Linear Programming Problem, Study notes of Mathematics

Download Linear Programming Problem and more Study notes Mathematics in PDF only on Docsity!

Linear Programming

UPSC Mathematics Optional

Asiya Shaikh

AROUSH ACADEMY

Linear Programming Problem (LPP)-Model Formulation

Definition

Linear programming (LP) is a mathematical optimization technique used to find the best outcome in a

mathematical model with linear relationships, subject to a set of constraints. The goal of linear programming is

to maximize or minimize a linear objective function while satisfying a system of linear inequalities or equations.

In simple terms, Linear Programming problem is a method to optimize a linear objective function subject to

constraints represented by linear equations or inequalities.

Key components of a linear programming problem:

1. Objective Function : It is a linear equation that represents what we want to optimize (either minimize or

maximize).

It is usually written in the form "𝑎

1

1

2

2

2

3

𝑛

𝑛

" where 𝑥

1

2

𝑛

are decision variables &

1

2

𝑛

are coefficients defining their impact on optimization.

2. Decision variables : These are the variables that we can control or adjust to achieve the desired outcome.

They are typically represented as 𝑥, 𝑦, 𝑧, etc., and can take on real values.

3. Constraints : These are a set of linear equations or inequalities that restrict the values of the decision

variables. Constraints represent limitations or requirements of the problem.

4. Non-negativity restriction: It refers to a constraint that specifies that decision variables must take on

non-negative values.

5. Feasible Region : The feasible region is the set of all possible combinations of values for the decision

variables that satisfy all the constraints. It is the region in which a solution to the problem must lie.

6. Optimization Goal : The ultimate goal of LPP is to find the values of the decision variables that either

maximize or minimize the objective function while staying within the feasible region. That is, maximize or

minimize some numerical value representing profit, cost, production quantity, etc.

Formulation of a Linear Programming Problem

The standard form of a linear programming problem is

Decision variables : 𝑥

1

2

3

𝑛

:(which need to be found)

Objective Function : Maximize / Minimize: 𝒵 = 𝑐

1

1

2

2

3

3

𝑛

𝑛

Constraints :

Subject to : 𝑎

11

1

12

2

13

3

1 𝑛

𝑛

1

21

1

22

2

23

3

2 𝑛

𝑛

2

𝑚 1

1

𝑚 2

2

𝑚 3

3

𝑚𝑛

𝑛

𝑚

Non negativity : 𝑥

1

2

3

𝑛

Here 𝑎

11

12

1 𝑛

21

22

2 𝑛

𝑚𝑛

are input output constraints and

1

2

3

𝑚

are capacities where ∀ 𝑏

𝑖

≥ 0 and

1

2

3

𝑚

are cost or profit coefficients.

Note : Subject to may be any one of the following ≤, =, 𝑜𝑟 ≥. And here 𝑛 = number of decision variables and

𝑚 = number of constraints.

The linear programming can be stated in matrix form as follows:

To find the Decision variables : 𝑥 1

2

3

𝑛

and to optimize the

Objective Function : Max/Min: 𝒵 = 𝐂𝓍

Subject to : 𝐀𝓍 (≤ = ≥ )𝐛

Non negativity : 𝓍 ≥ 𝟎

Where 𝐀 = [𝑎 𝑖𝑗

]

𝑚×𝑛

is called coefficient matrix, where 𝑎

𝑖𝑗

indicates the amount of 𝑖

𝑡ℎ

type of resource

necessary to manufacture one unit of product 𝑗, 𝐂 =

[

1

2

3

𝑚

]

are cost vector or sometime called as profit

matrix , 𝟎 is null matrix of type 𝑛 × 1 ,

B gives a return of 15% but has a risk factor of 8. Total amount invested is ₹ 5 , 00 , 000. Total minimum return on

investment should be 12%. Maximum combined risk should be more than 6. Formulate the LPP.

