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in the last class we saw the graphical method
to solve a linear programming problem we considered
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the example that is shown which is a maximization
problem that maximizes ten x one plus nine
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x two subject to three x one plus three x
two less than or equal to twenty one four
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x one plus three x two less than or equal
to twenty four x one x two greater than or
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equal to zero we do the graph corresponding
to the constraints we identified the feasible
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region with is shown in a different color
and we also computed the corner points and
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then we observed that it is enough to evaluate
up the corner point and we also found out
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that this corner point the corner point three
comma four the optimal solution lies in this
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when we evaluated the objective function value
at the four corner points zero comma zero
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gave us z equal to zero six comma zero gave
us z equal to sixty zero comma seven gave
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us z equal to sixty three and three comma
gave us z equal to sixty six since we are
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maximizing this function we said this is the
corner point so this is the corner point three
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comma four that has the best value of the
objective function and hence is the optimal
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solution now let us look at the same thing
from a slightly different way
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now what we do now is we draw objective function
lines i have shown a few objective function
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lines so what we have essentially done is
the objective function is ten x one plus nine
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x two and we keep ten x one plus nine x two
equal to a constant for example we could keep
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ten x one plus nine x two equal to fifty so
this line could represent this line could
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represent
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this line could represent ten x one plus nine
x two equal to fifty and this line is drawn
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this straight line is drawn and then we objective
function lines which are parallel to this
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for example this could represent ten x one
plus nine x two equal to sixty and so on so
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as we increase ten x one plus nine x two to
a constant and keep increasing this constant
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the objective function lines become parallel
to each other and move in this direction you
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can see that the objective function is moving
in this direction and as it moves the objective
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function line finally it touches the feasible
region at the point three comma four the corner
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point where the objective function lines last
touch before leaving the feasible region is
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this point
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is this point three comma four which is the
optimum solution
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so there are two ways of of getting to the
optimum solution one of which is to evaluate
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the four corner points in this example there
are four corner points in other examples there
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could be a different number of corner points
evaluate the objective function at all the
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corner points and find out which corner point
has the best value the other is to draw a
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series of objective function lines where the
objective function is a constant and if it
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is maximization increase the constant the
lines move in the direction as shown the lines
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move in this direction as shown
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the lines move in this direction and because
we are maximizing the lines finally leave
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the feasible region at this corner point which
happens to be the optimum solution now we
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will understand how we solve a minimization
problem using the graphical method the method
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is similar however we would explain it through
an example
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now we look at this minimization problem which
minimizes seven x one plus five x two subject
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to two constraints x one plus x two greater
than or equal to four and five x one plus
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two x two greater than or equal to ten x one
x two greater than or equal to zero we draw
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the graph as we did for the maximization problem
and then as since x one and x two are greater
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than or equal to zero we will look at only
the first quadrant we will not look at the
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other quadrants because in these quadrants
which are here in these quadrants which are
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here or here or in this place you'll get x
one or x two or both will become negative
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and therefore we will be looking at only this
quadrant which is the first quadrant and we
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also have x one here as the x axis and we
have x two here as the y axis we now draw
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x one plus x two greater than or equal to
four to do that we first draw x one plus x
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two equal to four so x one plus x two equal
to four is this line which passes through
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four comma zero and zero comma four which
are shown here so this is the first line is
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x one plus x two equal to four which passes
through four comma zero and zero comma four
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which are here and this line is shown here
x one plus x two equal to four
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now this line divides the graph into two regions
one of which is less than four and the other
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is greater than four so to find out which
one it is we look at a convenient point which
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is the origin you look at the origins which
is here and then check whether zero comma
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zero satisfies x one plus x two greater than
or equal to four or not now when we substitute
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zero comma zero into the left hand side of
this inequality to the left hand side of this
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inequality
left hand side of this inequality we realize
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the left hand side value is zero and therefore
it does not satisfy the inequality therefore
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the region that is x one plus x two greater
than four is the region on this side
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so all this region is the region x one plus
x two greater than or equal to four
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we now draw the next constraint which is five
x one plus two x two greater than or equal
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to ten so we first draw the line five x one
plus two x two equal to ten and that line
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is shown here so that line is shown here
so this is the line five x one plus two two
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x two equal to ten and to draw this line we
have two points which is two comma zero and
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zero comma five now once again this line divides
a graph into two regions one of which is less
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than ten and the other is greater than ten
so once again we look at the origin which
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is here and check whether the origin satisfies
