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in this class we will look at the algebraic
method to solve minimization problems we consider
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the same minimization problem that we used
the same problem that we solved using the
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graphical method so the problem minimizes
seven x one plus five x two subject to x one
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plus x two greater than or equal to four five
x one plus two x two greater than or equal
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to ten x one x two greater than or equal to
zero now once again we will follow the same
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procedure where we convert the inequalities
two equations and we try to solve the equations
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so as we did before we will consider the two
constraints we will at present leave out the
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objective function and the non negativity
so the two constraints are x one plus x two
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greater than or equal to four and five x one
plus two x two greater than or equal to ten
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now let's look at x one plus x two greater
than or equal to four and convert it to and
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equation and we do it as follows now this
constraint x one plus x two greater than or
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equal to four is now written as x one plus
x two minus x three equal to four now x one
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and x two are greater than or equal to zero
and we want x one plus x two to be either
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equal to four or more than four so we introduce
the variable x three such that if x one plus
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x two is more than four then x three would
take a positive value and then make it equal
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four
so that x one plus x two minus x three is
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equal to four if x one plus x two is exactly
equal to four than x three will take value
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zero now x one and x two have to be such that
they ordinarily should not be less than four
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therefore x three does not take a negative
value so x one plus x two greater than or
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equal to four is now rewritten as x one plus
x two minus x three equals to four so x one
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plus x two greater than or equal to four is
now written as x one plus x two minus x three
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equals to four now this x three is called
a surplus variable it is also a slack variable
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but because x three has a negative sign it
is also a called a negative slack variable
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or it's called a surplus variable so when
we have a greater than or equal to the constraint
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like this
now that would result in a negative slack
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or a surplus variable in a similar manner
five x one plus two x two greater than or
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equal to ten is written as five x one plus
two x two minus x four equal to ten now the
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minus x four comes because when five x one
plus two x two exceeds ten then x four will
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take a positive value to make it equal to
ten if five x one plus two x four is exactly
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equal to ten then x four will take zero we
also should have x one x two such that five
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x one plus two x two is greater than or equal
to ten five x one plus two x two is not less
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than ten
and therefore x four does not take a negative
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value so when we have a greater than or equal
to constraint we have a negative slack variable
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or surplus variable introduced x three and
x four have been introduced and importantly
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x three x four are greater than or equal to
zero and they will have a negative sign if
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we have a greater than or equal to constraint
in the earlier class when we had a less than
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or equal to constraint we introduced x three
and x four as positive slack variables again
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x three and x four where greater than or equal
to zero but x three and x four had a plus
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one sign appearing so when we convert an inequality
to an equation by adding a slack variable
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we could add a positives slack variable if
the constraint is less than or equal to type
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and a negative slack if it is greater than
or equal to type but what is important is
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that the slack variables are defined as greater
than or equal to zero in both the cases just
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as the decision variables are defined as greater
than or equal to zero so once again we observe
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like in the previous example that with the
addition of the two slack variables we now
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have two equations but we have four variables
we also know that if we have two equations
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we can only solve for two variables therefore
we follow a similar manner like what we did
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in the previous class to try and solve this
system that now has four variables and two
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equations just as in the previous example
the negative slack or surplus variables do
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not contribute to the objective function
and therefore they contribute a zero to both
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x three and x four we now go on to solve this
system that minimizes seven x one plus five
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x two plus zero x three plus zero x four subject
to x one plus x two minus x three equal to
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four five x one plus two x two minus x four
equal to ten x one x two x three and x four
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greater than or equal to zero now the linear
programming problem that we have to solve
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is to minimize seven x one plus five x two
plus zero x three plus zero x four subject
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to the two constraints and non negativity
since we have four variables and two equations
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we can only solve for two variables at a time
and these two variables from four given variables
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can be chosen in four c two or it can be chosen
in six ways so it can be chosen in so it be
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chosen in four c two ways or it can be chosen
in six ways
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now these six ways of choosing the two variables
are shown here for example one of them would
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be to solve for x three and x four another
one could be to solve for x one and x and
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three and so no now when we choose the two
variables in anyone out of the six ways we
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have the other two variables which we are
not solving for and these variables have to
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be fixed to some arbitrary value now they
can be actually fixed to any arbitrary value
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which means they can take infinite values
but as we did before in the previous example
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we fix those variables to zero
so these variables which we are fixing at
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zero or these variables which we are fixing
variables fix to zero are called non basic
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variables and those variables that we are
solving are called basic variables so at a
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time we take two variables out of the six
and solve for them by fixing the remaining
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two variables to zero so now let us go back
and look at all the six possible solutions
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so first let us look at the so the case where
we are going to solve for x three and x four
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by fixing x one and x two to zero now when
we solve when we fix x one and x two to zero
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i am kind of deleting x one and x two so we
