# Mastering Linear Programming: A Step-by-Step Guide to Graphical Solutions

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## Introduction

In this comprehensive guide to linear programming, we will focus on the graphical method as a powerful tool for solving optimization problems. Linear programming is widely used in various fields for decision-making processes, particularly when evaluating how to maximize or minimize an objective function while maintaining constraints. This discussion will include understanding the graphical solution method, identifying feasible regions, and interpreting slack variables.

### What is Linear Programming?

Linear programming involves the optimization of a linear objective function, subject to linear equality or inequality constraints. It’s used in production, transportation, finance, and many other sectors. The aim is often to maximize profits or minimize costs, establishing the conditions under which this can be achieved.

### The Graphical Method of Linear Programming

The graphical method is a visual way to analyze linear programming problems, particularly effective when there are only two decision variables. Let’s dive deeper into the steps involved in solving a maximization problem using this method.

## Step-by-Step Approach to Graphical Solutions

### 1. Formulating the Problem

To illustrate how to solve a linear programming problem graphically, we will consider the following example:

**Objective Function:**Maximize Profit = 10S + 90D**Constraints:**- $$\frac{7}{10}S + D \leq 630$$
- $$S + rac{2}{3}D \leq 708$$
- $$D \leq 135$$
- $$S, D \geq 0$$

Where S stands for quantity of one product, and D represents another.

### 2. Graphing the Constraints

#### Plotting Individual Constraints

- Start by plotting each constraint line by converting inequalities to equalities. For example, for the first constraint:

$$\frac{7}{10}S + D = 630$$

- Identify points by setting S or D to zero:
- If S=0, then D=630.
- If D=0, then S=900.

By repeating this process for each constraint, you get:

**First Constraint:**Line segments from (0, 630) to (900, 0)**Second Constraint:**From points derived by setting variables accordingly and calculating intersections.

### 3. Identifying the Feasible Region

The next step is to shade the feasible region based on the inequalities of the constraints. The feasible region includes all points that satisfy all constraints simultaneously.

#### Determining Feasibility

To identify which side of the line satisfies the inequality, pick test points. If the test point satisfies the inequality, shade the area where the corresponding values lie. For instance, if using the point (200, 200) satisfies a given constraint, shade that area.

### 4. Finding the Optimal Solution

Once all constraints are graphed and the feasible region is identified, we move to the objective function. The objective function line itself can be graphed. By selecting values for the profit level, you can draw parallel lines across the feasible region.

### 5. Trial and Error for Optimal Solutions

By systematically moving the profit line outward, you can identify the last point at which it remains within the feasible region. The corner points of this region often have potential as optimal solutions. The final optimal solution is achieved when the objective function's line is tangent to the feasible region boundary, essentially maximizing your output. In our case, substituting values from the derived corner points into the profit equation will give the highest profit achievable — for instance, achieving a profit of $7668 with particular amounts of S and D.

### Understanding Slack Variables and Constraints

**Slack Variables:**They represent unused resources in your constraints. For instance, if the sieving time has 120 hours not utilized, this represents a slack in this area. These give insights into constraints where resources aren't fully utilized.**Binding vs Non-Binding Constraints:**Binding constraints have no slack (fully utilized), while non-binding constraints allow for slack to exist. Recognizing these helps in understanding which constraints truly limit the maximum potential of your objective function.**Redundant Constraints:**Some constraints might not affect the feasible region but should be taken into account as they can become binding under different conditions.

### Conclusion

In summary, solving linear programming problems using graphical methods enables professionals to visualize complex situations and identify optimal solutions effectively. The integration of slack variables, binding constraints, and the identification of feasible regions is essential in reaching maximum outcomes. In forthcoming discussions, we will explore graphical calculators like Desmos and computational tools like Excel Solver to solve problems with greater complexity and efficiency.

