Linear optimization: Formulation and graphical solution of a LP 4
Inhaltsverzeichnis
Introduction
Linear optimization is a mathematical technique used in Operations Research to solve problems in which the objective function and the requirements are given as linear functions. This method generates, considering the constraints that are formulated as equalities or inequalities, the best outcome of the given mathematical model. The objective function includes two, three or more variables and can be either minimized or maximized.
Definition/ Formulation
objective function to maximize (or minimize): F(x1,...,xp) = c1 x1 + ... + cp xp
problem constraints:
- ai1 x1 + … + aip xp ≤ bi for i = 1,...,m1
- ai1 x1 + ... + aip xp ≥ bi for i = m1+1,...,m2
- ai1 x1 + ... + aip xp = bi for i = m2+1,...,m
consideration of the non-negative condition: xj ≥ 0 for all j = 1,...,p
Example
In the following part we consider only two variables. First we formulate the problem mathematically which includes the objective function and the restrictions. Then we solve it graphically.
Maximization
In the following example we want to produce clothes. Therefore we need cloth and machines. Goal is to maximize revenues.
- x1: sweat shirts
- x2: T-shirts
The table leads to the following model:
goal function: max F(x1, x2) = 40 x1 + 30 x2
restrictions:
- • x1 + x2 ≤ 70 restriction 1
- • 9 x1 + 5 x2 ≤ 450 restriction 2
- • x1 , x2 ≥ 0 non-negative condition
First we draw the restrictions. They show us the maximal feasible solution set. Afterwards we draw the goal function and implement a parallel translation to get as far as possible away from the beginning. That's how we get the optimal feasible solution.
The optimal solution is:
(x*) = (x1* = 25, x2* = 45)
The maximal profit is:
F (x*) = 40 • 25 + 30 • 45 = 2350 Euro
Minimization
In the following example we want to plant flowers. Therefore we need gardener and fertilizers. The costs shoud be minimal.
- x1: roses
- x2: sunflowers
The table leads tot he following model:
goal function: min F(x1, x2) = 2 x1 + 2 x2
restrictions:
- • 8 x1 + 4 x2 ≥ 32 restriction 1
- • 5 x1 + 5 x2 ≥ 25 restriction 2
- • x1, x2 ≥ 0 non-negativ condition
As we proceeded in the example above, we first draw the restrictions. But now the feasible solution set is different to the maximation example caused by the different restriction inequations. Then we draw the goal function again and shift it downwards as far as possible. This leads to the optimal solution.
The optimal solution is:
(x*) = (x1* = 3, x2* = 2)
The costs are:
F (x*) = 3 • 3 + 2 • 2 = 13 Euro