Linear Programming Part 2: Graphical Solution Method

Graphical solution method is suitable for Linear Programming with two decisions variables because it uses Cartesian coordinate system: Each axis represents three value of decision variable with straight lines represent both constraints and objective function. Recall the Mathematical model from Indemy Coffee Shop example:

(OF) max𝑥0 = 6𝑥𝑎 + 3𝑥𝑐

(1) 9𝑥𝑎 + 5𝑥𝑐 ≤ 450

(2) 50𝑥𝑎 ≤ 2000

(3) 0.1𝑥𝑐 ≤ 5

(4) 𝑥𝑎 + 𝑥𝑐 ≤ 70

(5) 𝑥𝑎 ≥ 0

(6) 𝑥𝑐 ≥ 0

Here are the steps of finding optimal solution with Graphical Solution Method:

1. Converts all constraints into equality equation

(1) 9𝑥𝑎 + 5𝑥𝑐 = 450

(2) 50𝑥𝑎 = 2000

(3) 0.1𝑥𝑐 = 5

(4) 𝑥𝑎 + 𝑥𝑐 = 70

(5) 𝑥𝑎 = 0

(6) 𝑥𝑐 = 0

2. Draw straight lines corresponds to each equation into the coordinate system. Let x-axis represent 𝑥𝑎 and y-axis represent 𝑥𝑐

Note that for constraints (5) and (6) corresponds to 𝑥𝑎-axis and 𝑥𝑐-axis respectively

3. Determine the feasible solution space

Feasible Solution space is identifiable by observing the inequality signs. Constraint (1) to (4) have “less than equals to” inequality signs, hence the feasible space is the left side (for y-axis) and bottom side (for x-axis) of each constraint lines.

On the other hand, constraint (5) and (6) have “more than equals to” signs, hence the feasible space right side and upper side from the constraint lines.

4. Draw arbitrary objective function line

Define 𝑥0 value arbitrarily for the objective functions. For example, let 𝑥0 have the value 120. Then draw the objective functions to graph.

It is clear that the arbitrary objective function value is in the feasible solution space, however the value is still un-ideal because other combinations of 𝑥𝑎 and 𝑥𝑐 value could produce higher objective function value. In graphical solution method, the optimal solution lies at the outermost point of the feasible solution space (marked with blank point on the graph).

5. Shift the objective functions line parallel to the outermost point of the feasible solution space and determine the optimal solution.

The objective function is now at the optimal solution since it intersects with the outermost point in the feasible solution space. Note that this point also the intersection between constraint (1) and (2). Since each equation consist of two variables then it can be solved mathematically and the result is 40 for 𝑥𝑎 and 18 for 𝑥𝑐.

Based on graphical solution, the owner should sell 40 affogato and 18 café latte each day two maximize profit that is 40 × €6 + 18 × €3 = €294. This method provides simple graphical representation for solving the problem and easy to implement. However, it becomes problematic when the problem has huge variables and constraints size. In the next part, we discuss another approach for solving this problem namely Simplex Method.