This step-by-step guide is intended for decision-makers and analysts looking to optimize their strategic planning and enhance their operations through linear programming. By following these methods, you can efficiently address real-world problems and maximize your organization’s potential profits.

ok a linear programming problem involving only two decision variables can be solved using a graphical solution

equal to 10 s plus 90 subjected to there are four constraint is there i will explain how to solve this problem

on the graph can be identified by the s and d values which indicate the position of the point

corresponds to a possible solution every point on the graph is called solution point the solution point where

because s and d must be non negative the graph in figure 1 only displays positive solutions

say s equal to two hundred d equal to eight hundred and if it is this point s equal to four

hundred and d equal to 300 this is the one solution point the next stage is plotting all the constraint

into the graph so let us see how to plot the constraint number one that is cutting and dyeing

so i have taken the constraint number one seven upon ten s plus one d less than or equal to six thirty

to show all solution point that satisfy this relationship we start by graphing the solution

it can be obtained by identifying two points that satisfy this equation and then drawing a line through the

so ok if you substitute is equal to zero say d equal to six thirty so we got one point zero comma six thirty

you have to substitute these values into this equation if this point is satisfied you have to

consider two point s equal to two hundred d equal to two hundred and s equal to six hundred and d equal

to five hundred what i have explained in the previous slides you can see from the figure two that the

first solution point is below the constraint line and the second point is above the constraint line which of these

hundred and d equal to two hundred we are getting in the values of three forty so this three forty is less than six

but the nine hundred and twenty is greater than six thirty that is violating our constraint so

if a solution point is not feasible for a particular constraint then all other solutions point on the

feasible for a particular constraint then all other solution points on the same side of the constraint line are

feasible for that constraint so what is the point here is suppose if you take any one point

if it is satisfying our constraint right so this is less than or equal to this equal to represents only the

straight line when you write less than or equal to this shaded line which is shaded in blue

so one has to evaluate the constraint function for only one solution point one solution point to determine which

we have to see which side of this constraint line is feasible whether it can be either either on the left hand

side or the right hand side in this figure 3 we indicate all the points satisfying the cutting and dying

constrained by the shadowed region so this region is the region which satisfy our constraint

the same way i have taken the another constraint that is sieving constraint so this less than or equal to six

hundred this less than or equal to 708 less than or equal to 135 so these points re satisfy our

kind of figure so the area which is common for all the constraint is called feasible region

any point if you take any point that point will satisfy all our constraint feasible solution and

feasible region the shaded region in this figure include every solution point that satisfy all

feasible solutions and the shaded region is called feasible solution region or simply the feasible

feasible solu feasible region is a feasible solution point so in this boundary this is a feasible

the z one because it is a maximization problem the z two is the point corresponding to z two

the difficulty with this approach is the infinite number of feasible solutions are possible so checking all the points

and verifying whether which is the optimal value is very difficult one because one cannot possibly evaluate an

infinite number of feasible solutions so this trial and error procedure cannot be used to identify the optimal solution

so we are going to find out an another method what we are going to do we are going to select some arbitrary value for

which feasible solution provide a profit of thousand eight hundred dollar so i have fixed suppose if i want to have my

profit contribution is thousand eight hundred dollar what should be what should be the value of s and d

these solutions are given by the values of s and d in the feasible region that will make the objective function

solution for an arbitrary value of thousand eight hundred dollar of your profit contribution

eight hundred dollar ok so this is the line of ten years plus ninety equal to thousand eight hundred

dollar if i want to make this much profit contribution i have to find out what is the value of

axis and y axis e i can find out where it is intersecting otherwise in this equation if i substitute s equal to 0 i

this expression in simply the equation of a line which one ten in ten years plus ninety equal to thousand 800

value of s and d will satisfy this our objective function it is thousand eight hundred the procedure for

graphing the profit or objective function line is the same what i have explained previously

so when you substitute s equal to zero the d will be two hundred so we got one point this point

zero comma two hundred similarly when you substitute d equal to zero ok we will get another point

these two points identifies all solution that have a profit contribution of 1800 so the graph of this profit line is

we have taken only one value similarly i can take another profit contribution value arbitrarily three thousand six

this line ten years plus ninety equal to three thousand six hundred 10 years plus 90 equal to 5400

because the objective function is to find the feasible solution yielding the largest profit contribution

let us proceed by selecting higher profit contribution and finding the solution yielding the

