This document provides an overview of operations research and linear programming techniques. It begins with an introduction to the graphical method for solving linear programming problems with two variables by plotting the feasible region defined by the constraints. It then defines key terms like feasible solutions and optimal solutions. The document provides examples of using the graphical method to find the optimal solution for both maximization and minimization problems. It also discusses special cases that can occur with linear programs, such as alternative optimal solutions, unbounded solutions, infeasible solutions, and degenerate solutions. Finally, it provides an introduction to the concept of duality in linear programming.

1. Operations
Research
Presented by,
Snehal S. Athawale
Ph.D 1st year
(Agricultural Economics)
School of Social Sciences
College of Post Graduate Studies in Agricultural
Sciences, Umiam, Meghalaya.
AEC-605
Linear Programming

2. Table of contents
Graphical
method
Duality
01 02

3. Graphical Method
For LP problems that have only two variables, it is possible that
the entire set of feasible solutions can be displayed graphically by
plotting linear constraints on a graph paper in order to locate the
best (optimal) solution. The technique used to identify the optimal
solution is called the graphical solution method (approach or
technique) for an LP problem with two variables.

4. IMPORTANT DEFINITIONS
Solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that satisfy the
constraints of an LP problem is said to constitute the solution to that LP
problem.
Feasible solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that
satisfy all the constraints and non-negativity conditions of an LP problem
simultaneously is said to constitute the feasible solution to that LP problem.
Infeasible solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that
do not satisfy all the constraints and non-negativity conditions of an LP problem
simultaneously is said to constitute the infeasible solution to that LP problem.

5. IMPORTANT DEFINITIONS
Optimum basic feasible solution : A basic feasible solution that optimizes
(maximizes or minimizes) the objective function value of the given LP problem
is called an optimum basic feasible solution.
Unbounded solution A solution that can increase or decrease infinitely the
value of the objective function of the LP problem is called an unbounded
solution

6. Extreme Point Solution Method
Refers to the corner of the feasible region (or space), i.e. this point lies
at the intersection of two constraint equations.
The steps of the method are summarized as follows:
Step 1 : Develop an LP model
Step 2 : Plot constraints on graph paper and decide the feasible region.
Step 3 : Examine extreme points of the feasible solution space to find an
optimal solution

7. Examples on Maximization LP Problem
Example
Maximize Z = 15x1 + 10x2
subject to the constraints (i) 4x1 + 6x2 ≤ 360,
(ii) 3x1 + 0x2 ≤ 180,
(iii) 0x1 + 5x2 ≤ 200
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint 4x1 + 6x2 = 360
We get, X2 = 60 , when X1=0
X1= 90 , when X2=0

8. Examples on Maximization LP Problem
Example
Maximize Z = 15x1 + 10x2
subject to the constraints (i) 4x1 + 6x2 ≤ 360,
(ii) 3x1 + 0x2 ≤ 180,
(iii) 0x1 + 5x2 ≤ 200
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint ) 3x1 + 0x2 =180,
We get, X2 = 0, when X1=0
X1= 60 , when X2=0

9. Examples on Maximization LP Problem
Example
Maximize Z = 15x1 + 10x2
subject to the constraints (i) 4x1 + 6x2 ≤ 360,
(ii) 3x1 + 0x2 ≤ 180,
(iii) 0x1 + 5x2 ≤ 200
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 3st constraint ) 0x1 + 5x2 = 200
We get, X2 = 40, when X1=0
X1= 0 , when X2=0

10. Examples on Maximization LP Problem
Since all constraints have been graphed,
the area which is bounded by all the
constraints lines including all the boundary
points is called the feasible region (or
solution space). The feasible region is
shown in fig by the shaded area OABCD
Note: feasible region is the overlapping area
of constraints that satisfies all of the
constraints on resources.

11. Examples on Maximization LP Problem
Objective function Z is to be maximized
from Table
maximum value of Z = 1,100 is achieved at
the point extreme B (60, 20).

12. Examples on Minimization LP Problem
Example
Minimize Z = 3x1 + 2x2
subject to the constraints (i) 5x1 + x2 ≥ 10,
(ii) x1 + x2 ≥ 6,
(iii) x1 + 4x2 ≥12
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint 5x1 + x2 = 10
We get, X2 = 10 , when X1=0
X1= 2 , when X2=0

13. Examples on Minimization LP Problem
Example
Minimize Z = 3x1 + 2x2
subject to the constraints (i) 5x1 + x2 ≥ 10,
(ii) x1 + x2 ≥ 6,
(iii) x1 + 4x2 ≥12
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint x1 + x2 = 6,
We get, X2 = 6, when X1=0
X1= 6 , when X2=0

14. Examples on Minimization LP Problem
Example
Minimize Z = 3x1 + 2x2
subject to the constraints (i) 5x1 + x2 ≥ 10,
(ii) x1 + x2 ≥ 6,
(iii) x1 + 4x2 ≥12
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 3st constraint x1 + 4x2 =12
We get, X2 = 3, when X1=0
X1= 12 , when X2=0

15. Examples on Minimization LP Problem
The coordinates of the extreme points of
the feasible region (bounded from below)
The value of objective function at each of
these extreme points is shown in Table

16. Examples on Minimization LP Problem
The minimum (optimal) value of the
objective function Z = 13 occurs at the
extreme point C (1, 5). Hence, the optimal
solution to the given LP problem is: x1 = 1,
x2 = 5, and Min Z = 13.

