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in this class we will study the algebraic
method to solve linear programming problems
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we will begin by considering a maximization
problem subject to less than or equal to constraints
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we will look at the same example that we used
to illustrate the graphical method so we go
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back to the same maximization problem which
maximizes ten x one plus nine x two subject
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to three x one plus three x two less than
or equal to twenty one and four x one plus
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three x two less than or equal to twenty four
x one x two greater than or equal to zero
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now in order to solve this algebraically let
us first look only at the constraints for
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for a moment let us set aside the objective
function and for a moment let us set aside
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the non negativity so we are looking at three
x one plus three x two less than or equal
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to twenty one and four x one plus three x
two less than or equal to twenty four so we
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have two inequalities so at the moment let's
again assume that we know to solve equations
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while we don't know at the moment to solve
inequalities so under the assumption that
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we know to solve equations let us see how
we can apply that idea to solving the two
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given inequalities so in order to solve them
as equations the first thing is to write them
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as equations from inequalities
so we have three x one plus three x two less
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than or equal to twenty one and x one x two
greater than or equal to zero so if we have
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to satisfy x one x two greater than or equal
to zero and three x one plus three x two less
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than or equal to twenty one it is now possible
to introduce another variable called x three
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and write three x one plus three x two less
than or equal to twenty one as three x one
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plus three x two plus x three equal to twenty
one which is what we have shown here we have
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written three x one plus three x two plus
x three equal to twenty one now since x one
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and x two have to be they have to be greater
than or equal to zero and they have to satisfy
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three x one plus three x two less than or
equal to twenty one therefore three x one
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plus three x two can either be equal to twenty
one in which case x three will be equal to
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zero or if three x one plus three x two is
less than twenty one then x three will take
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a positive value therefore x three will now
be defined as greater than or equal to zero
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so x three is greater than or equal to zero
so three x one plus three x two less than
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or equal to twenty one is now rewritten as
three x one plus three x two plus x three
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equal to twenty one its written as three x
one plus three x two plus x three equal to
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twenty one and x three greater than or equal
to zero
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in a similar manner four x one plus three
x two less than or equal to twenty four is
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written as four x one plus three x two so
it's written as four x one plus three x two
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plus x four equal to twenty four and x four
will be greater than or equal to zero now
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x four is greater than or equal to zero because
if four x one plus three x two is equal to
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twenty four then x four will become zero if
four x one plus three x two is less than twenty
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four then x four will take a positive value
therefore x four is greater than or equal
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to zero
so the given problem maximize ten x one plus
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nine x two subject to three x one plus three
x two less than equal to twenty one four x
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one plus three x two less than equal to twenty
four x one x two greater than or equal to
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zero is now written as maximize ten x one
plus nine x two plus zero x three plus zero
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x four subject to three x one plus three x
two plus x three equal to twenty one four
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x one plus three x two plus x four equal to
twenty four x one x two x three x four greater
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than or equal to zero now these variables
x three and x four that are written here
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these variables x three and x four that are
written here are called slack variables now
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these slack variables are used to convert
an inequality into an equation so this inequality
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is converted to an equation by the addition
of slack variables
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now we when we formulated the problem we said
the revenue associated with x one is ten the
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revenue associated with x two is nine so the
revenue or profit or money that is associated
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with x three and x four are zero the money
associated with x three and x four are zero
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and it is assumed that they do not contribute
to the objective function now we have converted
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a problem with inequalities into a problem
with equations and we now try to solve this
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problem now we started by saying that since
we know to solve equations we will now look
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at these two equations and try to solve them
so we have two equations but now with the
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addition of the two slack variables we now
have four variables x one x two x three and
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x four when we learnt methods to solve linear
equations we learnt methods where the number
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of variables that we solve is equal to the
number of equations but now in our example
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we have four variables and we have two equations
now when we have two equations we can solve
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only for two variables now we have to find