are going to solve for x three and x four
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so we solve for minus x three equals four
minus x four equals ten which gives us a solution
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x three equal to minus four and x four equal
to minus ten it gives us this solution now
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when we consider this solution where x one
equal to zero x two equal to zero x three
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equal to minus four and x four equal to minus
ten we have this solution where x three equal
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to minus four and x four equal to minus ten
violates the non negativity restriction and
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therefore this solution is infeasible so this
solution becomes infeasible because it violates
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the non negativity restriction
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now we look at the second instance where x
one and x three we are going to solve and
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we are going to fix x two and x four to zero
so we are going to if we take the second instance
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we are fixing x two and x four to zero and
we are going to solve for x one and x three
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since we are fixing x two and x four to zero
i am leaving them out so i am going to solve
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x one equal to four five x one is equal to
ten so five x one equal to ten x one minus
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x three is equal to four so five x one equal
to ten gives me x one equal to two gives me
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x one equal to two and when i substitute x
one equal to two i will get x three equal
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to minus two so that x one minus x three is
equal to four
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so we have a solution x one equal to two x
two equal to zero
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so x two equal to zero x three is equal to
minus two and x four is equal to zero now
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because x three is equal to minus two it violates
the non negativity restriction and therefore
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this solution also becomes infeasible now
we look at the third out of this six problems
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where we are going to solve for x one and
x four and we are going to fix x two and x
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three to zero since i am fixing x two and
x three to zero i am removing x two and x
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three from this solution
so this gives me a system where x one is equal
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to four this gives me a system where x one
is equal to four and five x one minus x four
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is equal to ten so x one is equal to four
is shown here that is the solution and when
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i substitute five x one five x one becomes
twenty therefore x four becomes ten so that
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five x one minus x four is equal to ten so
i get a solution x one equal to four x four
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equal to ten so x one equal to four x two
equal to zero x three equal to zero x four
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equal to ten satisfies the non negativity
restriction and therefore it is feasible it
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is a basic solution because i am solving for
two variables i am solving for two variables
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because i have two equations so this solution
is both basic and feasible therefore it is
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basic feasible
now i compute the value of the objective function
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for x one equal to four x four equal to ten
so seven into four plus five into zero gives
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z equal to twenty eight the value of the objective
function is twenty eight now we move on to
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the fourth out of the six solutions now in
this solution
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in this solution i am going to solve for x
two and x three and i am going to fix x one
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and x four to zero so i am going to fix x
one and x four to zero so this goes this goes
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this also goes so the second equation becomes
two x two equal to ten from which i get x
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two equal to five from which x two is equal
to five and if i substitute x two five i get
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five minus x three is equal to four therefore
x three equal to one
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so i have a solution x one equal to zero x
two equal to five x three equal to one x four
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equal to zero and this solution is feasible
because it satisfies the non negativity restriction
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it also satisfies the two equations it is
also a basic because i am solving for two
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variables x two and x three and i have two
equation so i have a basic feasible solution
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therefore i try to find out the objective
function value so the objective function value
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would be seven into zero plus five into five
which gives me z equal to twenty five
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now i move on to the fifth of the six problems
where i am going to fix x one and x three
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to zero and i am going to solve for x two
and x four so i fix x one and x three to zero
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x so x one and x three to zero so the first
equation gives me x two equal to four first
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equation gives me x two equal to four which
is shown here and when i substitute two equal
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to four eight minus x four is equal to ten
which gives me x four is equal to minus two
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now when x four is equal to minus two it violates
the non negativity restriction and therefore
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this solution is not feasible so it become
infeasible and this is written as in feasible
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now we go to the last of the six situations
now the last one i am going to solve for x
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one and x two i am going to fix x three and
x four to zero so i am going to fix x three
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and x four to zero so this goes this also
goes i have to solve for x one plus x two
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equal to four five x one plus x two x two
equal to ten i multiply the first equation
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by two to get two x one plus two x two is
equal to eight subtracting i will get three
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x one is equal to two x one is equal to two
by three i will get x one is equal to two
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by three which is shown here
now when i substitute x one is equal to two
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by three in x one plus x two equal to four
i will get x two is equal to ten by three
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so that x one plus x two is twelve by three
which is four now this satisfies the non negativity
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restriction and therefore it is feasible it
is also a basic because i am solving for two
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variables since i have two equations so it
is a basic feasible solution now among the
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i calculate the value of the objective function
for this
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so seven x one plus five x two will become
seven into two by three fourteen by three
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plus five into ten by three fifty by three
which will give me sixty four by three which
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is equal to twenty one point three three now
out of the three solutions which are basic
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feasible the one that has the lowest value
minimizing the objective function the one
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that has the lowest value is optimum it is
the best solution so the solution two by three
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comma ten by three is optimum to the minimization
linear programming problem in the next class
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we will see some more aspects of the algebraic
method and we will try to relate the algebraic
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method to the graphical method