Hello, welcome to the course on Mechanics
of Machining. Today, it is the 14th lecture,
and I will discuss about Optimization of Machining
Processes. Optimization of machining processes
is often called as economics of machining
in the books, although it is somewhat misnomer,
but they use word economics of machining,
optimization of machining is more appropriate
term.
Now, suppose you are doing some turning operation
like this, you have to decide that what parameter
I should take, what should be my feed, what
should be my cutting speed. Considering various
type of constraints, so that my cost of production
may be less, sometimes I may be worried that
my production time should be less, whatever
is the objective. So, this process is called
Optimization.
Now, there, in any optimization problem, first
thing is that you should know: what are your
objectives. So, certainly any optimization
problem will have at least one objective,
we can if we write in the form of mathematics,
we call it a objective function. So, we should
have one objective. We may have more than
one objective also, that is called multi objective
problem. Then we have to decide that how to
fulfill both the objectives, may be both objectives
may not be fulfilled to their fullest extent,
but you can decide what type of trade off
should I employ that means, at least this
objective should be satisfied 90 percent,
other objective should also be satisfied by
certain percentage like that.
So, here we have got various objectives, like
objective 1 can be, to produce a component
of required dimension and surface finish at
the minimum possible cost. So, my objective
is basically cost minimization, now this one
that is cost minimization is my objective.
So, here we say cost minimization, so that
can be your objective, cost minimization right.
And but, we have to also maintain proper dimensions
and surface finish, it should not be that
I minimize my cost, but then the surface finish
has become very poor. So, these are a type
of constraint type of thing.
Objective 2 can be to produce a component
of required dimension and surface finish at
the maximum possible production rate, that
means, my production rate should be very high,
so that means, maximizing production rate,
that can be objective function maximize, maximize
production rate or minimize production time.
If production time is less, then the production
rate will be high.
Objective 3 is that to produce a component
of required dimension and surface finish at
the maximum possible profit rate, that my
profit rate should be very good. Suppose,
why this is different from the other things,
profit is something different. Suppose, I
have produced a component, which is of very
low price, is it not in the market, it may
be available in 10 rupees, I am producing
in 5 rupees. So, in a way, I am getting 5
rupee as a profit, because I have reduced
my cost.
But, if production rate is very low, then
I am earning no doubt profit, but that is
in a very long time, that means, if I am producing
suppose only two components per day and I
am saving the cost 5 rupees in each component,
then total cost saving will become 10 rupees.
But, another person is actually, suppose he
is reducing the cost not by 5 rupee, but only
by 3 rupee, but he is making 10 components
in that day, so he has basically made a profit
of 30 rupees. So, per day profit has increased
in that case, so that is called the profit
rate in the second case is high. So, profit
rate for minimizing, maximizing the profit
rate, you have to worry about reducing the
cost as well as the production rate, so that
is why, profit rate is another objective.
And there can be several other objectives
also. Sometimes objective may be only to produce
a good surface finish, and other things can
be treated as a constraint, that can be also
done. But, mostly we talk about these three
objectives; one is minimization of the cost,
another objective is maximization of the production
rate, and the third objective is maximization
of the profit rate. Now, this factors that
can be varied during a machining operation,
they are called speed, cutting speed, you
have to decide what cutting speed I have to
choose. Then, it can be feed that means, in
turning operation how much millimeter the
tool will move in one revolution.
And then, it can be depth of cut. Then, it
can be tool and work material, means what
type of that. This is not any quantitative
thing, but it may be in the form of a decision
that means, I have to machine something, I
may have to take a decision, whether I should
use this tool for this work or I should use
another type of tool material. Then, tool
geometry, what should be my geometry, what
should be rake angle, what should be relief
angle or clearance angle. And cutting fluid
that means, what type of cutting fluid I should
be using or I should not use the cutting fluid
at all that type of decision also has to be
taken.
So, these things if we have to take decision
that means, these things are in our hand,
we can choose speed, we can use speed, we
can use depth of cut. And depending on what
decision I take, I will be getting a type
of may be a cost or I will get particular
production rate. So, objective is that I should
chose these parameters in such a way, I should
decide on these parameter in such a way, so
that I should get my desired objective.
So, therefore, these are called design variables
or decision variables. In management, mostly
we speak about these things as decision variables.
These are decision variables, because we are,
these are variables, speed is a variable,
but we are taking decision about that, sometimes
they are also called as design variable, means
I can design the process like that. So, decision
variables and design variables have to be
chosen depending on the objective that is
called the optimization.
Now, in this, total cost of producing a component
involves many factors. So, main factors are
cost of material. What is the cost of material?
Then cost of material handling, then cost
of material processing that means actual cost
spend in machining. Nature and effects of
cutting conditions are machining cost must
be analyzed. For example, when the cutting
speed is low, suppose I can keep the cutting
speed very very low, then I will get very
high tool life, my tool will not get worned
out, but the production rate will be very
low. On the other hand, suppose I increase
cutting speed, then production rate will be
very high, but then the tool life will be
very short, so that means, you have to do
some compromise, neither this is good nor
that is good, something in between is good,
and that you have to find out.
So, similarly higher values of speed and depth
of cut would decrease the machining time.