out a way by which we solve for two variables
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at a time when we have four variables and
two equations so let us now see how we solve
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this
so the problem that we have now it or we are
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now going to solve is given on the right hand
side the problem that we are going to solve
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is given here maximize ten x one plus nine
x two plus zero x three plus zero x four subject
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to subject to the first equation here second
equation here and son on so there are four
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variables and two equations and since we have
two equations we can solve only for two variables
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at a time so we choose two variables that
we are going to solve which means the other
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two variables that we are not going to solve
we have to give them some value for example
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if we assume that we are going to solve for
x one and x two we need to give some value
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for x three and x four if we give a value
of one for x three and x four then the equations
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will become three x one plus three x two equal
to twenty and four x one plus three x two
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equal to twenty three
if we give zero value for x three and x four
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it'll become three x one plus three x two
equal to twenty one and four x one plus three
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x two equal to twenty four so we have to give
some value for the two variables that we are
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not going to solve which are called fixed
variables we have to give them some value
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now we can give any value to x three and x
four except that we want x three and x four
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to be greater than or equal to zero therefore
fixing x three and x four or fixing any two
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variables we can do it in infinite possible
ways the easiest thing to do and perhaps the
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first thing we can try and do is to fix these
variables at zero so what we are going to
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do is at any point we are going to solve for
two variables by fixing two other variables
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to zero those variables that we are going
to solve are called basic variables and those
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variables that we are going to fix to zero
we are not going to solve for these they are
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called non basic variables
now we have a total of four variables in this
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formulation that we have on the right hand
corner so we have four variables in this formulation
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that we have there are four variables at any
point i am going to solve only for two variables
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and i am going to fix some other two variables
to zero so if i am going to solve for two
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variables out of four the two variables that
i am going to solve can be chosen in four
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c two ways or they can be chosen in six ways
so they can be chosen in four c two equal
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to six ways i can choose the variables
now these six ways by which i can choose them
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are written here from one to six so one pair
could be i am solving for x three and x four
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which means i am fixing x one x two to zero
expanded through zero i can solve for x one
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x three by fixing x two and x four to zero
i can solve for x one and x four by fixing
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x two x three to zero i can solve for x two
and x three by fixing x one x four to zero
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i can solve for x two and x four by fixing
x one x three to zero and i can solve for
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x one x two by fixing x three x four to zero
so i can solve six problems i can choose two
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variables out of four in six ways four c two
ways that have been shorn in this corner and
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i can solve the problem using these six pairs
variables in each of these six problems there
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are two basic variables the variables that
we are going to solve and we have the remaining
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two variables as non basic variables and we
solve we we fix the non basic vary variables
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to zero so those variables that are fixed
to zero are the non basic variables and those
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variables that we are going to solve are the
basic variables
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now let us look at each one of these six solutions
and see what happens when we actually solve
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them the first out of the six solutions we
are going to solve for x three and x four
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and we are going to fix x one and x two at
zero
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so when we solve for x three and x four we
are fixing x one and x two to zero so i am
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just striking x one x two off now we can easily
see that the solution is x three equal to
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twenty one x four equal to twenty four because
x one x two are fixed at zero they are not
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going to contribute so we will have x three
equal to twenty one x four equal to twenty
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four that is shown here as the first solution
now we move to the second solution coming
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back to the first one we said we fixed x one
x two to zero therefore we got the solution
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x three equal to twenty one x four equal to
twenty four which is written here x three
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equal to twenty one x four equal to twenty
four we now evaluate the objective function
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value for this because this solution is basic
feasible so let me explain what a basic feasible
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solution is now this solution is basic because
we have fixed x one x two two out of the four
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variables to zero and more importantly we
are solving for two variables x three and
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x four so we are solving for x three and x
four those variables are called basic variables
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non basic variables are fixed at zero we get
a solution x three equal to twenty one x four
