The document presents an overview of fractional and 0/1 knapsack problems. It defines the fractional knapsack problem as choosing items with maximum total benefit but fractional amounts allowed, with total weight at most W. An algorithm is provided that takes the item with highest benefit-to-weight ratio in each step. The 0/1 knapsack problem requires choosing whole items. Solutions like greedy and dynamic programming are discussed. An example illustrates applying dynamic programming to a sample 0/1 knapsack problem.

1. WELCOME
TO
OUR
PRESENTATION

2. OUR TOPIC IS FRACTIONAL AND 0/1
KNAPSACK.
SUBMITTED BY:
Lithy Ema Rozario-162-15-7989
Imran Hossain-162-15-7672
Tania Maksum-162-15-7698
Shekh Hasibul Islam-162-15-7748
Mohd Nasir Uddin-162-15-7797
Md Fazlur Rahman-162-15-7796
Salowa Binte Sohel-162-15-7820
Submitted to :
Rifat Ara
Shams

3. THE FRACTIONAL KNAPSACK PROBLEM
• Given: A set S of n items, with each item i having
• bi - a positive benefit
• wi - a positive weight
• Goal: Choose items with maximum total benefit but with weight at
most W.
• If we are allowed to take fractional amounts, then this is the fractional
knapsack problem.
• In this case, we let xi denote the amount we take of item i
• Objective: maximize
• Constraint:
Si
iii wxb )/(
ii
Si
i wxWx 
0,

4. EXAMPLE
• Given: A set S of n items, with each item i having
• bi - a positive benefit
• wi - a positive weight
• Goal: Choose items with maximum total benefit but with total weight at
most W.
Weight:
Benefit:
1 2 3 4 5
4 ml 8 ml 2 ml 6 ml 1 ml
$12 $32 $40 $30 $50
Items:
Value:
3($ per ml) 4 20 5 50
10 ml
Solution: P
• 1 ml of 5 50$
• 2 ml of 3 40$
• 6 ml of 4 30$
• 1 ml of 2 4$
•Total Profit:124$
“knapsack”

5. THE FRACTIONAL KNAPSACK ALGORITHM
• Greedy choice: Keep taking item with highest value (benefit to
weight ratio)
• Since
Algorithm fractionalKnapsack(S, W)
Input: set S of items w/ benefit bi and weight wi; max. weight W
Output: amount xi of each item i to maximize benefit w/ weight at most W
for each item i in S
xi  0
vi  bi / wi {value}
w  0 {total weight}
while w < W
remove item i with highest vi
xi  min{wi , W - w}
w  w + min{wi , W - w}
 

Si
iii
Si
iii xwbwxb )/()/(

6. THE FRACTIONAL KNAPSACK ALGORITHM
• Running time: Given a collection S of n items, such that each item i has
a benefit bi and weight wi, we can construct a maximum-benefit subset of
S, allowing for fractional amounts, that has a total weight W in O(nlogn)
time.
• Use heap-based priority queue to store S
• Removing the item with the highest value takes O(logn) time
• In the worst case, need to remove all items

7. KNAPSACK PROBLEM
Given a set of items, each with a mass and a value,
determine the number of each item to include in a
collection so that the total weight is less than or equal
to a given limit and the total value is as large as
possible.
 It derives its name from the problem faced by
someone who is constrained by a fixed-
size knapsack and must fill it with the most valuable
items.

8. KNAPSACK PROBLEM
• In a knapsack problem or rucksack problem,
we are given a set of 𝑛 items, where each item 𝑖
is specified by a size 𝑠𝑖 and a value 𝑣𝑖. We are
also given a size bound 𝑆, the size of our
knapsack.
Item # Size Value
1 1 8
2 3 6
3 5 5

9. KNAPSACK PROBLEM
There are two versions of the problem:
1. 0-1 Knapsack Problem
2. Fractional Knapsack Problem
i. Bounded Knapsack Problem
ii. Unbounded Knapsack Problem

10. SOLUTIONS TO KNAPSACK
PROBLEMS
Greedy Algorithm – keep taking most
valuable items until maximum weight is
reached or taking the largest value of each
item by calculating 𝑉𝑖 =
𝑣𝑎𝑙𝑢𝑒 𝑖
𝑠𝑖𝑧𝑒 𝑖
Dynamic Programming – solve each sub
problem once and store their solutions in
an array

11. EXAMPLE
Given:
𝑛 = 4 (# of elements)
𝑆 = 5 pounds (maximum size)
Elements (size, value) =
{ (1, 200), (3, 240), (2, 140), (5, 150) }

12. GREEDY ALGORITHM
1. Calculate Vi =
vi
si
for 𝑖 = 1,2, … , 𝑛
2. Sort the items by decreasing Vi
3. Find j, such that
𝑠1 + 𝑠2 + ⋯ + 𝑠𝑗 ≤ 𝑆 < 𝑠1 + 𝑠2 + ⋯ + 𝑠𝑗+1

