Linear Programming Problems (LPP) via Graphical Method, Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com

linear programming problem graphical
method in the previous video I have
explained the maximization problem with
two constraints in this video I am going
to explain minimization problem and
maximization problem in minimization
with two constraints in maximization
with three constraints let us see the
problem solve the following LPP using
graphical method minimize Z is equal to
4 X 1 plus 6 X 2 subject to X 1 plus X 2
greater than or equal to 8 6 X 1 plus X
2 greater than or equal to 12 X 1 and X
2
Karuna equal to 0 this is a minimization
problem our objective function is
minimize that is equal to 4 X 1 plus 6 X
2 so our final answer will be the
optimum value so let us see first you
need to convert the first constraint
into an equation X 1 plus X 2 is equal
to H okay have converted the bus to
concern into an equation so now we need
to find the value for X 1 and X 2 so
first you need to consider X 1 as 0 so
when X 1 is equal to 0 what is the value
for X 2 we need to find the value for x
2 when X 1 is 0 just substitute this
value in the first equation so X 1 value
is 0 plus X 2 is equal to 8 so X 2 is
equal to 8 so you have got X 2 value in
the same way when X 2 is equal to 0 what
is the value for X 1 we need to find the
value for X 1 now substitute this value
in the first equation so X 1 plus X 2
value is 0 is equal to 8 so X 1 is equal
to 8 so we have got the value for X 1 so
in the first thing X 1 X 2 okay so we
need to have two points to draw a line
in the
when X 1 is 0 X 2 values 8 when X 2
values 0 X 1 value is 8 so we have got 2
points with this we can draw a line now
look at this Akane 1 6 X 1 plus X 2
greater than or equal to 12
convert into an equation 6 X 1 plus X 2
is equal to 12 so second equation X 1 is
0 what is the value of 1 X 2 this
substitute 6 X 1 values 0 plus X 2 is
equal to 12 6 into 0 0 plus X 2 is equal
to 12 so X 2 is equal to 12 now when X 2
is equal to 0 what is the value for x 1
just substitute this value in the
equation 6 X 1 plus X 2 1 is 0 is equal
to 12 so 6 X 1 is equal to 12 X 1 is
equal to 12 divided by 6 so X 1 is equal
to 2 so we have got the value for X 1 X
1 is 2 when X 1 is 0 see when X 1 is 0 X
2 is 12 the same way when X 2 is 0 X 1
value is 2 so we have got 2 points and
so with this you can draw a line now I
am going to draw a line for first
equation see the value X 1 value is 0 X
2 value is 8 move X 1 axis this one is X
2 axis so first a point is X 1 0 X 2 8
second point this X 1 is 8 X 2 is 0 X 1
value is 8 X 2 van is 0 so this is the
point now we need to join the two points
now see the values for second equation X
1 is 0 X 2 is 12 and a second point is X
1 is 2 X 2 is Z
for second constrains bus to point is in
zero to L that is X 1 is 0 X 2 value is
12 this is the point look at this again
point to 0 that is X 1 value is 2 X 2
value is 0 so this is the point now we
need to join the two points see this
line represent the first constraint and
this line represents the second
constraint so now you need to find the
feasible region look at the first
constraint greater than or equal to 8
second one also greater than or equal to
12 when you find the feasible region
just check whether it is lesser than or
greater than
so both the constraints are greater than
which means the feasible region should
be above the two boundaries that is two
lines say both the constraints are
greater than or equal to so the feasible
region must be above the first
constraint as well as above the second
constraint this is the feasible region
see this is a feasible region here we
have three corner points this is the
first corner point second corner point
third corner by its value is 8 and 0
that is X 1 values 8 X 2 n is zero and
series value is X 1 values 0 X 2 value
is 12 and look at this this is the point
which intersect both the lines both was
to constrain and second constraint so
now we have to find the value for B C X
1 value is 0 point H X 2 value is 7
point one seven point two seven point
two a B C so in the feasible region we
have three corner points and the optimal
solution will be any one of the corner
points and let us find the optimum
