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in the second module we move on to study linear
programming solutions in the earlier module
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we looked at linear programming formulations
now we will study solution to linear programming
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problem specifically the graphical and algebraic
methods of solving linear programming problems
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we will first look at the graphical method
to solve a given linear programming problem
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and the problem that we will consider is a
maximization problem maximize ten x one plus
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nine 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
there where are two variables x one and x
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two and since there are two variables it is
possible to represent these two variables
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in a graph so we first try to represent the
constraints and the objective function using
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a graph so we first start drawing the graph
by defining the origin and by drawing the
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x and y axis x usually the x and y axis divides
the graph into four quadrants but when we
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solve linear programming problems we look
at only one out of the four quadrants because
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x one and x two have to be greater than or
equal to zero then x one and x two are greater
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than or equal to zero we look at the first
quadrant and we do not look at the other three
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quadrants which are here where atleast one
variable takes negative values so in all linear
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programming problems when we use the graphical
method we will be restricting ourselves only
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to the first quadrant where the two variables
are greater than or equal to zero
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now we try and draw the x axis will now represent
the first variable which is x one and the
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y axis will represent the second variable
which is x two we now try to draw the two
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constrains in to the graph so we take the
first constraint which is three x one plus
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three x two less than or equal to twenty one
which has an inequality so what we do is we
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draw first the corresponding equation and
then we find out that area in the graph which
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maps to the inequality so the first constraint
will be three x one plus three x two equal
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to twenty one and that is drawn this way so
the constraint or the line three x one plus
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three x two equal to twenty one is drawn here
we require two points to draw this line and
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these two points are seven comma zero and
zero comma seven and with these two points
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we can draw the line three x one plus three
x two equal to twenty one or x one plus x
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two equal to seven
no this line divides the graph into two regions
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one region which is below this line which
i am showing the other region is above this
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line which is this space now one of these
two is less than or equal to and the other
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will become greater than or equal to now to
find out which area is less than or equal
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to we take a convenient point which is the
origin and then we substitute into this equation
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and we realize that the left hand side is
zero and the right hand side is twenty one
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left hand side is less than the right hand
side therefore the origin is in the region
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which is less than or equal to twenty one
and therefore this triangle which is made
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out of these three points is the region corresponding
to three x one plus three x two less than
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or equal to twenty one
having drawn the first constraint we now draw
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the second constraint which is four x one
plus three x two less than or equal to twenty
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four and this is how the line will look like
first we will draw four x one plus three x
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two equal to twenty four and to draw that
we need to mark two points six comma zero
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and zero comma eight which satisfy four x
one plus three x two equal to twenty four
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so if the line four x one plus three x two
equal to twenty four is drawn once again this
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line will divide the region into two portions
one is this portion the other is the portion
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on the other side one of which will correspond
to four x one plus three x two less than or
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equal to twenty four and the other will correspond
to the greater than or equal to as we did
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before we take a convenient point which is
the origin and evaluate it
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so when we evaluate at zero comma zero which
is the origin four x one plus three x two
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becomes zero and is less than the right inside
value of twenty four therefore the region
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formed by these three points the triangle
formed by these three points zero eight six
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zero and zero zero forms the region four x
one plus three x two less than or equal to
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twenty four so now what we do is we now look
at this region which is common to both the
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regions we first saw that for the first constraint
this triangle was there it was the region
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between zero seven seven zero and zero zero
and for the second constraint it was the region
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between zero eight six zero and zero zero
and the common region which is now shaded
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and shown is the region where both these constraints
are satisfied since we want to satisfy both
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the constraints we are now looking at that
region where both the constraints are satisfied
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and the shaded region is where both the contains
are satisfied and we also have this point
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of intersection which is three comma four
and the shaded region is called feasible region
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we use the word feasible to represent that
the constraints are satisfied for example
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let us look at other points
if we look at the points seven comma zero
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the point seven comma zero satisfies the first
constraint but the point seven comma zero
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does not satisfy the second constraint and
therefore it is infeasible so any point which
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violates even one constraint and does not
satisfy even one constraint is called infeasible
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so seven comma zero is infeasible now zero
comma eight satisfies the second constraint
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but violates the first constraint therefore
zero comma eight is infeasible if we take
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a point which is somewhere here which could
be seven comma five or something like that
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which is somewhere here then it violates both
the constraints so any point other than that
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the point which is in that feasible region
violates atleast one constraint so all the
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points in this space are not feasible or infeasible
because they have negative values for x one
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so it's important to identify the feasible
region and we have identified the feasible
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region for this linear programming problem
now this feasible region is the shaded region
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that we have shown and any point inside the
feasible region including these corner points
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and the boundary points for example a point
here like three comma zero is a boundary point
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for example a point here which is zero comma
three is also a boundary point so all the
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points which are the in the corners in the
boundaries and inside these region are all
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feasible points they belong to the feasible
region and they are called feasible solutions
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when we use the word solution we use the word
solution to indicate values that the variables
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take for example zero comma zero is a solution
for example one comma one which would be somewhere
