Systems Analysis Methods SMU EMIS 5300/7300 NTU SY-521-N Systems Analysis Methods Dr. Jerrell T. Stracener, SAE Fellow LP Solutions The Graphical Method prepared and presented by Imran Ismail updated 10.11.01

Consider the following LP Maximize X + 2Y Subject to: X + Y <= 4 X - 2Y <= 2 -2X + Y <= 2 X, Y >= 0 This problem is in 2-dimensions and can be solved graphically

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.

The constraint lines can be constructed by changing all inequalities in to equal signs. i.e. X + Y <= 4 become X + Y = 4, the inequalities play an important role in defining the feasible region and are implemented later on.

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 XY plane

Similarly for constraint X-2Y <= 2, we get

And for constraint 3, -2X+Y <= 2: Since X and Y must be nonnegative, the feasible region for the LP is the bounded by the three lines described (and shown) above and the X and Y axes. Any point inside or on the boundary of the region described above is a feasible solution to the LP

Finding an Optimal Solution The objective function is given as an equation whose value has yet to be determined. The information provided in the LP gives us 2 clues: The function has to be maximized The function must satisfy all constraints i.e. it must lie in the feasible regions

In such an instance, we can start by assuming an objective function value of 0 (zero). This gives us the following equation X + 2Y = 0 The line above will provide us with a set of points that will have an objective function value of 0. When drawn on the graph, the segment of this line that intersects the feasible region will represent the set of feasible solutions with an objective function value of zero. In this case, the set is just the single point (0,0).

Similarly, the line X + 2Y = 2 describes the set of points that have an objective function value of two. So, the segment of this line that intersects the feasible region represents the set of feasible solutions with an objective function value of two. We can continue this method till we reach an objective function value such that it no longer intersects the feasible region.

By continuing on in this fashion, we can find an optimal solution for the LP by "pushing" the objective function line up until it last touches the feasible region.

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.

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

It is observed that objective functions with different slopes, while traveling through the feasible region might be maximized at different points. It is also important to note that the corner points of the feasible region are the only part of this region that do not allow the objective function to move further ahead while still remaining feasible! Therefore, if an LP has an optimal solution, it is guaranteed that this solution will occur at a corner or extreme point of the feasible region

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.

In our example we can identify the extreme points by labeling A, B, C, D and E, such that:

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 Solving these equations and plugging them into the objective function, we get:

Here, 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

Point A is also where the objective function line was graphically maximized. Therefore, we conclude that point A is the optimal extreme point of the feasible region. It: maximizes the value of the objective function and satisfies all constraints The (x,y) co-ordinates of point A are called the optimal decision variables or the optimal solution with optimal objective function value ZA

Conclusion The graphical method and the brute force method will always obtain the same result. In our example, it was found that the optimal solution is such that: Xopt=2/3 = 0.66 Yopt=10/3 = 3.33 Zopt = 22/3 = 7.333