If I take high value of feed and high value
of depth of cut, more material will be removed,
but would adversely affect the tool cost,
same thing happens. Moreover, suppose I have
to get a required dimension, then I may be
constraint to take particular depth also,
it is not that you can take any depth, feed
you can take of course anything, but you cannot
take the depth. I have to reduce suppose 10
mm to 8 mm, naturally I have to maximum depth
of cut I can take 1 mm, I can decide whether
I will do in one pass by taking 1 mm depth
of cut, so that 2 mm diameter gets reduced
or I can do in two steps, that means two passes.
So, too high value may increase tool changing
time drastically, and it will again further
it will take more time. So, for simplicity,
first let us only consider the case of a single
pass straight turning operation and obtain
the value of optimum cutting speed. Assume
that, the depth of feed and depth of cut is
a constant, it is kept constant, only we are
changing the cutting speed. So, it is a single
variable optimization problem, single variable
optimization, that means, my variable is only
cutting speed, and I have to do what, I have
to minimize may be cutting, may be cost.
So, this is so, let us first concentrate on
the machining cost. Machining cost can be
divided into non-productive cost, cutting
cost, and tool cost. So, in this, what are
the non-productive cost? We although it is
non, misnomer, because it is not non-productive,
it is also essential, but we are using the
term non-productive that means, they may not
be doing actual machining that time. So, non-productive
cost can be evaluated by adding all non-productive
time and multiplying it by cost rate, that
means, what is cost rate means, what is the
cost per hour, that means, you add non-productive
times.
So, relevant non-productive times are initial
setup time. You, occurring once per batch,
you have a batch of hundred components, you
need some preparation something it has to
be brought from some place, it has to be put
properly, all that type of things then machine
has to be cleaned, and started things have
to be put at appropriate place. Means, whatever
you do for initial setting up. You have to
put the tool also properly in correct position,
this is done once per batch that means, this
one. After that, for each piece, you do not
do that means, you already have brought may
be carton, in which there are hundred pieces,
again and again you do not have to bring it,
so that is one initial setup time.
Then loading and unloading time, that means,
you put the job in the chuck, then it will
take some time, you will climb it in three
jaw chuck, and then after the piece has been
made, then you will unclamp that, that is
called unloading. So, you will spend some
time in loading and unloading also, so that
time also has to be considered. So, this is
non-productive cost.
Then tool advance and withdrawal time, it
is also occurs once per piece. Previous thing
also loading and unloading is for each piece,
and tool advance that means, my tool is advancing,
it is come my job is here, but tool is at
this location, from here to here, it is approaching.
So, some distance it traverse in the air,
that time also has to be taken into account,
it is not actually doing machining, but it
is approaching, and then it has to be withdrawn.
So, for withdrawal, I am just moving this
much distance to tool. I have machine, I have
finished my machining, then my tool is reached
up to this. Again, I am bringing my tool to
the original position, in that also some time
will be lost, that will be called it occurs
once per piece, and that is called withdrawal
time.
And tool removing and replacing time. This
does not occur after every piece, but it occurs
once per tool regrind that means, suppose
my tool has become damaged, I have to take
it from the machine, and then that particular
tool has to be ground, and again it has to
be put in the proper position, so that is
called tool removing and replacing time, that
also has to be taken, but it may will be done
once per tool regrinding. May be after making
the fifty pieces, tool has to be reground
at so, therefore after fifty piece, there
will be tool removing and replacing time,
per piece time can be obtained by dividing
it by 50. And then, there will be some idle
time, that also can be per piece that is idle
time that is T, T i we can write, not accounted.
So, let us see that we will explain some mathematics
here, but let us be clear about the notations.
N B, I am calling as number of components
in a batch, suppose a batch of hundred. N
S is the number of components produced between
two tool regrinds, that means, between two
tool regrind. Suppose, a tool can machine
ten components before failing, means before
it is a tool life got gets exhausted, and
there is a need to regrind it again, so then
ten will be the number of components produced
between two tool regrinds.
And T P is the machining time per piece, per
piece this much actual machining. And R P
is the production rate. P R is the profit
rate. And I P can be called income per piece,
income per piece, how income you are making
or this one. Now, in this case, let us say
you so, we have defined the notations machining
yeah, now here, yeah these notations, income
per piece is also known that means, how much
income you generate by selling one piece.
And then, let us discuss that: what is the
sum of the total non-productive time; let
us call it T N. So, T N is equal to T S plus
T l plus T a into N B. T S is the setting
time, see T S here is the total setting time.
And T l plus T a is what, T l is the loading
time, and T a is the, was let us see the previous
this one, T l t T l was loading and unloading
time, and Ta was approach time. So, loading
plus approach that is for each piece; so,
for whole batch, the time will be T l plus
T a into N B, because in one batch, there
are N B component. But then, there are N B
component, and before new grinding, N g tools
number of pieces can be produced. So, total
number of regrinds will be N B divided by
N g ok, and then multiplied by tool replacement
time, so that means, if we are considering,
suppose, suppose there is a batch size of
100 and let us say, so we say that N B is
equal to 100, and then N g is equal to supposing
it is 50 that means, tool can machine up to
this one 50.