17. SPECIAL CASES IN LINEAR
PROGRAMMING
1) Alternative (or Multiple) Optimal Solution
2) Unbounded Solution
3) Infeasible Solution
4) Degenerate Solution

18. Alternative (or Multiple) Optimal Solution
In certain cases, a given LP problem may have more than one solution
yielding the same optimal objective function value. Each of such optimal
solutions is termed as alternative optimal solution.
Conditions
(i) The slope of the objective function should be the
same as that of the constraint forming the boundary of
the feasible solutions region.
(ii) The constraint should form a boundary on the feasible
region in the direction of optimal movement of the
objective function. In other words, the constraint
should be an active constraint.

19. Example on Alternative optimum solution
Example
Maximize Z = 20x1 + 10x2
subject to the constraints (i) 10x1 + 5x2 ≤ 50,
(ii) 6x1 + 10x2 ≤ 60,
(iii) 4x1 + 12x2 ≤ 48
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint 10x1 + 5x2 = 50,
We get, X2 = 10, when X1=0
X1= 5 , when X2=0

20. Example on Alternative optimum solution
Example
Maximize Z = 20x1 + 10x2
subject to the constraints (i) 10x1 + 5x2 ≤ 50,
(ii) 6x1 + 10x2 ≤ 60,
(iii) 4x1 + 12x2 ≤ 48
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint 6x1 + 10x2 = 60,
We get, X2 = 6, when X1=0
X1= 10 , when X2=0

21. Example on Alternative optimum solution
Example
Maximize Z = 20x1 + 10x2
subject to the constraints (i) 10x1 + 5x2 ≤ 50,
(ii) 6x1 + 10x2 ≤ 60,
(iii) 4x1 + 12x2 ≤ 48
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 3st constraint 4x1 + 12x2 = 48,
We get, X2 = 4, when X1=0
X1= 12 , when X2=0

22. Examples on Alternative optimum solution
The solution space is denoted by O(0,0),
B(5,0), C(3.6,2.8) and D(0,4). The objective
function value at the corner point B&C are
same and maximum among all values.
This indicates existence of multiple
combinations of value of the decision
variable for the same maximum objective
function.
Extrem
e Point
Coordinates
(X1, X2)
Objective Function Value
Z = 20x1 + 10x2
O (0,0) 20(0)+ 10(0) = 0
B (5,0) 20(5)+ 10(0) = 100
C (3.6,2.8) 20(3.6)+ 10(2.8) = 100
D (0,4) 20(0) + 10(4) =40
Objective function is parallel
to first constraint

23. Unbounded Solution
• Infinite solution is referred as an unbounded solution.
• It happens when value of certain decision variables and
the value of the objective function (maximization case)
are permitted to increase infinitely, without violating
the feasibility condition.

24. Example on Unbounded solution
Example
Maximize Z = 3x1 + 2x2
subject to the constraints (i) x1 - x2 ≥ 1,
(ii) x1 + x2 ≥ 3,
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint x1 - x2 = 1,
We get, X2 = -1, when X1=0
X1= 1, when X2=0

25. Example on Unbounded solution
Example
Maximize Z = 3x1 + 2x2
subject to the constraints (i) x1 - x2 ≥ 1,
(ii) x1 + x2 ≥ 3,
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint x1 + x2 = 3,
We get, X2 = 3, when X1=0
X1= 3, when X2=0

26. Example on Unbounded solution
Since the given LP problem is of
maximization, there exist a number
of points in the shaded region for
which the value of the objective
function is more than 8.
Extreme
Point
Coordinat
es (X1, X2)
Objective Function Value
Z = 3x1 + 2x2
A (0,3) 3(0)+ 2(3) = 6
B (2,1) 3(2)+ 2(1) = 8

27. Example on Unbounded solution
For example, the point (2, 2) lies in
the region and the objective function
value at this point is 10 which is more
than 8.
Thus, as value of variables x1 and x2
increases arbitrarily large, the value
of Z also starts increasing. Hence, the
LP problem has an unbounded
solution.

28. Infeasible Solution
• An infeasible solution to an LP problem arises when there is
no solution that satisfies all the constraints simultaneously.
• This happens when there is no unique (single) feasible
region.
• This situation arises when a LP model that has conflicting
constraints.
• Any point lying outside the feasible region violates one or
more of the given constraints.

