Linear Programming Problem

Definition

A linear programming problem is the problem of optimizing (maximizing or minimizing) a linear function (a function in the form p = ax + by ) which is subject to a set of conditions which are expressed as a system of linear inequalities or equations (the set of all points satisfying this set of conditions is called the feasible set or the feasible region). Each point of this set is called a feasible solution

Solution of a linear problem

1. If the linear problem has a solution, then it must be at a vertex (corner) feasible point of the feasible set.

2. If the linear problem has a solution at two adjacent vertices of the feasible set, then every point on the line segment joining these points is also a solution. How many solutions are these in this case?

1. If the feasible set is bounded then the objective function has both maximum and minimum on that set.

2. If the feasible set is empty, then the linear programming problem has no solution.

1. If the feasible set is unbounded then the objective function has no maximum but it has a minimum if:

a. The feasible set includes the inequalities: x ≥ 0 and y ≥ 0.

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b. The coefficients of both x and y in the formula of the objective function are positive

The feasible set & the existence of a max and or max for the objective function: p = ax + by

The graphical method to find the maximum (minimum)

1. Graph the feasible set ( the set resulting from graphing the conditions)

2. Find the coordinates of the vertices (the corners) of the feasible set, and then find the value of the objective function at each of these vertices.

3. Consider the vertex or vertices at which the objective function has the greatest (smallest) value.

a. If there is only one such vertex, then there is a unique solution to the maximizing problem if the feasible set is bounded ( a unique solution to the minimizing problem, if either the feasible set is bounded or the feasible set is in the first quadrant and the coefficients of the objective functions are both positive). Where does this solution occur?

b. If there is a maximum (minimum) at two adjacent (neighboring) vertices, then every point on the line segment joining them is a solution to the maximizing (minimizing) problem. How many solution do we have in this case?

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Example

How did we find the corner (3,2) ?

To find the corner, we find the point of intersection of two lines, which is the solution of the system of he equations of the lines:

2x + y = 8 and 2x + 3y = 12

Subtracting the first equation from the second, we get: 2y = 4 → y = 2

Substituting that in the first equation, we get:

2x + 2 = 8 → x = 3

Thus, the two lines intersect at the point (3,2)

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Example

How did we find the corners (40,20) and (20,50) ?

The feasible set has four corners. Two of them are obvious (which ones?)

The other corners are:One is the intersection of the (green) line

x + 2y = 80 and the (red) line 3x + 2y = 160

Solving these two equation, we get:2x = 80→x = 40. Substituting that in the first

equation, we get: 40 + 2y = 80 →y = 20This the point of the intersection of these two lines

is (40,20)

The other one is the intersection of the (green) linethe (red) line 3x + 2y = 160 and the (blue) line 5x + 2y = 200

Solving these two equation, we get:

2x = 40→x = 20. Substituting that in the first equation, we get: 60 + 2y = 160 →y = 50

This the point of the intersection of these two lines is (20,50)

Is there a point at which p is maximum?

Why not?

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Example

How did we find the corners (8,4)?

The feasible set has three corners. Two of them are obvious (which ones?)

The third corner is the intersection of the linesx - 4y = - 8 and x + 2y = 16

Solving these two equation, we get:6y = 24→y = 4. Substituting that in the first

equation, we get: x - 16 = - 8 →x = 8This the point of the intersection of these

two lines is (8,4)

Assignment

1. Find a formula for the set of all feasible points on which the maximum occurs?

2. Give examples of other solution points ( points on which the maximum occurs) other the one given in the answer.

Homework

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