So, what I am telling, that supposing I do
one grinding, suppose I have brand new tool
ok, so may be after, so I have produced 50
tools components, and then did one grinding;
and another 50 components I produced, I did
another grinding. Of course, that batch got
exhausted, but suppose you want to put the
tool in the means, Ok condition means same
condition, so that is why, you will do two
times grinding. Is it not, because you had
hundred pieces, so first these was so many
100 pieces, 50 pieces were made, then you
felt a need of regrinding, after again 50
pieces, again you felt the need of regrinding,
because other batch will come, for that batch
also you need the fresh tool.
So, basically we can say two times I have
done grinding. So, how it will come, 100 divided
by 50, that is why I am putting N B by N g.
So, you got two. And tool replace time is
T r, so that means, replace time includes
the time in grinding, so that means, 2 into
whatever time is going, that much you have
to multiply, and then you get this one, and
plus T I, T i is the idle time, I have combined
the idle time for whole batch. So, this way
I have got the total non-productive time.
Now, let us discuss about the cost thing.
C R is the total cost rate. And C L is the
labour cost rate, rate means, it can be in
per hour. And C O is the overhead cost rate,
and C D is the depreciation cost rate. Actually
overhead cost rate means, suppose the cost
of manager, salary of manager etc., that will
be on monthly basis, but you can decide that
in a month there are 25 days, then 25 days
multiply by 8, 8 hours, suppose are you say
that effectively say 6 hour, so that means,
it is about 25 into 6 means 150 hour, so that
means, whatever salary he is getting, that
you can divide by 150. So, this way, you have
to decide about that what will be per hour
overhead cost rate. And it is not only just
manager salary, other things also like rent
of the building, and all these things will
be included ok.
And C D is the depreciation cost rate, that
means, machine is depreciating each day. So,
suppose you have purchased machine of 5 lakh,
so if you consider very simple rate line depreciation
formula, so that machine which is purchased
in 5 lakh, thatís its cost has to be recovered
in 5 years, if suppose its life is 5 years,
so that means, per year 1 lakh depreciation
I have to pay account. So, in a year, how
many hours are there in which work take place,
so you divide by that, then you will say per
hour depreciation cost. Like that, you have
to find out per hour depreciation cost.
So, total cost rate will be actually combination
of this thing; labour cost rate that is means
hourly wage of the worker, and then overload,
overhead cost rate, and depreciation cost
rate will give you total cost rate. And then
the total non-productive cost per batch, then
will be what, C N will be C R into T N of
course. So, T Ns expression I have already
obtained here, just plug in the value of C
R from here, C R goes here, and T N goes here,
and then you get C N is equal to C L plus
C 0 plus C D, and that much expression ok.
I am not going to repeat, you can see on the
screen, that this is what you got the cost.
Then, we have to worry now about cutting cost,
that was non-productive cost. Actual cutting
cost C c means during the cutting operation
can be evaluated by multiplying tool cutting
time with cost rate, so that means, cutting
cost will be C L means time for loading unloading,
then C 0 plus C D, and this will be T C into
N B, because T C is the cutting time per component,
there are N B component in a batch. So, in
a batch, this much cost will be encountered.
So, C L is what, C O is what, overhead cost,
that means, overhead cost will be spend in
that period also in which this one. See overhead
cost is coming here also, in non-productive
time also, that means, when we are doing setting
up or when we are just understanding that
how a job will be done, that is also part
of the setting, so that overhead cost is here
also.
And but, but their loading cost is not there,
loading I am indicating by small, labour cost
is there ok. So, I have put that here labour
cost, and then overhead cost C L plus C O
and plus C D. C D is the depreciation cost
ok. So, here in the overhead cost, I have
put this one. So, here you have put C L plus
C O plus C D. And in this case, yes, just
see here that in this case, cost is C L plus
C O plus C D multiplied by T C into N B, where
T C is the cutting time. Now, loading and
unloading time, we have already incorporated.
Here, that in this case no, means it is T
C is the cutting time. So, in the cutting
time, we have to consider that means, we have
to consider the cost of actual cutting ok,
that means, I have to consider only the cutting
time that means, actual machining of that
thing.
So, tool cost will be what? This is the cutting
cost, and then we have to consider the third
thing tool cost. Tool cost is C T is the tool
cost including initial cost and regrinding
cost. Let us say C i is the initial cost of
the tool, and you C g is the cost of tool
regrinding. So, thus in this case, you have
to say C T is equal to C T is equal to C I
by R g that means, suppose I have done two
tool regrinds, so I am distributing that initial
cost in two parts. And plus C g, C g is the
cost of tool regrind that means, actually
this was the cost of regrind ok, and but there
were number of regrinds were R g. So, initial
cost I am distributing equally among all the
between all the regrind, so I divide it by
C i by R g plus C g into N B by N g that means,
N B by N g means, these many times I have
to do that.
So, N B by N g will be no doubt, it will be
R g, because N B by, N B by this one was there,
this was R g. So, if I do that like this,
then this will be R g. So, they basically
I am telling that you have to say C I, R g
R g gets canceled, and this will be C g into
it will be r g, so that means, suppose initial
cost was something, but I have ground it two
times, so the overall cost has become in that
particular batch. Suppose, in that batch,
you had to grind two times, then naturally
that it will be C i plus C g into R g, that
is the way you have to do that. If you are
throwing away, you are not grinding. So, there
is no cost involved in this one, then naturally
you do not have to consider this component,
but then you have to consider only the initial
cost of that tool, so this way.