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here is a solution now zero comma zero and
one comma one are feasible solutions because
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they lie in the feasible region seven comma
zero is also a solution it represents x one
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equal to seven x two equal to zero but it
is infeasible it is an infeasible solution
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since we are interested in solving the linear
programming problem using the graph we now
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want to have x one and x two which satisfy
the two given constraints and x one and x
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two to satisfy the non negativity now the
two axis basically represent the region where
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the non negativity is satisfied the first
quadrant and then we have added the two constraints
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into that first quadrant to get a feasible
region where are all the points in the feasible
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region satisfy the two given constraints as
well as satisfy the non negative restriction
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so anyone or everyone of the points in this
feasible region is a solution that satisfies
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all the constraints and we want to find out
that solution or that value of x one and x
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two which lies inside the feasible region
which maximizes ten x one plus nine x two
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when we take a close look at the feasible
region we quickly realize that there are infinite
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points in the feasible region and we should
now have to find the way by which we are able
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to get the best point out of infinite possible
solutions that are feasible now to get the
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best solution the value of x one and x two
that has the highest value of ten x one plus
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nine x two we now do a few more things
we first start by saying this every point
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inside the feasible region is dominated by
a boundary point so let me explain this now
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let us take a point which is inside the feasible
region and let this point be one comma one
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now the value of the objective function is
ten plus nine nineteen when x one is equal
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to one and x two is equal to one so if we
assume that we are going to make one unit
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of x one and one unit of x two the revenue
that we will get is nineteen now what i do
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is i can simply extend this point increase
the x one value upto it reaches a boundary
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point so here what will happen is i will be
getting something like five point five comma
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one or something like that this is the boundary
point which will be here so this boundary
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point when i evaluate ten x one plus nine
x two for this boundary point the value of
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ten x one plus nine x two for this boundary
point will be higher than the value of ten
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x one plus nine x two for one comma one so
if i move this point to to the right which
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means i keep increasing x one by keeping x
two the same and i keep moving this point
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to the right till it touches a boundary point
i realize that the value of the objective
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function or the revenue is increasing
in a similar manner if i keep this one comma
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one and i keep increasing the x two value
and push it upwards it will reach one comma
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six and at that point it would still be feasible
now when i take one comma six the point that
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is here which is a boundary point the value
of the objective function will be more than
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nineteen it would be sixty four in fact therefore
for every point inside this region there will
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be a boundary point that will give a better
value of the objective function i use the
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term better because if the objective function
was were to be minimize then when i move it
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leftwards i will get a better value if it
is to minimize i will move it down to get
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the better value but depending on whether
it is maximize or minimize depending on whether
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sum of these coefficients are positive or
negative it is always possible to move this
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point towards the boundary and we could do
that in many ways but clearly four ways by
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shifting it to the right by moving it to the
left by moving it to the top and moving it
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to the bottom and we will realize that out
of these four one or more will give better
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values of the objective function
therefore we summarize that i don't have to
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evaluate points strictly inside the feasible
region it is enough to evaluate points which
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are on the boundaries so it is enough to evaluate
points which are on the boundaries but then
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we realize that this is the boundary of this
region starting from zero zero to six zero
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to three four and zero seven but then we also
realize that there are infinite boundary points
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now we make another observation that every
boundary point is dominated by a corner point
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for example if i take the point one comma
six the point one comma six which is a boundary
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point has a value ten plus fifty four equal
to sixty four now i can try and move this
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boundary point along the boundary region and
try to come to either zero comma seven or
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three comma four if i do that i am still feasible
i am satisfying all the constraints so if
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i do that zero comma seven will give me sixty
three and three comma four will give me sixty
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six so by moving downwards towards three comma
four i will be able to get higher value of
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the objective function
so every boundary point is dominated by a
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corner point because when we have two variables
and two lines we can always move the boundary
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point in in along the boundary till it touches
a corner point and one of the corner points
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will give a better value if it is maximization
a higher value if it is minimization a lower
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value it will give a better value of the objective
function so every boundary point is now dominated
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by a corner point and therefore it is enough
to evaluate only the corner points now we
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realize that there are four corner points
zero comma zero six comma zero three comma
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four and zero comma seven so there are four
corner points in this and now we evaluate
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the objective function for the four corner
points
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the corner point zero comma zero z is the
value of the objective function so at x one
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equal to zero x two equal to zero ten x one
plus nine x two will be zero and therefore
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z is equal to zero for the corner point six
comma zero ten x one plus nine x two will
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be ten into six plus nine into zero which
is sixty for the corner point zero comma seven
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ten into zero plus nine into seven will gives
us sixty three and for three comma four we
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will get ten into three thirty plus nine into
four thirty six which is sixty six which is
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the best solution which is the one that gives
us the highest value of the objective function
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or the revenue now the best solution is called
the optimum solution
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so in summary for the graphical method we
first plot the constraints find the feasible
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region which is the common region find the
feasible region and evaluate find the corner
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points enough to evaluate the corner points
and that corner point which has the best value
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of the objective function is the best solution
or the optimum solution
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we will see more aspects of the graphical
method and some more aspect in the subsequent
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classes