M=20 (total capacity of bag)

N=4 (no of items)

Let’s find Pi/Wi (profit/weight) for each items:

P1/W1=3/5 = 0.60 (O1)

P2/W2=10/13= 0.76(O2)

P3/W3= 15/12 = 1.25 (O3)

P4/W4 = 5/8 = 0.62 (O4)

Where O is object no

Let’s arrange them in order: 1.25 >> 0.76 >> 0.62 >> 0.60

Now,

• Knapsack weight left to be filled is 8 kg but item-2 has a weight of 13 kg.
• Since in fractional knapsack problem, even the fraction of any item can be taken.
• So, knapsack will contain the following items-

< O3,(8/13)O2 >