B Com Exam > B Com Videos > Business Mathematics and Statistics > Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics
Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com
|
115 videos|142 docs
|
FAQs on Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics Video Lecture - Business Mathematics and Statistics - B Com
| 1. What is the Simplex Method in linear programming? |  |
The Simplex Method is an algorithm used to solve linear programming problems. It starts with an initial feasible solution and iteratively moves towards an optimal solution by improving the objective function value at each step. It does this by moving from one corner point (vertex) of the feasible region to another until the optimal solution is reached.
| 2. How does the Simplex Method work in solving linear programming problems? | |
The Simplex Method works by continuously improving the objective function value at each iteration. It starts with an initial feasible solution and identifies the pivot element, which is the most negative coefficient in the objective function. Then, it performs row operations to create a new feasible solution with a better objective function value. This process is repeated until an optimal solution is found.
| 3. What are the key steps involved in applying the Simplex Method to solve linear programming problems? | |
The key steps in applying the Simplex Method are as follows:
1. Formulate the linear programming problem by defining the objective function and the constraints.
2. Convert the problem into standard form by introducing slack or surplus variables as necessary.
3. Construct the initial feasible solution by setting the slack or surplus variables to zero.
4. Calculate the coefficients of the objective function and set up the initial Simplex tableau.
5. Identify the pivot element, which is the most negative coefficient in the objective function.
6. Perform row operations to obtain a new feasible solution with a better objective function value.
7. Repeat steps 5 and 6 until an optimal solution is reached, where all coefficients in the objective function are non-negative.
| 4. What is the role of the Simplex Method in business mathematics and statistics? | |
The Simplex Method plays a crucial role in business mathematics and statistics as it helps in optimizing business decisions. It is used to solve linear programming problems that involve resource allocation, production planning, inventory management, transportation, and other business-related optimization problems. By finding the optimal solution, the Simplex Method helps businesses maximize profits, minimize costs, and make efficient use of resources.
| 5. Are there any limitations or drawbacks of the Simplex Method in solving linear programming problems? | |
Yes, there are a few limitations and drawbacks of the Simplex Method. It may become computationally expensive and time-consuming for problems with a large number of variables and constraints. Additionally, it may not handle certain types of constraints, such as integer or non-linear constraints, efficiently. However, advancements in optimization algorithms have addressed some of these limitations, and alternative methods like interior-point methods have been developed to handle more complex problems efficiently.
Text Transcript from Video
in this video we are going to see linear
programming problem using Big M method
the Viggen method is a modified version
of simplex method only so we can use the
same simplex algorithm to solve this Big
M method there is no much difference
between simplex method and Big M method
in simplex method it contains a lesser
than or equal to constraint so in that
case we need to add slack variables that
is plus as this we have already
discussed in the simplex method in a
separate video now we are going to see
begin method which contains greater than
or equal to constraints so in case of
greater than or equal to we need to do
two things first thing we need to
subtract the surplus variables that is -
yes and we need to add artificial
variables that is plus a let us see the
problem using Bigham method
see the problem solve the following
linear programming problem using begum
method minimization set is equal to 7 X
1 plus 15 X 2 plus 20 X 3 subject to 2 X
1 plus 4 X 2 plus 6 X 3 greater than or
equal to 24 and second one 3 X 1 plus 9
X 2 plus 6 X 3 greater than or equal to
30 X 1 X 2 X 3 greater than or equal to
0 so these are the two constraints are
this one is objective function so this
problem is minimization problem using
Bigham method the first step is we need
to subtract surplus variables and we
need to add artificial variables in the
two constraints before that we need to
write the objective function that is
minimization Z is equal to 7 X 1 plus 15
X 2 plus 20 X 3 subject to the
constraint subject to the first
constraint is 2 X 1 plus 4 X 2 plus 6 X
3 so in order to balance the equation we
need to subtract a surplus variable 1
and
add artificial variable one in the
foster constrain is equal to 24 in the
same way in the next constrain is 3 X 1
plus 9 X 2 plus 6 X 3 so here also
greater than or equal to 9 so in order
to make it equal just subtract the
surplus variable 2 and add artificial
variable 2 is equal to 30 and what the
variables are then X 1 comma X 2 comma X
3 comma we have subtracted to surplus
variable now so s 1 comma s 2 and we
have added artificial variables 2 comma
a 1 and a 2 greater than or equal to 0
okay so here we have done two things
first thing is we have subtracted a
surplus variable and we have added
artificial variable the same thing need