the condition or violates the condition
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so when we substitute five x one plus two
x two to the left hand side of of this inequality
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we get zero on the left hand side which is
less than ten therefore the other one is the
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one that satisfies greater than or equal to
ten so now looking at this you will realize
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that the the region goes like this so all
of these is the region that we are talking
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about so the entire thing is the region that
we are talking about and that is shown the
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line is five x one plus two x two equal to
ten and this is the region which is now shown
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in a different color which
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now this is the region which is shown in a
in a light green color which is the feasible
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region for this particular example
we also observed that points in this space
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satisfy one of the constraints they satisfy
the first constraint but they violate the
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second constraint and therefore they do not
lie in the feasible region
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now our feasible region has three corner points
the first corner point is given here which
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is four comma zero the other corner point
is here and the third corner point is zero
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comma five so at four comma zero we evaluate
the objective function and the objective function
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value is given as z is equal to twenty eight
the next corner point is zero comma five which
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is here and when we evaluate the objective
function seven x one plus five x two we get
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z is equal to twenty five seven into zero
plus five into five gives us twenty five
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we now look at the third corner point which
is two by three comma ten by three now this
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two by three comma ten by three can be found
in the graph if we draw this graph to scale
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and we draw the graph correctly as we have
attempted to do now you realize that this
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point has an x coordinate of two by three
and has a y coordinate of ten by three so
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this point is two by three comma ten by three
so this is two by three comma ten by three
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that i have shown here and the objective function
value when we substitute we get seven into
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two by three which is fourteen by three plus
ten into five by three which is fifty by three
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so fourteen by three plus fifty by three is
sixty four by three which is what we have
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here
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now we have twenty eight for this corner point
we have twenty five for this corner point
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and we have sixty four by three which is twenty
one point three three for the other corner
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point which is two by three comma ten by three
therefore the optimum solution to this minimization
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problem is that corner point which has the
smallest value of objective function and out
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of these three we realize that the point two
by three comma ten by three with sixty four
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by three has the smallest value and therefore
is the optimum solution so this point is the
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optimum solution with value sixty four by
three
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like we did for the maximization problem we
can also draw what are called iso objective
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function lines where we fix seven x one plus
five x two to to a constant so we fix seven
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x one
so these are
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seven x one
seven x one plus five x two equal to a constant
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for example this would mean seven x one plus
five x two would be equal to about thirty
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and since it's a minimization problem we draw
the objective function lines in the other
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direction where seven x one plus five x two
gets progressively reduced
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so the direction of movement of the objective
function is this way for a minimization problem
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unlike in a maximization problem so it moves
in this direction and as it moves downwards
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or in the direction in this direction it finally
touches this corner point finally touches
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this corner point
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it finally touches this corner point before
it leaves the feasible region so the last
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feasible point is given by two by three comma
ten by three is given by two by three comma
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ten by three which is the optimum solution
to the minimization problem so this way we
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solve minimization problems we solve minimization
problems using the graphical method
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now the only difference between the maximization
and the minimization is many times it will
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happen that in a minimization problem the
feasible region will not be bounded as in
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this case even though i have shown a certain
portion of the feasible region one would realize
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that a point here a point here is also feasible
and a feasible region is not bounded or it's
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not a closed region whereas in a maximization
problem many times we would get a feasible
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region that is a closed region so that is
one of the things that we need to understand
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the other most important thing where the reason
for which we evaluate only the corner points
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is that we have shown that every point inside
the region there will be a corner point which
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will have a better value of the objective
function now that happened due to what is
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called the convexity property of the feasible
region if we take two points in the feasible
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region and join them by a line we will realize
that all the points in the line will lie in
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the feasible region and because this feasible
region is convex we we can easily show that
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for every point inside the feasible region
there is always a point on the boundary there
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is always a point on the boundary and then
there is always a corner point which has a
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better value of the objective function so
these are the basic ideas and principles that
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we use to solve the linear programming problem
using the graphical method while the graphical
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method is simple and easily understandable
we are limited by the fact that we can only
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represent two variables in the graphical method
the x one and x two we have already seen formulations
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that involve more than two variables and several
linear programming problems have a large number
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of variables so we need other methods other
than the graphical to solve linear programming
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problems when we have more than two variables
we will see the algebraic method in the next
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class