that are on the following lines so these are the two lines right we have to find the all the points which are in these

feasible region at least some points on the line are and it is therefore possible to obtain a

it is not covering see this is covering only from this point to this point this region is not covered by this line

but all the points in this region will satisfy your objective function can we find a feasible solution yielding an

even higher profit contribution please look at the figure 8 and see what general observations you

farther from the origin we are getting the higher profit line so when you go this way it may be five thousand

six hundred this may be something else so what are the two point all the objective functions are parallel

slope is minus 10 by 9. so when p equal to thousand eight hundred when you substitute in the

previous values you will getting minus ten by nine yes two hundred when you increase p equal to two three thousand

so what is the similarity you are saying even though the value of the objective function the right hand side is

increasing the slope remains constant minus ten by nine minus ten by nine minus ten by nine so this implies that

the objective function lines are parallel to each other so the slope minus ten by nine is the

same for each profit line because profit lines are parallel further we see that the d intercept

from the origin that we have understood algebraically next we will look at how to find out the

optimal solutions because the profit lines are parallel and higher profit lines are farther from

parallel lines and when we move away from this origins so there is a possibility the value of

our objective function will increase however at some point we will find that any further outward movement will place

the profit line completely outside the feasible region so when we move away from the origins

because solutions outside the feasible regions are unacceptable the point in the feasible region that

you have to draw different objective function line we have to keep on move away from the origin at the particular

as you can so what you can do you can move this way you can move this way you can move this way

extreme point is our optimal solution so what is this point how we are getting this point there are two constraints are

that two lines where it is intersecting that corresponding point is called your optimal solution when you substitute

so there are two constraints so wherever the right hand side is 630 which one that is a cutting dying

because i want to know the value of s and d i am taking equal to sign only not less than or equal to because otherwise

intersecting is your finishing constraint so i have taken one by one s plus two upon three d seven zero eight

d plus two upon three d so d is common you take on the left hand side the remaining is minus ten upon

twenty one d equal to minus one hundred ninety two so the d value at the end you are getting two fifty two

so i have solved the two constraint to get this point so 540 minus comma 252 when you substitute this value

50 units of the professional model for its own sales force and expects sales of the professional

so what the constraint enforcing this requirement is p minus 50 because this 50 is consumed

assistant tablet so less than or equal to 1 by 2 a when you simplify this will be getting 2 p minus 100 less than or

plot it so substitute p equal to zero will getting a equal to minus hundred so this is the point

so when you substitute a equal to zero you will get p equal to fifty so this is another point so what is the point you

you may get your point that point may lie on the negative sides but at the time of considering the

so if r equal to hundred t also hundred that is the meaning that is a way of drawing the constraint

d equal to 252. so in the constraint number one when you substitute it the resources used is 630

but the hours available is six thirty so all the resources are fully utilized that is unused resources

so this slack variable for this first constraint is zero look at the second constrained sieving

consumed is 480 but the available resources is 600. so 100 120 unit hours not utilized so here the slack

the complete solution tells management that the production of 540 standard bags and 252 deluxe bags will require

so that means there won't be any unutilized resources so in these department all the resources

135 minus 117 so 18 hours of inspection packaging time will remain unused so what we are understanding out of this

for another two constraint we have some positive slack variables if it is a positive slack variables in

slack for the two department in linear programming terminology any unused capacity for a constraint less

so in a linear programming model if the constraint is the type of less than or equal to type

the corresponding unutilized resources called slack variable often variables called slack variables

are added to the formulation of a linear programming problem to represent the slack or ideal capacity

so that the slack unutilized resources has to be added in the linear programming problem so

is zero so in our objective function the coefficient is zero s one zero s two zero s three and zero s four

what is the binding constraint we can see that the cutting and dying and the finishing constraint restrict or

finishing constraint and cutting and dying constrain so this is restricting our optimal solutions beyond which it

what are they finishing and cutting and dying in other words the graph shows as that the cutting and dying and the

on other hand sieving and inspection packaging constraint are not binding the feasible region see the

solution which means we can expect some unused time or slog or slack for these two operations