The document provides information about linear programming techniques: - Linear programming is a mathematical modeling technique used to optimize allocation of scarce resources among competing demands. It involves linear objectives and constraints. - The goal is to maximize or minimize some quantity by determining an optimal feasible solution given problem constraints. A feasible solution satisfies all constraints while an optimal solution results in the largest/smallest objective function value. - Several examples of linear programming problems are provided to maximize revenue, profit or resources by determining optimal production levels given processing times, availability of resources, and sales prices of products. Solutions involve setting up objective functions and constraints then solving using the simplex method.

1. Forest Business Management
NRM 357
Credit hour (2+0)
B.Sc Forestry (Fifth Semester)
Prabin Pandit
Lecturer
Purbanchal University
College of Environment and Forestry (PUCEF)
foresterpandit@gmail.com

2. Linear Programming
• Linear programming is a widely used mathematical modelling technique to determine the
optimum allocation of scarce resources among competing demands.
• Resources typically include raw materials, manpower, machinery, time, money and space.
• The technique is very powerful and found especially useful because of its application to
many different types of real business problems in areas like finance, production, sales and
distribution, personnel, marketing and many more areas of management.
• As its name implies, the linear programming model consists of linear objectives and linear
constraints, which means that the variables in a model have a proportionate relationship.
For example, an increase in manpower resource will result in an increase in work output.

3. Maximization and Minimization
• The maximization or minimization of some quantity is the objective in all
linear programming problems.
• All LP problems have constraints that limit the degree to which the objective
can be pursued.
• A feasible solution satisfies all the problem's constraints.
• An optimal solution is a feasible solution that results in the largest possible
objective function value when maximizing (or smallest when minimizing).

4. Linear Programming (Example)
1. A company makes two products; Chair and Table both require processing on
Band saw and Sander machine. Table takes 10 and 15 minutes on Band saw
and Sander machine per unit respectively. Where as Chair takes 22 and 18
minutes on Band saw and Sander machine per unit respectively. Both the
machines are available for 2640 minutes per week. The products; Chair and
Table are sold for Rs 280 and Rs175 respectively per unit. Formulate a linear
programming to maximize revenue?

5. Linear Programming
Solution:
Tabulated form of the given problem is;
Products
Processing Time Required (minutes)
Selling price (Rs)
Band Saw Sander
Chair 10 15 200
Table 22 18 175
Availability of Machines 2640 min/week 2640 min/week Z
Decision variables;
• Let x be the number of chairs to be produced and y
be the number of tables to be produced.
Objective Functions;
Z = 200 x + 175 y
Constraints;
10x + 22y ≤ 2640
15x + 18y ≤ 2640
Now we can write the problem as follow;
Maximize:
Z = 200 x + 175 y
Subject to:
10x + 22y ≤ 2640
15x + 18y ≤ 2640
Where, x,y≥ 0

6. Linear Programming
2. A Factory can make Particle board and Plywood. They use two glues; phenol-
formaldehyde and urea-formaldehyde. To make one unit of Particle board, they need 3
liter of phenol-formaldehyde and 4 liter of urea-formaldehyde. Similarly, To make one
unit of Plywood, they need 2 liter of phenol-formaldehyde and 5 liter of urea-
formaldehyde. They have 30 liter of phenol-formaldehyde and 25 liter of Urea-
formaldehyde. Particle board and plywood are sold at Rs 800 and 900 per unit
respectively. find the best product need to show that total revenue will be maximum

7. Linear Programming
Products
Resources (Liter) Selling price
per unit (Rs)
phenol-formaldehyde urea-formaldehyde
Particle board 3 4 800
Plywood 2 5 900
Availability of
resources
30 25

8. Linear Programming
3. A furniture company produces tables and chairs. Each table requires 4 hours of Carpentry and two
hours of Painting work while each chair requires 3 hours of carpentry work and 1 hour of painting
work. During the current production period, 240 hours of carpentry time are available and 100 hours
of painting time are available. Each table sold at a profit of rupees 70 while each chair sold yields a
profit of Rs. 50. formulate a linear programming model for this problem.

