WEBVTT
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in this lecture we continue the discussion
on operator and task assignment in the previous
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lectures we have seen some aspects of operator
and task assignment we have seen some aspects
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of rabbit chasing and dedicating machines
to operators so we have seen rabbit chasing
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and we have seen some aspects of dedicating
machines to operators we have also seen dedicating
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both these are seen under the context of labor
intensive cells
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and we have seen dedicating operator dedicating
machines to operators in the context of machine
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intensive cells in the context of machine
intensive cells we assume that while the operation
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is being carried out or the manufacturing
takes place the operator need not be there
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with the machine which means the operator
does only load and unload and while the actual
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manufacturing or machining is happening the
operator can attend to some other machine
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and can load or un load parts on other machine
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so far we have seen three models basically
where one is the rabbit chasing model and
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the dedicating model where in the context
of labor intensive cell the operator is physically
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present when the actual machining takes place
we also saw that rabbit chasing gives better
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output and can increase the productivity but
has its own concerns such as limit on the
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distance that people walk ability of the operate
all the operators to handle all the machines
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and so on
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so the next model that we will see a model
where we have a single cell and we do multiple
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products so we have a single cell and multiple
products under the context of rabbit chasing
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and dedicating
so let us assume that there are two products
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this cell makes this cell has five machines
and this is the direction of the flow
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the process times are twelve eight ten nine
and six for the first product and the process
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times are nine four six eight and fifteen
for the second product so the sum of the processing
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times is twenty thirty thirty nine plus six
forty five for the first product and nine
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plus four thirteen nineteen twenty seven plus
fifteen forty two for the second product
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now if we have two operators if we have two
operators the first product we can expect
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a piece every twenty two point five time units
the second one we can expect every twenty
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one time units if we have three operators
the first one we could expect forty five by
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three fifteen fifteen is more than twelve
eight ten nine and six so we would expect
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fifteen that forty two divided by three is
fourteen and this fifteen is more than fourteen
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so the output will come every fifteen minutes
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so we could do this analysis for two operators
three operators and so on and if we do rabbit
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chasing we can compute what is the expected
output at steady state then we dedicate the
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machines to operators we have already seen
that the best output in this case for two
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operators would be
one and two which gives us twenty ten nineteen
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plus six twenty five so two operators we would
get twelve plus eight twenty and ten plus
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nine plus six therefore the best value is
twenty five is when is the steady state output
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of this system
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now when we take second product we have nine
plus four thirteen plus six nineteen fifteen
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plus eight twenty three so for the second
product we would get nineteen and twenty three
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so the output will be twenty three for two
operators the only thing we understand here
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is we are having two operators and if we are
dedicating machines to operators under the
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context of a labor intensive cell then we
observe that for the first product the operator
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one would do twelve and eight which is twenty
while operator two would take twenty five
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time units while for the second product the
first operator will do the first three machines
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will do nineteen and the other operator would
do twenty three
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so we have to look at two things one is that
for different products when we dedicate we