to be added in the objective function ok
so now the objective function is
minimization set is equal to 7 X 1 plus
15 X 2 plus 20 X 3 and we need to add
that is the four surplus variable we
need to give positive 0 ok the value is
positive 0 so 0 s 1 plus 0 s 2 4
artificial variable we need a you
maximum positive value that is a Yong
Yong a 1 plus young a 2 m denotes a
maximum positive value say these
artificial variable are included in the
model just to solve the model ok and in
the final solution these artificial
variables should not be available let us
say the first initial simplex table let
us say the initial table for matter be V
stands for basic variables R that is in
this problem we have these variables x1
x2 x3 s1 s2 a1 and a2 so these are the
basic variables
CJ stands for coefficient of objective
function and CB stands for coefficient
of basic variables so this one is Row
one and second one row two
we need to write solution and finally a
ratio so now let us see how I am framing
these things so BB is a basic variables
are these are the basic variables that
we have already written
CJ stands for coefficient of objective
function this we can write from the
objective function look at the objective
function 7 X 1 so X 1 is there now right
715 X 2 X 2 is there right 15 next 120 X
3 xx 0 s 1 0 0 s 2 0 and for artificial
variable the most positive value that is
young for a 1 young for a 2 also young
so these are the objective function that
is coefficient of objective function
these are the basic variables in this
column we need to write the two
artificial constraints that is a 1 and a
2 so write a 1 and a 2 what is the
coefficient of the basic variables look
at the objective function for a 1 m4a2
also M so write m4a1 and m4a2 also so
now write the first row values so look
at the first constraint to X 1 for X - 6
X 3 minus s 1 plus a 1 is equal to 24 -
X 1 for X - 6 X 3 s 1 that is minus s 1
so s 1 is the inner right minus 1 so in
the first constraint there is no as to
put 0 and in the first constraint a 1 is
there so write plus 1 know a 2 so write
0 is equal to 24
in the same way fill all these things
for the row 2 for Row 2 3 X 1 3 9 X - 4
X - 9 4 X 3 6 and there is no s 1 so 4 0
s 2 is there
- 1 a 1 is now then so 0 a 2 is then so
plus 1 is equal to 13 so this is the way
to frame initial table so now we need to
proceed further the next step is we need
to find Z J okay so in order to find the
Z J this is the formula to find Z J so
instead of using this formula let me
explain the simple method so first we
need to take coefficient of basic
variable of r1 into the first element
plus coefficient of basic variable of r2
into first element then you will be
getting ZJ take the coefficient of a
basic variable into the first element
plus the second row coefficient of basic
variable into first element will be
getting ZJ that is M into 2 plus M into
3 in the same way for the next value M
into 4 plus M into 9 m into 6 plus M
into 6 for the fourth value M into minus
1 plus M into 0 in this way we need to
find the ZJ let us see the calculation
for the first one M into two - yum plus
M into three Liam 2 plus 3 Phi M for the
next value M into 4 for young plus m
into 9 9 young so totally 13 M for the
next one M into 6 plus M into 6 so 12 m
next value M into -1 - y M plus M into 0
0 so value minus young next one M into 0
0 plus M into -1 - y M
next one yum into 1 M plus M into 0 0 so
plus young next one M into 0 0 plus M
into 1 plus M mix 1 M into 24 24 M plus
M into 30 30 M totally 54 young ok so
this is the way to find
is each a so after finding ZJ the next
step is a CJ - ZJ so we need to compare
CJ and ZJ and we need to find the
difference as it is a the formless CJ -
ZJ so look at the first value CJ is 7
minus Z J's Phi M so answer 7 minus 5m
next value CJ is 15 ZJ is 13 M so CJ
minus ZJ is equal to 15 minus 13 young
next value 20 minus 12 young next 1 0
minus young already - is there so 0 plus
young so answer is young next 0 minus
minus young so minus minus plus plus
young next one young - young 0 M minus
young 0 after finding CJ - ZJ we need to
check the optimality so what is
optimality condition for minimization
problem this is the optimality condition
that is CJ - ZJ must be greater than or
equal to 0 so all the CJ - ZJ value must
be 0 or positive value that is greater
than or equal to 0 look at the answer CJ
- ZJ we have got a negative value for
three things the first one is 7 minus
Phi M here 15-13 IAM and here 20 minus 2
a yam so we did not reach the optimality
so in order to reach the optimality we
need to proceed further
in order to proceed further the first
step is we need to find the key column
for K column we need to find most
negative value from the CJ - ZJ so look
at these three values which one is most
negative - 30 mm that is 15 minus 13 M
is
most negative value so this is the key
column so this column is a key column
that is x2 column is a K column okay so
after finding key column the next one is
we need to find key row for that we need
to find the ratio in ratio column for 24
divided by 4 is equal to 6 for the
second row the solution is 13 house or
30 divided by 9 is equal to 10 by 3 in
order to find the key row we need to
select at least a positive value so
which one is at least a positive value
10 by 3 is at least a positive value
than the 6 so this row is a key row this
row is the key row ok so now we have got
a key column x2 is a key column and a 2
is a key row and the intersecting point
is called key element so the key element
is 9 and this one is key row and this
one is key column so after finding key