So, machining cost per batch of components,
excluding the material cost will be this,
so if it C B is equal to this much. So, this
is the cost in the manufacturing. And no matter,
whatever process we employ, the material cost
will be same. So, therefore, right now I need
not consider the material cost in my optimization
problem, means I just have added non-productive
cost, cutting cost, and tool cost.
So, I get this type of expression, C B is
equal to this much, I just rearranged everything.
And machining cost per piece C P is C P divided
by is equal to C B divided by N B, I divide
it by N B. So, it is clear, that the machining
cost can be reduced by decreasing the non-productive
time. In this regard, the use of jigs and
fixture, improved tool holder design could
be very effective. And a variety of criteria
have been used for optimization.
These include, suppose we say minimum cost
per component, another can be maximum production
rate, and the third can be maximum profit
rate criteria. So, from the manufacturerís
point of view, the maximization of profit
rate without increasing the price of the commodity
may be of very great interest. Now, time T
B required for machining of a batch of components
is non-productive plus cutting time together
is given by this.
This we have already discussed in which everything
has in included, even the idle time has been
included here. And N g is equal to T divided
by, this one, means N g is equal to T B divided
by this one, I think this, I had written N
g means number of regrinds that means, number
of regrinds, so that is this one N B by N
g was equal to r, so that is this one. So,
number of components produced in each regrind;
so, it should be actually r g type of ok ok,
N g, suppose this is T, this is N g, so number
of components produced between each tool regrind
ok.
So, in this case that, if T is the tool life,
suppose T is the tool life and my, this is
the total time required for machining of a
batch ok. So, therefore number of components
required between this may be N g may be T
B divided by T. Suppose, my total time required
for machining of a batch of components is
actually suppose 100, and then here, this
tool life is may be 100, so therefore, number
of regrinds will be only 1 ok. So, number
of so this is what, that this one.
So, in this case, now suppose you have T B,
so T B is equal to T s, T l plus T a multiplied
by N B this is T r. And now, I am writing
here that T c, T c was what? T c is the cutting
T c is just understand this thing, forget
about this equation, just concentrate on equation
9.12 ok. So, in equation 9.12, see here it
is written that this one is T r, N B by N
g. N B is what number of components in a batch,
N g is that number of components between two
regrinds ok, so N B by N g ok correct. I am
putting N B by N B as it is T r that is T
r is the tool replacement time that how much
time will be needed in grinding and replacing.
But, here N B instead of N g, I am writing
N B by N g that means, 1 by N g is equal to
I am writing T c divided by T. T c is the
actual time spent in machining, so that means,
I am writing instead of 1 by N g, it is this
thing or basically I am writing N g is equal
to T by T c, T by T c. So, this would be I
am making it more clear now. This is this
should be N g is equal to T by T c, T by T
c that means, number of components produced
in one regrind or in one life of the tool.
If T is the tool life and T c is the machining
time, let us say the machining time is same
as the tool life, then N g will be one only,
because in the one life of the tool only you
will be able to make one component. But, suppose
the tool life is 20 minute and machining time
is only this one, machining time is only 10,
so this will be then, then you will be able
to make two components. So, this is about
that this one, so that means, machining time
and tool life. Tool life is more, then you
will be this one.
So, if we got to, we go back to our N g is
the number of components produced between
two regrinds, so that is why between two regrinds
you will produce these many. And T c is actually,
then T c must be equal to cost, it should
be cutting time per component, it should be
then, you should use the cutting time per
component. You should not do the mistake of
taking T c as the cutting time per batch.
Here T B is of course, the cutting time for
entire batch. So, T c is this one, that is
why you keep understanding each and everything
properly, we are doing just arithmetic, but
you do not anything any formula you do not
learn by wrote.
So, you see that now it is clear, that T s
T B whole batch. For whole batch, I am finding
out the time. So, this is T s and whole and
T l plus T a into N B, then T r replacement
time, T c by T means these are the number
of components between two regrinds that is
means this whole thing is 1 by N g. And then,
this is N B by N g, so that means, this takes
care about this. This is idle time for whole
batch, then this is T c into N B. N B is the
number of components in a batch, and per piece
is T c time, so that is why T c into N B you
have to do. And then, T is the tool life,
I am not using any suffix here.
So, thus machining time per piece is now I
divide the whole thing by T B, so that means,
I divide sorry, I divide the whole thing by
N B, then I say T p is equal to this much,
N B is equal to this much. And inverse of
T p gives the production rate that is number
of pieces produced per unit time is equal
to this much.
So, production rate can be increased by reducing
non-productive time. And profit rate can be
expressed as I p, suppose I am getting the
income per piece excluding material cost,
this much I am generating income minus C p
is the machining cost per piece. So, I p minus
C p by and T is what? T is the suppose machining
time per piece. So, T is the so, it is per
piece I am having I p minus C p divided by
T p, actually it is T p. Here it has shifted
in power point, so it is looking this.
I am writing again P r P r I p means per piece
how much is the income, excluding the material
portion and means how much I am generating,
suppose somebody has given me material, he
just ask me to produce that component, and
I said give me 100 rupee, but then my actual
expenditure, was cost was 80 rupee that is
cost. So, income is what we get and cost is
that why we spend.
