what is linear programming,primal and dual problems,methods to solve -graphical and simplex

2.  Many real-life problems consist of maximizing
or minimizing a certain quantity subject to
some constraints.
 Linear programming is one approach to this
kind of problem.
 We will see examples in which we are
maximizing or minimizing a linear expression
in any number of variables subject to some
linear constraints.
 This technique is applied to a wide variety of
problems in industry and science.

3.  LP is the mathematical modelling technique
useful for the allocation of “scarce or limited”
resources such as labor ,material , machine
,time ,warehouse space etc…,to several
competing activities such as product , service ,
job, new equipments ,projects ,etc..on the
basis of given criteria of optimality.
A mathematical technique used to obtain an
optimal solution of resources allocation
problems,such as product planning.

4. .
it is a mathematical tool or technique or
technique for efficient or effective
utilization of limited resources to achieve
organisation objectives ( maximise profit or
minimum cost)
When solving a problem using linear
programming ,the program is put into
number of linear inequalities and then an
attempt is made to maximise (or minimise )
the input.

5. • There must be a well defined objective
function.
• There must be a constraint on the amount.
• There must be alternative course of action.
• The decision variable should be interrelated
and non negative .
• The source must be limited in supply.

6. General problem : Given a linear expression z=ax+by in two variables x and
y, find values of x and y that either maximize or minimize z subject to the
linear constraints:
And x>0,y>0
 The function z in the above problem is called the objective function.
 If (x, y) satisfies all the constraints, then it is called a feasible solution.
 The set of all feasible solutions is a subset of the xy-plane called
the feasible region.

7. .
• Note that each constraint of the form ax+by=c represents a straight line
in the xy-plane, whereas each constraint of the form defines a half-
plane that includes its boundary line
• The feasible region is then an intersection of finitely many lines and half-
planes. If the feasible region can be enclosed in a sufficiently large
circle, it is called bounded; otherwise it is called unbounded.
• If a feasible region is empty (contains no points), then the constraints
are inconsistent and the problem has no solution. The extreme points of
a feasible region are those boundary points that are intersections of the
straight-line boundary segments of the region. In a linear programming
problem, the following theorem tells us when we will be successful and
what points to search for.

9. • It helps in attaining optimal use of productive factors.
• It improves the quality of decision.
• It improve better tools for meeting the changing
condition.
• for large problems the computation difficulties are
enormous.
• It is only applicable to static situation.
• LP deals with the problems with single objective.

11. • The graphical method is limited to LP problems
involving two decision variables and a limited
number of constraints due to the difficulty of
graphing and evaluating more than two decision
variables.
• This restriction severely limits the use of the
graphical method for real-world problems
• Also the graphical method is simple and very
easy to understand

12. Constructing the LP problem requires four steps:
Step 1. Define the decision variables- describe the decisions to be made
Step 2. Define the objective function- define the goal we want to achieve
Step 3. Determine the constraints- some limitation under which the enterprise mus
operate
Step 4. Declare sign restrictions- Can the decision variable assume only nonnegati
values, or is it allowed to assume both positive and negative values?
The objective function deals with two types of objectives:
Maximization of such things as profits, revenue, or productivity
Minimization of such things as cost, time, or scrap

13. • Product mix problem- Beaver Creek Pottery company
• How many bowls and mugs should be produced to
maximize profits given labor and materials constraints?
• Product resource requirements and unit profit:

14. Resource 40 hrs of labor per day
Availability: 120 lbs of clay
Decision x1= number of bowls to produce per day
Variables: x2= number of mugs to produce per day
Objective Maximize z = $40x1 + $50x2
Function Where z= profit per day
Resource 1x1 + 2x2 ≤ 40 hours of labor
Constraints : 4x1 +3x2≤ 120 pounds of clay
Non negativity x1 ≥0 x2 ≥0
Constraints :

15. • Maximize z = $40x1 + $50x2
• Subject to : x1 + 2x2 ≤ 40
4x1 +3x2 ≤ 120
x1 ≥0 x2 ≥0

16. • A feasible solution does not violate any of the constraints
• Example: x1 = 5 bowls
x2 = 10 mugs
z = $40x1 + $50x2 = 700
• Labor constraint check: 1(5) + 2(10)=25 < 40 hours
• Clay constraint check : 4(5) + 3(10) =70 < 120 pounds

17. • An infeasible solution violates atleast one of the
constraints
• Example: x1 = 10 bowls
x2 = 20 mugs
z = $40x1 + $50x2 = 1400
• Labor constraint check: 1(10) + 2(20)=50 > 40 hours

