© 5/7/2002 V.S. Subrahmanian1 Knapsack Problem Notes V.S. Subrahmanian University of Maryland.

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© 5/7/2002 V.S. Subrahmanian1 Knapsack Problem Notes V.S. Subrahmanian University of Maryland

© 5/7/2002 V.S. Subrahmanian2 Knapsack Problem You have a knapsack that has capacity (weight) C. You have several items I 1,…,I n. Each item I j has a weight w j and a benefit b j. You want to place a certain number of copies of each item I j in the knapsack so that: –The knapsack weight capacity is not exceeded and –The total benefit is maximal.

© 5/7/2002 V.S. Subrahmanian3 Example ItemWeightBenefit A260 B375 C490 Capacity = 5

© 5/7/2002 V.S. Subrahmanian4 Key question Suppose f(w) represents the maximal possible benefit of a knapsack with weight w. We want to find (in the example) f(5). Is there anything we can say about f(w) for arbitrary w?

© 5/7/2002 V.S. Subrahmanian5 Key observation To fill a knapsack with items of weight w, we must have added items into the knapsack in some order. Suppose the last such item was I j with weight w i and benefit b i. Consider the knapsack with weight (w- w i ). Clearly, we chose to add I j to this knapsack because of all items with weight w i or less, I j had the max benefit b i.

© 5/7/2002 V.S. Subrahmanian6 Key observation Thus, f(w) = MAX { b j + f(w-w j ) | I j is an item}. This gives rise to an immediate recursive algorithm to determine how to fill a knapsack.

© 5/7/2002 V.S. Subrahmanian7 Example ItemWeightBenefit A260 B375 C490

© 5/7/2002 V.S. Subrahmanian8 f(0), f(1) f(0) = 0. Why? The knapsack with capacity 0 can have nothing in it. f(1) = 0. There is no item with weight 1.

© 5/7/2002 V.S. Subrahmanian9 f(2) f(2) = 60. There is only one item with weight 60. Choose A.

© 5/7/2002 V.S. Subrahmanian10 f(3) f(3) = MAX { b j + f(w-w j ) | I j is an item}. = MAX { 60+f(3-2), 75 + f(3-3)} = MAX { , } = 75. Choose B.

© 5/7/2002 V.S. Subrahmanian11 f(4) f(4) = MAX { b j + f(w-w j ) | I j is an item}. = MAX { 60 + f(4-2), 75 + f(4-3), 90+f(4-4)} = MAX { , 75 + f(1), 90 + f(0)} = MAX { 120, 75, 90} =120. Choose A.

© 5/7/2002 V.S. Subrahmanian12 f(5) f(5) = MAX { b j + f(w-w j ) | I j is an item}. = MAX { 60 + f(5-2), 75 + f(5-3), 90+f(5-4)} = MAX { 60 + f(3), 75 + f(2), 90 + f(1)} = MAX { , , 90+0} = 135. Choose A or B.

© 5/7/2002 V.S. Subrahmanian13 Result Optimal knapsack weight is 135. Two possible optimal solutions: –Choose A during computation of f(5). Choose B in computation of f(3). –Choose B during computation of f(5). Choose A in computation of f(2). Both solutions coincide. Take A and B.

© 5/7/2002 V.S. Subrahmanian14 Another example Knapsack of capacity items –Item 1 has weight 10, benefit 60 –Item 2 has weight 20,benefit 100 –Item 3 has weight 30, benefit 120.

© 5/7/2002 V.S. Subrahmanian15 f(0),..,f(9) All have value 0.

© 5/7/2002 V.S. Subrahmanian16 f(10),..,f(19) All have value 10. Choose Item 1.

© 5/7/2002 V.S. Subrahmanian17 f(20),..,f(29) F(20) = MAX { 60 + f(10), f(0) } = MAX { 60+60, 100+0} =120. Choose Item 1.

© 5/7/2002 V.S. Subrahmanian18 f(30),…,f(39) f(30) = MAX { 60 + f(20), f(10), f(0) } = MAX { , , 120+0} = 180 Choose item 1.

© 5/7/2002 V.S. Subrahmanian19 f(40),…,f(49) F(40) = MAX { 60 + f(30), f(20), f(10)} = MAX { , , } = 240. Choose item 1.

© 5/7/2002 V.S. Subrahmanian20 f(50) f(50) = MAX { 60 + f(40), f(30), f(20) } = MAX { , , } = 300. Choose item 1.

© 5/7/2002 V.S. Subrahmanian21 Knapsack Problem Variants 0/1 Knapsack problem: Similar to the knapsack problem except that for each item, only 1 copy is available (not an unlimited number as we have been assuming so far). Fractional knapsack problem: You can take a fractional number of items. Has the same constraint as 0/1 knapsack. Can solve using a greedy algorithm.