Knapsack dynamic programming formula top down (1)

0/1 KNAPSACK PROBLEM
DYNAMIC PROGRAMMING
APPROACH
Dr. P. Subathra
Prof/ IT
KAMARAJ College of Engg. & Tech
(AUTONOMOUS)
Madurai
Tamil Nadu
India

2
Dr. P. Subathra, KAMARAJ College of
Engineering & Tech (AUTONOMOUS),
Madurai, Tamil Nadu, India

Dynamic Programming
3

Dynamic Programming – Top Down
4

Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4
5
Dynamic Programming – Top Down

KNAPSACK PROBLEM – Dynamic – Top Down
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4
6
max { F(i-1, Cj), (Vi+F(i-1,Cj-wi)) } ; if Cj-Wi >=0
F(i,Cj) =
F(i-1, Cj) ; if Cj-Wi = 0

F(i,Cj) = max { F(i-1, Cj), (Vi+F(i-1,Cj-wi)) }
F(4,5)
7
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(4,5)=max { F(4-1, 5), (15+F(4-1,5-2)) }
8
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(4,5)=max { F(4-1, 5), (15+F(4-1,5-2)) }
= max{F(3,5),(15+F(3,3)}
9
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(4,5)= max{F(3,5),(15+F(3,3)}
10
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,5)=max { F(3-1, 5), (15+F(3-1,5-3)) }
11
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,5)=max { F(3-1, 5), (15+F(3-1,5-3)) }
= max{F(2,5),(15+F(2,2)}
12
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,5)= max{F(2,5),(15+F(2,2)}
13
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,3)=max { F(3-1, 3), (20+F(3-1,3-3)) }
14
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,3)=max { F(3-1, 3), (20+F(3-1,3-3)) }
= max{F(2,3),(20+F(2,0)}
15
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(3,3) = max{F(2,3),(20+F(2,0)}
16
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,5)=max { F(2-1, 5), (10+F(2-1,5-1)) }
17
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,5)=max { F(2-1, 5), (10+F(2-1,5-1)) }
= max{F(1,5),(10+F(1,4)}
18
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,5) = max{F(1,5),(10+F(1,4)}
19
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,3)=max { F(2-1, 3), (10+F(2-1,3-1)) }
20
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,3)=max { F(2-1, 3), (10+F(2-1,3-1)) }
= max{F(1,3),(10+F(1,2)}
21
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,3)= max{F(1,3),(10+F(1,2)}
22
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,2)=max { F(2-1, 2), (10+F(2-1,2-1)) }
23
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,2)=max { F(2-1, 2), (10+F(2-1,2-1)) }
= max{F(1,2),(10+F(1,1)}
24
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,2) = max{F(1,2),(10+F(1,1)}
25
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,2)= max{F(1,2),(10+F(1,1)}
26
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,0)=max { F(2-1, 0), (10+F(2-1,0-1)) }
27
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,0)=max { F(2-1, 0), (10+F(2-1,0-1)) }
= max{F(1,0),(10+F(1,-1)} =max{F(1,0),10+0}
28
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,0)=max { F(2-1, 0), (10+F(2-1,0-1)) }
= max{F(1,0),(10+F(1,-1)} =max{F(1,0),10+0}
29
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(2,0)=max { F(2-1, 0), (10+F(2-1,0-1)) }
= max{F(1,0),(10+F(1,-1)} =max{F(1,0),10+0}
30
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0
12 2 1
10 1 2
20 3 3
15 2 4

F(0,5)=max { F(0-1, 5), (12+F(0-1,5-2)) }
= max{0,(12+0} =max{F(0,12} =12
31
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0 12
12 2 1
10 1 2
20 3 3
15 2 4

F(0,5)=12
32
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0 12
12 2 1
10 1 2
20 3 3
15 2 4

F(0,4)=max { F(2-1, 0), (10+F(2-1,0-1)) }
= max{F(1,0),(10+F(1,-1)} =max{F(1,0),10+0}
33
Value/
Profit
Of Item
(Vi)
Weight
of Item
(Wi)
Capacity
of Bag
(Cj)
Item
No. (i)
0 1 2 3 4
5
(Max.)
0 0 0 12
12 2 1
10 1 2
20 3 3
15 2 4

KNAPSACK PROBLEM – Dynamic – Top Down
36
Items Selected : 1, 2 & 4
Total Cost = 12+10+15
= 37
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Knapsack dynamic programming formula top down (1)

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