solution by substituting the three
values in the objective function see
these are the three corner points value
is given now we need to substitute this
value in
the objective function our objective
function is minimizes that is equal to 4
X 1 plus 6 X 2 now we need to substitute
these values in the X 1 and X 2 let us
see the first one is value is H and 0 X
1 value is 8 X 2 value is 0 Z a is equal
to 4 X 1 value 8 plus 6 X 2 value 0 4
into 8 32 plus 6 into 0 0 so answer is
32 in the same way find the value for B
also
so such V is equal to 4 into X 1 value
is 0.8 plus 6 X 2 value 7.2 okay 4 into
0.8 3.2 plus 6 into 7.2 43.2 so what is
answer 3.2 plus forty three point two is
40 6.0 in the same way find the value
for C also 4 into X 1 value is 0 plus 6
into X 2 value is 12 so 4 into 0 0 plus
6 into 12 72 so what is answer 72 okay
now we have got the value for 3 corner
points for the first one is 32 4 B 46.4
4 c 72 okay so since the type of
objective function is minimum the
solution corresponding to the minimum Z
values to be selected as optimum
solution so look at which one is least
value 32 or 46.4 or 72 32 is the least
value so this point is a optimum point
is the optimum value so optimum solution
is a optimum solution is
a point a this corner is the optimal
solution so X 1 is equal to 8 and X 2 is
equal to 0 Z optimum value how much 32
organism optimal solution here I am
going to explain third model in LPP
graphical method so this is a
maximization problem with three
constraints take the first constraint
make it as a equation 5 X 1 plus 10 X 2
is equal to 50 so find the value for x1
and x2 when x1 is equal to 0 what is the
value for X 2 so just apply this value
in the first equation 5 X 1 values 0
plus 10 X 2 is equal to 50 so 5 into 0 0
so we have only 10 x2 is equal to 50 so
x2 is equal to 50 divided by 10 is equal
to 5 so when X 1 when X 1 is 0 X 2 value
is 5 in the same way we need to find the
value for x1 when X 2 is 0 then X 2 is
equal to 0 what is the value for x1 okay
just apply this value in the first
equation 5 X 1 plus 10 X 2 value is 0 is
equal to 50 so 10 into 0 0 we have only
5 X 1 so 5 X 1 is equal to 15 X 1 is
equal to 50 divided by 5 so what is a
value 10
so of x1 values when X 2 is 0 X 1 value
will be 10 see the second constrain 8 x1
+ 2 x2 greater than or equal to 16 make
it as an equation 8 x1 + 2 X
is equal to 16 so we need to find the
value for x1 and x2 when x1 is 0 what is
the value for X - 8 X 1 values 0 + 2 x2
is a part of 1618 - 0 0 so we have only
2 x2 is equal to 16 so x2 is equal to 16
by 2 so what is the value 8 6 - value is
8 in the same way when x2 is equal to 0
what is the value for x1 8 x1 + 2 x2
value is 0 is equal to 16 so 8 x1 - 2 0
0 is equal to 16 so x1 is equal to 16
divided by a what is a value - 6 1 value
is 2 so when X 1 value is 0 X 2 is 8
when X 2 is 0 X 1 value will be 2 so we
have got the value for second equation
also in the same way find the value for
third equation also that is 3 X 1 minus
2 X 2 greater than or equal to 6 3 X 1
minus 2 X 2 is equal to 6 so find the
value for x1 and x2 when X 1 is equal to
0 what is the value for x 2 this apply
this value in that equation 3 0 - 2 X 2
is equal to 6 3 into 0 0 - 2 X 2 is
equal to 6 so X 2 is equal to 6 divided
by minus 2 so X 2 is equal to minus 3
when X 1 is equal to 0 X 2 value is
minus 3 in the same way when X 2 is
equal to 0 what is the value for x 1 so
when X 2 is equal to 0 what is the value
for x 1 3 X 1 minus 2 X 2 values 0
zero is equal to 6 so 2 into 0 0 so we
have 3 X 1 is equal to 6 so X 1 is equal
to 6 divided by 3 2 so we have got two
points when X 1 is equal to 0 X 2 value
is minus 3 when X 2 is 0 X 1 value is 2
so with this you can plot the line in
the graph sheet see in the first
equation we have got two points now when
X 1 is equal to 0 X 2 value is 5 ok when
X 2 is 0 X 1 value is 10 so if this true
point you can draw a line when X 1 0 X 2
value is 5 so 0 + 5 6 1 value 0 extra
value 5 this is the first point see if
it's a cane pointer X 1/10 X 2 is 0 X 1
10 X 2 value 0 so this is the point X 1
10 X 2 0 so we have got two points with
this you can draw a line first line join
the two points this line represent the
first constraint look at the second