Solution :

Decision variables are:

1

= Amount invested in A

2

= Amount invested in B

The Objective function is

Maximize 𝒵 = 0. 09 𝑥

1

2

Constraints

Related to total investment: 𝑥

1

2

Related to return: 0. 09 𝑥

1

2

= 0. 12 × 5 , 00 , 000 = 60 , 000

Related to Risk: 5 𝑥

1

2

= 6 × 5 , 00 , 000 = 30 , 00 , 000

Non negativity : 𝑥

1

2

Inspection Problem

Q : In the context of quality control inspection, a company employs two grades of inspectors, Grade I and Grade

II. The task is to perform quality control inspections, with a requirement of inspecting at least 1 , 500 pieces in

an 8 - hour workday. Grade I inspectors can examine 20 pieces per hour with an accuracy rate of 96%, while

Grade II inspectors can check 14 pieces per hour with an accuracy rate of 92%.

The company incurs different wages for these inspectors, with Grade I inspectors earning ₹ 5 per hour & Grade

II inspectors earning ₹ 4 per hour. Additionally, any inspection error by an inspector result in a cost of ₹ 3 to the

company. Given that the company employs 10 Grade I inspectors and 15 Grade II inspectors, the objective is to

determine the optimal assignment of inspectors to minimize the daily inspection cost.

Solution :

Hourly cost of a Grade I inspector: ₹( 5 + 3 × 0. 04 × 20 ) = ₹ 7. 40

Hourly cost of a Grade II inspector: ₹( 4 + 3 × 0. 08 × 14 ) = ₹ 7. 36

Decision variables are:

1

= Number of Grade I inspector

2

= Number of Grade II inspector

The Objective function is

Minimize 𝒵 = 7. 40 𝑥

1

2

: daily inspection cost

Constraints

Number of Grade I inspector: 𝑥

1

Number of Grade I inspector: 𝑥

2

Number of pieces Inspected daily: 160 𝑥

1

2

Non negativity : 𝑥

1

2

Trim Loss Problem:

Q: A manufacturer of cylindrical containers receives tin sheets in widths of 30 cm and 60 cm. These sheets need

to be cut into three different widths of 15 cm, 21 cm, & 27 cm to manufacture containers. The production plan

involves creating 400 containers of 15 cm width, 200 containers of 21 cm width, & 300 containers of 27 cm width.

The manufacturer purchases bottom plates and top covers separately, and there are no limitations on the

lengths of standard tin sheets. The goal is to formulate a Linear Programming Problem (LPP) to determine the

production schedule that minimizes trim losses.

Solution :

Decision variables : Let 𝑥

𝑖𝑗

represent the cutting combinations where each combination results in a

certain amount of trim loss.

Possible cutting combinations for both types of sheets are:

Width 30 cm 60 cm

11

12

13

21

22

23

24

25

26

Loss cm 0 9 3 0 9 3 3 12 6

The Objective function is

Minimize 𝒵 = 0 𝑥

11

12

13

21

22

23

24

25

26

Constraints

11

21

22

23

24

21

22

24

25

13

23

25

26

Non negativity : 𝑥

11

12

13

21

22

23

24

25

26

Media Selection Problem

Q : An advertising agency is planning to launch an advertising campaign across various media platforms, including

T.V., Radio, & Newspaper. Each medium has different reach potential & costs associated with it. The agency has

specific requirements, including reaching a minimum of 1 , 00 , 00 , 000 households through T.V. advertising, keeping

newspaper expenditure below ₹ 10 , 00 , 000 , & staying within a total advertising budget of ₹ 20 million.

Medium Cost per Unit Reach per Unit

TV 2 , 00 , 000 20 , 00 , 000

Radio

Newspaper 40 , 000 2 , 00 , 000

Solution :

Decision variables are:

1

= Number of units of TV ads

2

= Number of units of Radio Ads

3

= Number of units of Newspaper

The Objective function is

Maximize 𝒵 = 20 , 00 , 000 𝑥

1

2

3

: daily reach

Constraints

Total cost budget: 2 , 00 , 000 𝑥

1

2

3

On TV for reaching units: 20 , 00 , 000 𝑥

2

On Newspaper expenditure: 40000 𝑥

2

Non negativity : 𝑥

1

2

3

Diet Problem

Q : The diet problem involves selecting foods to meet specific nutritional requirements while minimizing the cost

of the diet. In this particular scenario, we are dealing with vitamins 𝐵 1

and 𝐵

2

, which are found in two foods, 𝐹

1

and 𝐹 2

. The nutritional content and costs of these foods are given as follows:

▪ 1 unit of 𝐹

1

contains 3 units of 𝐵

1

and 4 units of 𝐵

2

▪ 1 unit of 𝐹

2

contains 5 units of 𝐵

1

and 3 units of 𝐵

2

The minimum daily prescribed consumption of 𝐵 1

and 𝐵

2

is 50 units & 60 units, respectively. The cost per unit of

1

and 𝐹

2

is ₹ 6 and ₹ 3 , respectively.

