knapsack: 0-1 Knapsack Problem

In adagio: Discrete and Global Optimization Routines

0-1 Knapsack Problem

Description

Solves the 0-1 (binary) single knapsack problem.

Usage

knapsack(w, p, cap)

Arguments

w	integer vector of weights.
p	integer vector of profits.
cap	maximal capacity of the knapsack, integer too.

Details

knapsack solves the 0-1, or: binary, single knapsack problem by using the dynamic programming approach. The problem can be formulated as:

Maximize sum(x*p) such that sum(x*w) <= cap, where x is a vector with x[i] == 0 or 1.

Knapsack procedures can even solve subset sum problems, see the examples 3 and 3' below.

Value

A list with components capacity, profit, and indices.

Author(s)

HwB email: <[e-mail]>

References

Papadimitriou, C. H., and K. Steiglitz (1998). Combinatorial Optimization: Algorithms and Complexity. Dover Publications 1982, 1998.

Horowitz, E., and S. Sahni (1978). Fundamentals of Computer Algorithms. Computer Science Press, Rockville, ML.

See Also

knapsack::knapsack

Examples

# Example 1
p <- c(15, 100, 90, 60, 40, 15, 10,  1)
w <- c( 2,  20, 20, 30, 40, 30, 60, 10)
cap <- 102
(is <- knapsack(w, p, cap))
# [1] 1 2 3 4 6 , capacity 102 and total profit 280

## Example 2
p <- c(70, 20, 39, 37, 7, 5, 10)
w <- c(31, 10, 20, 19, 4, 3,  6)
cap <- 50
(is <- knapsack(w, p, cap))
# [1] 1 4 , capacity 50 and total profit 107

## Not run: 
##  Example 3: subset sum
p <- seq(2, 44, by = 2)^2
w <- p
is <- knapsack(w, p, 2012)
p[is$indices]  # 16  36  64 144 196 256 324 400 576

##  Example 3': maximize number of items
#   w <- seq(2, 44, by = 2)^2
#   p <- numeric(22) + 1
#  is <- knapsack(w, p, 2012)

## Example 4 from Rosetta Code:
w = c(  9, 13, 153,  50, 15, 68, 27, 39, 23, 52, 11,
       32, 24,  48,  73, 42, 43, 22,  7, 18,  4, 30)
p = c(150, 35, 200, 160, 60, 45, 60, 40, 30, 10, 70,
       30, 15,  10,  40, 70, 75, 80, 20, 12, 50, 10)
cap = 400
system.time(is <- knapsack(w, p, cap))  # 0.001 sec

## End(Not run)

adagio documentation built on Oct. 26, 2023, 5:08 p.m.