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will have situations where operators will
handle different machines or there is different
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machine allocation when multiple products
are considered this also brings us to the
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requirement that the operators should be able
to handle all the machines we have a situation
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where operator one would be doing only one
and two and operator two would be doing three
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four and five when product one is being made
while operator one will have to do one two
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and three when product two is made
so this necessitates that operator one should
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be capable of handling one two and three while
operator two should be capable of handling
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three four and five when we have multiple
products being made in the same cell we can
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have another situation which necessitates
that all all operators should be able to handle
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all the machines
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now let us assume that we are doing product
number two while we do product number two
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and we dedicate operators to machines when
we do product number two the first operator
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would handle one two and three to have a time
of nineteen the second operator would have
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eight plus fifteen which is twenty three and
the output would be every twenty three time
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units
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now the out since the operator one takes up
the things first and then operator two does
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it and if we have a situation where the operator
one takes a piece finishes this keeps it here
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goes back takes another piece finishes keeps
here and operator two takes these two then
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if operator one is busy all the time then
there will be a buildup of inventory at this
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point because operator one would have the
raw material or entering material inventory
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here and here if we have unlimited supply
or if the entire production batch is available
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here then operator one would be taking a piece
he would finish these three leave it here
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go back take another piece and every nineteen
time units the pieces will get built up here
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operator two would be requiring twenty three
time units for every piece and this buildup
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is going to be replenished at the at one piece
per twenty three time units so there will
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be an inventory buildup if operator one continues
to work all the time we have already we should
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not allow or we should not plan inventory
buildup or excess inventory buildup in the
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context of modern manufacturing methods and
therefore the only way by which we can prevent
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inventory buildup from happening here is to
control the amount of inventory that is entering
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here
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so this principle is a fundamental principle
in just in time manufacturing which we would
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be seeing later in this lecture series but
if we control the inventory here then this
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person will be walking on a piece every nineteen
minutes while this person would take twenty
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three minutes or time units while working
on this piece
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if we control the inventory at this point
what will happen is that so this operator
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whom we call operator one will be idle for
four time units while this operator would
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be working all the time so when we have a
situation where while dedicating the there
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is a difference between these two outputs
or there is a difference between these two
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outputs twenty and twenty five which gives
us twenty five for product one which simply
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means if we make a table we say operator one
operator two product one product two operator
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one product one is twenty time units operator
two product two is twenty five time units
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so operator two operate operator two product
one is twenty five time units so steady state
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output is twenty five time units when we consider
the second product nine plus four thirteen
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plus six nineteen this is twenty three so
steady state output is twenty three if we
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permanently dedicate according to this then
the first operator will be doing these two
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for product one and these three for product
one second operator will do these three for
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product one and these two for product two
and if in both the instances it happens that