Solution :

Decision variables are:

1

= Number of units of food 𝐹

1

consumed per day.

2

= Number of units of food 𝐹

2

consumed per day.

The Objective function is

Minimize 𝒵 = 6 𝑥

1

2

: the cost of diet

Constraints

For 𝐵

1

consumption: 3 𝑥

1

2

≥ 50 (Minimum daily consumptions)

For 𝐵

2

consumption: 4 𝑥

1

2

≥ 60 (Minimum daily consumptions)

Non negativity : 𝑥

1

2

Blending Problem

Q : In the blending problem, a manager seeks to optimize the blend of two processes, 𝑃 1

and 𝑃

2

, to maximize

profit while accounting for input constraints and minimum gasoline demand. This is formulated as a Linear

Programming Problem (LPP) with provided data including crude oil inputs, gasoline outputs, profit per operation,

and availability constraints.

LPP Solutions

Convex Set

It refers to a set of points in which any line segment connecting two points within the set is also entirely

contained within the set.

In mathematical terms, A subset 𝑺 of ℝ

𝑛

space is said to be convex set if for all pairs 𝑥, 𝑦 ∈ 𝑺 any linear

combination { 𝑚𝑦 + ( 1 − 𝑚)𝑥 ; ( 0 ≤ 𝑚 ≤ 1 )}, is also contained in 𝑺. This property is called convexity of the set.

i.e. ∀ 𝑥, 𝑦 ∈ 𝑺 ⟹ 𝑚𝑦 + ( 1 − 𝑚)𝑥 ∈ 𝑺

Convex Region

It is the special type of convex set that is bounded by a finite number of hyperplanes. Otherwise it is concave.

Solution in Linear Programming

A solution to a linear programming problem is defined as a set of variables {𝑥 1

2

3

𝑛

} that adheres to &

satisfies the given set of constraints.

Feasible region

Refers to the set of all points that satisfy the constraints of the problem. In other words it is the region in

the scope of the decision variables where the solution to the problem can lie. For an LPP it is always a convex

region.

Feasible Solution

A set of values for the decision variables that satisfies all the constraints of the problem. In other words it is

a solution that lies within the feasible region.

Infeasible solution

An infeasible solution in linear programming refers to a solution that does not satisfy one or more of the

problem's constraints, making it impossible to implement within the given constraints.

Basic Solution

A basic solution is a solution in an LP problem where a subset of decision variables (equal to the number of

constraints) is set to zero, while the remaining variables are determined based on the constraints. This results

in a unique solution that lies on the boundary of the feasible region.

The variables whose values are set to zero in the solution and do not directly affect the objective function's

value are called Non-basic variables. Basic variables are those whose values are determined by the solution to

the system of equations, representing feasible solutions.

Note : In the system of simultaneous linear equations 𝐀𝓧 = 𝐛, where 𝐀 is an 𝑚 × 𝑚 matrix (rank (𝐀)= 𝑚), if 𝑘 =

𝑛 − 𝑚 variables not associated with the 𝑚 × 𝑚 matrix are set to zero, the resulting solution is referred to as

the basic solution

Basic Feasible Solution

A basic feasible solution is a basic solution in which all the basic variables are set to non-negative values , and

it satisfies all the constraints of the LP problem. Basic feasible solutions are typically located at the vertices

or extreme points of the feasible region.

“Basic” because it satisfies a subset of the constraints in the problem and “feasible” because it satisfies all the

constraints.

Non-Basic Feasible Solution:

A non-basic feasible solution is a solution in which not all decision variables are set to zero , and it still

satisfies all the constraints of the LP problem. This type of solution lies within the feasible region but not

necessarily at a vertex.