So, I made pure profit of I p minus C p means
20 rupees. So, 20 rupees I made profit and
divide it by T p that means, suppose this
took, suppose T p is 2 hour, so 20 rupees
I made in 2 hours. So, my profit rate has
become 10 rupee per hour. So, like that, it
has to be done, that profit rate can be expressed
as this one, because one components thing
has been made, and I am doing that one component
as a time, at time of course, I am not talking
about parallel machines and all these things,
otherwise those things you can calculate later
on. Right now, I am going to see that my,
in a single machine, how my profit will be
more. So, now what happens that this type
of. So, these are the basic equations. Now,
you have to choose whole parameters, you have
to minimize these things these functions.
So, here you can use calculus and I am illustrating
by one example, suppose you have optimum cutting
speed thing you have to find out, so use Taylorís
tool life equation. VT to the power n is equal
to C; T is the tool life, n and C are material
dependent constants. So, T is equal to this
much, and V is the cutting speed; n and C
are constants for a workpiece material for
a particular set of workpiece material constants,
and also the combination of the tool and workpiece,
means it is the constants for a tool, but
for a particular workpiece material for a
particular set of values of feed and depth
of cut.
T c is the cutting time for a cylindrical
workpiece of length L. Suppose, its length
is L diameter is this, then T c will be pi
D L divided by f V. You can easily see that,
T c will come out to be pi D L divided by
this one, because it will be basically you
can say L divided by f N ok. N may be spindle
rpm, f, f is the feet per millimeter per revolution.
And L you can you can use another formula
V is equal to pi D N, pi D N type of thing,
then cutting speed. Then from that, you replace
N by this, you get pi D L f V. Then, T c can
be written as some constant K 1 by V, because
for the same job D is fixed, so and f is also
suppose fixed, so T c is inversely proportional
to V, and where K 1 constant is pi D L by
f, that expression is there.
So, this cutting speed for minimum cost can
be written how? That I can put the value of
T c in this whole thing, and I get this type
of expression. Similarly, for T c by T t,
T c I have written. And T i can write in terms
of T i can write in terms of this one, C by
V 1 by n, so I replaced it here. So, I got
this type of expression, this type of lengthy
expression, this you have got. Now, since
N g is equal to T by T c that is also expressed
in this one in this form, where K 2 is equal
to this thing, so I have got this thing. I
am writing everything trying to write in the
form of velocity.
So, whole thing is now in the form of velocity,
other things are constant, and I have to minimize
this cost. So, what should I do, I should
do dC B by dV equal to 0, and I will be getting
that expression. So, this I will not be doing
for that big expression, because it will be
here very big expression, but you can also
once you get the correct values, you can always
minimize. You can just do dC B by dV equal
to 0, and then solve that equation, you get
a particular value of V.
Let us see graphically, what happens? That
suppose, generally, suppose you have I plot
cutting speed on x axis and cost per piece
on y axis, so I am doing this thing cutting
speed and cost per piece. Non-productive cost
and non-productive time remains constant,
so that is why; here I am having non-productive
cost as a constant. But, the cutting cost,
this one if I say that cutting cost will keep
on decreasing that means, tool cost and tool
changing time will keep on increasing, so
tool cost will keep on increasing, but cutting
cost and cutting time will decrease, so that
is why, so one factor it decreases. So, overall
that summation if I make the total cost, so
total cost actually, initially it decreases,
and then increases that means, there is some
optimum cutting speed in between at which
the cost will be minimum.
So, here at any location suppose at this point,
here the non-productive cost is fix only,
cutting cost is this much, but you see that
tool cost has increased considerably, because
you have now you are operating at a high cutting
speed. So, cutting cost is less, because you
are not spending much time on the machine,
but then the tool has to be replaced again
and again, that is why tool cost has increased.
In this location, here your non-productive
cost is same, but cutting cost is little bit
higher than this, but the tool cost has reduced.
In this particular situation, suppose I make
it here. Here, cutting speed non-productive
cost is same, and cutting cost is actually
very, very high, because you are operating
at a very slow speed, so that is why cutting
cost is of course, the tool cost is very very
low, almost 0, because the tool is not getting
failed, so that is why, this thing is also
there. So, this type of curve is there.
And same thing about cutting time, suppose
it is non-productive time. Now, if I operate
at this one, if I increase my cutting speed,
my cutting time will decrease, but the tool
changing time will increase. So, total time
will increase. Here also the total time will
increase. Why? Because here the cutting time
will be very high, because you are operating
at a low speed. Of course, the tool will not
be changed, so tool changing time is almost
0. So, here also you have some optimum. So,
you get V 1 star is cost per piece, optimum
speed for cost per piece, and V 2 star is
the optimum speed for minimum time.
And V 2 star will always be V 1, more than
V 1 star, this you can even understand from
you intuition, because cost for maximum production
rate, this is speed for maximum production
rate. You will get only more production rate,
if the cutting speed is somewhat high also,
but of course, it should not be too high,
otherwise then this one. But, still it is,
it is V 2 star, because here my concern is
to minimize the time of machining, means increase
the production rate, so this V 2 star will
always be greater than V 1 star. But, what
may happen, that difference between these
two may be very small, and profit rate value
will fall somewhere between V 1 and V 2, so
that is the thing.
So, these factors give optimum cutting values
for cutting speed for minimum cost and minimum
machining time or maximum production rate.