23. )
• If an LPP has many constraints, then it may be
long and tedious to find all the corners of the
feasible region. There is another alternate and
more general method to find the optimal solution
of an LP, known as 'ISO profit or ISO cost method„
• Example-
Suppose the LPP is to optimize z= ax+by subject to
the constraints
a1x + b1y ≤ (or ≥) c1
a2x + b2y ≤ (or ≥) c2
X≥0
Y≥0

24. • This method of optimization involves the following method.
• Step 1: Draw the half planes of all the constraints
• Step 2: Shade the intersection of all the half planes which is the feasible
region.
• Step 3: Since the objective function is Z = ax + by, draw a dotted line for the
equation ax + by = k, where k is any constant. Sometimes it is convenient to
take k as the LCM of a and b.
• Step 4: To maximise Z draw a line parallel to ax + by = k and farthest from the
origin. This line should contain at least one point of the feasible region. Find
the coordinates of this point
• To minimise Z draw a line parallel to ax + by = k and nearest to the origin. This
line should contain at least one point of the feasible region. Find the co-
ordinates of this point by solving the equation of the line on which it lies.
• Step 5: If (x1, y1) is the point found in step 4, then
x = x1, y = y1, is the optimal solution of the LPP and
Z = ax1 + by1 is the optimal value.

25. Example: Solve the following LPP graphically
using ISO- profit method.
maximize Z =120x + 100y
Subject to the constraints
10x + 5y <= 80
6x + 6y <=66
4x + 8y >= 24
5x + 6y <= 90
x>=0 , y>=0

26. Identify all the half planes of the constraints. The
intersection of all these half planes is the feasible
region as shown in the figure.

27. • Give a constant value 600 to Z in the objective function, then we have an equation
of the line
120x + 100y = 600 or 6x + 5y= 30 …. (1)
P1Q1 is the line corresponding to the equation 6x + 5y = 30.
• P2Q2 is a line parallel to P1Q1 and has one point 'M' which belongs to feasible
region and farthest from the origin. If we take any line P3Q3 parallel to P2Q2 away
from the origin, it does not touch any point of the feasible region.
• The co-ordinates of the point M can be obtained by solving the equation 2x + y = 16
and x + y =11 which give
x = 5 and y = 6
• The optimal solution for the objective function is x = 5 and y = 6
• The optimal value of Z
120 (5) + 100 (6) = 600 + 600
= 1200

29. The theory of duality is a very elegant and
important concept within the field of
operations research. This theory was first
developed in relation to linear
programming, but it has many applications,
and perhaps even a more natural and
intuitive interpretation, in several related
areas such as nonlinear programming,
networks and game theory.

30. The notion of duality within linear programming
asserts that every linear program has associated
with it a related linear program called its dual. The
original problem in relation to its dual is termed the
primal.
it is the relationship between the primal and its
dual, both on a mathematical and economic level,
that is truly the essence of duality theory.

31. There is a small company in Melbourne which has
recently become engaged in the production of office
furniture. The company manufactures tables, desks
and chairs. The production of a table requires 8
kgs of wood and 5 kgs of metal and is sold for $80;
a desk uses 6 kgs of wood and 4 kgs of metal and is
sold for $60; and a chair requires 4 kgs of both
metal and wood and is sold for $50. We would like to
determine the revenue maximizing strategy for this
company, given that their resources are limited to
100 kgs of wood and 60 kgs of metal.

32. max
x
Z x x x80 60 501 2 3
8 6 4 100
5 4 4 60
0
1 2 3
1 2 3
1 2 3
x x x
x x x
x x x, ,

33. Now consider that there is a much bigger
company in Melbourne which has been the
lone producer of this type of furniture for
many years. They don't appreciate the
competition from this new company; so
they have decided to tender an offer to buy
all of their competitor's resources and
therefore put them out of business.

34. The challenge for this large company then is
to develop a linear program which will
determine the appropriate amount of money
that should be offered for a unit of each type
of resource, such that the offer will be
acceptable to the smaller company while
minimizing the expenditures of the larger
company.

