MBA Semester linear analysis and programming function

Linear Programming
Vikash Singh
Senior Data Scientist, MBA, BSc (Statistics), UGC Net
18+ years of experience in DS, ML, AI and Strategy

Session 7
• Introduction to Linear Programming Problems
• LPP Formulations
• Graphical Solution
Session 8
• Simplex method
• Training on Excel
What we will cover:

• Linear Programming Problems (LPP): Linear programming or linear optimization is a process
which takes into consideration certain linear relationships to obtain the best possible solution
to a mathematical model. It is also denoted as LPP.
• Linear programming is a process that is used to determine the best outcome of a linear
function. It is the best method to perform linear optimization by making a few simple
assumptions.
• The linear function is known as the objective function.
• Real-world relationships can be extremely complicated. However, linear programming can be
used to depict such relationships, thus, making it easier to analyze them.
LPP

• Simplex Method is one of the most powerful & popular
methods for linear programming.
• The simplex method is an iterative procedure for getting the
most feasible solution. In this method, we keep transforming
the value of basic variables to get maximum value for the
objective function.
Simplex Method

Simplex Solution
Maximize P: 6x + 5y + 4z
subject to 2x + y + z <= 180
x + 3y + 2z <= 300
2x + y + 2z <= 240
x, y, z >=0

Simplex Solution - Steps
x + 3y + 2z <= 300
2x + y + 2z <= 240
x, y, z >=0
1 Slack Variables
2x + y + z + u = 180
x + 3y + 2z + v = 300
2x + y + 2z + w = 240
2 Rewrite the objective function
P - 6x - 5y - 4z = 0
177 < 180
177 + 3 = 180

1. Simplex Table
x y z u v w P C
2 1 1 1 0 0 0 180
1 3 2 0 1 0 0 300
2 1 2 0 0 1 0 240
-6 -5 -4 0 0 0 1 0
x + 3y + 2z <= 300
2x + y + 2z <= 240
x, y, z >=0
1 Slack Variables
2x + y + z + u = 180
x + 3y + 2z + v = 300
2x + y + 2z + w = 240
2 Rewrite the objective function
P - 6x - 5y - 4z = 0

x y z u v w P C
2 1 1 1 0 0 0 180
1 3 2 0 1 0 0 300
2 1 2 0 0 1 0 240
-6 -5 -4 0 0 0 1 0
2. Pivot Column
Pivot column is the one with smallest value which is -6 so this becomes my
pivot column.

2. Pivot Row
Divide the constants by corresponding values of the pivot column.
Coefficients and constants
x y z u v w P C
2 1 1 1 0 0 0 180 180/2
1 3 2 0 1 0 0 300 300/1
2 1 2 0 0 1 0 240 240/2
-6 -5 -4 0 0 0 1 0

Pivot Row
Divide the constants by corresponding values of the pivot column.
x y z u v w P C
2 1 1 1 0 0 0 180 180/2 90
1 3 2 0 1 0 0 300 300/1 300
2 1 2 0 0 1 0 240 240/2 120
-6 -5 -4 0 0 0 1 0
this is smallest,
so becomes
pivot row.
2 is the pivot cell

Multiply the pivot row so that the pivot cell becomes 1
x y z u v w P C
2 1 1 1 0 0 0 180 180/2
1 3 2 0 1 0 0 300 300/1
2 1 2 0 0 1 0 240 240/2
-6 -5 -4 0 0 0 1 0
x y z u v w P C
1/2 R1 1 1/2 1/2 1/2 0 0 0 90
1 3 2 0 1 0 0 300
2 1 2 0 0 1 0 240
-6 -5 -4 0 0 0 1 0
No change anything in 2nd and 3rd row

Make the other elements of the pivot column to zero
x y z u v w P C
1/2 R1 1 1/2 1/2 1/2 0 0 0 90
1 3 2 0 1 0 0 300
2 1 2 0 0 1 0 240
-6 -5 -4 0 0 0 1 0

