Linear programming is the minimization (or maximization) of a linear objective subject to linear constraints. There are several widely adopted schemes for representing linear programming problems. Here we adopt a scheme corresponding where the linear objective is written in terms of decision variables $x_1, x_2, \ldots, x_N$ as
\begin{align} \min_{x_1, x_1, \ldots, x_N} c_1x_1 + c_2x_2 + \cdots + c_Nx_N \end{align}subject to
\begin{align} x_i \geq 0 & \qquad i=1,\ldots,N\quad\mbox{lower bounds on decision variables}\\ \sum_{j=1}^N a^{ub}_{ij}x_j \leq b^{ub}_i & \qquad i=1,\ldots,M_{ub}\quad\mbox{upper bound constraints} \\ \sum_{j=1}^N a^{eq}_{ij}x_j = b^{eq}_i & \qquad i=1,\ldots,M_{eq}\quad\mbox{equality constraints}\\ \end{align}The notation can be simplified by adopting a matrix/vector formulation where
\begin{align} \min_{x\geq 0} c^T x \end{align}subject to
\begin{align} A_{ub} x \leq b_{ub} \\ A_{eq} x = b_{eq} \end{align}where $c$, $A_{ub}, b_{ub}$, and $A_{eq}, b_{eq}$, are vectors and matrices of coefficients constructed from the linear expressions given above.
The decision variables are
| Variable | Description | Units |
|---|---|---|
| $x_1$ | crude #1 | bbl/day |
| $x_2$ | crude #2 | bbl/day |
| $x_3$ | gasoline | bbl/day |
| $x_4$ | kerosine | bbl/day |
| $x_5$ | fuel oil | bbl/day |
| $x_6$ | residual | bbl/day |
The overall objective is to maximize profit
\begin{align} \mbox{profit} & = \mbox{income} - \mbox{raw_material_cost} - \mbox{processing_cost} \end{align}where the financial components are given by
\begin{align} \mbox{income} & = 72x_3 + 48x_4 + 42x_5 + 20x_6 \\ \mbox{raw_material_cost} & = 48x_1 + 30x_2 \\ \mbox{processing_cost} & = 1 x_1 + 2x_2 \end{align}Combining these terms, the objective is to maximize
\begin{align} f & = c^T x = - 49 x_1 - 32 x_2 + 72 x_3 + 48x_4 + 42x_5 + 20x_6 \end{align}The material balance equations are
\begin{align} \mbox{gasoline: } x_3 & = 0.80 x_1 + 0.44 x_2 \\ \mbox{kerosine: } x_4 & = 0.05 x_1 + 0.10 x_2 \\ \mbox{fuel oil: } x_5 & = 0.10 x_1 + 0.36 x_2 \\ \mbox{residual: } x_6 & = 0.05 x_1 + 0.10 x_2 \end{align}Production limits
\begin{align} \mbox{gasoline: } x_3 & \leq 24,000 \\ \mbox{kerosine: } x_4 & \leq 2,000 \\ \mbox{fuel oil: } x_5 & \leq 6,000 \end{align}\begin{align} \underbrace{\left[\begin{array}{cccccc} 0.80 & 0.44 & -1 & 0 & 0 & 0 \\ 0.05 & 0.10 & 0 & -1 & 0 & 0 \\ 0.10 & 0.36 & 0 & 0 & -1 & 0 \\ 0.05 & 0.10 & 0 & 0 & 0 & -1 \end{array}\right]}_{A_{eq}} \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6 \end{array}\right] & = \underbrace{\left[\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \end{array}\right]}_{b_{eq}} \end{align}
from scipy.optimize import linprog
c = [49, 32, -72, -48, -42, -20]
A_ub = [[0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 1, 0]]
b_ub = [24000, 2001, 6000]
A_eq = [[0.80, 0.44, -1, 0, 0, 0],
[0.05, 0.10, 0, -1, 0, 0],
[0.10, 0.36, 0, 0, -1, 0],
[0.05, 0.10, 0, 0, 0, -1]]
b_eq = [0, 0, 0, 0]
results = linprog(c, A_ub, b_ub, A_eq, b_eq)
results
p0 = 573517.24
print('additional profit', -results.fun - p0)
additional profit 175.035862069
print(results.message)
if results.success:
for k in range(len(results.x)):
print('x[{0:2d}] = {1:7.1f} bbl/day'.format(k+1, results.x[k]))
Optimization terminated successfully. x[ 1] = 26199.3 bbl/day x[ 2] = 6910.3 bbl/day x[ 3] = 24000.0 bbl/day x[ 4] = 2001.0 bbl/day x[ 5] = 5107.7 bbl/day x[ 6] = 2001.0 bbl/day