

- For one item there are two choices, either to select or reject. For 2 items we have four choices:
- Select both items

- Reject both items

- Select first and reject second

- Reject first and select second

- In general, for n items, knapsack has 2
^{n}choices. So brute force approach runs in O(2^{n}) time. - We can improve performance by sorting items in advance. Using merge sort or heap sort, n items can be sorted in O(nlog
_{2}n) time. Merge sort and heap sort are non-adaptive and their running time is the same in best, average and worst case. - To select the items, we need one scan to this sorted list, which will take O(n) time.
- So the total time required is

T(n) = O(nlog_{2}n) + O(n) = O(nlog_{2}n).

## Examples of Fractional Knapsack

**Problem: Consider the following instances of the fractional knapsack problem: n = 3, M = 20, V = (24, 25, 15) and W = (18, 15, 20) find the feasible solutions.**

**Solution:**

Let us arrange items by decreasing order of profit density. Assume that items are labeled as X = (I_{1}, I_{2}, I_{3}), have profit V = {24, 25, 15} and weight W = {18, 15, 20}.

Item (x_{i}) | Value (v_{i}) | Weight (w_{i}) | p_{i} = v_{i} / w_{i} |

I_{2} | 25 | 15 | 1.67 |

I_{1} | 24 | 18 | 1.33 |

I_{3} | 15 | 20 | 0.75 |

We shall select one by one item from Table. If the inclusion of an item does not cross the knapsack capacity, then add it. Otherwise, break the current item and select only the portion of item equivalent to remaining knapsack capacity. Select the profit accordingly. We should stop when knapsack is full or all items are scanned.

Initialize, Weight of selected items, SW = 0,

Profit of selected items, SP = 0,

Set of selected items, S = { },

Here, Knapsack capacity M = 20.

**Iteration 1 :** SW= (SW + w_{2}) = 0 + 15 = 15

SW ≤ M, so select I_{2}

S = { I_{2} }, SW = 15, SP = 0 + 25 = 25

**Iteration 2 :** SW + w_{1 }> M, so break down item I_{1.}

The remaining capacity of the knapsack is 5 unit, so select only 5 units of item I_{1}.

*frac* = (M – SW) / W[i] = (20 – 15) / 18 = 5 / 18

S = { I_{2, }I_{1 }* 5/18 }

SP = SP + v_{1} * *frac* = 25 + (24 * (5/18)) = 25 + 6.67 = 31.67

SW = SW + w_{1} * *frac* = 15 + (18 * (5/18)) = 15 + 5 = 20

The knapsack is full. Fractional Greedy algorithm selects items { I_{2, }I_{1 }* 5/18 }, and it gives a profit of 31.67 units.

**Problem: Find the optimal solution for knapsack problem (fraction) where knapsack capacity = 28, P = {9, 5, 2, 7, 6, 16, 3} and w = {2, 5, 6, 11, 1, 9, 1}.**

**Solution:**

Arrange items in decreasing order of profit to weight ratio

Item | Profit p_{i} | Weight w_{i} | Ratio v_{i}/w_{i} |

I_{5} | 6 | 1 | 6.00 |

I_{1} | 9 | 2 | 4.50 |

I_{7} | 3 | 1 | 3.00 |

I_{6} | 16 | 9 | 1.78 |

I_{2} | 5 | 5 | 1.00 |

I_{4} | 7 | 11 | 0.64 |

I_{3} | 2 | 6 | 0.33 |

Initialize, Weight = 0, P = 0, M = 28, S = { }

Where S is the solution set, P and W is profit and weight of included items, respectively. M is the capacity of the knapsack.

**Iteration 1**

(Weight + w_{5}) ≤ M, so select I_{5}

So, S = { I_{5} }, Weight = 0 + 1 = 1, P = 0 + 6= 6

**Iteration 2**

(Weight + w_{1}) ≤ M, so select I_{1}

So, S = {I_{5} ,I_{1} }, Weight = 1 + 2 = 3, P = 6 + 9= 15

**Iteration 3**

(Weight + w_{7}) ≤ M, so select I_{7}

So, S = {I_{5}, I_{1}, I_{7} }, Weight = 3 + 1 = 4, P = 15 + 3= 18

**Iteration 4**

(Weight + w_{6}) ≤ M, so select I_{6}

So, S = {I_{5}, I_{1}, I_{7}, I_{6} }, Weight = 4 + 9 = 13, P = 18 + 16= 34

**Iteration 5**

(Weight + w_{2}) ≤ M, so select I_{2}

So, S = {I_{5}, I_{1}, I_{7}, I_{6}, I_{2} }, Weight = 13 + 5 = 18, P = 34 + 5= 39

**Iteration 6**

(Weight + w_{4}) > M, So I_{4} must be broken down into two parts x and y such that x = capacity left in knapsack and y = I_{4} – x.

Available knapsack capacity is 10 units. So we can select only (28 – 18) / 11 = 0.91 unit of I_{4}.

So S = {I_{5}, I_{1}, I_{7}, I_{6}, I_{2}, 0.91 * I_{4} }, Weight = 18 + 0.91*11 = 28, P = 39 + 0.91 * 7= 45.37

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