knapsack | R Documentation |

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

```
knapsack(w, p, cap)
```

`w` |
integer vector of weights. |

`p` |
integer vector of profits. |

`cap` |
maximal capacity of the knapsack, integer too. |

`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.

A list with components `capacity`

, `profit`

, and `indices`

.

HwB email: <hwborchers@googlemail.com>

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.

`knapsack::knapsack`

```
# 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)
```

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