Types of Basic Solution:

There are two main types of basic solutions in LP:

a) Degenerate Basic Solution:

In a degenerate basic solution, at least one of the basic variables takes on the value zero. This

situation often occurs when the current solution is at a corner or vertex of the feasible solution region.

b) Non-Degenerate Basic Solution:

In a non-degenerate basic solution, all the basic variables have positive values. These solutions are

typically found in the interior of the feasible region and do not involve zero values for any basic

variables.

Optimum or Optimal Solution

An optimal solution in linear programming (LP) is a feasible solution that maximizes (or minimizes) the objective

function while satisfying all constraints. It represents the best possible outcome within the problem's

constraints. It is also called Optimal feasible solution.

Types of Optimal Solutions in LP

▪ Unique Optimal Solution:

There is only one feasible solution that optimally maximizes (or minimizes) the objective function.

▪ Multiple Optimal Solutions:

Multiple feasible solutions yield the same optimal objective function value within the problem's constraints.

Unbounded Solution

An unbounded solution in linear programming occurs when there is no finite upper or lower limit to the objective

function value. In other words, the objective function can be continuously increased (for maximization) or

decreased (for minimization) without violating any constraints because the feasible region extends infinitely in

one or more directions.

This situation indicates that the LP problem does not have a finite optimal solution

Corner Point Solution

A corner point solution in LP is a feasible solution that lies at one of the vertices or extreme points of the

feasible region. These solutions are characterized by having a subset of the decision variables fixed at their

extreme values (usually zero or a positive value) while the rest are determined by the constraints.

3

2

4

1

3

1

3

3

1

3

Yes No

No

3

4

1

2

1

2

2

1

2

Yes No

Yes

The basic solutions are: ( 0 , 0 , 3 , 2 ), ( 0 , 6 , 0 , 2 ), (− 12 , 0 , 0 , 8 ), ( 0 , 0 , 4 , 4 ) & ( 4 , 8 , 0 , 0 )

The basic feasible solutions are: ( 0 , 0 , 3 , 2 ), ( 0 , 6 , 0 , 2 ), ( 0 , 0 , 4 , 4 ) & ( 4 , 8 , 0 , 0 )

The non generated basic feasible solutions are: ( 0 , 0 , 3 , 2 ), ( 0 , 6 , 0 , 2 ), ( 0 , 0 , 4 , 4 ) & ( 4 , 8 , 0 , 0 )

It has no degenerated basic feasible solution.

The optimal basic feasible solutions are: ( 0 , 6 , 0 , 2 ) & ( 4 , 8 , 0 , 0 )

Initial basic feasible solution

An initial basic feasible solution serves as the initial configuration in algorithms such as the Simplex algorithm,

which aims to determine the optimal solution for a Linear Programming Problem (LPP). This solution must adhere

to two essential properties:

a) It must be feasible, and

b) It should have a number of basic variables equal to the number of constraints in the problem.

Example : Maximize: 𝒵 = 3 𝑥 + 4 𝑦

Subject to: 𝑥 + 𝑦 ≤ 5

Where 𝑥

1

2

One possible initial basic feasible solution is 𝑥 = 0 and 𝑦 = 0 , with 𝑠 1

= 5 and 𝑠

2

= 8. This solution satisfies the

constraints and has two basic variables, 𝑠 1

and 𝑠

2

, which corresponds to two constraints.

Note :

▪ A solution in which the vectors associated with 𝑚 variables are linearly independent and remaining 𝑘 = 𝑛 −

𝑚 are zero are called as basic solution.

▪ If 𝐵 is the matrix of 𝑚 linearly independent vectors of 𝐀 and 𝓍 𝐵

is the corresponding variables or basic

variables then the basic solution is given by 𝐵𝓍

𝐵

= 𝐛 or 𝓍

𝐵

− 1

Theorem

A necessary and sufficient condition for the existence and non-degeneracy of all basic solutions of the linear

system 𝐀𝓍 = 𝐛 is that every set of 𝑚 columns of the augmented matrix 𝐀𝐛 =

[

]

is linearly independent.

Corollary

The necessary and sufficient condition for any given basic solution 𝓍 𝐵

− 1

𝐛 to be non-degenerate is the linear

dependence of 𝐛 and every (𝑚 − 1 ) columns from 𝐵.