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the first operator is spending less time than
the second operator
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while in rabbit chasing if we have two operators
both of them are going to work about the same
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time and a steady state output will be twenty
two point five for each at steady state and
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fifteen for each if we have three or twenty
one for both the operators if we have two
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operators for the second product while rabbit
chasing utilizes them equally dedicating does
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not utilize them equally if we want to utilize
them equally then we end up building inventory
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at some point
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if we control the inventory then we would
have less utilization of operator one and
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more utilization of operator two so operator
to be spending more time with thee with the
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machines than operator one so in order to
balance that what we could do is if the same
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two products are made lets say in two consecutive
shifts then the person who becomes operator
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one will be operator one in one shift and
operator two in the second shift
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so the people can change that the dedicating
machines will remain the same so if we have
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a person called m and a woman called f which
are our two operators then m will be called
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operator one and say f is called operator
two so the person m in the first shift will
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be operator one and will be doing these two
when the first product is being made and these
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three when the second product is being made
so the person number one who will be m will
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now have utilizations of twenty and nineteen
in the first shift while the person f who
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is now called operator two she will have utilizations
of twenty five and twenty three in the first
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shift
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now when the second shift if we interchange
them and the person f takes up these two for
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product one and these three for product two
and so on now this person will now the one
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who had twenty and nineteen in the first shift
will now have twenty five and twenty three
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as the time spent so that way in one shift
one person will be utilized more the other
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will be utilized less in the next shift this
person who has utilized more will now be utilized
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less
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so this way we could ensure that one person
is not overworked in the cell but if we have
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to do that again both m and f should be capable
of handling all the five machines so both
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in rabbit chasing as well as in dedicating
it is necessary that operators are multi skilled
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they are trained sufficiently so that they
can handle a variety of machines so operator
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training cross functional training is a very
very important aspect in the context of cellular
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manufacturing
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now lets continue further and look at single
product with multiple cells so let us look
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at a a situation where lets look at a situation
where three cells feed to the assembly so
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this will be called assembly
this is a this is b and this is c all of them
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feed to the assembly and the final product
comes out of the assembly now let us assume
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that each cell cell one has five machines
and the process times are twelve eight ten
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nine and six now cell two has five machines
nine four six eight and fifteen and cell three
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also has five machines seven nine twelve ten
and six
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now if we each each one has five machines
as described and the product flow will be
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like this now when we do rabbit chasing we
actually dont need all these numbers we just
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need two numbers which is the total which
is forty five and the maximum which is twelve
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here the total is forty two and the maximum
is fifteen and here the total is forty four
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and the maximum is twelve
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now what happens is if we assign now from
this we know that if we have one operator
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then the output is forty five if we have two
operators its twenty two point five if we
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have three operators it is fifteen four operators
it is twelve and so on so here again if we
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have one operator its forty two two operators
its twenty one three operators its fifteen
