Method of Semi Averages, Business Mathematics and Statistics Video Lecture | SSC CGL Tier 2 - Study Material, Online Tests, Previous Year

in this video I am going to explain
linear programming problem by using
graphical method so before you go for
the problem let me explain the basic
concept of linear programming problem
see L P that is linear programming is a
mathematical programming technique to
optimize the performance generally in an
organization there are two major
objectives first one is a minimization
and second one is maximization so
minimization of what minimization offer
total cost of production that is a main
objective and second important objective
is a maximization of their profit so
these are the major objective of any
organization okay so here in graphical
method there are different models or
there okay so with the help of LPP
graphical method you can able to find
the optimum solution in ltp we have
different methods for example graphical
method simplex method dual simplex
method Bigham method so these are the
different methods are available but the
graphical method is applicable only when
we have only two variables if you have
more than two variable means the
graphical method is not suitable so here
I am going to explain three different
models of graphical methods the first
one is maximization problem and second
one is minimization problem okay but in
the first two methods we have two
constraints so this is the objective
function subject to two constraints are
given in the first problem
the second problem was the same thing
but the first one is maximization
problem and second one for minimization
problem so I am going to explain two
different models maximization model and
minimization model and the third one is
a maximization but here we have three
constraints in the above two problems we
have only two constraints but in the
third problem we have three constraints
so constraints with three equations are
they okay so here I am going to explain
three
in the first video I am going to explain
the first one the remaining second enter
will see in the separate video let us
see the first problem say solve the
following LPP using graphical method
this is a problem
okay maximize Z is equal to 12 X 1 plus
16 X 2 this is an objective function of
the problem subject to two constraints
are given first one 10 X 1 plus 20 X 2
less than or equal to 120 this is the
first constraint and second constraint
is 8 X 1 plus 8 X 2 less than or equal
to 80 so now we have two constraints our
objective function is maximization
problem okay you need to find the value
for x1 and x2 the value should be
greater than or equal to 0 which means
positive value only greater than or
equal to 0 means 0 or more than 0 so
first step is just convert the first
concern into in equation so write the
first one 10 X 1 plus 20 X 2 is equal to
120 this is the first equation so now we
have to find the value for x1 x2 we need
two points to plot in the graph sheet
okay so first thing you need to take X 1
as 0 see when X 1 is equal to 0 what is
the value for x 2 so when X 1 is 0 what
is the value for x 2 let's apply this in
the first equation 10 X 1 is 0 plus 20 X
2 is equal to 120 10 into 0 0 we have
only 20 X 2 so 20 X 2 is equal to 120 so
X 2 is equal to 120 divided by 20 is
equal to 6 X 2 value is 6 when X 1 is 0
X 2 will be 6 the same way now you have
to consider X 2
is equal to zero then what is the value
for x 1 when x 2 is equal to 0 what is
the value for x 1 let's apply this in
the same equation 10 X 1 plus 20 X 2
value is 0 no put 0 here 1 2 in chain so
20 into 0 is 0 we have only 10 X 1 so 10
X 1 is equal to 120 X 1 is equal to 120
divided by 10 6 1 is equal to 12 when X
2 is equal to 0 X 1 value will be 12 so
now we have got two points with this you
can plot into the graph sheet in order
to draw a line we need to have two
points okay first one laser X 1 0 X 2 6
this is the first point and second point
is X 1 12 X 2 0 this axis X 1 axis and
this one next X 2 axis so plot the line
first 1 X 1 0 X - 6 X 1 0 X 2 6 0 comma
6 X 1 0 X 2 6 a second point X 1 12 X 2
0 X 1 12 X 2 0 so this is the point X 1
12 X 2 0 we have got two points just
join the two points to make a line this
line represents the first constraint now
we have to find the second line ok look
at the problem look at the second
constraint 8 X 1 plus 8 X 2 less than or
equal to 8 8 second equation 8 X 1 plus
8 X 2 is equal to 80 so second equation
in the same way we need to find the
value for x1 and x2 so in order to draw
a line we need to have two points x1 x2
