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Linear Programming

 Primal Linear Programming Problem

Text Box: Linear Programming (LP) Problem A linear programming problem is one in which we are to find the maximum or minimum value of a linear expression ax + by + cz + . . . (called the objective function), subject to a number of linear constraints of the form Ax + By + Cz + . . . N or Ax + By + Cz + . . . N. The largest or smallest value of the objective function is called the optimal value, and a collection of values of x, y, z, . . . that gives the optimal value constitutes an optimal solution. The variables x, y, z, . . . are called the decision variables.  

 

 

 

 

 

 

 

 

 

 

 

 

Text Box: Example Here is an example of an LP problem: Find the maximum value of p = 3x - 2y + 4z subject to 4x + 3y - z 3 x + 2y + z 4 x 0, y 0, z 0 The objective function is p = 3x - 2y + 4z. The constraints are 4x + 3y - z 3 x + 2y + z 4 x 0, y 0, z 0. Q Wait a minute! Why can't I simply choose, say, z to be really large (z = 1,000,000 say) and thereby make p as large as I want? Because: 1. That will make P too large! 2. You have to change x first 3. It will violate the first constraint. 4. It will violate the first constraint

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Sketching the Solution Set of a Linear Inequality

To sketch the region represented by a linear inequality in two variables:

A. Sketch the straight line obtained by replacing the inequality with an equality.

B. Choose a test point not on the line ((0,0) is a good choice if the line does not pass through the origin, and if the line does pass through the origin a point on one of the axes would be a good choice).

C. If the test point satisfies the inequality, then the set of solutions is the entire region on the same side of the line as the test point. Otherwise it is the region on the other side of the line. In either case, shade out the side that does not contain the solutions, leaving the solution region showing.

Example

To sketch the linear inequality

3x - 4y 12,

first sketch the line 3x - 4y = 12.

 

Next, choose the origin (0, 0) as the test point (since it is not on the line). Substituting x=0, y=0 in the inequality gives

3(0) - 4(0) 12.

 

 

Since this is a true statement, (0, 0) is in the solution set, so the solution set consists of all points on the same side as (0, 0). This region is left unshaded, while the (grey) shaded region is blocked out.

 

 

 

Feasible Region

The feasible region determined by a collection of linear inequalities is the collection of points that satisfy all of the inequalities.

To sketch the feasible region determined by a collection of linear inequalities in two variables: Sketch the regions represented by each inequality on the same graph, remembering to shade the parts of the plane that you do not want. What is unshaded when you are done is the feasible region.

 

 

 

 

 

Example

The feasible region for the following collection of inequalities is the unshaded region shown below (including its boundary).

3x - 4y 12,
x + 2y 4
x 1
y 0.

 

 

 

Graphical Method

The graphical method for solving linear programming problems in two unknowns is as follows.

A. Graph the feasible region.
B. Compute the coordinates of the corner points.
C. Substitute the coordinates of the corner points into the objective function to see which gives the optimal value.
D. If the feasible region is not bounded, this method can be misleading: optimal solutions always exist when the feasible region is bounded, but may or may not exist when the feasible region is unbounded. The textbook shows a straightforward way for determining whether optimal solutions exist in the case of unbounded feasible regions

Example

Minimize C = 3x + 4y subject to the constraints

3x - 4y 12,
x + 2y 4
x 1,   y 0.

The feasible region for this set of constraints was shown above. Here it is again with the corner points shown.

 

The following table shows the value of C at each corner point:

Point

C = 3x + 4y

(1, 1.5)

3(1)+4(1.5) = 9

minimum

(4, 0)

3(4)+4(0) = 12

 

Therefore, the solution is x = 1, y = 1.5, giving the minimum value C = 9.

 
 
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