9. Linear Programming
Products
Tasks (hour) Selling price
per unit (Rs)
Carpentry Painting
Tables 4 2 70
Chairs 3 1 50
Availability of
resources
240 100

10. Linear Programming
4. Consider a chocolate manufacturing company which produces two types of chocolate Kit Kat and
dairy milk. Both the chocolates requires milk and Choco only. To manufacture each unit of Kit Kat
and Dairy milk, the following quantities are required;
• Each unit of Kitkat requires 1 unit of milk and 3 unit of Choco
• Each unit of Diary milk requires 1 unit of milk and 2 unit of Choco.
The Company Kitchen has a total of 5 units of milk and 12 units of choco. On each sale the Company
makes a profit of Rs 6 per unit Kitkat sold and a profit of Rs. 5 per unit Diary milk sold. Now, the
company wishes to maximizing the profit. How many limits of Kitkat and Diary Milk Should it
Produces respectively?

11. Linear Programming
5. Using Simplex method, Maximize Z = 5x + 7y
subject to;
2x + 3y ≤ 13
3x + 2y ≤ 12 Where, x,y ≥ 0
Solution;
Let, S1 and S2 be the non negative slack variables. The linear programming problem (LPP) can be
written as;
2x + 3y + S1 + 0*S2 = 13
3x + 2y + 0*S1 + S2 = 12
- 5x - 7y + Z + 0*S1 + 0* S2 = 0
Where; x, y, S1 & S2 ≥ 0

12. Linear Programming
2x + 3y + S1 + 0*S2 = 13
3x + 2y + 0*S1 + S2 = 12
- 5x - 7y + 0*S1 + 0* S2 + Z = 0
Where; x, y, S1 & S2 ≥ 0
Simplex Tableau
X Y S1 S2 Z RHS Ratio
2 3 1 0 0 13
3 2 0 1 0 12
-5 -7 0 0 1 0

13. Linear Programming
X Y S1 S2 Z RHS Ratio
2 3 1 0 0 13
3 2 0 1 0 12
-5 -7 0 0 1 0
Here, -7 is the most negative entry in the last row. So, 2nd column is the pivot column.
13/3 = 4.3
12/2 = 6
Since, 13/3 = 4.3, 12/2 =6 and 4.3 is smaller than 6 so, 1st row is the pivot row.
Intersection of pivot row and pivot column is 3. So, 3 is the pivot number.

14. Linear Programming
X Y S1 S2 Z RHS Ratio
2 3 1 0 0 13 4.3
3 2 0 1 0 12 6
-5 -7 0 0 1 0
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
3 2 0 1 0 12
-5 -7 0 0 1 0
Operation: Applying R1 R1* 1/3
First off all we have to make pivot element 1
Y

15. Linear Programming
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
3 2 0 1 0 12
-5 -7 0 0 1 0
We have to make remaining value in pivot column (2 & -7) to 0
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
3-4/3 = 5/3 0 -2/3 1 0
12-26/3 =
10/3
-5+14/3 =
-1/3
0 7/3 0 1
13*7/3 =
91/3
Operation: Applying R2 R2 - 2*R1 Applying R3 R3+ 7*R1
Y
Y

16. Linear Programming
This is not the optimal solution, there is Still Negative value in Last Row.
Again we have to remove the negative value in last row.
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
5/3 0 -2/3 1 0 10/3
-1/3 0 7/3 0 1 91/3
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3 13/3*3/2 = 13/2
5/3 0 -2/3 1 0 10/3 2
-1/3 0 7/3 0 1 91/3
Here, -1/3 is the most negative entry in the last row. So, 1st column is the pivot column.
Since, 2 is smaller than 13/2 so, 2nd row is the pivot row
Intersection of pivot row and pivot column is 5/3. So, 5/3 is the pivot number.
Y
Y

17. Linear Programming
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3 13/3*3/2 = 13/2
5/3 0 -2/3 1 0 10/3 2
-1/3 0 7/3 0 1 91/3
Operation: Applying R2 R2* 3/5
First off all we have to make pivot element 1
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
1 0 -2/5 3/5 0 2
-1/3 0 7/3 0 1 91/3
Y
X
Y

18. Linear Programming
X Y S1 S2 Z RHS Ratio
2/3 1 1/3 0 0 13/3
1 0 -2/5 3/5 0 2
-1/3 0 7/3 0 1 91/3
We have to make remaining value in pivot column (2/3 & -1/3) to 0
Operation: Applying R1 R1 – 2/3*R2 Applying R3 R3+ 1/3*R2
X Y S1 S2 Z RHS Ratio
0 1
(1/3)-(2/3*-
2/5) = 3/5
-2/5 0 3
1 0 -2/5 3/5 0 2
0 0 11/5 1/5 1 31
X
Y
Y
X