when x1 is 0 what is the value
for x2 in the same way when x2 is 0 what
is the value for x1 when X 1 is equal to
0 what is the value for X 2 you can just
apply this in equation 8 X 1 that is 8 X
1 plus 8 X 2 is equal to 18 ohm so 8
into 0 X 1 value is 0 no plus 8 X 2 is
equal to 88 into 0 0 we have only Inc X
- 8 X 2 is equal to 80 so X 2 is equal
to 80 divided by 8 so what is the value
10 10 in the same way
find the value for x1 when X 2 is equal
to 0 what is the value for x1 what is
the equation 8 x1 no so H X 1 plus a x2
values 0 is equal to 88 into 0 0 we have
only eight X 1/8 X 1 is equal to 80 so
x1 is equal to 80 divided by 8 so value
is 10 so x1 is equal to 10 when x2 is
equal to 0 X 1 value will be 10 with
this now you can plot the second
constraint X 1 0 X 2 10 this point 0
comma 10 18 1 X 1 10 X 2 0 X 1 10 X 2 0
now join the two points
this line represent the second
constraint now you need to find the
feasible region in order to find the
feasible region just see the two
constraints that is two equation just
check whether it is less than or equal
to or greater than or equal to because
based on that only can able to find the
feasible region so look at the first
constraint less than or equal to second
constraint also less than or equal to
which means you have to find the
feasible region the feasible region
should be below the line which you have
drawn in the graph shader so this line
represents the first constraint and this
line represents the second constraint
okay see the shaded portion is the
feasible region ok so in order to find
the feasible region both the constraints
are less than or equal to so the
feasible region should be below the two
lengths so if we select this one this
will be under second constraint but
above the first constraint if we select
this region this will be under faster
region and above the second region so
these are the two region which is not
feasible so this is the feasible region
why because this particular shaded
portion is below both the lines we have
four corners this is the first corner
second corner third corner fourth corner
so these are the four corners in the
feasible region know this UV a and this
is B this corner minus B this is C and
this one is D the closed polygon a B C D
is the feasible region okay so after
finding the four points so you have to
find the value what is the value for a 0
and 0 x1 also 0 x2 also 0
what about B B value is X 1/10 X 2 0
look at the C point X 1 H X 2 2 X 1/8 X
2 2 just see
eight and two okay the same way D D
value already is a zero six that is x1
is 0 X 2 6 so we have got four values
for Coke or NASA that is after finding
the feasible region now we need to
substitute these values in the objective
function what is our objective function
so this is the objective function that
is maximization that is equal to 12 X 1
plus 16 X 2 so we have got four points
no so we have a x1 value x2 value in
each points okay so now you need to
substitute those values in this
objective function in order to find the
optimum point there is optimum corner so
let us see how I am going to work out
this thing now I am going to substitute
this value that is a value x1 value is 0
x2 value also 0 so now I am going to
substitute this value in the objective
function see this is the objective
function maximization of Z is equal to
12 X 1 plus 16 X 2 sales value 0 0 no so
x10 12 X 1 value 0 plus 16 X 2 value
also 0 so what is answer 12 into 0 0
plus 16 into 0 0 so 0 plus 0 0 then in
the same way we need to find the value
for being seeing and D also in order to
find the optimal point look at the bees
x1 x2 value 10 0 x1 value 10 x2 value 0
so 12 X 1 value 10 plus 16 X 2 value 0
so 12 into 10 1 20 plus 16 into 0 0 1 20
a B is over now look at the C X 1 is 8 X
2 is 2 4 C 12 X minus h plus 16 X 2 is 2
12 - 8 9
6 plus 1680 232 so 96 plus 32 is equal
to 128 ABCs are over now look at the
fourth one Dee Dee's value is X 1 value
is 0 X 2 value is 6 1412 X 1 value is 0
plus 16 X 2 value is 6 gives 12 into 0 0
plus 16 into 6 96 so 0 plus 96 96 now we
have what is that value for ABCD since
it is a maximization problem we need to
find the maximum pointer so look at the
four answers which one is highest value
0 120 128 96 or so use the highest
maximum amount at C the Z value is
maximum in the corner point of see now
is that value is maximum for the corner
Point C so this is the highest two
points no C is the highest point optimum
point so X 1 value is 8 X 2 values to
optimal visit is 128 hope you all
understand the sponsor thank you