The fractional knapsack problem involves selecting items to fill a knapsack of limited capacity (M=20 in this case) in such a way as to maximize the total value.

Here’s how to solve it:

Sort the items in decreasing order of their value per unit weight (i.e. Pi/Wi).

Item 1: 3/5 = 0.6

Item 2: 10/13 = 0.769

Item 3: 15/12 = 1.25

Item 4: 5/8 = 0.625

Take the items in the order of the sorted list until the knapsack is full. If a complete item cannot fit, take a fraction of it.

Item 3 (15/12) can fit completely, so add it to the knapsack and reduce the capacity to 20-12=8.

Item 2 (10/13) can fit completely, so add it to the knapsack and reduce the capacity to 8-13= -5.

Item 1 (3/5) cannot fit completely, so add 3/5 of it to the knapsack and reduce the capacity to 0.

Item 4 (5/8) is not considered since the knapsack is full.

The total value of the items in the knapsack is 15 + 10 + (3/5)*3 = 22.

So the optimal solution is to choose items 3, 2 and a fraction of item 1, with a total value of 22.

The fractional knapsack problem involves selecting items to fill a knapsack of limited capacity (M=20 in this case) in such a way as to maximize the total value.

Here’s how to solve it:

So the optimal solution is to choose items 3, 2 and a fraction of item 1, with a total value of 22.