row and key column and gear element we
need to prepare the first iteration in
the first iteration X 2 will be the
entering variable and eight will be the
leaving variable in the first iteration
BB and CJ remains same ok so now we have
to fill row 3 and Row 4 ok so in the
previous initial table so which variable
is entering variable in which variables
leaving variable X 2 is a entering
variable and a 2 is a leaving variable
okay in the first iteration instead of a
to unity write X 2 X 2 what is the cost
for x2 15 right 15 here okay and a 2 is
a leaving variable no so in the a two
column you need not find anything
support - okay now we need to find the
new value in the first iteration okay in
order to find a new
value just take the old value divided by
K element okay for the leaving variable
okay you need to find the old value
divided by the King element that is
three by nine nine by nine 6x9 0 by 9
minus 1 by 9 0 by 9 1 by 9 and 30 by 9
in this way you can find a new value for
Row 4 so first 1 3 by 9 that is 1 by 3 9
by 9 1 6 by 9 is equal to 2 by 3 0 by 9
is equal to 0 minus 1 by 9 is equal to
minus 1 by 9 0 by 9 is equal to 0 okay
and 30 by 9 is equal to 10 by 3 okay we
have got the new value for Row 4 ok this
is the way to find new value for Row 4
now we need to find the values for Row 3
in order to find values for 203 there is
a formula the formula is a old value
minus K column value into Kiro value
noted by key value ok so look at the
table initial table for example for the
first value old value is 2 okay so 2
minus corresponding column values for
okay into correspondent key row value is
3 divided by key value is 9 okay this is
the formula to find the new value so
instead of this formula you can find
this formula that is old value minus K
column value into row 4 value okay this
we have already applied in the previous
step that is a K Rho value divided by K
value okay this is denotes r4 so you can
simply take them or 4 value let me
explain with the problem old value minus
key column value into row 4 value look
at the initial table old value is 2
minus key column value is 4 into row 4
values 1 by 3 again I repeat old values
2 so 2 minus
and Kay column value is 4 so 4 into or 4
value 1 by 3 2 - 4 into 1 by 3 is equal
to 2 minus 4 by 3 is equal to 3 6 minus
4 is equal to 2 by 3 so the new value is
2 by 3 2 by 3 in the same way for the
next value old values for corresponding
column value is also 4 and the new value
row 4 value is 1 so formula is old value
4 - correspondent column value for n 2
or 4 values 1 so 4 minus 4 is equal to
0-0
in the same way for the next value is
the old value is 6 so 6 minus
corresponding column value is 4 into our
4 value 2 by 3 6 minus 4 into 2 by 3 so
6 minus photos are 8 8 by 3 LCM 18 minus
8 is equal to 10 by 3 new value is 10 by
3 10 by 3 ok for the next value old
value is minus 1 minus K column value 4
into Rho value row value is 0 minus 1
minus 4 into 0 is equal to minus 1 minus
1 for the next value old value is 0 so 0
minus 4 into minus 1 by 9 is equal to
plus 4 by 9 for the next value old value
is 1 1 minus column value 4 into new row
value 0 1 minus 4 into
zero is equal to one one okay for the
solution all the valleys 24 - : value
for into new row value ten by three so
24 minus 4 into 10 40 by 3 so LCM 3 3 to
24 72 minus 40 is equal to 30 2 by 3 30
2 by 3 see here the value is the basic
variable remains same and the
coefficient of basic variables are in
mind same so this is the way to frame
first iteration tables after framing all
these things the next step is you need
to find the ZJ okay in order to find the
ZJ you can follow the same procedure
what we did in the initial table for the
first value M into 2 by 3 plus 15 into 1
by 3 the answer is 2 by 3 M plus 5 okay
for the second value M into 0 0 plus 15
into 1 15 next value M into 10 by 3 10
by 3 M plus 15 into 2 by 3 plus 10 next
value M into minus 1 plus 15 into 0 so
minus M next value M into 4 by 9 so 4 by
9 M plus 15 into minus 1 by 9 so minus 5
by 3 next value M into 1 M plus 18 into
0 0 the answer is young okay the same
way for solution M into 32 by 3 plus 15
into 10 by 3 the answer is a 32 by 3 M
plus 50 after finding ZJ the next step
is we need to find CJ minus
J so look at the first value the formula
CJ - ZJ so first value is 7 7 minus 2 by
3 M plus 5 so you cannot subtract this
value from 7 because it's a here we have
a young value so we can subtract only 5
from 7 ok the formula is CJ - ZJ so we
will be getting minus 2 by 3 M plus 2
that is 7 minus 5 will be getting 2 for
the next value 15 minus 15 will be
getting zero for the next value 20 minus
10 by 3 M plus 10 the same way we cannot
subtract this value okay from 20 we can
select only 10 with the for my CJ - ZJ
no the answer is minus 10 by 3 M plus 10
that is 20 minus 10 we'll be getting 10
the next value is 0 minus minus young so
we'll be getting plus young next value 0
minus 4 by 9 M minus PI by 3 the same
thing is applicable here also because we
have a young value here so the answer is
minus 4 by 9 M plus 5 by 3 okay again I
repeat 0 minus 4 by 9 M minus PI by 3 so
minus into minus VB getting Plus value
the next value M minus young 0
after finding CJ minus ZJ we need to
check the optimality condition what is
the optimality condition for
minimization problem all the CJ minus ZJ
value must be greater than or equal to 0
which means we need to get a to 0 or
positive value but in this iteration we
have got negative values for three
things in so we didn't attain the
optimality so we need to proceed further
in order to proceed further we need to
