Note

This is only for educational purposes, all credits go towards the authors of the book

Currently Included Content

• Formation of LP of form $Ax\text{ ? }b$ with objective function vector and equality constraints ($?$ can denote $=$, $\geq$, or $\leq$)
• Canonical form computation for given LP and proposed basis
• Simplex computation for given LP
• Two-phase Simplex computation for given LP
• Duality transformation for given LP

To be added

• complementary slack conditions
• other algorithms included in the textbook (?)

Example Usage

formulating an LP simple case:

$$\max\left\{z(x)=v^{\top} x:Ax=b,x\geq 0 \right\}$$

where

$$A=\begin{pmatrix} 1&1&-3&1&2\\ 0&1&-2&2&-2\\ -2&-1&4&1&0 \end{pmatrix},\quad b=\begin{pmatrix} 7\\ -2\\ -3 \end{pmatrix},\quad v=\begin{pmatrix} -1\\ 0\\ 3\\ 7\\ -1 \end{pmatrix}$$

A <- rbind(
  c(1,1,-3,1,2),
  c(0,1,-2,2,-2),
  c(-2,-1,4,1,0)
)
b <- c(7,-2,-3)
v <- c(-1,0,3,7,-1)
LP <- form.LP(b=b, A=A, v=v)

formulating an LP with basis $B=\{1,4\}$ and initial $z$ value of $3$:

$$\begin{aligned} &\max\qquad 3 + \begin{pmatrix}0 & -1 & -2 & 0 & -3\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} 1 & -2 & 1 & 0 & 2\\ 0 & 1 & -1 & 1 & 3 \end{pmatrix}x=\begin{pmatrix} 2\\ 4 \end{pmatrix}\\ &\quad x \geq 0 \end{aligned}$$

B <- c(1,4)
A <- rbind(
  c(1,-2,1,0,2),
  c(0,1,-1,1,3)
)
b <- c(2,4)
v <- c(0,-1,-2,0,-3)
z <- 3
LP <- form.LP(B, b, A, v, z)

optimal solution exists for raw Simplex with given basis $B=\{3,5\}$:

$$\begin{aligned} &\max\qquad \begin{pmatrix}-1 & 3 & -5 & 9 & 3\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} 4 & 1& 3 & -1 & -2\\ 3 & 1 & 2 & 0 & -1 \end{pmatrix}x=\begin{pmatrix} 2\\ 3 \end{pmatrix}\\ &\quad x \geq 0 \end{aligned}$$

B <- c(3,5)
A <- rbind(
  c(4,1,3,-1,-2),
  c(3,1,2,0,-1)
)
b <- c(2,3)
v <- c(-1,3,-5,9,3)
LP <- form.LP(B, b, A, v)
simplex(LP)

unbounded LP under raw Simplex with basis $B=\{2,3,5,6\}$:

$$\begin{aligned} &\max\qquad \begin{pmatrix}0 & 7 & -8 & -2 & -4 & -6\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} 0 & -2 & 2 & 1 & 1 & 1 \\ 2 & -3 & 3 & 0 & 2 & 4 \\ 4 & -2 & 4 & -2 & 1 & 2 \\ 3 & 4 & -3 & -4 & -2 & -1 \end{pmatrix}x=\begin{pmatrix} 1\\ 9\\ 6\\ 2 \end{pmatrix}\\ &\quad x \geq 0 \end{aligned}$$

B <- c(2,3,5,6)
A <- rbind(
  c(0,-2,2,1,1,1),
  c(2,-3,3,0,2,4),
  c(4,-2,4,-2,1,2),
  c(3,4,-3,-4,-2,-1)
)
b <- c(1,9,6,2)
v <- c(0,7,-8,-2,-4,-6)
LP <- form.LP(B, b, A, v)
simplex(LP)

optimal solution exists under two-phase Simplex:

$$\begin{aligned} &\max\qquad \begin{pmatrix}2 & -1 & 2\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} -1 & -2 & 1 \\ 1 & -1 & 1 \end{pmatrix}x=\begin{pmatrix} -1\\ 3 \end{pmatrix}\\ &\quad x \geq 0 \end{aligned}$$

A <- rbind(
  c(-1,-2,1),
  c(1,-1,1)
)
b <- c(-1,3)
v <- c(2,-1,2)
LP <- form.LP(b=b, A=A, v=v)
twophase(LP)

finding dual of given LP simple case:

$$\begin{aligned} &\max\qquad \begin{pmatrix}28 & -7 & 20\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} 6 & -2 & 1 \\ -1 & 1 & 2 \\ 4 & -1 & 3 \end{pmatrix}x\leq\begin{pmatrix} 3\\ 3\\ 3 \end{pmatrix}\\ &\quad x \geq 0 \end{aligned}$$

A <- rbind(
  c(6,-2,1),
  c(-1,1,2),
  c(4,-1,3)
)
b <- c(3,3,3)
v <- c(28,-7,20)
LP <- form.LP(b=b, A=A, v=v, A.c=rep('<=',3))
dual(LP)

finding dual of given LP complex case:

$$\begin{aligned} &\min\qquad \begin{pmatrix}53 & 52 & 51 & 54 & 55\end{pmatrix}x\\ \text{s.t.}\qquad&\\ &\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \end{pmatrix}x\text{ }\begin{matrix} =\\ \geq\\ \geq \end{matrix}\begin{pmatrix} 9\\ 15\\ 29 \end{pmatrix}\\ &\\ &\quad x_1\geq0,x_2\geq0,x_3\leq0,x_4\leq0,x_5\text{ free} \end{aligned}$$

A <- matrix(1:15, ncol=5, nrow=3, byrow=T)
b <- c(9, 15, 29)
v <- c(53, 52, 51, 54, 55)
LP <- form.LP(b=b, A=A, v=v, A.c=c('=','>=','>='), x.c=c('>=0','>=0','<=0','<=0','free'))
dual(LP, opt=0)