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and so on fifteen comes because forty two
by three is fourteen fourteen is smaller than
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fifteen so this will be fifteen
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here with one operator its forty four two
operators its twenty two three operators is
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fourteen point six six and four operators
its twelve so if we have if we assign in this
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case if we assign two operators to this and
if we assign three operators to this and if
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we assign two operators to this so if we do
two comma three comma two this will be producing
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a piece at the rate of twenty two point five
time units this will be producing a piece
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at the rate of fifteen time units this will
be producing a piece at the rate of twenty
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two time units and the steady state output
from the assembly is the maximum of these
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three numbers which is twenty two point five
time units
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if we do three two and three then by assigning
three operators to the first one we would
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get fifteen by assigning two operators to
this one we would get twenty one by assigning
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three up by assigning again two operators
to this one we would get twenty two and the
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best one will be twenty two
if we do two two three it would be twenty
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two point five for two it will be twenty one
and for three it will be fourteen point six
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six and the best value will be twenty two
point five
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so if we assume that we have seven operators
with us and if we are trying to allocate these
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seven operators to these three cells in three
different ways which is two three two three
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two two and two two three now we observe that
the output we get this twenty two point five
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twenty two and twenty two point five so the
problem is how do we allocate a given number
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of operators to these three cells such that
we maximize the productivity or minimize the
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maximum of the times that we take the maximum
of the times is the rate at which it is coming
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we want it to be as small as possible
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so this problem is called the operator allocation
problem when we have multiple cells feeding
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to assembly now if we assume rabbit chasing
so let us assume that x one is the number
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of operators allocated to the first cell x
two is the number of operators allocated to
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the second cell and x three is the number
of people allocated to the third cell so when
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we have x one people coming in the time taken
is forty five by x one the time taken is forty
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two by x two and the time taken is forty four
by x three
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we also know that its not always forty five
by x one it is forty five by x one up to this
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and this so right now we say that the number
of operators that i wish to assign to the
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first cell cannot exceed four because there
is even if i increase it beyond four i am
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not going to have any improvement in this
because by the very we saw in a previous lecture
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that even if we have five operators here the
bottleneck time or the maximum time will take
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over and this will only be twelve and it will
not be more than twelve
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so there is a restriction that x one is less
than or equal to four x two is less than or
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equal to three and x three is less than or
equal to four so we have these restrictions
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which is there now lets assume that we have
eight operators therefore we have a constraint
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in addition to this x one plus x two plus
x three is less than or equal to eight now
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what do we want to do so if we assign x one
operators to the first one we are going to
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assume that the output is forty five by x
one forty two by x two forty four by x three
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and therefore we want to minimize the maximum
of these three values
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so we will say minimize u we will minimize
the maximum of the three values minimize u
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is not maximized you minimize u
so let u be the maximum of the three values
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now we want to minimize the maximum of these
three values so we minimize v and define v
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as the maximum of the three values so since
v is the maximum of these three values v is
26:26.690 --> 26:33.340
greater than or equal to forty five by x one
v is greater than or equal to forty two by