For obtaining optimum cutting speed V 1 star
for minimum cost per batch of component, equation
must be differentiated with respect to V and
equated to 0. So, we do dC B by dV is equal
to this thing. So, we get this type of complicated
expression, but here easily it looks only
complicated. But, if you do with pen and paper,
it is not difficult, you just say V will come
this side, and ultimately you get some expression
V 1 star is equal to this much. So, this expression
is there. n is the Taylorís tool life exponent
that is also known to me, and all these things
are known to me, these are cost data.
And similarly, cutting speed for maximum production
rate is given like this. And R p is equal
to this much. For obtaining the optimum cutting
speed for maximum production rate, for a batch
components, equation must be differentiated.
And then, we did this we get this type of
thing. So, this is basically optimum this
thing, V 2 star. Now, you can see mathematically
also that V 1 star is smaller, and then V
2 star is slightly more. And cutting speed
for maximum profit rate, maximum profit rate
is given by equation this one.
What we have already discussed, and cost per
piece has come like this T p was this so I
p minus this, divided by this.
And now again you differentiate it with respect
to time, so you say dP R by dV. And you get
this type of expression, but the solution
of this equation gives the value of V 3 star.
Here the solution is a bit difficult, may
not be what you can, of course get or you
get just use any numerical method like bisection
method. And we get the cutting speed for maximum
profit, not profile, it is maximum profit
ok. So, we get maximum profit ok right.
So, this is, now these are things, which we
have considered, but, we in actual scenario,
we also have to consider number of constraints.
Suppose, I get cutting speed of something,
but can I operate this thing. May be, I am
getting minimum cost, but I may not be getting
minimum, may not be getting good surface finish.
So, you have to satisfy number of constraints
also in actual practice.
So, these may be like this, maximum power
restriction your machine is old you cannot
operate at a high speed. So, power required
during a cutting operation is a function of
the cutting condition and expressed as P w
is equal to B w V may be proportional to V
f times some exponent m, d times m 2 that
depth of cut. So, P w is B w prime into V
and B w prime is this much.
So, this type of thing will be there. So,
this would of course mean a change in the
maximum cost or maximum production rate or
maximum profit rate ok. So, here, this one,
so, so suppose you, you implement that one
may be that optimum rate is cutting speed
is coming something, but you have restriction
because of the power. Then you may have another
type of speed restriction. Most machine tools
have cutting speeds available in steps over
a certain range.
Suppose, you have 400 after that, it may be
only 320, your answer is coming 360. So, you
cannot choose 360, you have to choose only
say 320, whatever is available on the machine,
machine may be having discrete speed. On such
machine, step closes to the optimum value
should be used. And optimum speed must not
exceed the upper limit of the available machine
cutting speed. Lower limit of the cutting
speed is generally limited by the formation
of built-up edge. Suppose, you take very low
cutting speed, if your, by mathematical formula
it comes out like this, that very low speed
is the optimum speed, but you may not able
to use it, because it will create built up
edge.
Then, we have force and vibration restrictions.
Because, machine components are designed for
the maximum permissible load, beyond which
the tool deflections are excessive. So, job
accuracy will be affected. So, cutting force
should also be restricted. And cutting force
generally can be a function of feed and depth
of cut and some constant B w, B w may include
effect of the cutting speed.
Cutting speed has very little effect on cutting
force generally. In fact, in most of the cases,
if the cutting speed increases, then cutting
force decreases. So, therefore, it is safe
to neglect that effect. There is no restriction
on cutting speed in terms of force, but this
will limit the maximum values of feed and
depth of cut that can be used. Excessive vibrations
and chatter also puts restrictions on cutting
condition.
Then, surface finish restriction. You cannot
keep very high feed, because the, your surface
finish will be deteriorated. Surface finish
depends on tool work material, tool geometry,
process geometry, cutting conditions and the
type of coolant. If feed is more, then the
surface finish is poor, because roughness
is more. So, for a given operation the surface
finish restriction may be put like this, that
h m is the maximum permissible. Surface, so
that means, h m must be greater or equal to
B s f to the power something, V to the power
something. Of course, with increase in cutting
speed, sometimes the surface roughness decreases,
means surface finish improves. So, cutting
speed affects the surface finish due to the
formation of built-up-edge that gets smaller
and smaller and disappears at sufficiently
high speed, and improves the surface finish.
So, these constraints also have to be taken
together with that objective. Then comparison
of the three criteria, a comparison of the
three optimization criteria considered here
is of considerable practical interest. We
have to consider this one. So, maximum production
rate criteria is the simplest one to use,
since we do not need it any information regarding
various cost rates required. So, you can safely
say, I need maximum production rate and minimum
cost speed is not much different from maximum
production rate.
Minimum cost criteria is slightly more complex
and maximum profit rate criteria involves
solution of complex equation, there you may
have to use numerical technique; so, each
of these criteria, yield different values
of optimum cutting speed. The optimum cutting
speed for minimum cost is always less than
that for maximum production rate. This I have
told already, again I am emphasizing that
optimum cutting speed for minimum cost is
always less than that for maximum production
rate.
V 1 star and V 2 star may not maximize the
profit rate, which depend on the margin between
the selling price and the cost of production
as well as the rate of production. At the
optimum value V 1 star, the production rate
may be low while that speed V 2 the cost of
production may be high. So, both of these
conditions may lead to low profit margin.