Make the other elements of the pivot column to zero
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
R2 - 1 R1 0 5/2 3/2 -1/2 1 0 0 210
R3 - 2 R1 0 0 1 -1 0 1 0 60
R4 + 6 R1 0 -2 -1 3 0 0 1 540
x y z u v w P C
1/2 R1 1 1/2 1/2 1/2 0 0 0 90
1 3 2 0 1 0 0 300
2 1 2 0 0 1 0 240
-6 -5 -4 0 0 0 1 0

x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
0 5/2 3/2 -1/2 1 0 0 210
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540
Now which column is the pivot column????
Now this is the new table.

x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
0 5/2 3/2 -1/2 1 0 0 210
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540
we have two negatives, and the largest negative is -2. so that becomes now
a pivot column.
Now this is the new table.

Find the pivot row???
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90 90/(1/2)
0 5/2 3/2 -1/2 1 0 0 210 210/(5/2)
0 0 1 -1 0 1 0 60 60/0 this does not count, so 84 is the smallest.
0 -2 -1 3 0 0 1 540
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
0 5/2 3/2 -1/2 1 0 0 210
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540

Make pivot cell equal to 1
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
0 5/2 3/2 -1/2 1 0 0 210
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
2/5 R2 0 1 3/5 -1/5 2/5 0 0 84
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540

Make other cells of pivot row zero
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
2/5 R2 0 1 3/5 -1/5 2/5 0 0 84
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540
x y z u v w P C
R1 - 1/2 R2 1 0 1/5 3/5 -1/5 0 0 48
0 1 3/5 -1/5 2/5 0 0 84
0 0 1 -1 0 1 0 60
R4 + 2 R2 0 0 1/5 13/5 4/5 0 1 708

Make other cells of pivot row zero
x y z u v w P C
1 1/2 1/2 1/2 0 0 0 90
2/5 R2 0 1 3/5 -1/5 2/5 0 0 84
0 0 1 -1 0 1 0 60
0 -2 -1 3 0 0 1 540
x y z u v w P C
R1 - 1/2 R2 1 0 1/5 3/5 -1/5 0 0 48
0 1 3/5 -1/5 2/5 0 0 84
0 0 1 -1 0 1 0 60
R4 + 2 R2 0 0 1/5 13/5 4/5 0 1 708
Now we have no negative cells in last row

Final simplex matrix
x y z u v w P C
1 0 1/5 3/5 -1/5 0 0 48
0 1 3/5 -1/5 2/5 0 0 84
0 0 0 -1 0 1 0 60
0 0 1/5 13/5 4/5 0 1 708
• In a simplex matrix, the variables that have a unit column are called basic variables.
• A unit column is a column that is all zeros except for a single one.
• The variables that are not a unit column, i.e. a column of junk, are called non-basic variables.

x y z u v w P C
1 0 1/5 3/5 -1/5 0 0 48
0 1 3/5 -1/5 2/5 0 0 84
0 0 0 -1 0 1 0 60
0 0 1/5 13/5 4/5 0 1 708
• In a simplex matrix, the variables that have a unit column are called basic variables.
• A unit column is a column that is all zeros except for a single one.
• The variables that are not a unit column, i.e. a column of junk, are called non-basic variables.
• The non-basic variables act like parameters and thus in the simplex process, we set their values to
zero.

Maximize P: 6x + 5y + 4z: This is happening at x = 48, y = 84, z =0, P = 708
x y z u v w P C
1 0 1/5 3/5 -1/5 0 0 48
0 1 3/5 -1/5 2/5 0 0 84
0 0 0 -1 0 1 0 60
0 0 1/5 13/5 4/5 0 1 708
x 48 u 0
y 84 v 0
z 0 w 60
max 708
x + 3y + 2z <= 300
2x + y + 2z <= 240
x, y, z >=0

Formulation of LPP in MS-Excel
Problem:
The World Light Company produces two light fixtures (product 1 and 2) that require both metal frame parts and
electrical components. Management wants to determine how many units of each product to produce so as to maximize
profit. For each unit of the product 1, 1 unit of frame parts and 2 units of electrical components are required. For each
unit of product 2, 3 units of frame parts and 2 units of electrical components are required. The company has 200 units
of frame parts and 300 units of electrical components. Each unit of product 1 gives a profit of $1, and each unit of
product 2, upto 60 units, gives a profit of $2. Any excess over 60 units of product 2 brings no profit, so such an
excess has to be ruled out.
(a) Formulate the LP model for this problem
(b) Use the graphical method to solve this model. What is the resulting total profit?
(c) Solve this using MS-Excel solver, Python and R-programing (optional).