35. 8 5 80
6 4 60
4 4 50
0
1 2
1 2
1 2
1 2
y y
y y
y y
y y,
min
y
w y y100 601 2

36. a x a x a x b
a x a x a x b
a x a x a x b
x x x
n n
n n
m m mn n m
n
11 1 12 2 1 1
21 1 22 2 2 2
1 1 2 2
1 2 0
...
...
... ... ... ...
... ... ... ...
...
, ,...,
max
x
j j
j
n
Z c x
1

37. a y a y a y c
a y a y a y c
a y a y a y c
y y y
m m
m m
n n mn m n
m
11 1 21 2 1 1
12 1 22 2 2 2
1 1 2 2
1 2 0
...
...
... ... ... ...
... ... ... ...
...
, ,...,
min
y
i
i
m
iw b y
1

38. z Z cx
s t
Ax b
x
x
*: max
. .
0
w* : min
x
w yb
s.t.
yA c
y 0

39. 1. The number of variables in the dual problem is equal to the number of
constraints in the original (primal) problem. The number of constraints in
the dual problem is equal to the number of variables in the original
problem.
2. Coefficient of the objective function in the dual problem come from the
right-hand side of the original problem.
3. If the original problem is a max model, the dual is a min model; if the
original problem is a min model, the dual problem is the max problem.
4. The coefficient of the first constraint function for the dual problem are the
coefficients of the first variable in the constraints for the original problem,
and the similarly for other constraints.
5. The right-hand sides of the dual constraints come from the objective
function coefficients in the original problem.

40. 1. The dual of the dual problem is again the primal
problem.
2. Either of the two problems has an optimal solution
if and only if the other does; if one problem is
feasible but unbounded, then the other is
infeasible; if one is infeasible, then the other is
either infeasible or feasible/unbounded.

41. The graphical method is useful only for
problems involving two decision variables and
relatively few problem constraints.
What happens when we need more decision
variables and more problem constraints?
We use an algebraic method called the simplex
method, which was developed by George B.
DANTZIG (1914-2005) in 1947 while on
assignment with the U.S. Department of the air
force.

42. A linear programming problem is said to be a standard
minimization problem in standard form if its mathematical model
is of the following form:
Minimize the objective function
Subject to problem constraints of the form
With non-negative constraints
1 1 2 2 ... , 0n na x a x a x b b
1 2, ,..., 0nx x x
Zmin = c1x1 + c2 x2 + ……..+ cn xn

43. Basic variables are selected arbitrarily with the restriction that
there be as many basic variables as there are equations.
The remaining variables are non-basic variables.
This system has two equations, we can select any two of the four
variables as basic variables. The remaining two variables are
then non-basic variables. A solution found by setting the two
non-basic variables equal to 0 and solving for the two basic
variables is a basic solution. If a basic solution has no negative
values, it is a basic feasible solution.
1 2 1
1 2 2
2 32
3 4 84
x x s
x x s

44. • Optimal solution(x*) is the best solution i.e. the
value of x for which objective function(Z) is minimum
or maximum.
• To find x*, simplex method must decide which
component “enters” by becoming positive and
which component “leaves” by becoming zero.
• This exchange is chosen so as to lower the total cost
or to increase the profit.

45.  Minimize the cost
c.x = 3x1 + x2 + 9x3 + x4
Constraints: x>=0
Equations Ax=b :
x1 + 2x3 + x4 = 4
x2 + x3 - x4 = 2

46. Step 1.Write equations in terms of basic variable and z
in terms of non basic variables,
x1 = 4 – 2x3 – x4
x2 = 2 – x3 + x4
c.x = 3 ( 4 – 2x3 – x4 ) + ( 2 – x3 + x4 ) + 9x3 + x4
c.x = 14 + 2x3 – x4

47. c.x = 14 + 2x3 – x4
Step 2.As x3 and x4 are non basic variables,
x3=x4=0. To minimize c.x either x3 should be
decreased or x4 should be increased.
So, entering variable is x4.
x1 = 4 – 2x3 – x4 <- binding
equation
x2 = 2 – x3 + x4
Leaving variable is x1.

48. Step 1.Write equations in terms of basic variable
(x2,x4) and z in terms of non basic variables
(x1,x3) as they are zero.
x4 = 4 – 2x3 – x1
x2 = 2 – x3 + (4 – 2x3 – x1 )
x2 = 6 – 3x3 – x1
c.x = 14 + 2x3 – (4 – 2x3 – x1 )
c.x = 10 + 4x3 + x1

49. • As, x3 and x1 are zero so they cannot be
reduced further.
• Therefore cost c.x cannot be minimized
further.
• Z=10 is the minimum cost.
• Value of constraints are
x1=0, x2=6, x3=0, x4=4
• The optimal solution x* = ( 0, 6, 0,4 )