# simplex

##### Simplex Method for Linear Programming Problems

This function will optimize the linear function `a%*%x`

subject
to the constraints `A1%*%x <= b1`

, `A2%*%x >= b2`

,
`A3%*%x = b3`

and `x >= 0`

. Either maximization or
minimization is possible but the default is minimization.

- Keywords
- optimize

##### Usage

```
simplex(a, A1 = NULL, b1 = NULL, A2 = NULL, b2 = NULL, A3 = NULL,
b3 = NULL, maxi = FALSE, n.iter = n + 2 * m, eps = 1e-10)
```

##### Arguments

- a
A vector of length

`n`

which gives the coefficients of the objective function.- A1
An

`m1`

by`n`

matrix of coefficients for the \(\leq\) type of constraints.- b1
A vector of length

`m1`

giving the right hand side of the \(\leq\) constraints. This argument is required if`A1`

is given and ignored otherwise. All values in`b1`

must be non-negative.- A2
An

`m2`

by`n`

matrix of coefficients for the \(\geq\) type of constraints.- b2
A vector of length

`m2`

giving the right hand side of the \(\geq\) constraints. This argument is required if`A2`

is given and ignored otherwise. All values in`b2`

must be non-negative. Note that the constraints`x >= 0`

are included automatically and so should not be repeated here.- A3
An

`m3`

by`n`

matrix of coefficients for the equality constraints.- b3
A vector of length

`m3`

giving the right hand side of equality constraints. This argument is required if`A3`

is given and ignored otherwise. All values in`b3`

must be non-negative.- maxi
A logical flag which specifies minimization if

`FALSE`

(default) and maximization otherwise. If`maxi`

is`TRUE`

then the maximization problem is recast as a minimization problem by changing the objective function coefficients to their negatives.- n.iter
The maximum number of iterations to be conducted in each phase of the simplex method. The default is

`n+2*(m1+m2+m3)`

.- eps
The floating point tolerance to be used in tests of equality.

##### Details

The method employed by this function is the two phase tableau simplex method. If there are \(\geq\) or equality constraints an initial feasible solution is not easy to find. To find a feasible solution an artificial variable is introduced into each \(\geq\) or equality constraint and an auxiliary objective function is defined as the sum of these artificial variables. If a feasible solution to the set of constraints exists then the auxiliary objective will be minimized when all of the artificial variables are 0. These are then discarded and the original problem solved starting at the solution to the auxiliary problem. If the only constraints are of the \(\leq\) form, the origin is a feasible solution and so the first stage can be omitted.

##### Value

An object of class `"simplex"`

: see `simplex.object`

.

##### Note

The method employed here is suitable only for relatively small
systems. Also if possible the number of constraints should be reduced
to a minimum in order to speed up the execution time which is
approximately proportional to the cube of the number of constraints.
In particular if there are any constraints of the form ```
x[i] >=
b2[i]
```

they should be omitted by setting `x[i] = x[i]-b2[i]`

,
changing all the constraints and the objective function accordingly
and then transforming back after the solution has been found.

##### References

Gill, P.E., Murray, W. and Wright, M.H. (1991)
*Numerical Linear Algebra and Optimization Vol. 1*. Addison-Wesley.

Press, W.H., Teukolsky, S.A., Vetterling, W.T. and Flannery, B.P. (1992)
*Numerical Recipes: The Art of Scientific Computing (Second Edition)*.
Cambridge University Press.

##### Examples

```
# NOT RUN {
# This example is taken from Exercise 7.5 of Gill, Murray and Wright (1991).
enj <- c(200, 6000, 3000, -200)
fat <- c(800, 6000, 1000, 400)
vitx <- c(50, 3, 150, 100)
vity <- c(10, 10, 75, 100)
vitz <- c(150, 35, 75, 5)
simplex(a = enj, A1 = fat, b1 = 13800, A2 = rbind(vitx, vity, vitz),
b2 = c(600, 300, 550), maxi = TRUE)
# }
```

*Documentation reproduced from package boot, version 1.3-24, License: Unlimited*