Problem Definition
The WorldLight Company produces two light fixtures (product 1 and 2) that require both metal frame parts and
electrical components. Management wants to determine how many units of each product to produce so as to maximize
profit.
Number of units of Product 1 and Product 2.
Let X1, be the number of units of Product 1
X2, be the number of units of Product 2
Identifying the variables

Problem Definition
Each unit of product 1 gives a profit of $1, and each unit of product 2, upto 60 units,
gives a profit of $2.
Objective of the problem is to maximize the profit
Max Z = P = 1 * X1 + 2 * X2
5 units of pr 1 = x1 = 5
4 units of pr 2 = x2 = 4
P = 5 * 1 + 4 * 2 = 13dollars
Identify the objective of the problem:

Problem Definition
For each unit of the product 1, 1 unit of frame parts and 2 units of electrical components are required. For each unit of
product 2, 3 units of frame parts and 2 units of electrical components are required. The company has 200 units of
frame parts and 300 units of electrical components.
Any excess over 60 units of product 2 brings no profit, so such an excess has to be ruled out.
Constraints:
Constraints Product 1 Product 2 Availabilit
y
Raw material
constraint:
Metal frame parts 1 3 200
Electrical
components
2 2 300
Production
constraint:
<=60

Problem Definition
Raw material constraint:
For Metal frame:
1X1 + 3X2 <=200
For Electrical component:
2X1 + 2X2 <=300
Production constraint:
For Product 2:
X2<=60
Constraints:
Constraints Product 1 Product 2 Availability
Raw material
constraint:
Metal frame parts 1 3 200
Electrical
components
2 2 300
Production
constraint:
<=60
Non-negativity condition:
X1,X2>=0

Problem Definition
Objective function:
Max Z = 1X1 + 2X2
Subject to the constraint:
Raw material constraint:
For Metal frame:
1X1 + 3X2 <=200
For Electrical component:
2X1 + 2X2 <=300
Production constraint:
For Product 2:
X2<=60
Non-negativity condition:
X1,X2>=0
Problem:

Problem Definition – In Excel
Problem:

Solver in MS-Excel:

Solver Parameters:
Objective function
Decision variables
Constrains
Solving method

Problem solving – In Excel
Solver in MS-Excel:
Objective function
Decision variables
Constrains
Solving method

Problem solving – In Excel
Problem:

Solution– In Excel
Problem:
Hence the company should produce 1 unit of X1 and 2 units of X2 to make the maximum profit of 180.
This is achieved by using 180 units of metal frame and 300 units of electrical component.
The number of units of product 2 has not exceeded 60 units as defined in the constraint

MBA Semester linear analysis and programming function

• https://www.cuemath.com/algebra/linear-programming/
• https://www.analyticsvidhya.com/blog/2017/02/lintroductory-guide-on-linear-programming-
explained-in-simple-english/
• https://www.spiceworks.com/tech/it-strategy/articles/linear-
programming/#:~:text=Linear%20programming%20(LP)%20uses%20many,restrictions%20of%20sup
plies%20and%20personnel.
• https://www.youtube.com/watch?v=rQt_SWrOktg&ab_channel=TimMelvin
• https://www.youtube.com/watch?v=y9WWTi5sffo&ab_channel=DrD%E2%80%99sMathHelp
• Study material- https://drive.google.com/drive/folders/1J_eV95KDwMDlGvvlGkCpEZM2Taial-Db
References

MBA Semester linear analysis and programming function

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