MatheAss 9.0Linear Algebra

Linear Optimization

The task of optimization is to determine the optimal value for a given goal, whereby given conditions must be observed.
The program determines the optimal solution for a two-variable objective function with linear inequalities as boundary conditions.

Example 1 (Maximization)

A factory produces two different mobile phones. X devices of type A and y devices of type B are to be completed daily.

Boundary conditions:
The individual departments have the following production capacities per day:
  1. The assembly line for Type A can produce a maximum of 600 devices.
    x ≤ 600
  2. The assembly line for type B can produce a maximum of 700 devices.
    y ≤ 700
  3. The plastics department produces a maximum of 750 cases in total.
  4. x + y ≤ 750
  5. The electrical department produces a maximum of 400 type A devices or 1200 type B devices or a combination thereof. This means that 1/400 of the total time is required per device of type A and 1/1200 per device of type B.
    1/400·x + 1/1200·y ≤ 1   or   3·x + 1·y ≤ 1200;
Objective function:
How many devices of each must be produced per day in order to achieve maximum profit if the profit for type A is $140 per device and for type B $80.
Objective function:   
  ƒ(x,y) = 140·x + 80·y → Maximum

Constraints:
  x ≥ 0
  y ≥ 0
  x ≤ 600
  y ≤ 700
  x + y ≤ 750
  3·x + y ≤ 1200

Maximum
  x = 225   y = 525
  ƒ(x,y) = 73500

The maximum profit of $73500 is therefore achieved if 225 type A and 525 type B devices are manufactured daily.

Example 2 (Minimization)

Objective function:   
  ƒ(x,y) = 140·x + 80·y → Minimum 

Constraints:
  x ≥ 0
  y ≥ 0
  x ≥ 160
  y ≥ 80
  x + y ≥ 750
  3 x + y ≥ 1200

Minimum
  x = 225   y = 525
  ƒ(x,y) = 73500

See also:

Wikipedia: Linear programming
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