13. GREEDY ALGORITHM
Sample Problem
𝑉𝑖 =
𝑐𝑜𝑠𝑡𝑖
𝑤𝑒𝑖𝑔ℎ𝑡 𝑖
A B C D
cost 200 240 140 150
weight 1 3 2 5
value 200 80 70 30
𝑉𝑖 =
𝑣𝑎𝑙𝑢𝑒 𝑖
𝑠𝑖𝑧𝑒 𝑖
=
𝑐𝑜𝑠𝑡 𝑖
𝑤𝑒𝑖𝑔ℎ𝑡 𝑖

14. GREEDY ALGORITHM
 The optimal solution to the fractional
knapsack
 Not an optimal solution to the 0-1
knapsack

15. DYNAMIC PROGRAMMING
Recursive formula for sub problems:
𝑉 𝑘, 𝑠
=
𝑉 𝑘 − 1, 𝑠 𝑖𝑓𝑠 𝑘 > 𝑠
max 𝑉 𝑘 − 1, 𝑠 , 𝑉 𝑘 − 1, 𝑠 − 𝑠 𝑘 + 𝑣 𝑘 𝑒𝑙𝑠𝑒

16. DYNAMIC PROGRAMMING
for 𝑠 = 0 to 𝑆
V[ 0, s ] = 0
for 𝑖 = 1 to 𝑛
V[𝑖, 0 ] = 0

17. DYNAMIC PROGRAMMING
for 𝑖 = 1 to 𝑛
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else V i, s = V[i − 1, s]

18. EXAMPLE
Given:
𝑛 = 4 (# of elements)
𝑆 = 5 (maximum size)
Elements (size, value) =
{ (2, 3), (3, 4), (4, 5), (5, 6) }

19. EXAMPLE
is 0 1 2 3 4 5
0
1
2
3
4

20. EXAMPLE
for 𝑠 = 0 to 𝑆
V[ 0, 𝑠 ] = 0
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1
2
3
4

21. EXAMPLE
for 𝑖 = 0 to 𝑛
V[𝑖, 0 ] = 0
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0
2 0
3 0
4 0

22. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0
2 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

23. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3
2 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

24. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3
2 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

25. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3
2 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

26. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

27. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

28. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

29. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

30. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

31. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

32. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

33. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝐕 𝐢, 𝒔 = 𝒗𝒊 + 𝑽 𝒊 − 𝟏, 𝒔 − 𝒔𝒊
else
V i, s = V[i − 1, s]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

34. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
𝑉 𝑖, 𝑠 = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
𝑽 𝒊, 𝒔 = 𝑽[𝒊 − 𝟏, 𝒔]
else 𝑉 𝑖, 𝑠 = 𝑉[𝑖 − 1, 𝑠]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

35. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

36. EXAMPLE
if 𝑠𝑖 ≤ 𝑠
if 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖 > 𝑉[𝑖 − 1, 𝑠]
V i, s = 𝑣𝑖 + 𝑉 𝑖 − 1, 𝑠 − 𝑠𝑖
else
V i, s = V[i − 1, s]
else 𝐕 𝐢, 𝒔 = 𝐕[𝐢 − 𝟏, 𝐬]
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

37. EXAMPLE
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝑖 = 𝑖 − 1, 𝑘 = 𝑘 − 𝑠𝑖
else
𝑖 = 𝑖 − 1
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

38. EXAMPLE
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝑖 = 𝑖 − 1, 𝑘 = 𝑘 − 𝑠𝑖
else
𝒊 = 𝒊 − 𝟏
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

39. EXAMPLE
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝑖 = 𝑖 − 1, 𝑘 = 𝑘 − 𝑠𝑖
else
𝒊 = 𝒊 − 𝟏
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

40. EXAMPLE
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝒊 = 𝒊 − 𝟏, 𝒌 = 𝒌 − 𝒔𝒊
else
𝑖 = 𝑖 − 1
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

41. EXAMPLE
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝒊 = 𝒊 − 𝟏, 𝒌 = 𝒌 − 𝒔𝒊
else
𝑖 = 𝑖 − 1
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

42. EXAMPLE
𝑖 = 𝑛, 𝑘 = 𝑆
while 𝑖, 𝑘 > 0
if V 𝑖, 𝑘 ≠ 𝑉 1 − 𝑖, 𝑘
𝑖 = 𝑖 − 1, 𝑘 = 𝑘 − 𝑠𝑖
else
𝑖 = 𝑖 − 1
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)

43. EXAMPLE
The optimal knapsack should
contain {1,2} = 7
is 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 3 3 3 3
2 0 0 3 4 4 7
3 0 0 3 4 5 7
4 0 0 3 4 5 7
Items (𝑠𝑖, 𝑣𝑖)
1: (2, 3)
2: (3, 4)
3: (4, 5)
4: (5, 6)