Example : To demonstrate that the feasible solution 𝑥 1

2

3

= 1 and 𝒵 = 6 to the system of equations:

1

2

3

1

2

3

For 𝑥

1

2

3

which minimizes 𝒵 = 2 𝑥

1

2

3

, is not basic.

Solution :

Here we have 𝑛 = 3 variables and 𝑚 = 2 constraints.

In the given feasible solution, there are only two (𝑚 = 2 ) non-zero variables, namely 𝑥

1

and 𝑥

3

and 𝑘 =

𝑛 − 𝑚 = 3 − 2 = 1 variable 𝑥

2

is zero.

Let the vectors associated with these variables be 𝑣

1

= [

] and 𝑣

2

= [

].

The determinant of column vectors corresponding to variables 𝑥

1

and 𝑥

3

is |

This implies that the vectors 𝑎

1

and 𝑎

2

are linearly dependent.

Hence the solution is not a basic solution.

Q : Find all the basic solution of the following system and prove that they are non-degenerate.

1

2

3

= 4 and 2 𝑥

1

2

3

Solution : The given system of equation can be expressed in matrix form as 𝐀𝓍 = 𝐛 where matrix 𝐀 = (𝑣 1

2

3

So that 𝑣

1

= [

], 𝑣

2

= [

] and 𝑣

3

= [

].

Here, 𝑛 = 3 variables and 𝑚 = 2 constraints resulting

3

2

= 3 basic solutions. Consequently, there are 2

basic variables and 1 non-basic variable, which is set to zero.

Now, we have three sets of two vectors formed:

1

= [𝑣

2

3

] = [

], 𝐵

2

= [𝑣

1

3

] = [

], and 𝐵

3

= [𝑣

1

2

] = [

],

All of these matrices have non-zero determinants, indicating that each set of two vectors is linearly

independent. Consequently, all three basic solutions exist. The corresponding basic variables are

calculated as follows:

𝐵

1

1

− 1

𝟏

𝟗

[

] [

] = [

], 𝓍

𝐵

2

2

− 1

𝟏

𝟑

[

] [

] = [

] & 𝓍

𝐵

3

3

− 1

𝟏

𝟑

[

] [

] = [

]

In basis 𝐵

1

, the basic vectors are 𝑣

2

3

and non-basic variable is 𝑣

1

resulting in the basic solution

( 0 , 5 / 3 , 2 / 3 ). Similarly other basic solutions are ( 2 , 1 , 0 ) & ( 5 , 0 , − 1 )

In all of these basic solutions, none of the basic variables are zero, demonstrating that they are all non-

degenerate.

There are two main types of graphical methods to solve LPP:

1. Corner Point Method (Graphical Method)

This is the most common type of graphical method used to solve LPPs. It involves identifying and evaluating

the objective function at the corner points of the feasible region.

Procedure to solve LPP by Corner Point Method

▪ Formulate the LPP problem.

▪ Express all constraints as equations and create a graph.

▪ Determine the feasible region.

▪ Find the coordinates of each vertex (corner point) within the feasible region, either by inspection or by

solving the intersecting line equations.

▪ Evaluate the objective function using these corner points.

▪ For maximization, select the highest value among these results as the optimum. For minimization, choose

the lowest value as the optimum.

2. Iso-Profit (Iso-Cost) Line Method

The Iso-Profit Line Method is used when the goal is to find the profit-maximizing (or cost-minimizing)

solution given a fixed profit (or cost) value. It involves drawing a line parallel to the objective function line

at a specified profit (or cost) level and finding the point where it intersects the feasible region. This

method is particularly useful for sensitivity analysis when you want to explore how changes in profit (or

cost) affect the optimal solution.

Procedure to solve LPP by Iso-Profit (Iso-Cost) Line Method

▪ Formulate the LPP problem.

▪ Express all constraints as equations and create a graph.

▪ Determine the feasible region. The feasible region, satisfying all constraints and non-negativity, forms a

convex set.

▪ Identify the extreme points of this feasible region.

▪ Assign a value, 𝑘, to the objective function 𝒵 and draw the corresponding line in the 𝑥𝑦-plane.

▪ For maximization , draw lines parallel to 𝒵 = 𝑘, finding the farthest line from the origin with at least

one common point in the feasible region.

