Knapsack Problem - Introduction to Operations Research - Lecture Slides, Slides of Operational Research

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Knapsack Problem

• There are many other versions

max

x n n N n

Z = w x

= ∑ 1

v x V x

n n N n n = ∑

1

Interpretation

• n = 1,2,..., N - Item-type
• V = Volume of knapsack
• v n

= Volume of an item of type n

• w n

= Weight of an item of type n

• x n

= Number of items of type n put in the

knapsack

DP Strategy

• Let,

f(s):= maximum weight of knapsack

with available volume of s units,

s=0,1,2,...,V.

• We are interested in f(V).
• Clearly, f(0)=
• Clearly, f(s) =0 for all s smaller than

the smallest item (volume-wise).

• Clearly, if f(s)>0, then f(s) = f(s-v

n

) + w

n

for some item-type n.

• Obvious question:

What is the best n?

• Answer:

The n that maximizes f(s-v

n

) + w

n

(NILN)

V

s

n

f(s) =?

(NILN)

V

s

n

s-v

n

f(s) = f(s-v

n

) + w

n

V

s

n

f(s) =?

Notation

• As we solve the functional equation

for s=0,1,2,...,V - in this order (why?)

we store the optimal decisions. ie.

N ( s ) := n * : f ( s ) = w

n *

+ f ( s − v

n *

• Since there is no item-type with v n

we introduce a dummy variable n=4,

with w

4

=0 and v

4

n 1 2 3

w

n

v

n

9.2.2 Example

Procedure

• f(0)= f (^) { w f v (^) } { w f v (^) } w f v f n v n n n n n n ( ) max ( ) max ( ) ( ) ( ) { } 1 1 1 1 0 0 0 1 4 = + − = + − = 4 + − 4 = + = ≤ ∈

N(1)={4}

f (^) { w f v (^) } { w f v (^) } w f v f n v n n n n n n ( ) max ( ) max ( ) ( ) ( ) { } 2 2 2 2 0 1 0 2 4 = + − = + − = 4 + − 4 = + = ≤ ∈

N(2)={4}

n 1 2 3 4

wn

vn

{ } { } f w f v w f v w f v w f v n v n n n n n n ( ) max ( ) max ( ) max { ( ), ( )} max { , } { , } 3 3 3 3 3 5 0 0 0 5 3 2 4 2 2 4 4 = + − = + − = + − + − = + + = ≤ ∈

N(3)={2}

f (5) = max n vn ≤ 5 w n

• f (5 − v n { )} =^ max n ∈{2,3, 4} w n
• f (5 − v n { )} = max { w 2
• f (5 − v 2 ), w 3
• f (5 − v 3 ), w 4
• f (5 − v 4

= max {5 + f (2),6 + f (1),0 + f (4)} = max{5 + 0,6 + 0, 0 + 6} = 6

N(5)={3,4}

Recovery of Optimal Solution

• Using the table we select the optimal items

from N(s) and then go to N(s-v

n

• x (0)

=(0,0,0,0), s=V=

• x (1)

=(0,0,1,0) (N(13)={2,3}), next s = s-v

3

• x (2)

=(0,1,1,0) (N(9)={2}, next s = s-v

2

• x (3)

=(0,2,1,0) (N(6)={2}, next s = s-v

2

• x (4)

=(0,3,1,0) (N(3)={2}, next s = s-v

2

• x=(0,3,1,0)*