26:33.340 --> 26:43.049
x two and v is greater than or equal to forty
four by x three
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so the problem now would be to minimize this
so we rewrite this first and then rewrite
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it as now we have x one plus x two plus x
three less than or equal to eight because
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we have a maximum of eight operators now we
at the moment ignore these three we will use
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them when required so this constraint is written
now these things have to be written differently
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so this will become x one is greater than
or equal to forty five by v x two is greater
27:24.409 --> 27:34.700
than or equal to forty two by v and x three
is greater than or equal to forty four by
27:34.700 --> 27:49.260
v and we minimize v we minimize v
27:49.260 --> 27:59.400
now this is still not a linear programming
problem because v is a variable x one is also
27:59.400 --> 28:04.799
a variable so this is not a linear programming
problem so we will now define u is equal to
28:04.799 --> 28:17.960
one by v and then we say that minimize
one by u u is equal to one by v so minimize
28:17.960 --> 28:26.001
one by u subject to x one greater than or
equal to forty five u x two greater than or
28:26.001 --> 28:35.610
equal to forty two u and x three greater than
or equal to forty four u plus the constraint
28:35.610 --> 28:39.480
x one plus x two plus x three less than or
equal to eight
28:39.480 --> 28:50.300
now minimizing one by u is the same as maximizing
u so now this becomes a linear programming
28:50.300 --> 28:56.370
problem that maximizes u subject to x one
greater than forty five u x two greater than
28:56.370 --> 29:01.830
equal to forty two u x three greater than
or equal to forty four u x one plus x two
29:01.830 --> 29:07.970
plus x three less than or equal to eight and
x j greater than or equal to zero so this
29:07.970 --> 29:17.809
becomes a linear programming problem and when
we solve this problem we get x one equal to
29:17.809 --> 29:26.600
three x two equal to two we get a solution
x one equal to three x two equal to two and
29:26.600 --> 29:32.840
x three equal to three we get a solution this
way
29:32.840 --> 29:39.909
now this would mean that for eight operate
with eight operators we allocate them in the
29:39.909 --> 29:49.520
order three comma two comma three so when
we have three this person would be producing
29:49.520 --> 30:01.779
at fifteen when we have two this one would
be producing at twenty one
30:01.779 --> 30:10.510
and then we have three operators the the third
cell will be producing at fourteen point six
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six and the maximum of that will be twenty
one which will be the which will be the steady
30:21.080 --> 30:29.250
state output of the assembly
30:29.250 --> 30:34.300
now when we look at twenty one as the steady
state output of the assembly then we also
30:34.300 --> 30:42.020
have to look at one more aspect here so we
are assign three operators here in the end
30:42.020 --> 30:49.720
we have two operators here four eight and
then we have three operators here for the
30:49.720 --> 30:58.620
third cell now what happens is the if we allow
this many operators to work this person is
30:58.620 --> 31:06.059
going to produce at the rate of fifteen minutes
per piece twenty one minutes per piece and
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fourteen point six six minutes per piece steady
state output from this system is twenty one
31:12.860 --> 31:18.409
minutes per piece or twenty one time units
per piece
31:18.409 --> 31:27.340
now if we allow these number of operators
to work continuously then what happens is
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this is the bottleneck this is the cell thats
creating the twenty one time units so these
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two cells will over produce so either we have
to make sure that they dont produce more or
31:42.360 --> 31:47.880
we will end up building inventory in these
two places if we dont want to build inventory
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in these two places then we have to control
the amount of material that comes into this
31:55.640 --> 32:02.890
and comes into this as previously mentioned
they lead to just in time manufacturing models
32:02.890 --> 32:09.659
where we control the amount of material that
come into these two the net effect of it would
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be that the operators here would be underutilized
and will not be utilized fully
32:16.169 --> 32:24.210
so once again we will get back to shifting
operators so that the the person the roster
32:24.210 --> 32:31.380
will shift operators from one shift to another
so that the same person does not get overworked
32:31.380 --> 32:36.660
this again leads us to the fact that all the
operators will have to be cross trained and
32:36.660 --> 32:43.529
be capable of working on all the machines
in this context another possibility also is
32:43.529 --> 32:50.710
that if building a little bit of inventory
is permitted then what we can do is let them
32:50.710 --> 32:58.770
work at their full speed build some inventory
here and then after some time shift an operator
32:58.770 --> 33:05.910
from here to this or from here to this because
this cell can have a maximum of three beyond
33:05.910 --> 33:10.240
which it is not producing more than the rate
of fifteen
33:10.240 --> 33:15.549
so after some inventory is built up one person
can move and work here so that this twenty
33:15.549 --> 33:23.080
one becomes fifteen this will slow down because
this from from fifteen will become twenty
33:23.080 --> 33:33.890
two point five so after some time we can adjust
it and make it as two three three with output
33:33.890 --> 33:45.730
twenty two point five fifteen and fourteen
point six six even though this is producing
33:45.730 --> 33:51.580
at twenty two point five there is some inventory
thats built up here which can be consumed
33:51.580 --> 33:56.950
and now the steady state output will come
to fifteen
33:56.950 --> 34:03.059
then after some time this inventory will come