29. Example on Infeasible solution
Example
Maximize Z = 6x1 – 4X2
subject to the constraints (i) 2x1 + 4x2 ≤ 4,
(ii) 4x1 + 8x2 ≥ 16,
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint 2x1 + 4x2 = 4, ,
We get, X2 = 1, when X1=0
X1= 2, when X2=0

30. Example on Infeasible solution
Example
Maximize Z = 6x1 – 4X2
subject to the constraints (i) 2x1 + 4x2 ≤ 4,
(ii) 4x1 + 8x2 ≥ 16,
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint 4x1 + 8x2 = 16, ,
We get, X2 = 2, when X1=0
X1= 4, when X2=0

31. Example on Infeasible solution
The constraints are plotted on graph as
usual as shown in Fig. Since there is no
unique feasible solution space,
therefore a unique set of values of
variables x1 and x2 that satisfy all the
constraints cannot be determined.
Hence, there is no feasible solution to
this LP problem because of the
conflicting constraints.

32. Degeneracy
A basic feasible solution is called degenerate if value of at least one
basic variable is zero.
Example
Maximize Z = 100x1 + 50x2
subject to the constraints (i) 4x1 + 6x2 ≤ 24,
(ii) x1 ≤ 4,
(iii) x2 ≤ 4/3
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 1st constraint 4x1 + 6x2 = 24
We get, X2 = 4, when X1=0
X1= 6, when X2=0

33. Example on Degeneracy
Example
Maximize Z = 100x1 + 50x2
subject to the constraints (i) 4x1 + 6x2 ≤ 24,
(ii) x1 ≤ 4,
(iii) x2 ≤ 4/3
Non-negativity x1, x2 ≥ 0
Solution : Compute the coordinates on X1X2 plane.
From the 2st constraint x1 = 4
From the 3st constraint x2 = 4/3
by plotting these coordinates on graph,

34. Example on Degeneracy
The Closed polygon ABCD is
the feasible region. The
intersection of two lines will
define a corner point of feasible
solution. But at corner point C,
three lines intersect. This shows
the presence of degeneracy in
the problem.
The coordinates the corner
point C are (4. 4/3). Hence,
Z(A) = 0 Z(B)= 400
Z(C)= 1400/3 Z(D)= 66.67
= 466.67

35. Duality
In linear programming, duality implies that each linear
programming problem can be analyzed in two different ways but
would have equivalent solutions. Any LP problem (either
maximization and minimization) can be stated in another
equivalent form based on the same data. The new LP problem is
called dual linear programming problem or in short dual.

36. Duality
Every Linear Programming problem is associated with another linear
programming problem called the Dual of the problem.
The Dual problem of the LPP is obtained by,
1) Transposing the coefficient matrix
2) Interchanging the role of constant terms and the coefficient of the
objective function
3) Reverting the inequalities
4) Minimization of objective function instead of maximizing it
Primal : Maximize Z(P) = CX
constraints Ax ≤ b
non-negativity x ≥0
Dual Problem : Minimize Z(D) = b’w
constraints A’w ≤ c’
non-negativity w ≥0
Where, w = (w1, w2, w3……wm)

37. Duality
Rules for Constructing the Dual from Primal
1. A dual variable is defined corresponds to each constraint in the primal LP
problem and vice versa. Thus, for a primal LP problem with m constraints and n
variables, there exists a dual LP problem with m variables and n constraints and
vice-versa.
2. The right-hand side constants b1, b2, . . ., bm of the primal LP problem
becomes the coefficients of the dual variables y1, y2, . . ., ym in the dual
objective function Z y. Also the coefficients c1, c2, . . ., cn of the primal variables
x1, x2, . . ., xn in the objective function become the right-hand side constants in
the dual LP problem.
3. For a maximization primal LP problem with all ≤ (less than or equal to) type
constraints, there exists a minimization dual LP problem with all ≥ (greater than
or equal to) type constraints and vice versa. Thus, the inequality sign is reversed
in all the constraints except the non-negativity conditions.

38. Duality
Rules for Constructing the Dual from Primal
4. The matrix of the coefficients of variables in the constraints of dual is
the transpose of the matrix of coefficients of variables in the constraints of
primal and vice versa.
5. If the objective function of a primal LP problem is to be maximized, the
objective function of the dual is to be minimized and vice versa.
6. If the ith primal constraint is = (equality) type, then the ith dual variables
is unrestricted in sign and vice versa.

39. Duality
The primal-dual relationships may also be memorized by using the
following table

40. Duality
Example
Maximize Z = 4x1 + 10x2 + 25x3
subject to the constraints (i) 2x1 + 4x2 + 8x3 ≤ 25,
(ii) 4x1 + 9x2 + 8x3 ≤ 30,
(iii) 6x1 + 8x2 + 2x3 ≤ 40
Non-negativity x1, x2, x3 ≥ 0
Solution :Let Yi be the dual variable associated with the ith constraints of
the primal problem.
Minimize Z’= 25Y1 + 30Y2 + 40Y3
subject to the constraints (i) 2Y1 + 4Y2 + 6Y3 ≥ 4,
(ii) 4Y1 + 9Y2 + 8Y3 ≥ 10,
(iii) 8Y1 + 8Y2 + 2Y3 ≥ 25
Non-negativity Y1, Y2, Y3 ≥ 0

41. A summary of the general relationships between primal and dual
LP problems is given in Table
Duality

42. t
You can find me at – snehalathawale98@gmail.com