Linear Programming
Linear Programming is a particular case of constrained optimization problems. What sets
the linear programming aside is that optimal values are sought for a linear function subject to
linear constraints. To have a general formulation let's assume that we are given
- an m×n matrix A
- an m×1 (column) vector b
- a 1×n (row) vector c
The role of the unknown is played by a n×1 (column) vector x which is required to satisfy
two constraints:
- Ax = b, and
- x
0,
the latter means that all the components of x are nonnegative. Vectors that satisfy constraints
1 and 2 are called feasible. We are interested in the situations where the set of feasible
vectors is not empty.
Finally, the cost or objective function f is given by f(x) = c·x, where
c·x is the scalar product of two vectors c and x: c·x = c1x1 + ... + cnxn. In linear
programming, one is requested to maximize (or minimize) the cost function f subject to constraints 1 and 2:
| (P) |
Maximize f(x) = c·x subject to Ax = b and x 0 |
Example
Assume that we have decided to position defenders of a square castle according to the following plan:
so that the total number of defenders is 4(p+q) while (2p+q) fighters face the enemy on every side. Let's
denote p = x1 and q = x2. Let also A = (4 4), be a 1×2 matrix,
and c = (2 1), a 1×2 row vector. Assume that a 1×1 vector b is equal to (28). The linear programming problem (P) is then interpreted as:
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How to position 28 fighters according to the symmetric arrangement above so as to have the
maximum number of defenders on each side?
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In a more formal manner, the problem is to maximize f(x) = 2x1 + x2 subject to
4(x1 + x2) = 28 and x1 0 and x2 0.
Geometrical solution to the problem is easy and illuminating.
The set F of feasible vectors is defined by
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F = {x: x1 + x2 = 7 and x1 0 and x2 0}.
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In the plane with orthogonal axes x1 and x2, F is a straight line segment between two points: (0,7) and (7,0).
The function f depends on two variables: x1 and x2. Instead of drawing its graph
in the three-dimensional space, we'll draw a series of its level lines in the plane. The function is
constant on each of the lines as shown on the diagram at the left. The function attains the maximum
value 14 on the line 2x1 + x2 = 14 that passes through one of the
endpoints (7,0) of the interval F. This is always the case in linear programming problems - solution is found
at one (or more) vertices of the set of feasible vectors. Note that if we were looking to minimize f the
solution would be located at the other endpoint (0,7) of F. Thus for the arrangement (*) we have two extreme
configurations: a minimum and a maximum:
References
- J.L.Casti, Five Golden Rules, John Wiley & Sons, 1996
- J.A.Paulos, Beyond Numeracy, Vintage Books, 1992
- G.Strang, Introduction to Applied Mathematics, Wellesley-Cambridge Press, MA, 1986.

Copyright © 1996-2008 Alexander Bogomolny
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