Description Usage Arguments Value Note Author(s) References See Also Examples

Solves a linear programming problem,

*min(sum(Cost_i x_i))*

subject to

*Ex=f*

*Gx>=h*

*x_i>=0*

(optional)

This function provides a wrapper around `lp`

(see note)
from package lpSolve, written to be consistent with the functions
`lsei`

, and `ldei`

.

It allows for the x's to be negative (not standard in lp).

1 2 |

`E ` |
numeric matrix containing the coefficients of the equality
constraints |

`F ` |
numeric vector containing the right-hand side of the equality constraints. |

`G ` |
numeric matrix containing the coefficients of the inequality
constraints |

`H ` |
numeric vector containing the right-hand side of the inequality constraints. |

`Cost ` |
numeric vector containing the coefficients of the cost function;
if |

`ispos` |
logical, when |

`int.vec ` |
when not |

`verbose ` |
logical to print error messages. |

`... ` |
extra arguments passed to R-function |

a list containing:

`X ` |
vector containing the solution of the linear programming problem. |

`residualNorm ` |
scalar, the sum of absolute values of residuals of equalities and violated inequalities. Should be very small or zero for a feasible linear programming problem. |

`solutionNorm ` |
scalar, the value of the minimised |

`IsError ` |
logical, |

`type ` |
the string "linp", such that how the solution was obtained can be traced. |

If the requirement of nonnegativity are relaxed, then strictly speaking the problem is not a linear programming problem.

The function `lp`

may fail and terminate R for very small problems that
are repeated frequently...

Also note that sometimes multiple solutions exist for the same problem.

Karline Soetaert <[email protected]>

Michel Berkelaar and others (2007). lpSolve: Interface to Lpsolve v. 5.5 to solve linear or integer programs. R package version 5.5.8.

`lp`

the original function from package lpSolve

`Blending`

, a linear programming problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 | ```
#-------------------------------------------------------------------------------
# Linear programming problem 1, not feasible
#-------------------------------------------------------------------------------
# maximise x1 + 3*x2
# subject to
#-x1 -x2 < -3
#-x1 + x2 <-1
# x1 + 2*x2 < 2
# xi > 0
G <- matrix(nrow = 3, data = c(-1, -1, 1, -1, 1, 2))
H <- c(3, -1, 2)
Cost <- c(-1, -3)
(L <- linp(E = NULL, F = NULL, Cost = Cost, G = G, H = H))
L$residualNorm
#-------------------------------------------------------------------------------
# Linear programming problem 2, feasible
#-------------------------------------------------------------------------------
# minimise x1 + 8*x2 + 9*x3 + 2*x4 + 7*x5 + 3*x6
# subject to:
#-x1 + x4 + x5 = 0
# - x2 - x4 + x6 = 0
# x1 + x2 + x3 > 1
# x3 + x5 + x6 < 1
# xi > 0
E <- matrix(nrow = 2, byrow = TRUE, data = c(-1, 0, 0, 1, 1, 0,
0,-1, 0, -1, 0, 1))
F <- c(0, 0)
G <- matrix(nrow = 2, byrow = TRUE, data = c(1, 1, 1, 0, 0, 0,
0, 0, -1, 0, -1, -1))
H <- c(1, -1)
Cost <- c(1, 8, 9, 2, 7, 3)
(L <- linp(E = E, F = F, Cost = Cost, G = G, H = H))
L$residualNorm
#-------------------------------------------------------------------------------
# Linear programming problem 3, no positivity
#-------------------------------------------------------------------------------
# minimise x1 + 2x2 -x3 +4 x4
# subject to:
# 3x1 + 2x2 + x3 + x4 = 2
# x1 + x2 + x3 + x4 = 2
# 2x1 + x2 + x3 + x4 >=-1
# -x1 + 3x2 +2x3 + x4 >= 2
# -x1 + x3 >= 1
E <- matrix(ncol = 4, byrow = TRUE,
data =c(3, 2, 1, 4, 1, 1, 1, 1))
F <- c(2, 2)
G <- matrix(ncol = 4, byrow = TRUE,
data = c(2, 1, 1, 1, -1, 3, 2, 1, -1, 0, 1, 0))
H <- c(-1, 2, 1)
Cost <- c(1, 2, -1, 4)
linp(E = E, F = F, G = G, H = H, Cost, ispos = FALSE)
``` |

limSolve documentation built on Aug. 14, 2017, 3:01 p.m.

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