Optimum cutting speed V 3 star for maximum
profit rate will, therefore, be different
and will lie between V 1 star and V 2 star.
When the optimum speeds are adjusted to take
into account the limited number of speed steps
and other restrictions, the final value of
V 1 and V 2 and V 3 star may not be widely
different, because suppose in between there
is difference of suppose V 1 is, suppose V
1 is, say 330, and V 2 may come suppose 370,
and V 3 profit rate thing in between may be
340, but your speed is one speed is 320, after
that you have 400.
So, what is the use, you can have to you have
to choose either 320 or 400, so that is why
sometimes, we do not bother much about that.
But, nowadays modern CNC machines of course,
they have variable speed, continuous speed,
there this becomes important. And it is often
suggest, so other cases it is often suggested
that the cutting conditions should be selected
between maximum cost per component and maximum
production rate.
Now, this was the analysis in which we did
a single variable optimization. We fixed f
and d, and we only adjusted basically the
cutting speed, but now we have to consider
the effect of feed and depth of cut also.
So, we use Taylorís extended tool life equation
in which we have T v p f q d r equal to C.
And for fixed d, suppose d is sometimes this
one you have to do single pass machining.
And d is fixed, then you can write this type
of equation.
If we do this, then we substitute you do the
same type of analysis, but now v and f both
will come into the picture in the objective
function. So, you have to do two variable
optimization. So, for multi variable optimization,
you can have partial derivative, suppose objective
function is O. So, del O by del v equal to
0, and del O by del f equal to 0. So, you
will be getting two equations. And these two
equations can be solved, and you can get the
value of f and v. This type of thing has been
discussed in the book of Ghosh Mallik and
these expressions, but those lengthy expressions.
We are not writing here; I am just, means
it can be easily done.
Now, coming to the multi-pass optimization,
we talk till now about single-pass optimization.
If I am doing the multi-pass optimization,
then we have to say that in each pass. Suppose,
I am taking say m number of passes, in that
m number of passes so we have to write these
type of expressions. And we have to say, that,
cutting time, we can find out by considering
these things, total cost, this is C T is the
total cost. And we are having for each pass,
and we are having then, we can write these
expressions.
I am not going to explain this line by line,
but this you can get, actually this is, these
type of expressions are available in research
papers, but you see that here, we are considering
these things for each one, we are considering
for each pass. Suppose, suppose t s is the
setting time. So, suppose you are having m
passes; so, m plus 1 and C 0 into t s, setting
in 1. And then you are having this type of
equations, suppose I am taking equal depth
of cut. So, you have to say like this, this
type of equation you will make, D 0 was your
original diameter, and let us see that 2 m
d R, minus 2 m d R, suppose depth of cut in
each roughing pass is minus 2, 2 d m, 2 m
d r, so that means, if you take depth of cut
d R, your diameter will reduce by 2 d R in
one pass, but there are m passes m roughing
passes. So, you are getting 2 m d R and minus
finishing pass is separate that why here m
plus 1. So, 2 d F that is 2 d F and minus
D L means final diameter, so of that 1, this
should be 0 that means, equation should be
satisfied or I can say D L is equal to D 0
minus 2 m d R minus 2 d F that has to be this
one done.
And in multi-pass optimization, you have total
cost is basically roughing cost plus finishing
cost, and then you can find out preparation
time cost and tool reset time cost this type
of thing to put. So, this is roughing cost,
finishing cost, and then preparation time.
Preparation time is t p, and then corresponding
cost is C 0 by that rate, overhead cost and
this is overhead here C 0, t st means tool
reset time cost and this way, it will go.
Now, here it will be like this, that here
roughing time is actually like this, suppose
I am finding per piece thing. So, roughing
time is I am having suppose in one pass the
roughing time is T r. So, there are m passes,
you add all roughing time together multiplied
by C 0 and that overhead portion of the cost.
And then, you have to say, i is equal to 1
to m, then now here it is due to machining
that tool changing time cost has come here,
this is and then this is C t, that means,
here the cutting cost C 0 into t r, t r is
the tool replacement. And C t was actually
the cutting this one and t R i by this one.
So, roughing, roughing cost is there.
And then, unequal depth of strategy; if you
imply, then you can say in each pass I have
unequal depth, instead of equal. Many people
have done equal depth of cut strategy, so
it will be like that. So, this expression
will be t R i can be written in this term
first pi L by D 0. D 0 is the initial diameter,
then diameter has become D 1. D 1 is equal
to what? D 0 minus d R 1 that means, in first
2 into 2 d R 1, d R 1 is the depth of cut.
So, this has come like this, then finishing
cost is C 0 t f plus C 0 tool changing time
etc., t r by t f T. And then, you have C t
t f by T f that means, it is the tool cost,
and this C 0 t f that means, it is basically
the total operating cost. It is not only the
overhead, but it is operating cost. So, C
0 into t f, this is actually that one and
this is the tool cost. And this was the operating
cost, but because of the tool replacement
time got lost.
So, like that you make the expression. And
total machining time for roughing pass is
this, and where D i minus 1 is equal to this,
this formula is there.
And then, you can actually get, now you can
make the strategy. Now, equal you, you, suppose
you can make equal depth of cut strategy or
unequal. I have seen one research paper, in
that, he has taken unequal depth of cut strategy.