constraint values X 1 is 0 X 2's H X 1 0
X 2 8 so this is the point X 1 0 X 2 a
this is a cluster point X 1 is 2 X 2 is
0 X 1 is 2 X 2 is 0
this is the second point X 1 is 2 X 2 is
0 so now you need to join the two points
this line represent the second
constraint look at the third equation
values X 1 s 0 X 2 is minus 3 so the
first point is X 1 0 X 2 s minus 3 this
is the point say X 1 0 X 2 minus 3 this
is the first point look at this again
point X 1 is 2 X 2 is
0 X 1 2 X 2 is 0 so this is the point
that is 2 comma 0
this is the endpoint now you need to
join the two points see this line
represent third constraint so we have
got three lines now you need to find the
feasible region so before you find the
peaceable region just see the three
constraints check whether it is a
greater than or less there based on that
only can able to find the feasible
region look at the first equation that
is less than or equal to 50 okay so less
than or equal to means the feasible
region must be below the first line and
that is first one is below and look at
the second equation here greater than or
equal to so greater than or equal to
means the feasible region should be
above look at the third one that is
greater than or equal to 6 so the third
line is greater than no so about the
feasible region should be above the
third line okay the first one is less
than or equal to so the feasible region
should be below the first line second
one greater than or equal to no so the
feasible region should be above the
second line and third one also greater
than so the feasible region should be
above the third line so with this you
can able to find the feasible region
this is the first line ok the feasible
region should be below this line just
shade like this so the feasible region
should be below the first line ok now go
for the second line look at the second
equation condition that is the feasible
region should be above this line see the
second line the second constraint rule
is the feasible region should be above
the second line so arise the below
shaded portion see the first constraint
condition has below the line the
feasible region is below the line
second constraint condition is above the
line the feasible area will be above the
line so two conditions are satisfied now
go for the third one look at the third
constraint the condition is above ok the
feasible region should be above this
line this is third constraint ok the
feasible region should be above this
line that is above this line no so this
is the inner set this one is outer side
ok the feasible region should be above
this line so I write this part third
constraint condition is the feasible
region should be above the third line so
this is the final feasible region this
feasible region you have found based on
the three condition first condition is a
it should be below the first line above
the second line and above the third line
this is a feasible region in this
feasible region we got three corners a B
and C now find the x1 x2 value for each
corners for a x1 is 2 x2 is 0 for B X 1
is 10 x2 is 0 for C X 1 is 4 X 2 is 3 so
X 1 value is 4 X 2 is 3 now substitute
three corners of the x1 and x2 values in
the objective function to find optimal
solution objective function is maximized
that is equal to hundred x1 + 16 X 2 so
these are the three corners x1 and x2
values now substitute these value in the
objective function Z a is equal to
hundred x1 value is 2 + 16 X 2 value is
0 what is the answer 100 into 2 200 +
6200 so unser is 204 B hundred into x1
value is 10 + 16 X 2 value is 0 100 into
10,000 plus 16 into 0 0
answer is thousand for seeing 100 in 2 X
1 value is 4 plus 16 2 X 2 value is 300
2 for 400 plus 16 2 3 180 total 5 80
since the type of the objective function
is maximization the solution
corresponding to the maximum Z value is
to be selected as the optimal solution
look at the Z values a value is 200 B
value is thousand say values by 80 so B
value is the maximum value hence the
corresponding solution is X 1 is 10 X 2
is 0 set optimum value is thousand in
this video we have discussed a
minimization problem with two
constraints and maximization problem
with three constraints hope you all
understand this concept thank you