Welcome back, so we were discussing about
the inventory management in our last session, and we started discussions about 2 stage inventory
management in the supply chain. And, we discussed that how directly at stage
2. We can apply EOQ model because of this sawteeth
pattern, but directly, we saw that we have this type of step pattern at stage 1. And, therefore we cannot apply that EOQ model
at stage 1 in this supply chain. But, then immediately it was told to us, that
there is a concept known as Echelon inventory stocks. The Echelon inventory stocks where, you have
physical inventory at stage 1 + physical inventory at stage 2. And, when we used this Echelon concept then
we saw in the last session, that we can have these types of dotted slanted lines. And as usual what we discuss in the last class,
that these vertical lines are representing the replenishment stocks. The stocks when it is coming to you, the replenishment
stocks. And these slanted lines, the line with slopes
these are representing the consumption lines. So, same way, now we have in our stage 1 also,
these vertical lines as our replenishment lines, and these virtual lines, these dotted
lines are the consumption line. And, therefore you have the sawteeth pattern
at installation1 also, and now because of the sawteeth pattern at installation1, you
can apply EOQ model at installation1 also. So, now this agreement is being made. That we can apply EOQ model at both these
installation 1, and installation 2. Now, we are told that how to apply EOQ for
a single stage, for a single installation. So, now let us start doing that only, and
as we have discussed that all the assumptions which we take for developing basic EOQ model
will apply here also. The only additional assumption, the only additional
condition which we discussed in the last session. That, because of the value addition as we
are discussing in a supply chain environment the h2 the holding cost which we are incurring
at our right hand side of the supply chain, will be more than the holding cost which we
incur at the left hand side. So, as we are coming from left to right, the
holding cost values keep on increasing. We will discuss one more concept about handling
this increasing holding cost because, now we have discuss about the echelon stocks. So, the concept of echelon stock will help
us in understanding the combined optimisation also. But, before we start this combined optimisation
let me go with the separate optimisation. When, we handle the inventories at installation1
and installation2 in separate manners. And that is how we do the inventory management,
when we are handling 2 cases separately, then at installation1 let us say the total cost
of inventory, total variable cost of inventory is being represented by C1, and now C1 is
composed of ordering cost or the setup cost and the holding cost. So, the ordering cost is the number of times
you give the order. Your total requirement is d, which is at the
end of the supply chain. This is the total requirement d, and each
time okay, let us discuss first for installation2, so C2 is our, because we have discussed that
our Q1 means nQ2. So, right now we are discussing the inventory
management for our second installation that we are starting this process. Because, here it was very clear that how we
can apply that basic EOQ model. So, for that purpose we are starting with
our installation 2, so, at installation2, C2 the cost is d by Q2. And, the cost of order per order is k2. So, dk2 upon Q2, is the ordering of setup
cost. Then the second cost is the holding cost. And, EOQ development we have discussed that
it is always paid on the average inventory stock you have. So, Q2 is the total, 0 is the least, so, Q2/2
is the average inventory and the holding cost is h2 per unit per year. So, this becomes the expression for calculating
the total variable cost, and we already know to determine the value of Q2. You need to differentiate it and the apply
principles of maxima minima ,equate that differential equals to 0. And, then you get the value of Q2. So, this will result in to Q2*, that is under
root 2dK2 upon h2. This is the economic order quantity value,
which we should have at installation2 and, this Q2 value if I substitute back into this. If I put Q2 back into this. This will result the C2min. Which will be I request all participants that
they should do this, and put this value of Q2 in this equation. And, then see that what C2min is coming. So, I wait here for a minute, and you can
practice, and you can check that your result should be under root 2dk2 upon h2, so I am
giving you directly the results. But, I request all the participants that if
you substitute these values of Q2, you get this C2. This is at stage 2. Now coming to
second level, we want to determine the value of Q1 also. The Q1 is nQ2 and, for that purpose I write
expression for C1 also. That is the total variable cost at installation1. d because, that is the rate of consumption
of your supply chain so d will remain as it is k2 will be change by k1, and Q2 will be
change by Q1. So, this is the ordering cost or setup cost
at installation 1. Now, comes the part of holding cost. Q2 by 2h2, now when you see you are procuring
the inventory immediately you are left with Q1-Q2. In this case we took Q1 equals to 3Q2. So, as soon as you procure the inventory your
1Q2. You say if it is nQ2. So, out of nQ2, your 1Q2 is gone to second
stage. So, you are left with nQ2 –Q2. You are left with at stage 1. You are left with nQ2 – Q2. And, then finally you reaches to the 0 level. So, the average inventory is at stage 1 is
nQ2 - Q2 divided by 2 that is the average inventory you have. And, that we will write here that is nQ2-Q2
divided by 2. This is the average inventory we are going
to have at stage 1 multiplied by the holding cost of stage 1 h1. And, again now you see, the first expression
is in Q1, the second step of this expression is in Q2. So, we can convert the whole expression in
terms of Q2 only. So, I want to remove this Q1 from my expression. So, it can be written as dK1 upon nQ2 + n
-1 Q2 by 2 h1. So, this is the expression I have for the
total variable cost at installation1. Now, out of this total variable cost Q2, I
have already determined here, so I need to determine only unknown quantity. So, can you have a guess, the only unknown
quantity you can see in this expression is this F. So, I will differentiate this total
variable cost at installation 1 with respect to n. Because, I want to determine the value of
n now, and this when I am putting the value of Q2 what I have received from here, that
I am going to put here. So, n will come, if I calculate the n will
come, and after rearranging my various mathematical terms. n will come k1upon k2 into h2 upon h1. So, this value of n will come, and here though
in this expression and in this model everywhere. We have assumed that the value of n will be
an integer 1, but here when I am doing this differentiation, when I am doing this calculation. I am not taking that integer aspect into consideration. Here, at this moment or I can write this n
as n*. So, when I am writing this n*, this n* is
any kind of fractional values. This can be any kind of fractional value. And, there is a procedure we can follow, since
we have taking this assumption that n will be exactly integer. Now, why n should be in an integer, so that
is every obvious question, and it must come to our mind, that why we need to have n exactly
an integer. The reason for n to be an integer is that
if n is an integer so, our procurement cycle at stage 1. Can be matched with procurement cycle at stage
2. If n is not an integer what may happen that
the procurement cycle at stage 1, will differ in it is timing, then the procurement cycle
at this stage 2. What may happen that either this right now,
you see this straight line is exactly matching with this straight line like this. So, you see as soon as you receive this stock
here, you pass Q2 portion to stage 2, and, this is happening only because n is an integer
in this case. n is an integer in this case only because
of that it is happening here, but if n is not an integer, what will happen in that case,
if n is not an integer in that case, this line will not match with this line. Either this line will be slightly ahead or
it will be slightly backward. That is the problem with this. How, that is be going to the problem. Because, if n is not an integer, let us say
it is 1.5, so you have 1.5 inventory. Which you are ordering if n is 1.5, so you
will order Q1 which is 1.5 times of Q2. And, for that reason if this is how your Q2
cycles are moving, and we have this other pen to show our Q1 cycles. So, here Q1 is 1.5 times of Q2. So, you received Q1 and out of that a portion
of Q1 is gone, so, you are left with Q1-Q2. Which is equals to 0.5 Q2. So, you are left with only 0.5 Q2 here, you
are going here. Now, by the time you order next round of inventory,
so you will go up to 2Q2 this level. Then, you will given of that to the installation2,
and therefore you will be either having more than required inventory yet sometime, and
sometime you will have less than required inventory. So, either you will fulfil the demand of this
by just 0.5Q. Half the requirement of second stage is being made, by the available stock
at installation 1. Or, if you are not willing to do that, then
you will have excess inventories at your installation1. So, because of these problems of mismatch
of inventory management at 1 and 2. We always keep n as integer , n has to be
integer this sometime we call as relaxation of original problem. A language of mathematics we say that we have
done the relaxation of the original problem. Original problem means by doing the differentiation
doing the maxima, minima, and * can be any fractional value, any continuous value. But, since we require only integer value,
so we will see that how do we follow a process to convert these n* into a proper integer. So, now let us see that part where we will
convert n* into a proper integer, so that this model can be properly used. Now, if the value of n*. If n* is less than 1, so your n is 1. But, if n* more than 1, in case of n* is more
than 1 than how to select the values of how to round off, the meaning is how to round
off, the n. So, that we can get the proper scientific
answer of doing the integer values. And, for that purpose we take this value. We consider this value, which is the most
possible, the nearest highest integer which is less than this which is less than this
is the highest possible integer which is less than this. The meaning is if the value of n* if n* let
us say is 2.57, so in that case n* in this bracket becomes 2, and then we also have
this type of relation. So you conceive this type of relationship
between these three values and with the help of comparison of these three values, you decide
whether we are going to have higher integer values or lower integer values. So, with this you will come to know. We will do the comparison with the help of
some numerical data in our next class and with that, we will come to know with that
how this model can be used. Now this is the case when we have done separate
optimisation, this is the case where we have done this separate optimisation and in this
separate optimisation we have only considered one factor that Q1 is dependent on calculation
of Q2, and then we determine the values of n. And, by doing a proper procedure of rounding
off of which we will discuss. We will have Q1 equal to nQ2 and that is and
of the separate optimisation separate calculation of inventory values. But you will see we have not taken the benefits
of supply chain environment in this particular case. We consider two entities separately, and for
any reason there can be a relation of Q1 equals nQ2 and this by doing this mathematical jugglery
we found out this particular process. Now when I go for simultaneous optimisation
I will use the concept of this echelon stock to determine whether I can use this supply
chain environment for my inventory management or not. The point I am trying to say the total cost,
total variable cost of inventory. In case of this inventory is C that is C1
+C2. This is the total cost of supply chain in
a separate optimisation, separate calculation of inventory values. Now, for simultaneous optimisation. I will take you to slightly different calculation
which is embedded in the concept of this echelon stocks. So, now we are going for that simultaneous
calculation. And in this simultaneous calculation, you
will see that we will define earlier the holding cost h was define for the installations. Now, we will define the holding cost for the
echelons, so now the holding cost as we are moving in the supply chain from stage 1 to
stage 2. We said the inventory holding cost h1 and
h2, so h2 will be more than h1, because you are doing some kind of value addition. Now we are defining echelon holding cost which
is represented bye. Now, when I am talking of h2, so already I
had some value addition up to stage 1 and on that value addition up to stage 1. I have already paid holding cost h1, so it
looks logical that I should pay at stage 2, the holding cost only for the amount of value
addition which is being done from the stage 1 to stage 2, So this concept will help me
in giving the concept of echelon holding cost and echelon holding cost e1 since this is
the starting stage of my supply chain. So, even is equals to h1, the holding cost
h1 is coming because of all the values which are added at stage1, but the echelon holding
cost at installation2 e2, will be the result of difference of value you are adding from
2 to 1. And therefore e2 will be h2 - h1, so this
new cost e2 and e1 will help me in getting a more robust model where I will use supply
chain environment for the inventory management. So, now when I have this concept of e1 and
e2. I will like to rewrite the total cost of inventory
in a single expression, I wrote cost of inventory very well cost of inventory in two separate
expressions C2 and C1 and then finally I combined as C equals to C1 + C2. Now, when I have define this two things I
am doing the simultaneous calculation simultaneous optimisation, I will like to write total C
in a single expression and let us write that total C. When we write total C, so obviously there
will be 2 expressions for the holding cost, and there will be 2 expressions for the ordering
cost, and let us see what happens, and how these types of results will give much. Now, when we are doing that, you will see
that we have total cost Which, is the result of whatever, you are
ordering at installation 2, . d by Q2 and k2 + dk2 upon n Q2 So, this relation will
be from the ordering cost. + The holding cost. Holding cost now will not be paid for the
installation. Rather we will have the holding cost given
by the echelon holding concepts. And, for that purpose Q2 by 2 that is e2 +
Q1 by 2 e1. Now, Q1 by 2 can be replaced. One it is nQ2, so this can be written as n-1
by 2 Q1, e1. So, this is the or I can write in a slightly
arranged manner, that it becomes d by Q2 , from these 2 expressions. k2 + k1 by n+ from these 2 steps, you see
you can take sorry, this is Q2 here. This Q2 by 2 you can take out, and this remains
e2 + n-1 e1. So, this is the expression, and when you see
this expression you can very well understand. That this holding cost right now is being
written as e2 and e1 and e1 here. It is actually h2 and h1. And, it will h2, h1. So, it will like this, and then you will have
slightly rearrangement of this expression, it will be Q2 by 2. h2 + n h1 – h1. And, this can be rewritten as Q2 by 2, h2
– h1, will account for e2. So, these 2 expressions will give me my e2,
and h1 is same as e1. So this will become ne1, so my expression
if I simply summarise will be if I just for the sake for clarity, remove these 2 lines. You can combine the expression like this. It is d by Q2, K2 + K1 by n + Q2 by 2 e2 +
ne1. Now, if I see any of my total cost expressions,
and you see this expression. So, this d by Q2 in the number of orders. I am placing again here, this becomes cost
of each order. This expression with in bracket K2 + K1 by
n. This becomes the cost of placing an order. Similarly, Q2 by 2 is the average inventory
level, And, the expression with in this bracket e2+
ne1. This is actually the holding cost for which
I am taking, so now based on our knowledge of EOQ models which we discussed, now if I
rearrange these to get the value of Q2. So, the final expression for my Q2 in case
of simultaneous calculation will be you all can also practise with me, you all can also
write with me, that Q2 will be now under root to d. Then here comes the cost of ordering that
is K2 +K1 by n divided by the holding cost. That holding cost is this e2 + ne1.so, this
is the new economic order quantity at Q2 level, and this takes here, because of the echelon
concept this takes here, of the supply chain environment. So, we will see the use of this formula with
the help of a numerical data, in our next class, and I request you to please do the
differentiation of this expression, those who are interested in mathematics. Do the differentiation of this expression
with respect to Q2, and try to get this formula of Q2. Though we have used our past one is of EOQ
write in this formula, but you can try a fresh, so that you can get this formula. Thank you very much.