to zero so we will shift this operator back
34:03.059 --> 34:08.490
by which time some inventory could be built
up here and so shift one person here so we
34:08.490 --> 34:16.819
will then move to three three and two two
would mean that this is this person will produce
34:16.819 --> 34:22.219
only twenty two will take twenty two time
units per piece but there is some inventory
34:22.219 --> 34:23.469
built up here
34:23.469 --> 34:30.169
so with three three and two will be operating
at fifteen fifteen and twenty two but that
34:30.169 --> 34:36.009
twenty two is offset by the fact that there
is some inventory and so on so if we keep
34:36.009 --> 34:42.690
doing this then typically we are solving essentially
like a linear programming problem so the solution
34:42.690 --> 34:51.409
is two point seven four eight x one is equal
to two point seven four eight x two is equal
34:51.409 --> 35:01.910
to two point five six four and x three is
equal to two point six eight seven
35:01.910 --> 35:12.450
this is like it it on an average we are working
at this way the three and three at some point
35:12.450 --> 35:18.359
the operator will move from here to here which
means the two will increase this will come
35:18.359 --> 35:23.479
down at some point one operator moves from
this to this so this will work at the steady
35:23.479 --> 35:28.539
state two point five six four well the other
thing will work at two point seven four eight
35:28.539 --> 35:35.150
two point six eight seven now these numbers
come because forty five by this will be equal
35:35.150 --> 35:39.140
to forty two by this will be equal to forty
four by this
35:39.140 --> 35:45.940
so at steady state the outputs from all the
three cells will be equal provided we allow
35:45.940 --> 35:51.479
a small amount of inventory to be built up
here and here and then shift these operators
35:51.479 --> 35:58.829
so it becomes what is called a dynamic way
of reallocating operators building a some
35:58.829 --> 36:04.349
amount of inventory so that the operators
are not underutilized and most importantly
36:04.349 --> 36:12.380
by doing so we are able to increase this output
from twenty one it will move to a balance
36:12.380 --> 36:17.380
which is equal to the same thing which will
be forty five by two point seven four eight
36:17.380 --> 36:23.229
forty two by two point five six four and forty
four by two point six eight seven
36:23.229 --> 36:31.209
so we will move to that number which is less
than twenty one time units per piece now another
36:31.209 --> 36:36.700
way of looking at this problem is when we
actually solved this there are two other things
36:36.700 --> 36:45.069
that we will have to look at one is in a way
we are making a certain approximation because
36:45.069 --> 36:51.969
this forty five by x one forty two by x two
forty four by x three are not exactly the
36:51.969 --> 37:02.829
same because the output is not forty five
by four if we put four values for x one the
37:02.829 --> 37:07.720
output is not forty four its not eleven point
two five but it is twelve
37:07.720 --> 37:17.479
so forty five by x one will hold only when
x one is between one and three when x one
37:17.479 --> 37:25.059
if x one is zero then this will start behaving
in a very very funny manner but the very problem
37:25.059 --> 37:31.089
the very nature of the allocation problem
will ensure that we do not allocate zero to
37:31.089 --> 37:37.130
any cell so we could put another condition
here such that x j is greater than or equal
37:37.130 --> 37:44.230
to one so that we do not allocate zero because
if we allocate zero forty five by x one will
37:44.230 --> 37:53.000
go to infinity as far as the linear programming
problem is concerned if x one is equal to
37:53.000 --> 37:58.640
four this would take the steady state output
as forty five by four and will not take it
37:58.640 --> 37:59.989
as twelve
37:59.989 --> 38:05.390
so we have to be conscious we have to be aware
of this fact and so we first solve this problem
38:05.390 --> 38:11.460
once then see what kind of an answer we get
and then progressively do it by adding these
38:11.460 --> 38:20.079
types of constraints and then by adjusting
the values at some point so there is a kind
38:20.079 --> 38:29.019
of a iterative i approach there is a kind
of a redoing this problem depending on what
38:29.019 --> 38:36.420
kind of a solution we get one way of overcoming
that is by taking the linear programming solution
38:36.420 --> 38:38.789
and by working with it
38:38.789 --> 38:44.119
so if we take the linear programming solution
the linear programming solution is two point
38:44.119 --> 38:53.400
seven four eight two point five six four and
two point six eight seven with eight operators
38:53.400 --> 39:01.509
so first thing what we do is you allocate
one operator to each cell so when you have
39:01.509 --> 39:08.549
one operator to each cell in fact from the
linear programming solution we could do this
39:08.549 --> 39:15.779
what we could do is the upper the closest
integer value for this is three the closest
39:15.779 --> 39:22.410
integer value for this is three the closest
integer value for this is also three so we
39:22.410 --> 39:23.839
cannot have three three and three
39:23.839 --> 39:32.339
at the same time we will have to find out
which variable has the largest fractional
39:32.339 --> 39:39.599
value so this one has the largest fractional
value so two point seven is approximated to
39:39.599 --> 39:49.709
three the next one that has largest fractional
value is this so this is approximated to three
39:49.709 --> 39:56.220
and since there is a total of eight this will
have to take a value two which was our optimum
39:56.220 --> 39:58.920
solution with three two and three
39:58.920 --> 40:07.690
so many times the lp optimum suitably rounded
suitably rounded i i say suitably because
40:07.690 --> 40:12.849
if we actually round it we will get three
three and three two point five six four will
40:12.849 --> 40:18.209
be rounded to three but then we dont have
nine operators so will suitably round them
40:18.209 --> 40:25.680
based on the largest fractional component
the variable that has the largest fraction
40:25.680 --> 40:31.799
first becomes takes its upper integer value
and this is progressively done till we utilize
40:31.799 --> 40:36.670