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equal to twenty four x one equal to zero x
two equal to zero now this solution x one
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equal to zero x two equal to zero x three
equal to twenty one x four equal to twenty
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four satisfies the non negativity restriction
of x one x two x three x four greater than
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or equal to zero and also satisfies these
two equations which are now the constraints
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because this solution satisfies all the constraint
it becomes feasible because it satisfies all
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the constraints it becomes feasible and since
it is a basic solution it becomes basic feasible
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so again a solution is basic when we solve
for as many variables as the number of equations
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there are two equations two constraints we
solve for two variables the basic variables
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here are x three and x four the non basic
variables are set to zero x one and x two
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are set to zero so when we solve we get a
solution x three equal to twenty one x four
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equal to twenty four now this is a basic solution
and since this basic solution is also a feasible
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it satisfies this constraint this and the
non negativity restriction it becomes a basic
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feasible solution once we have a basic feasible
solution we find the value of the objective
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function the objective function is zero because
both x one and x two are zero x three and
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x four do not contribute to the objective
function so the value is ten into zero plus
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nine into zero which is zero
we now move to the second of the six solutions
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we now move to the second of the six solutions
now in this second solution
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we are going to solve for x one and x three
by keeping x two and x four to zero since
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i am keeping x two and x four to zero i am
writing these off x two and x four are not
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here so directly i get four x one equal to
twenty four from the second equation that
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gives me x one equal to six and once i substitute
x one equal to six three x one is eighteen
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therefore x three is equal to three so this
basic solution where we are solving for two
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variables x one and x three fixing the remaining
variables to zero gives us x one equal to
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six and x three equal to three this is also
basic feasible because x two equal to zero
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x four equal to zero x one equal to six x
three equal to three satisfies the non negativity
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restriction and therefore this is basic feasible
now we evaluate the objective function value
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so x one equal to six x two equal to zero
would give us ten x one plus nine x two equal
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to ten into six plus nine into zero which
is sixty so we get the second solution which
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is here
we now look at the third solution
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in the third solution we are going to solve
for x one and x four so when we solve for
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x one and x four we have x two and x three
equal to zero so x two and x three are now
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kept equal to zero so from the first equation
we will get three x one equal to twenty one
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we get x one equal to seven so when we substitute
x one equal to seven in the second equation
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we get twenty eight plus x four equal to twenty
four x four is minus four and therefore this
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solution is infeasible and does not satisfy
the non negativity restriction so we write
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it as infeasible and we do not evaluate the
objective function
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we now go to the fourth one where we are going
to solve for x two and x three we are going
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to solve for x two and x three so we will
eliminate x one and x four here so we will
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eliminate x one and x four here so this would
give us three x two equal to twenty four and
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then we would get x two equal to eight which
is shown here now when we substitute x two
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equal to eight in the first equation we get
twenty four plus x three equal to twenty one
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we get x three equal to minus three and this
solution is infeasible because it violates
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the non negativity restriction
we now move on to the fifth solution where
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we solve for x two and x four where we solve
for x two and x four and therefore we will
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eliminate x one and x three we will eliminate
x one and x three from the first equation
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we get three x two equal to twenty one so
x two is equal to seven now we substitute
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in the second equation seven into three twenty
one plus three equal to twenty four x four
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equal to three so this is basic feasible because
it satisfies the non negativity restriction
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the value of the objective function is nine
into seven which is sixty three
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the six and the last one is we solve for x
one and x two here by eliminating x three
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and x four so we solve for x one and x two
by eliminating x three and x four and when
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we solve for three x one plus three x two
equal to twenty one four x one plus three
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x two equal to twenty four we get a solution
x one equal to three x two equal to four which
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is also a basic feasible so we evaluate the
objective function to get sixty six which
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is shown here and sixty six is the best solution
that we have out of the basic feasible solutions
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and hence is the optimum solution to this
problem
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in the next class we will revisit this algebraic
method and we will try to get more insights
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into the algebraic method