find key column K row and key element
for the key column we need to find most
negative value so which one is most
negative value this one is most negative
value so this column is a key column
okay after finding key : then we need to
find the ratio in order to find key row
for the first row 32 by 3 divided by 10
by 3 which means 3 by 10 the answer is
16 by 5 okay for the next value 10 by 3
divided by 2 by 3 that is 3 by 2 the
answer is 5 in order to find the key row
we need to select at least a positive
value so which one is at least positive
value so this value is the least
positive value when compared to 5 so 16
by 5 is the least positive and
compactify so this row is a king row
after finding Cairo and K :
this intersection point is the key
element so x3 is a key column and a 1 is
a Cairo and 10 by 3 is the key element
okay so according to this I tration x3
is the entering variable and a1 is the
leaving variable
so these other two things is very
important in order to frame iteration to
say the iteration to basic variable and
CJ values or remind same now we need to
frame our five that is row five occasion
the previous iteration so which one is
entering and which one is leaving
variable x3 is entering variable and a1
is a leaving variable so instead of a
one we need to write X 3 and X 3 value
is 20 here we need to write 20 and here
we need to write X 3 for these value
take the old value / key element that is
2 by 3 / 10 by 3 0 / 10 by 3 10 by 3 /
10 by 3 minus 1 divided by 10 by 3 4 by
9 / 10 by 3 for a 1 you need not write
anything why because a 1 is a leaving
variable you can put - 4 solution 32 by
3 / 10 by 3 so this is the procedure to
find the row 5 values 405 entering
variable is X 3 okay the cost is 20 okay
for the
you value you can apply the simple
formula that is older value divided by K
value 2x3 rotor by 10 by 3 is equal to 1
by 5 then 0 divided by 10 by 3 0 then 10
by 3 divided by 10 by 3 1 minus 1
divided by 10 by 3 minus 3 by 10 then 4
by 9 divided by 10 by 3 2 by 15 for a 1
a 2 - 4 solution 30 2 by 3 divided by 10
by 3 is equal to 16 by 5 okay so now we
have got row 5 values now we need to
find row 6 values so for row 6 basic
variable is X 2 as it is the coefficient
of x 2 is 15 so X - 15 now we need to
find a new value for row 6 so there is a
formula to find out row 6 look at the
formula row 6 value is equal to old
value minus K column value into Rho Phi
value so for the first value old value
is 1 by 3 - a key column value 2 by 3
into Rho Phi value 1 by 5 1 by 3 minus 2
by 3 into 1 by 5 1 7 1 by 3 minus Phi 3
so 15 2 into 1 2 so LCM 15 5 times 1
into 5 5 5 minus 2 is equal to 3 by 15
is equal to 1 by 5 so new value is 1 by
5 for the next value old value is 1 1
minus corresponding column value same
thing 2 by 3 into our five value that is
0 1 minus 2 by 3 into 0 so 2 by 3 into 0
will get 0 so our answer is 1 so new
value 1 for the next value old values 2
by 3 minus corresponding column value
same thing 2 by 3 into row five values
1 2 by 3 minus 2 by 3 into 1 so 2 by 3
into 1 2 by 3 2 by 3 minus 2 by 3 is
equal to 0 ok for the next value old
values 0 minus correspond and column
value 2 by 3 into Rho Phi value minus 3
by 10 older values 0 minus column value
2 by 3 in turn row 5 value minus 3 by 10
so 3 3 cancel 1 by 5 1 by 5 next one old
value is minus 1 by 9 minus column value
2 by 3 into Row 5 value 2 by 15 minus 1
by 9 minus 2 by 3 into 2 by 15 so minus
1 by 9 is equal to 3 into 15 45 minus 1
45 2 into 2 4 so LCM 45 5 times so 1
into 5 5 minus 5 minus 4 minus 9 by 45
is equal to minus 1 by 5 the answer is
minus 1 by 5 ok for a-1 a-2 need not
calculate because these two variables
are left a 1 year to put - for solution
old value is 10 by 3
- column value 2 by 3 into rho v value
16 by 5 10 by 3
- column value 2 by 3 into rho v value
16 by 5 is equal to 10 by 3 minus 5 into
3 15 6 into 2 32 so I will seem 15 5
times 5 into 10 50 - 30 - 50 - 32 is
equal to 18 by 15 Oh
6x5 answer is six by five now we have
bought row five values and row six
values now we need to find ZJ so in
order to find ZJ just multiply
coefficient of basic variable into each
element plus a coefficient of next row
variable into H element n so 20 into 1
by Phi plus 15 into 1 by 5 answer is 7
next 20 into 0 0 plus 15 into 1 15 20
into 1 20 plus 15 into 0 0 answer 20 and
20 in 2 minus 3 by 10 plus 15 into 1 by
5 answer minus 3 then 22 2 by 15 plus 15
into minus 1 by 5 answer is minus 1 by 3
the next value is 20 into 16 by 5 plus
15 into 6 by 5 answer 82 after finding
ZJ the next step is we need to find CJ
minus ZJ ok CJ is 7 7 minus 7 0 15 - 15
0 20 minus 2000 - minus 3 plus 3 0 minus
minus 1 by 3 plus 1 by 3 that's all okay
now check the optimality so what is
optimality condition for minimization
problem all the CJ minus J must be
greater than or equal to 0 which means
we need to get a 0 or positive value so
in this iteration we have got 0 0 0 3 &
1 by 3 so we have reached the optimality
so this is the final iteration we have
got the optimality solution the solution
is 82 ok so now we need to write the X 1
value X 2 value and X 3 value okay so
the minimization minimization is that
is equal to the final answer is solution
is 82 okay now just apply we have got x2
value x2 value is 6 by 5 so 6 by 5 x3
value is 16 by 5 16 by 5 so x1 value
take 0 there is no x1 we have got only
x2 and x3 so put 0 for X 1 now apply
these value in the objective function in
order to ensure the solution look at the
objective function minimization set is
equal to 7 x1 + 15 X 2 plus 20 X 3 7 X 1
s 0 + 15 X 2 is 6 by 5 plus 20 x3 value
is 16 by 5 is equal to 82 we need to get