For minimization , draw lines parallel to 𝒵 = 𝑘, finding the nearest line to the origin with at least one

common point in the feasible region.

▪ The common points obtained represent the optimal solution of the LPP.

Guidelines for Defining Feasible Regions

For the constraint 𝑎𝑥 + 𝑏𝑦 ≤ 𝑐 (where 𝑐 > 0 ), draw the line 𝑎𝑥 + 𝑏𝑦 = 𝑐 by selecting two suitable points

on it. This line divides the 𝑥𝑦-plane into two sections. The inequation 𝑎𝑥 + 𝑏𝑦 ≤ 𝑐 signifies the part of the 𝑥𝑦-

plane where the origin resides.

Similarly, for the constraint 𝑎𝑥 + 𝑏𝑦 ≥ 𝑐 (where 𝑐 > 0 ), draw the line 𝑎𝑥 + 𝑏𝑦 = 𝑐 using two points. This

line also divides the 𝑥𝑦-plane into two sections. The inequation 𝑎𝑥 + 𝑏𝑦 ≥ 𝑐 represents the part of the 𝑥𝑦-plane

where the origin is not located.

Example : Solve by graphical method the linear programming problem.

Maximization 𝒵 = 100 𝑥

1

2

Subject to: 6 𝑥

1

2

1

2

Where: 𝑥

1

2

Solution : Solving using Corner Point Method

1. Determination of axis: X-axis = 𝑥 1

& Y-axis = 𝑥

2

2 ) Finding the coordinates of constraint lines to represent a graph. Consider equality

𝟏

𝟐

𝟏

𝟐

1

2

1

2

3 ) Plot the constraints on the graph and determine the feasible region

4 ) To find the optimal solution, Consider all the corner points of the feasible region

Vertex Coordinates Value of 𝒵 = 100 𝑥

1

2

O

A ( 0 , 1000 ) 80000

B

C

Maximum optimal value is at B so Optimal profit = 𝑀𝑎𝑥 𝒵 = 128000

Example : Solve by graphical method the linear programming problem.

Minimization 𝒵 = 40 𝑥

1

2

Subject to: 72 𝑥

1

2

1

2

1

2

4. To find the optimal solution, Consider all the corner points of the feasible region

Vertex Coordinates Value of 𝒵 = 160 𝑥

1

2

A

B ( 4 , 6 ) 2080

C

Maximal optimal value is at A so Optimal profit = 𝑀𝑎𝑥 𝒵 = 2560

Example : Solve by graphical method the linear programming problem.

Minimization 𝒵 = 50 𝑥

1

2

Subject to: 𝑥

1

2

1

2

1

2

Where: 𝑥

1

2

Solution : Solving using Corner Point Method

1. Determination of axis: X-axis = 𝑥 1

& Y-axis = 𝑥

2

2. Finding the coordinates of constraint lines to represent a graph. Consider equality

𝟏

𝟐

𝟏

𝟐

𝟏

𝟐

1

2

1

2

1

2

3. Plot the constraints on the graph and determine the feasible region

4. To find the optimal solution, Consider all the corner points of the feasible region

Vertex Coordinates Value of 𝒵 = 40 𝑥

1

2

A

B

C

D

Minimum optimal value is at B so Optimal cost = 𝑀𝑖𝑛 𝒵 = 10500

Special Cases in LPP

▪ Infeasible solution (Infeasibility)

Infeasible means not possible. Infeasible solution is a solution that does not satisfy the constraints of the

problem, meaning it is not a valid or acceptable solution.

▪ Unbounded solution (Unboundedness)

Unbounded mean infinite solution. A solution which has infinity answer is called unbounded solution.

▪ Redundant constraints (Redundancy)

A constraint is called redundant when it does not affect the solution. The feasible region does not depend

on that constraint. Even if we remove the constraint from the solution the optimal answer is not affected.

▪ Alternative optimal solution (Multiple optimal solution)

Alternate or multiple optimal solution means a problem has more than one solution which gives optimal

answer. There are two or more sets of solution values which gives maximum profit or minimum cost.