19. Linear Programming
Since all the values in the last row are non negative. So, optimal solution is obtained
Therefore;
Maximize Z = 31 at
Y = 3 and
X = 2
Basic variables X Y S1 S2 Z RHS
y 0 1
(1/3)-
(2/3*-2/5)
= 3/5
-2/5 0 3
x 1 0 -2/5 3/5 0 2
0 0 11/5 1/5 1 31

20. Linear Programming (Example)
1. A company makes two products; Chair and Table both require processing on
Band saw and Sander machine. Table takes 10 and 15 minutes on Band saw
and Sander machine per unit respectively. Where as Chair takes 22 and 18
minutes on Band saw and Sander machine per unit respectively. Both the
machines are available for 2640 minutes per week. The products; Chair and
Table are sold for Rs 280 and Rs175 respectively per unit. Formulate a linear
programming to maximize revenue?

21. Linear Programming
Solution:
Tabulated form of the given problem is;
Products
Processing Time Required (minutes)
Selling price (Rs)
Band Saw Sander
Chair 10 15 200
Table 22 18 175
Availability of Machines 2640 min/week 2640 min/week Z
Decision variables;
• Let x be the number of chairs to be produced and y
be the number of tables to be produced.
Objective Functions;
Z = 200 x + 175 y
Constraints;
10x + 22y ≤ 2640
15x + 18y ≤ 2640
Now we can write the problem as follow;
Maximize:
Z = 200 x + 175 y
Subject to:
10x + 22y ≤ 2640
15x + 18y ≤ 2640
Where, x,y≥ 0

22. Linear Programming
Solution;
Let, S1 and S2 be the non negative slack variables. The linear programming problem (LPP) can be written as;
10x + 22y + S1 + 0*S2 = 2640
15x + 18y + 0*S1 + S2 = 2640
- 200x - 145y + 0*S1 + 0* S2+ Z = 0
Where; x, y, S1 & S2 ≥ 0
X Y S1 S2 Z RHS Ratio
10 22 1 0 0 2640 264
15 18 0 1 0 2640 176
-200 -145 0 0 1 0
Here, -200 is the most negative entry in the last row. So, 3rd column is the pivot column. 176 is less than 264, so 2nd row is
the pivot row and 15 is the pivot number

23. Linear Programming
X Y S1 S2 Z RHS Ratio
10 22 1 0 0 2640 264
15 18 0 1 0 2640 176
-200 -145 0 0 1 0
Operation: Applying R2 R2* 1/15
X Y S1 S2 Z RHS Ratio
10 22 1 0 0 2640
1 6/5 0 1/15 0 176
-200 -145 0 0 1 0
X

24. Linear Programming
X Y S1 S2 Z RHS Ratio
10 22 1 0 0 2640
1 6/5 0 1/15 0 176
-200 -145 0 0 1 0
X
Operation: Applying R1 R1 – 10*R2 Applying R3 R3+ 200*R2
X Y S1 S2 Z RHS Ratio
0 10 1 2/3 0 880
1 6/5 0 1/15 0 176
0 65 0 40/3 1 35200
X

25. Since all the values in the last row are non negative. So, optimal solution is obtained
Therefore;
Maximize Z = 35200 at
Y = 0 and
X = 176

26. Thank You

27. Linear Programming
Using Simplex method, Maximize Z = 2*X1 - X2 + 2* X3
Subject to;
2*X1 + X2 ≤ 10
X1 + 2* X2 - 2* X3 ≤ 20
X2 + 2* X3 ≤ 5 Where, X1, X2 and X3 ≥ 0
Solution;
Let, S1, S2 and S3 be the non negative slack variables. The linear programming problem (LPP) can
be written as;
2*X1 + X2 + S1 + 0S2 + 0S3 = 10
X1 + 2* X2 - 2* X3 + 0S1 + S2 + 0S3 = 20
X2 + 2* X3 + 0S1 + 0S2 + S3 = 5
-2*X1 + X2 - 2* X3 + 0S1 + 0S2 + 0S3 + Z = 0
Where; x, y, S1 & S2 ≥ 0

28. Linear Programming
Solution;
Let, S1, S2 and S3 be the non negative slack variables. The linear programming problem (LPP) can be written
as;
2*X1 + X2 + S1 + 0S2 + 0S3 = 10
X1 + 2* X2 - 2* X3 + 0S1 + S2 + 0S3 = 20
X2 + 2* X3 + 0S1 + 0S2 + S3 = 5
-2*X1 + X2 - 2* X3 + 0S1 + 0S2 + 0S3 + Z = 0
Where; x, y, S1 & S2 ≥ 0
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10
1 2 -2 0 1 0 0 20
0 1 2 0 0 1 0 5
-2 1 -2 0 0 0 1 0

29. Linear Programming
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10
1 2 -2 0 1 0 0 20
0 1 2 0 0 1 0 5 5/2
-2 1 -2 0 0 0 1 0
• Here, -2 is the most negative entry in the last row. So, 3rd column (Also can select 1st column) is the pivot
column, row 3 is the pivot row and 2 is the pivot number.