82 by applying these things just find 7
into 0 0 + 15 into 6 by 5 so 6 into 3 18
+ 4 into 16 64 so answer is 82 so we
have attained the optimality okay this
is the way to prepare LPP model using
Bigham method hope you all understood
this concept thank you
programming problem using Big M method
the Viggen method is a modified version
of simplex method only so we can use the
same simplex algorithm to solve this Big
M method there is no much difference
between simplex method and Big M method
in simplex method it contains a lesser
than or equal to constraint so in that
case we need to add slack variables that
is plus as this we have already
discussed in the simplex method in a
separate video now we are going to see
begin method which contains greater than
or equal to constraints so in case of
greater than or equal to we need to do
two things first thing we need to
subtract the surplus variables that is -
yes and we need to add artificial
variables that is plus a let us see the
problem using Bigham method
see the problem solve the following
linear programming problem using begum
method minimization set is equal to 7 X
1 plus 15 X 2 plus 20 X 3 subject to 2 X
1 plus 4 X 2 plus 6 X 3 greater than or
equal to 24 and second one 3 X 1 plus 9
X 2 plus 6 X 3 greater than or equal to
30 X 1 X 2 X 3 greater than or equal to
0 so these are the two constraints are
this one is objective function so this
problem is minimization problem using
Bigham method the first step is we need
to subtract surplus variables and we
need to add artificial variables in the
two constraints before that we need to
write the objective function that is
minimization Z is equal to 7 X 1 plus 15
X 2 plus 20 X 3 subject to the
constraint subject to the first
constraint is 2 X 1 plus 4 X 2 plus 6 X
3 so in order to balance the equation we
need to subtract a surplus variable 1
and
add artificial variable one in the
foster constrain is equal to 24 in the
same way in the next constrain is 3 X 1
plus 9 X 2 plus 6 X 3 so here also
greater than or equal to 9 so in order
to make it equal just subtract the
surplus variable 2 and add artificial
variable 2 is equal to 30 and what the
variables are then X 1 comma X 2 comma X
3 comma we have subtracted to surplus
variable now so s 1 comma s 2 and we
have added artificial variables 2 comma
a 1 and a 2 greater than or equal to 0
okay so here we have done two things
first thing is we have subtracted a
surplus variable and we have added
artificial variable the same thing need
to be added in the objective function ok
so now the objective function is
minimization set is equal to 7 X 1 plus
15 X 2 plus 20 X 3 and we need to add
that is the four surplus variable we
need to give positive 0 ok the value is
positive 0 so 0 s 1 plus 0 s 2 4
artificial variable we need a you
maximum positive value that is a Yong
Yong a 1 plus young a 2 m denotes a
maximum positive value say these
artificial variable are included in the
model just to solve the model ok and in
the final solution these artificial
variables should not be available let us
say the first initial simplex table let
us say the initial table for matter be V
stands for basic variables R that is in
this problem we have these variables x1
x2 x3 s1 s2 a1 and a2 so these are the
basic variables
CJ stands for coefficient of objective
function and CB stands for coefficient
of basic variables so this one is Row
one and second one row two
we need to write solution and finally a
ratio so now let us see how I am framing
these things so BB is a basic variables
are these are the basic variables that
we have already written
CJ stands for coefficient of objective
function this we can write from the
objective function look at the objective
function 7 X 1 so X 1 is there now right
715 X 2 X 2 is there right 15 next 120 X
3 xx 0 s 1 0 0 s 2 0 and for artificial
variable the most positive value that is
young for a 1 young for a 2 also young
so these are the objective function that
is coefficient of objective function
these are the basic variables in this
column we need to write the two
artificial constraints that is a 1 and a
2 so write a 1 and a 2 what is the
coefficient of the basic variables look
at the objective function for a 1 m4a2
also M so write m4a1 and m4a2 also so
now write the first row values so look
at the first constraint to X 1 for X - 6
X 3 minus s 1 plus a 1 is equal to 24 -
X 1 for X - 6 X 3 s 1 that is minus s 1
so s 1 is the inner right minus 1 so in
the first constraint there is no as to
put 0 and in the first constraint a 1 is
there so write plus 1 know a 2 so write
0 is equal to 24
in the same way fill all these things
for the row 2 for Row 2 3 X 1 3 9 X - 4
X - 9 4 X 3 6 and there is no s 1 so 4 0
s 2 is there
- 1 a 1 is now then so 0 a 2 is then so
plus 1 is equal to 13 so this is the way
to frame initial table so now we need to
proceed further the next step is we need
to find Z J okay so in order to find the
Z J this is the formula to find Z J so
instead of using this formula let me
explain the simple method so first we
need to take coefficient of basic
variable of r1 into the first element
plus coefficient of basic variable of r2
into first element then you will be
getting ZJ take the coefficient of a
basic variable into the first element
plus the second row coefficient of basic
variable into first element will be
getting ZJ that is M into 2 plus M into
3 in the same way for the next value M
into 4 plus M into 9 m into 6 plus M
into 6 for the fourth value M into minus
1 plus M into 0 in this way we need to
find the ZJ let us see the calculation