Example of Infeasible Solution

Maximization 𝓩 = 𝟓𝒙

𝟏

𝟐

Subject to: 𝟒𝒙

𝟏

𝟐

𝟏

𝟐

Where : 𝒙

𝟏

𝟐

Example of unbounded Solution

Maximization 𝓩 = 𝟒𝟎𝒙

𝟏

𝟐

Subject to: 𝟕𝟐𝒙

𝟏

𝟐

𝟏

𝟐

𝟏

𝟐

Where : 𝒙

𝟏

𝟐

Example of Redundant Constraints

Maximization 𝓩 = 𝟓𝒙

𝟏

𝟐

Subject to : 𝟑𝒙

𝟏

𝟐

𝟏

𝟐

𝟏

Where : 𝒙

𝟏

𝟐

Limitations of the Graphical Method:

This approach is applicable to problems with only two variables, but many practical scenarios involve more than

two variables. As a result, it is not a robust tool for solving Linear Programming problems.

Note :

▪ 𝒵 is of maximum or minimum type

▪ Inequalities are = type and all decision variables are ≥ 0

𝑖

Conversion to standard form and initial basic feasible solution

Example : Maximize 𝒵 = 80 𝑥 1

2

Such that 𝑥

1

2

1

2

Standard form is

Maximize 𝒵 = 80 𝑥

1

2

1

1

(−𝑚 is large penalty here)

Such that 𝑥

1

2

1

1

1

2

1

1

So initial basic feasible solution is 𝑥

1

2

1

= 25 and 𝐴

1

Note :

▪ In case of minimization problem we write 𝑚.

▪ If 𝑥 is unrestricted variable then we write 𝑥 = 𝑥

′

′′

where 𝑥

′

′′

≥ 0 and thus fit into standard form.

▪ Whenever decision variable appears only once we use that as artificial variable by adjusting its coefficient

to 1.

Example : 𝑥 1

2

1

2

3

Standard form: 𝑥

1

2

3

1

1

2

1

1

2

2

3

1

2

So 𝑥

3

is the artificial variable here and initial basic feasible solution is 𝑥

1

2

2

3

1

Row Method of Simplex Method

Procedure for Maximization Problems:

1. Formulate the objective function and inequality constraints.
2. Transform inequalities into equations by introducing slack variables.
3. Create the initial simplex tableau with the objective function as the bottom row.
4. Identify the pivot element or column as the most negative entry in the bottom row.
5. Calculate quotients by dividing the far-right column by the identified column (ignore zeros, negatives, or

zero denominators).

6. Select the row with the smallest quotient, and the intersection with the identified column from step 4

becomes the pivot element.

7. Perform pivoting using the Jordan method to make all other entries in the column zero.
8. If no more negative entries exist in the bottom row, process is complete; otherwise, repeat from step 4.

Example 1 : Niki has two part-time jobs, Job I and Job II, with a weekly time limit of 12 hours. She requires 2

hours of preparation for every hour at Job I and 1 hour for Job II, with a maximum of 16 hours for

preparation. Job, I pays $40 per hour, and Job II pays $30 per hour. To maximize her income, how many hours

should she allocate to each job per week?

Solution :

Step 1 : Formulate the LPP program

1

≔ Number of work hours at Job I

2

≔ Number of work hours at Job II

Maximize 𝒵 = 40 𝑥

1

2

Subject to: 𝑥

1

2

1

2

≤ 16 , where 𝑥

1

2

Step 2 : Convert the inequalities into equations

1

2

1

1

stack variables

1

2

2

2

stack variables

Rewrite objective as: − 40 𝑥

1

2

• 𝓏 = 0 where 𝑥

1

2

Step 3 : Construct initial tableau of simplex i.e. the augmented matrix

1

2

1

2

𝒵 C BV

1

2

Arbitrarily assigning values to some variables results in the initial basic solution.

For 𝑥

1

2

1

2

Step 4 : Identify the most negative entry in the bottom row as pivot column and calculate the quotients by

dividing the far-right column C by the pivot column.

Step 5 : Perform pivoting to make all the other entries in this column zero.

Row operation :

1

2

× 𝑅

2

1

2

1

2

C BV

1

2

Row operation : 𝑅

1

1

2

3

3

2

1

2

1

2

C

This solution is not optimal.

II: Step 6 : As last row contains non-negative now repeat the procedure from step 4.

Row operation : 2 × 𝑅

2