State-Space Tree Knapsack Problem | Branch & Bound Technique | L103 | Design & Analysis of Algorithm

[@psycogaming.1725]

Tq so much madam ❤ got a clarity after a long search

[@manum5032]

Really useful vedio mam

[@monyraesha1644]

the profit value should be 65 right? as we add the value of item 1 and 3. how is it 69

[@damnn5492]

why do we add the item of 1 and 3?

[@HellFir-q5x]

same doubt 🙄
In the end the ub is calculated without 4 right?
That means ud = 65...😶

[@irshad5961]

I'm grateful

[@basavadarshanbs9906]

Optimal solution 65 not 69

[@teamrdx8091]

How bro 65?

[@sukhi9065]

65+(1)(0) in last without 4th item step =65 ​@@teamrdx8091

[@dheeraj9619]

Bro ub value is 69 , optimal solution that is most profit is 65 , mam didn't say clearly that's it , both u should mention

[@hrushikesh6040]

Tqsm mam I am very grateful

[@MOHITKUMARRATHOD]

good explanation

[@VICKY_GUNDEBOINA-u1z]

Brilliant Medam

[@Lakshmi-il1vz]

Thankyou mam

[@rcbchallengersfans1412]

Tq mam

[@AnuragSingh-xe1nm]

ubber bound is 65.

[@teamrdx8091]

How it is 65

[@littleboy007-]

@@teamrdx8091 😂

[@shreyam2223]

Y can't we continue in without 1

[@stalinseif6982]

You can

[@BharathrajPraveenchandra]

Among with 1 and without 1,with 1 has the highest upper bound value so we should continue with 'with 1' and drop the without 1