for the first one M into two - yum plus
M into three Liam 2 plus 3 Phi M for the
next value M into 4 for young plus m
into 9 9 young so totally 13 M for the
next one M into 6 plus M into 6 so 12 m
next value M into -1 - y M plus M into 0
0 so value minus young next one M into 0
0 plus M into -1 - y M
next one yum into 1 M plus M into 0 0 so
plus young next one M into 0 0 plus M
into 1 plus M mix 1 M into 24 24 M plus
M into 30 30 M totally 54 young ok so
this is the way to find
is each a so after finding ZJ the next
step is a CJ - ZJ so we need to compare
CJ and ZJ and we need to find the
difference as it is a the formless CJ -
ZJ so look at the first value CJ is 7
minus Z J's Phi M so answer 7 minus 5m
next value CJ is 15 ZJ is 13 M so CJ
minus ZJ is equal to 15 minus 13 young
next value 20 minus 12 young next 1 0
minus young already - is there so 0 plus
young so answer is young next 0 minus
minus young so minus minus plus plus
young next one young - young 0 M minus
young 0 after finding CJ - ZJ we need to
check the optimality so what is
optimality condition for minimization
problem this is the optimality condition
that is CJ - ZJ must be greater than or
equal to 0 so all the CJ - ZJ value must
be 0 or positive value that is greater
than or equal to 0 look at the answer CJ
- ZJ we have got a negative value for
three things the first one is 7 minus
Phi M here 15-13 IAM and here 20 minus 2
a yam so we did not reach the optimality
so in order to reach the optimality we
need to proceed further
in order to proceed further the first
step is we need to find the key column
for K column we need to find most
negative value from the CJ - ZJ so look
at these three values which one is most
negative - 30 mm that is 15 minus 13 M
is
most negative value so this is the key
column so this column is a key column
that is x2 column is a K column okay so
after finding key column the next one is
we need to find key row for that we need
to find the ratio in ratio column for 24
divided by 4 is equal to 6 for the
second row the solution is 13 house or
30 divided by 9 is equal to 10 by 3 in
order to find the key row we need to
select at least a positive value so
which one is at least a positive value
10 by 3 is at least a positive value
than the 6 so this row is a key row this
row is the key row ok so now we have got
a key column x2 is a key column and a 2
is a key row and the intersecting point
is called key element so the key element
is 9 and this one is key row and this
one is key column so after finding key
row and key column and gear element we
need to prepare the first iteration in
the first iteration X 2 will be the
entering variable and eight will be the
leaving variable in the first iteration
BB and CJ remains same ok so now we have
to fill row 3 and Row 4 ok so in the
previous initial table so which variable
is entering variable in which variables
leaving variable X 2 is a entering
variable and a 2 is a leaving variable
okay in the first iteration instead of a
to unity write X 2 X 2 what is the cost
for x2 15 right 15 here okay and a 2 is
a leaving variable no so in the a two
column you need not find anything
support - okay now we need to find the
new value in the first iteration okay in
order to find a new
value just take the old value divided by
K element okay for the leaving variable
okay you need to find the old value
divided by the King element that is
three by nine nine by nine 6x9 0 by 9
minus 1 by 9 0 by 9 1 by 9 and 30 by 9
in this way you can find a new value for
Row 4 so first 1 3 by 9 that is 1 by 3 9
by 9 1 6 by 9 is equal to 2 by 3 0 by 9
is equal to 0 minus 1 by 9 is equal to
minus 1 by 9 0 by 9 is equal to 0 okay
and 30 by 9 is equal to 10 by 3 okay we
have got the new value for Row 4 ok this
is the way to find new value for Row 4
now we need to find the values for Row 3
in order to find values for 203 there is
a formula the formula is a old value
minus K column value into Kiro value
noted by key value ok so look at the
table initial table for example for the
first value old value is 2 okay so 2
minus corresponding column values for
okay into correspondent key row value is
3 divided by key value is 9 okay this is
the formula to find the new value so
instead of this formula you can find
this formula that is old value minus K
column value into row 4 value okay this
we have already applied in the previous
step that is a K Rho value divided by K
value okay this is denotes r4 so you can
simply take them or 4 value let me
explain with the problem old value minus
key column value into row 4 value look
at the initial table old value is 2
minus key column value is 4 into row 4
values 1 by 3 again I repeat old values
2 so 2 minus
and Kay column value is 4 so 4 into or 4
value 1 by 3 2 - 4 into 1 by 3 is equal
to 2 minus 4 by 3 is equal to 3 6 minus
4 is equal to 2 by 3 so the new value is
2 by 3 2 by 3 in the same way for the
next value old values for corresponding
column value is also 4 and the new value
row 4 value is 1 so formula is old value
4 - correspondent column value for n 2
or 4 values 1 so 4 minus 4 is equal to
0-0
in the same way for the next value is
the old value is 6 so 6 minus
corresponding column value is 4 into our
4 value 2 by 3 6 minus 4 into 2 by 3 so
6 minus photos are 8 8 by 3 LCM 18 minus
8 is equal to 10 by 3 new value is 10 by
3 10 by 3 ok for the next value old
value is minus 1 minus K column value 4
into Rho value row value is 0 minus 1
minus 4 into 0 is equal to minus 1 minus
1 for the next value old value is 0 so 0
minus 4 into minus 1 by 9 is equal to
plus 4 by 9 for the next value old value