So, suppose two passes are there, and he has
taken unequal depth of strategy. And his answer
is coming something; he has given this type
of table also. But, if you see very carefully,
if I have to let us say, I have to reduce
that, roughing depth, suppose, suppose I have
to reduce by depth of cut is 3.5 ok. So, suppose,
suppose it is like this, that here 2.51. And
this is, these are say cost per pieces in
dollar. So, what may happen that, it is suppose,
I have, I have taken, suppose I have chosen
2 mm, this one. And if I say that, I take
equal depth of cut strategy. So, I take corresponding
to 2 mm depth of cut, in roughing pass, the
cost is 0.551. If I remove 4 mm depth of cut
that means, I do two process, then my cost
will be 0.551 into 2 that means, it will be
2 this one point say, 1.102 like that.
Now, here that data is not there, because
the table was very big. And I have not seen
put it here, that if I do the same job in,
let us say that, if I take 4 mm, if I do the
same thing 4 mm ok. If I do that, let us say
that, if you, if this was done by taking,
yes say 2, 2 into 2 that this one. But, I
can do the same thing like this instead of
taking 2 mm in both, I can take in 1 pass
as 2.5. And another I can take 1.5 in another
this one, of course that data is not here.
So, 2.5 data is here. So, it may be that this
also comes out to be almost same, in most
of the, that means, there are multiple solution.
So, many times people have used this equal
depth of, unequal depth of cut strategy. And
the in order to get a regulate model, but
really that difference is not very appropriate
in most of the places equal depth of cut strategy
may be very very appropriate. And it may be
may be very pertinent, so that is what that
this point has to be seen into, taken into
consideration right, that means ok, I can
say, suppose I have 2.5. Let us see, if I
take 2.5, and then I am having 0.611. And
if I do that take 2.5 two times, so this is
2.2 this one 6 say, so 2, 2 and 2, 6, 12.
So, my cost is coming this much dollar, 1.2226,
12, 12, 1.22. And if I decide, no I will not
take 2.5 in each pass, because I have to remove
5 mm, I have to say take total 5 mm depth
of cut. So, first time I did what? I took
2.5 mm depth of cut. So, I did this operation
two times, my cost was this.
Another person says I will take one time 2.4,
another time I will take 2.6. So, he will
do what? 0.597, and then it has come 0.639.
So, it has come 976, no no sorry, this was
0.625, 0.625 so, 5, 7, 12; 9, 2 12; 6, 5 11
12. So, you see it also came exactly same
that means, in a even if that mathematically,
you have predicted that solution also may
be you have done some way. And you may get
a solution in which instead of some person
has given a solution 2.5, some person has
given a solution, in which says no, in first
you take 2.4, next you take 2.6 or vice versa.
First you take 2.6 depth of cut, then 2.4.
Both are taking total depth of cut as 5 mm,
but both are getting the same type of thing.
On, they may not realize that there are multiple
solution.
So, in many cases, I have seen in the research
papers that actually there are multiple solution.
At many times, what happens that this area
is actually very much investigated, lots of
papers have been produced. And sometimes this
optimization of machining has acted as a benchmark
to test different algorithms that which algorithm
is better that type of thing. So, from academic
point of view it is very good, but I have
noted that even the heuristic approach works.
What is that this one? That try to keep as
high depth of cut as possible. Most of the
time, when optimum solution is there, people
have take high depth of cut without violating
the constraint of course that means, there
should not be chatter or excessive vibrations.
And then try to have feed, such that it does
not violate surface roughness constraint.
So, you try to have feed, but you limit it
by this one. Because, if you take more feed,
then surface roughness will be decreased and
then, then after that you optimize for cutting
speed, then you will get this one. So, this
is a heuristic type of thing means I first
try to keep this one more depth of cut. In
fact, I should try to do the operation if
possible in one pass only, then I should go
to other two passes.
So, all these type of heuristics themselves
give that enough idea for that. So, heuristic
based optimization methods have been also
developed, and those things are also there
means applying this method or that method
will actually may be very good for academic
point of view. And it says that ok, I am getting
the solution, but many times that particular
solution will only predict the same thing
that which I know by even common sense that
ok, this feed can be taken this much high.
So, you take that one, and then it becomes
and then accordingly and then after that cutting
speed is only optimized.
So, otherwise in general, many people have
solved it as a four decision variable problem
that means, depth of cut is one variable,
no not only depth of cut as one variable.
They say feed, then they say cutting speed,
then may say ok. Suppose, they follow equal
depth of strategy then d r, then for finishing
they will say d F. And then, they will say
number of passes m. So, you see here itself
there are 1, 2, 3, 5 variables are there and
out of that one variable m is integer. So,
it becomes like a integer type of programming
problem.
So, no doubt that mathematically this is interesting,
and you can work out that, but you know that
in most of the cases, answer will this come
this only that no keep m is equal to 1, do
not do many passes. And then take depth of
cut more like that, so that is how that in
these practical aspects should also be seen
together with the mathematics.
And then, you can consider the cost of cutting
fluid. Here, I have not considered the cost
of cutting fluid; cost of cutting fluid can
also be included in the objective function.
So, this can be done.
So, this much for today, we will meet in next
class.
