Матеріали курсу SY CE/CSE ISE even 2026 та відповіді на тести | moodle.spit.ac.in

Consider the

0/1 Knapsack Problem with knapsack capacity m=15. The four items having weights

as (w

₁,w₂,w₃,w₄

)=(2,4,6,9) and

profits (p

₁,p₂,p₃,p₄

)=(10,10,12,18).

The cost

c and upper bound u of an intermediate node in

one of the step solving the 0/1 Knapsack Problem using LC Branch and Bound are

-38 and -38 respectively. The cost

c and upper bound u of next level of nodes in

State Space Tree are

First

Node (c,u)=(-36,-36) and

Second Node (c,u)=(-18,-18)

25%

First

Node (c,u)=(-24,-30) and

Second Node (c,u)=(-32,-32)

50%

First Node (c,u)=(-38,-38) and Second Node (c,u)=(-20,-20)

First

Node (c,u)=(-28,-28) and

Second Node (c,u)=(-22,-22)

25%

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