30. Linear Programming
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10
1 2 -2 0 1 0 0 20
0 1 2 0 0 1 0 5 5/2
-2 1 -2 0 0 0 1 0
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10
1 2 -2 0 1 0 0 20
0 1/2 1 0 0 1/2 0 5/2
-2 1 -2 0 0 0 1 0
Operation: Applying R3 R3* 1/2
X3

31. Linear Programming
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10
1 2 -2 0 1 0 0 20
0 1/2 1 0 0 1/2 0 5/2
-2 1 -2 0 0 0 1 0
Operation: Applying R3 R3* 1/2
X3
X1 X2 X3 S1 S2 S3 Z RHS Ratio
2 1 0 1 0 0 0 10 5
1 3 0 0 1 1 0 25 25
0 1/2 1 0 0 1/2 0 5/2
-2 2 0 0 0 1 1 5
Operation: Applying R2 R2 + 2*R3 Applying R4 R4+ 2*R3
There is still negative number in last row. Here, -2 is the most negative entry in the last row. So, 1st column is the pivot
column. 5 is less than 25, so 1st row is the pivot row and 2 is the pivot number
X3

32. Linear Programming
X1 X2 X3 S1 S2 S3 Z RHS Ratio
1 1/2 0 1/2 0 0 0 5 5
1 3 0 0 1 1 0 25 25
0 1/2 1 0 0 1/2 0 5/2
-2 2 0 0 0 1 1 5
Operation: Applying R1 R1*1/2
X3
X1
Operation: Applying R2 R2 - R1 Applying R4 R4+ 2*R3
X1 X2 X3 S1 S2 S3 Z RHS Ratio
1 1/2 0 1/2 0 0 0 5
0 5/2 0 -1/2 1 1 0 20
0 1/2 1 0 0 1/2 0 5/2
0 3 0 1 0 1 1 15
X1
X3

33. Since all the values in the last row are non negative. So, optimal solution is obtained
Therefore;
Maximize Z = 15 at
X1 = 5
X2 = 0
X3 = 5/2
Z = 2*X1 - X2 + 2* X3
15= 2*5 -0 +2*5/2
15= 15

34. Linear Programming
Solve the following linear programming problem by simplex method:
Z = 30x + 40y
Subject to;
2x + y ≤ 90
X + 2y ≤ 80
x + y ≤ 50
Where, x and y ≥ 0 ( PUCEF 2021)
Solution;
Let, S1, S2 and S3 be the non negative slack variables. The linear programming problem (LPP) can
be written as;
2x + y + S1 + 0S2 + 0S3 = 90
x + 2y + 0S1 + S2 + 0S3 = 80
X + y + 0S1 + 0S2 + S3 = 50
-30x - 40y+ 0S1 + 0S2 + 0S3 + Z = 0
Where; x, y, S1, S2 & S3 ≥ 0

35. Thank You

36. Linear Programming
Solve the following linear programming problem by simplex method:
Z = x + y + 3z
Subject to;
2x + y +3z ≥ 6
X + 2y+4z ≥ 8
3x + y – 2z ≥ 4
Where, x and y ≥ 0 ( Minimization Problem)
Solution;
X y z Constant
2 1 3 6
1 2 4 8
3 1 -2 4
1 1 3 0
Next, we form the
transpose of this
matrix by
interchanging its
rows and columns
X1 X2 X3 C
2 1 3 1
1 2 1 1
3 4 -2 3
6 8 4 0

37. Linear Programming
Solution………….
X1 X2 X3 C
2 1 3 1
1 2 1 1
3 4 -2 3
6 8 4 0
• Finally, we interpret the new matrix as a maximization problem as
follows.
• To do this, we introduce new variables, x1, x2, and x3. We call this
corresponding maximization problem the dual of the original
minimization problem.
• This gives the following dual problems;
Z = 6X1 + 8X2 + 4X3
Subject to;
2X1 + X2 + 3X3 ≤ 1
X1 + 2X2 + X3 ≤ 1
3X1 + 4X2 - 2X3 ≤ 3
Where, x and y ≥ 0