is 1 1 minus column value 4 into new row
value 0 1 minus 4 into
zero is equal to one one okay for the
solution all the valleys 24 - : value
for into new row value ten by three so
24 minus 4 into 10 40 by 3 so LCM 3 3 to
24 72 minus 40 is equal to 30 2 by 3 30
2 by 3 see here the value is the basic
variable remains same and the
coefficient of basic variables are in
mind same so this is the way to frame
first iteration tables after framing all
these things the next step is you need
to find the ZJ okay in order to find the
ZJ you can follow the same procedure
what we did in the initial table for the
first value M into 2 by 3 plus 15 into 1
by 3 the answer is 2 by 3 M plus 5 okay
for the second value M into 0 0 plus 15
into 1 15 next value M into 10 by 3 10
by 3 M plus 15 into 2 by 3 plus 10 next
value M into minus 1 plus 15 into 0 so
minus M next value M into 4 by 9 so 4 by
9 M plus 15 into minus 1 by 9 so minus 5
by 3 next value M into 1 M plus 18 into
0 0 the answer is young okay the same
way for solution M into 32 by 3 plus 15
into 10 by 3 the answer is a 32 by 3 M
plus 50 after finding ZJ the next step
is we need to find CJ minus
J so look at the first value the formula
CJ - ZJ so first value is 7 7 minus 2 by
3 M plus 5 so you cannot subtract this
value from 7 because it's a here we have
a young value so we can subtract only 5
from 7 ok the formula is CJ - ZJ so we
will be getting minus 2 by 3 M plus 2
that is 7 minus 5 will be getting 2 for
the next value 15 minus 15 will be
getting zero for the next value 20 minus
10 by 3 M plus 10 the same way we cannot
subtract this value okay from 20 we can
select only 10 with the for my CJ - ZJ
no the answer is minus 10 by 3 M plus 10
that is 20 minus 10 we'll be getting 10
the next value is 0 minus minus young so
we'll be getting plus young next value 0
minus 4 by 9 M minus PI by 3 the same
thing is applicable here also because we
have a young value here so the answer is
minus 4 by 9 M plus 5 by 3 okay again I
repeat 0 minus 4 by 9 M minus PI by 3 so
minus into minus VB getting Plus value
the next value M minus young 0
after finding CJ minus ZJ we need to
check the optimality condition what is
the optimality condition for
minimization problem all the CJ minus ZJ
value must be greater than or equal to 0
which means we need to get a to 0 or
positive value but in this iteration we
have got negative values for three
things in so we didn't attain the
optimality so we need to proceed further
in order to proceed further we need to
find key column K row and key element
for the key column we need to find most
negative value so which one is most
negative value this one is most negative
value so this column is a key column
okay after finding key : then we need to
find the ratio in order to find key row
for the first row 32 by 3 divided by 10
by 3 which means 3 by 10 the answer is
16 by 5 okay for the next value 10 by 3
divided by 2 by 3 that is 3 by 2 the
answer is 5 in order to find the key row
we need to select at least a positive
value so which one is at least positive
value so this value is the least
positive value when compared to 5 so 16
by 5 is the least positive and
compactify so this row is a king row
after finding Cairo and K :
this intersection point is the key
element so x3 is a key column and a 1 is
a Cairo and 10 by 3 is the key element
okay so according to this I tration x3
is the entering variable and a1 is the
leaving variable
so these other two things is very
important in order to frame iteration to
say the iteration to basic variable and
CJ values or remind same now we need to
frame our five that is row five occasion
the previous iteration so which one is
entering and which one is leaving
variable x3 is entering variable and a1
is a leaving variable so instead of a
one we need to write X 3 and X 3 value
is 20 here we need to write 20 and here
we need to write X 3 for these value
take the old value / key element that is
2 by 3 / 10 by 3 0 / 10 by 3 10 by 3 /
10 by 3 minus 1 divided by 10 by 3 4 by
9 / 10 by 3 for a 1 you need not write
anything why because a 1 is a leaving
variable you can put - 4 solution 32 by
3 / 10 by 3 so this is the procedure to
find the row 5 values 405 entering
variable is X 3 okay the cost is 20 okay
for the
you value you can apply the simple
formula that is older value divided by K
value 2x3 rotor by 10 by 3 is equal to 1
by 5 then 0 divided by 10 by 3 0 then 10
by 3 divided by 10 by 3 1 minus 1
divided by 10 by 3 minus 3 by 10 then 4
by 9 divided by 10 by 3 2 by 15 for a 1
a 2 - 4 solution 30 2 by 3 divided by 10
by 3 is equal to 16 by 5 okay so now we
have got row 5 values now we need to
find row 6 values so for row 6 basic
variable is X 2 as it is the coefficient
of x 2 is 15 so X - 15 now we need to
find a new value for row 6 so there is a
formula to find out row 6 look at the
formula row 6 value is equal to old
value minus K column value into Rho Phi
value so for the first value old value
is 1 by 3 - a key column value 2 by 3
into Rho Phi value 1 by 5 1 by 3 minus 2
by 3 into 1 by 5 1 7 1 by 3 minus Phi 3
so 15 2 into 1 2 so LCM 15 5 times 1
into 5 5 5 minus 2 is equal to 3 by 15
is equal to 1 by 5 so new value is 1 by
5 for the next value old value is 1 1
minus corresponding column value same
thing 2 by 3 into our five value that is
0 1 minus 2 by 3 into 0 so 2 by 3 into 0
will get 0 so our answer is 1 so new
value 1 for the next value old values 2
by 3 minus corresponding column value
same thing 2 by 3 into row five values
1 2 by 3 minus 2 by 3 into 1 so 2 by 3
into 1 2 by 3 2 by 3 minus 2 by 3 is
equal to 0 ok for the next value old
