1. Linear Programming in Two Dimensions
updated 22 January 2009

2. Consider the following LP
LP1: Maximize X + 2Y
Subject to:
X + Y  4 (1)
X - 2Y  2 (2)
-2X + Y  2 (3)
X  0 (4)
Y  0 (5)
This problem is in two dimensions and can be solved graphically.

3. Finding the Feasible Region
We begin by graphing the constraints on an X-Y coordinate system to determine the set of all points that satisfy all the constraints. This set of points is known as the feasible region for the LP.
Since both variables must be non-negative, we know that the feasible region must be within the first quadrant.

4. Graphing Constraints (1), (4), and (5)
The set of solutions to the equation X + Y = 4 can be represented by the straight line passing through the points (4,0) and (0,4) in the X-Y plane.
The shaded triangle on the next slide represents the set of points satisfying the constraint X + Y  4, X  0, Y  0.

5. Feasible Region With Respect to Constraints (1), (4), and (5)

6. Feasible Region With Respect to Constraints (1), (2), (4), and (5)

7. Feasible Region for LP1

8. Isoprofit Lines
The line X + 2Y = 0 describes the set of points that have an objective function value of 0.
The line X + 2Y = 2 describes the set of points that have an objective function value of 2
The line X + 2Y = 4 describes the set of points that have an objective function value of 4
The line X + 2Y = 6 describes the set of points that have an objective function value of 6
The line X + 2Y = 8 describes the set of points that have an objective function value of 8

9. Graphing the Isoprofit Lines

10. Graphing the Isoprofit Lines
6 < maximum Z < 8

11. Finding an Optimal Solution Graphically

12. The Optimal Solution to LP1
This occurs when we graph the line
X + 2Y = 22/3 = 7.333
which intersects the feasible region at the point (2/3, 10/3). Since there are no feasible solutions with a greater objective function value than 22/3, we say that X = 2/3, Y = 10/3 is an optimal solution and that 22/3 is the optimal value for the objective function.

13. It is important to investigate how other objective functions might behave given the same feasible region.

14. Interior-Point Solutions
Z(1,3) = 7
Z(1,2) = 5
Z(1,1) = 3

15. Corner-Point Solutions
Z(2/3,10/3) = 7 1/3
Z(1,3) = 7
Z(4/3,1/3) = 4 2/3

16. The “Brute Force” Method
Using the fact that if an LP has an optimal solution, then it has an extreme point solution, we can use a "brute force" method to find an optimal solution by: testing each extreme point to see if it is feasible and then comparing the objective function values.
Recall that an extreme, or corner-point, solution is formed by the intersection of two of the constraints.

17. In our example we can identify the extreme points by labeling them A, B, C, D and E, as shown below.

18. Feasible Corner Points
Point A is formed by the intersection of: X + Y = 4 and –2X + Y = 2
and similarly:
B: X - 2Y = 2 and X + Y = 4
C: X - 2Y = 2 and Y = 0
D: X = 0 and Y = 0
E: -2X + Y = 2 and X = 0

19. The co-ordinate points of A give us the maximum objective function value.
X-co-ord
Y co-ord
Obj. Val A
.66
3.33
7.33 B
4.66 C 2 D E 4

20. Unique Optimal Solution

21. Other Special Cases of LP
Adding the constraint Y  4 to the original LP makes the problem infeasible.
Removing the constraint X + Y  4 from the original LP makes the problem unbounded.
Changing the objective function of the original LP to maximize X + Y gives an LP with multiple optimal solutions.

22. An Infeasible LP

23. An Unbounded LP

24. An LP With Multiple Optima