all the operators that we have
40:36.670 --> 40:43.479
so that is one way to do this the other way
to do this is to start with one operator for
40:43.479 --> 40:58.799
each cell and if we do that our output will
be forty five forty two and forty four now
40:58.799 --> 41:08.709
this is the slowest give one more person to
the slowest so you get two one one now this
41:08.709 --> 41:16.089
becomes twenty two point five this becomes
twenty one this becomes forty two this becomes
41:16.089 --> 41:24.199
forty four this is the slowest one so you
give two one two so this is twenty two point
41:24.199 --> 41:34.239
five this is forty two this is twenty two
this is the slowest you get to two and two
41:34.239 --> 41:41.069
now this is twenty two point five this is
twenty one this is twenty two this is the
41:41.069 --> 41:47.319
slowest give one more person three two two
41:47.319 --> 41:55.099
now this becomes fifteen this becomes twenty
one this becomes twenty two so this is the
41:55.099 --> 42:02.859
slope this is the slowest give one more person
you get three two three my eight people are
42:02.859 --> 42:12.109
exhausted therefore i get this as a solution
so this heuristic solution almost in all cases
42:12.109 --> 42:22.829
gives the optimum solution another way to
do is this is my lp optimum so put everything
42:22.829 --> 42:33.829
to its corresponding upper integer value so
i would get three three and three with nine
42:33.829 --> 42:40.529
whereas i have only eight operators so with
three three and three my output will be fifteen
42:40.529 --> 42:53.839
here my output will be fifteen here and my
output there is fourteen point six six here
42:53.839 --> 42:56.359
now you have to reduce one
42:56.359 --> 43:07.599
so reduce from the fastest the reduce from
the fastest so if you do that if you reduce
43:07.599 --> 43:15.299
from the fastest you will get this is the
fastest fourteen point six six is the fastest
43:15.299 --> 43:23.769
so reduce from the fastest you get three three
and two which would give you fifteen fifteen
43:23.769 --> 43:29.559
and twenty two so you dont get the optimum
solution you get twenty two as a steady state
43:29.559 --> 43:35.609
output for this heuristic solution increase
everything to its upper value and then take
43:35.609 --> 43:45.950
away till you get the number of operators
that you have so you could do this one also
43:45.950 --> 43:52.640
so these are the several ways by which we
look at what is called operator allocation
43:52.640 --> 43:58.319
problems to manufacturing cells we what we
have actually done is we have actually done
43:58.319 --> 44:18.089
two aspects of this problem one is called
one is what we call as a static operator allocation
44:18.089 --> 44:26.289
and we have looked at what is called dynamic
operator allocation
44:26.289 --> 44:32.369
so in the static operator allocation case
where we could implement this solution which
44:32.369 --> 44:48.979
is three two and three which would give us
an output of twenty one time units but as
44:48.979 --> 44:55.279
i said this will be coming out at fifteen
one piece per fifteen time units one piece
44:55.279 --> 45:01.990
per twenty one time units one piece from fourteen
point six six time units
45:01.990 --> 45:11.219
so once again either in the static case either
there will be inventory or there will be under
45:11.219 --> 45:20.979
utilization if we control the number here
if we control the inventory here and let this
45:20.979 --> 45:27.299
produce at the rate of twenty one we this
will be producing at the rate of twenty one
45:27.299 --> 45:32.060
time units per piece we control inventory
and make this produce at the rate of twenty
45:32.060 --> 45:39.349
one per piece the steady state output will
be twenty one now in order to ensure do rotation
45:39.349 --> 45:46.609
so that the roster will ensure that no operator
is unfairly over utilized
45:46.609 --> 45:52.190
the other way as i said is to try and build
the inventory and then move operators from
45:52.190 --> 45:59.400
one cell to another dynamically so we would
start with three two three we would start
45:59.400 --> 46:04.759
with three two three static would be three
two three dynamic we would start with three
46:04.759 --> 46:10.709
two three and then after a while we build
up inventory now out of the three this is
46:10.709 --> 46:17.680
the fastest this for three people is fifteen
fourteen point six six and twenty one so this
46:17.680 --> 46:23.269
is the fastest this is also building inventory
this is also building inventory so then one
46:23.269 --> 46:31.299
person will come from here to this so we will
operate at three three two for some time and
46:31.299 --> 46:38.589
then we realize that this is getting a little
faster both are depleting at a faster rate
46:38.589 --> 46:42.769
because with three people here twenty one
also starts coming to fifteen
46:42.769 --> 46:50.010
so all the inventories that are built up is
getting little depleted this is producing
46:50.010 --> 46:55.539
at fifteen this is producing at fifteen this
is producing at twenty two but its getting
46:55.539 --> 47:03.269
depleted very quickly so then after a while
we would do two three three and then we will
47:03.269 --> 47:12.529
come back to three two three we will do this
which means at steady state we will be it
47:12.529 --> 47:23.910
is equivalent of two point seven four eight
two point five six four and two point six
47:23.910 --> 47:27.499
eight seven
47:27.499 --> 47:35.009
now in this model we have assumed that there
is only one product and all these three manufacturing
47:35.009 --> 47:42.519
cells will feed to the assembly so a very
logical extension would be when we have multiple
47:42.519 --> 47:50.749
products now when each cell makes lets say
the same product but two variants of the same
47:50.749 --> 47:58.219
product the two variants again may require
three parts or three sub assemblies each is
47:58.219 --> 48:04.140
made in a cell and then its brought into the
assembly then we also have to look at operator
48:04.140 --> 48:10.529
allocation in the context where we have multiple
cells making multiple products the model is
48:10.529 --> 48:16.219
very similar to what we have seen now with
a very slight difference and we will look
48:16.219 --> 48:18.490
at that model in the next lecture