values 0 minus correspond and column
value 2 by 3 into Rho Phi value minus 3
by 10 older values 0 minus column value
2 by 3 in turn row 5 value minus 3 by 10
so 3 3 cancel 1 by 5 1 by 5 next one old
value is minus 1 by 9 minus column value
2 by 3 into Row 5 value 2 by 15 minus 1
by 9 minus 2 by 3 into 2 by 15 so minus
1 by 9 is equal to 3 into 15 45 minus 1
45 2 into 2 4 so LCM 45 5 times so 1
into 5 5 minus 5 minus 4 minus 9 by 45
is equal to minus 1 by 5 the answer is
minus 1 by 5 ok for a-1 a-2 need not
calculate because these two variables
are left a 1 year to put - for solution
old value is 10 by 3
- column value 2 by 3 into rho v value
16 by 5 10 by 3
- column value 2 by 3 into rho v value
16 by 5 is equal to 10 by 3 minus 5 into
3 15 6 into 2 32 so I will seem 15 5
times 5 into 10 50 - 30 - 50 - 32 is
equal to 18 by 15 Oh
6x5 answer is six by five now we have
bought row five values and row six
values now we need to find ZJ so in
order to find ZJ just multiply
coefficient of basic variable into each
element plus a coefficient of next row
variable into H element n so 20 into 1
by Phi plus 15 into 1 by 5 answer is 7
next 20 into 0 0 plus 15 into 1 15 20
into 1 20 plus 15 into 0 0 answer 20 and
20 in 2 minus 3 by 10 plus 15 into 1 by
5 answer minus 3 then 22 2 by 15 plus 15
into minus 1 by 5 answer is minus 1 by 3
the next value is 20 into 16 by 5 plus
15 into 6 by 5 answer 82 after finding
ZJ the next step is we need to find CJ
minus ZJ ok CJ is 7 7 minus 7 0 15 - 15
0 20 minus 2000 - minus 3 plus 3 0 minus
minus 1 by 3 plus 1 by 3 that's all okay
now check the optimality so what is
optimality condition for minimization
problem all the CJ minus J must be
greater than or equal to 0 which means
we need to get a 0 or positive value so
in this iteration we have got 0 0 0 3 &
1 by 3 so we have reached the optimality
so this is the final iteration we have
got the optimality solution the solution
is 82 ok so now we need to write the X 1
value X 2 value and X 3 value okay so
the minimization minimization is that
is equal to the final answer is solution
is 82 okay now just apply we have got x2
value x2 value is 6 by 5 so 6 by 5 x3
value is 16 by 5 16 by 5 so x1 value
take 0 there is no x1 we have got only
x2 and x3 so put 0 for X 1 now apply
these value in the objective function in
order to ensure the solution look at the
objective function minimization set is
equal to 7 x1 + 15 X 2 plus 20 X 3 7 X 1
s 0 + 15 X 2 is 6 by 5 plus 20 x3 value
is 16 by 5 is equal to 82 we need to get
82 by applying these things just find 7
into 0 0 + 15 into 6 by 5 so 6 into 3 18
+ 4 into 16 64 so answer is 82 so we
have attained the optimality okay this
is the way to prepare LPP model using
Bigham method hope you all understood
this concept thank you
Related Exams
About this Video
|
4.92/5 Rating |
|
Nov 20, 2024 Last updated |
Video Description: Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics for B Com 2024 is part of Business Mathematics and Statistics preparation.
The notes and questions for Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics have been prepared according to the B Com exam syllabus.
Information about Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics covers all important topics for B Com 2024 Exam.
Find important definitions, questions, notes, meanings, examples, exercises and tests below for Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics.
Introduction of Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics in English is available as part of our Business Mathematics and Statistics for B Com & Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics in
Hindi for Business Mathematics and Statistics course. Download more important topics related with notes, lectures and mock test series for
B Com Exam by signing up for free.
Description
Video Lecture & Questions for Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com - B Com full syllabus preparation | Free video for B Com exam to prepare for Business Mathematics and Statistics.
Information about Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics
Here you can find the meaning of Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics defined & explained in the simplest way possible.
Besides explaining types of Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics theory, EduRev gives you an ample number of questions to practice Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics tests,
examples and also practice B Com tests.
|
115 videos|142 docs
|
|
Explore Courses for B Com exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
ppt
,Important questions
,Exam
,Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com
,Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com
,study material
,practice quizzes
,shortcuts and tricks
,past year papers
,Objective type Questions
,Viva Questions
,Extra Questions
,Previous Year Questions with Solutions
,Linear Programming Problems (LPP) via Simplex Method
,Linear Programming Problems (LPP) via Simplex Method
,mock tests for examination
,video lectures
,Summary
,Linear Programming Problems (LPP) via Simplex Method
,Free
,MCQs
,Sample Paper
,Business Mathematics and Statistics Video Lecture | Business Mathematics and Statistics - B Com
,Semester Notes
;
Study Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics on the App
Students of B Com can study Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Linear Programming Problems (LPP) via Simplex Method, Business Mathematics and Statistics is prepared as per the latest B Com syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup