Graphical Method: Owing to the importance of linear programming models in various industries, many types of algorithms have been developed over the years to solve them. Some famous mentions include the Simplex method, the Hungarian approach, and others. Here we are going to concentrate on one of the most basic methods to handle a linear programming problem i.e. the graphical method.

In principle, this method works for almost all different types of problems but gets more and more difficult to solve when the number of decision variables and the constraints increases. Therefore, we’ll illustrate it in a simple case i.e. for two variables only. So let’s get started with the graphical method!

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## The Graphical Method

We will first discuss the steps of the algorithm:

### Step 1: Formulate the LP (Linear programming) problem

We have already understood the mathematical formulation of an LP problem in a previous section. Note that this is the most crucial step as all the subsequent steps depend on our analysis here.

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### Step 2: Construct a graph and plot the constraint lines

The graph must be constructed in ‘n’ dimensions, where ‘n’ is the number of decision variables. This should give you an idea about the complexity of this step if the number of decision variables increases.

One must know that one cannot imagine more than 3-dimensions anyway! The constraint lines can be constructed by joining the horizontal and vertical intercepts found from each constraint equation.

### Step 3: Determine the valid side of each constraint line

This is used to determine the domain of the available space, which can result in a feasible solution. How to check? A simple method is to put the coordinates of the origin (0,0) in the problem and determine whether the objective function takes on a physical solution or not. If yes, then the side of the constraint lines on which the origin lies is the valid side. Otherwise it lies on the opposite one.

### Step 4: Identify the feasible solution region

The feasible solution region on the graph is the one which is satisfied by all the constraints. It could be viewed as the intersection of the valid regions of each constraint line as well. Choosing any point in this area would result in a valid solution for our objective function.

### Step 5: Plot the objective function on the graph

It will clearly be a straight line since we are dealing with linear equations here. One must be sure to draw it differently from the constraint lines to avoid confusion. Choose the constant value in the equation of the objective function randomly, just to make it clearly distinguishable.

### Step 6: Find the optimum point

## Optimum Points

An optimum point always lies on one of the corners of the feasible region. How to find it? Place a ruler on the graph sheet, parallel to the objective function. Be sure to keep the orientation of this ruler fixed in space. We only need the direction of the straight line of the objective function. Now begin from the far corner of the graph and tend to slide it towards the origin.

- If the goal is to minimize the objective function, find the point of contact of the ruler with the feasible region, which is the closest to the origin. This is the optimum point for minimizing the function.
- If the goal is to maximize the objective function, find the point of contact of the ruler with the feasible region, which is the farthest from the origin. This is the optimum point for maximizing the function.

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### Step 7: Calculate the coordinates of the optimum point.

This is the last step of the process. Once you locate the optimum point, you’ll need to find its coordinates. This can be done by drawing two perpendicular lines from the point onto the coordinate axes and noting down the coordinates.

Otherwise, you may proceed algebraically also if the optimum point is at the intersection of two constraint lines and find it by solving a set of simultaneous linear equations. The Optimum Point gives you the values of the decision variables necessary to optimize the objective function.

To find out the optimized objective function, one can simply put in the values of these parameters in the equation of the objective function. You have found your solution! Worried about the execution of this seemingly long algorithm? Check out a solved example below!

## Solved Examples for You

**Question 1: A health-conscious family wants to have a very well controlled vitamin C-rich mixed fruit-breakfast which is a good source of dietary fibre as well; in the form of 5 fruit servings per day. They choose apples and bananas as their target fruits, which can be purchased from an online vendor in bulk at a reasonable price.**

**Bananas cost 30 rupees per dozen (6 servings) and apples cost 80 rupees per kg (8 servings). Given: 1 banana contains 8.8 mg of Vitamin C and 100-125 g of apples i.e. 1 serving contains 5.2 mg of Vitamin C.**

**Every person of the family would like to have at least 20 mg of Vitamin C daily but would like to keep the intake under 60 mg. How much fruit servings would the family have to consume on a daily basis per person to minimize their cost?**

**Answer :** We begin step-wise with the formulation of the problem first. The constraint variables – ‘x’ = number of banana servings taken and ‘y’ = number of servings of apples taken. Let us find out the objective function now.

- Cost of a banana serving = 30/6 rupees = 5 rupees. Thus, the cost of ‘x’ banana servings = 5x rupees
- Cost of an apple serving = 80/8 rupees = 10 rupees. Thus the cost of ‘y’ apple servings = 10y rupees

Total Cost C = 5x + 10y

Constraints: x ≥ 0; y ≥ 0 (non-negative number of servings)

Total Vitamin C intake:

8.8x + 5.2y ≥ 20 (1)

8.8x + 5.2y ≤ 60 (2)

### Now let us plot a graph with the constraint equations-

To check for the validity of the equations, put x=0, y=0 in (1). Clearly, it doesn’t satisfy the inequality. Therefore, we must choose the side opposite to the origin as our valid region. Similarly, the side towards origin is the valid region for equation 2)

Feasible Region: As per the analysis above, the feasible region for this problem would be the one in between the red and blue lines in the graph! For the direction of the objective function; let us plot 5x+10y = 50.

Now take a ruler and place it on the straight line of the objective function. Start sliding it from the left end of the graph. What do we want here? We want the minimum value of the cost i.e. the minimum value of the optimum function C. Thus we should slide the ruler in such a way that a point is reached, which:

1) lies in the feasible region

2) is closer to the origin as compared to the other points

This would be our Optimum Point. I’ve marked it as P in the graph. It is the one which you will get at the extreme right side of the feasible region here. I’ve also shown the position in which your ruler needs to be to get this point by the line in green.

Now we must calculate the coordinates of this point. To do this, just solve the simultaneous pair of linear equations:

*y = 0*

8.8x + 5.2y = 20

We’ll get the coordinates of ‘P’ as (2.27, 0). This implies that the family must consume 2.27 bananas and 0 apples to minimize their cost and function according to their diet plan.

**Question 2: What is the purpose of a graphical method?**

**Answer:** We use a graphical method of linear programming for solving the problems by finding out the maximum or lowermost point of the intersection on a graph between the objective function line and the feasible region.

**Question 3: How do you solve the LPP with the help of a graphical method?**

**Answer:** We can solve the LPP with the graphical method by following these steps:

**1st Step:** First of all, formulate the LP problem.

**2nd Step:** Then, make a graph and plot the constraint lines over there.

**3rd Step:** Determine the valid part of each constraint line.

**4th Step:** Recognize the possible solution area.

**5thStep:** Place the objective function in the graph.

**6th Step:** Finally, find out the optimum point.

**Question 4: Define the graphical method for the simultaneous equations?**

**Answer:** For graphically solving a pair of simultaneous equations, firstly we have to draw a graph of both the equations simultaneously. We have 2 straight lines crossing each other at a common point which provides the solution of this pair of equations.

**Question 5: What is a graphical interpretation?**

**Answer:** A graphical interpretation proposes a number of valuable problem-solving methods. For example, finding the greatest value of a nonstop differentiable function ‘f(x)’ defined in some interval ‘an ≤ x ≤ b’.