LOT SIZE, REORDER POINT SYSTEMS (Q-R
POLICIES)
(Lecture held on 23/10/2002)
Assumptions for modeling
this system are as follows: -
1) The system is in continuous review.
2) Demand is
random and stationary, in other words the demand is independent of time and it
fluctuates around a stationary mean, hence there are no trends or
seasonalities.
3) There is a fixed positive lead time t for placing an order (random lead time can be other
assumption).
4) The costs are as follows: -
a) Set up cost at $ k per order.
b) Holding cost at $ h per unit held per year.
c) Stock out cost
at $ p per unit of unsatisfied demand.
It is noteworthy to mention
here that the concept of stock out costs comes into picture because the demand
is purely random and therefore there is always some probability of unusual
large demand, which might lead to unsatisfied demand as well.
Let the variable D
corresponds to the random demand during the lead time. It must be kept in mind
that D is demand only during the lead time and not for the entire cycle.
Let E(D) =
μ and variance (D) =
σ2
The variation of inventory
at any time t versus time is shown below:
Inventory R i i + 1 t p
Time (t)
Q
R is the reorder level of
the inventory. Whenever the inventory decreases to R, an order Q is placed.
Since the lead time is t, the stock Q comes at time t after the placement of the order. p
is one such reorder point corresponding to the ith period. Since the
demand during the lead time is random, the consumption of stock during the lead
time will be different for every period and therefore, the maximum inventory
will vary. The upper bound on the inventory is Q+R, and it occurs when the
demand during lead time is zero.
The aim of the whole
analysis is to find the best value of R and best value of Q. In other words we
would like to know when to order and how much to order.
Let E(D)
be the consumption during the lead time.
This
is also known as safety stock, i.e., the stock sufficient to prevent against
stock loss.
Let
us look at a particular cycle and calculate E(D), R,
the best values of Q & R.
If
Consumption
during the lead time =
Let
S be the remaining inventory when the order comes.
i.e. S = R – D
S
can be positive as well as negative,
Let
S+ is the value of the positive inventory, when
the order comes.
Mathematically,
S+ = max (S, 0)
i.e.,
if the inventory is positive then S+ = S, otherwise S+ =zero.
Clearly S+
Similarly
S – is the magnitude of the negative inventory
(stock loss), when the stock comes.
Mathematically,
S – = max (-S, 0).
Also,
S –
If
you look into the figure, you will find that the inventory for a particular
period i+1 varies from Q +
Let us suppose that the
stock outs are kept very small, then S is negative very rarely and therefore we
can safely assume that
E(S+)
It can be easily shown that
this the lower bound on the average inventory.
Average
inventory =
and,
E(S+)
Hence, making this
assumption is actually equivalent of getting the lower bound on the average
inventory, which is typically very close to the exact value.
Total cost in a given cycle
= Set up cost+ Holding cost + Penalty cost
= k +
h[
Penalty or backordering cost = p E(S –) = p
n(R)
Where, n(R) = E(S –) =
As it has been mentioned
earlier that S – = max (-S, 0),
In
other words S – =
Here,
f(x) is the pdf (probability distribution function) of demand during the lead
time.
So
we know both the expected value and the pdf of demand during the lead time, and
therefore the following relation should hold.
E(D) =
For
e.g., if the demand follows the Poisson distribution,
then, f(x) =
and,
E(D) =
Amount
of backorder in this case will be =
Our
next step is to find average cost per unit time, which is equal to the cost per
cycle divided by the cycle length. In order to calculate the cycle length, we
consider n (a large number) cycle lengths T1, T2, ………, Tn. We must appreciate it that though the
amount of stock consumed in a particular cycle is not Q, but if we consider a
large number of cycles (n), then the total amount of stock consumed in these n
cycles must be approximately equal to nQ.
i.e.,
or,
i.e,
the average cycle length =
Expected cost per unit time
= Total cost in a given cycle / Cycle length
To get the optimal Q we set,
And,
P (D
Where
F (R) is the cumulative distribution function (cdf) of D.
Suppose,
the stock out is just not possible, i.e., p
Q* =
[n (R) must tend to zero otherwise G ( Q,R) will become
infinite]
These
two equations together form a recursive equation, and therefore one has to
iterate forth and back to get the value of Q*. The following methodology can be
adopted in this case:
a) Find Q0 = EOQ =
b) Find R0 from 1 – F (R) =
c) Use Ri to
find Qi+1
d) Use Qi+2 to find Ri+1
In a few iterations, Qi will converge to Q* and Ri to R*.
In this way, we can
calculate the optimal reorder point as well as the reorder quantity. In step
(b) we have to use the value of n (Ri)
to get the value of Qi+1, but for that we nneed the pdf of the demand during the
lead time (D). One practical assumption would be to take demand as normally
distributed. This can be explained by the theorem of probability, which states
that if there are n independent random variables then the pdf of their sum is
normally distributed.
n (R) =
f(x) is the pdf of the demand during the lead time and according
to our assumption, it is normally distributed with mean μ and variance
σ2 .
i.e. f(x) =
then, we may re-express n (R) as
n (R) =
Put
t =
We
get, n (R) =
Where,
L(r) =
We
can get the value of L(r) for a particular value of r and vice-versa from the
standard tables (given at the back of any probability book). Hence, to
calculate n (R) we should first calculate
Annual interest rate = 20%
Loss of goodwill for every
jar of mustard not supplied (stock loss) = Rs 25/-
Fixed cost = Rs 50/-
Lead time = t = 6 months
Given E(D)
= 100 jars, σD = 25 and λ =
200jars per year
Also assume that D is
normally distributed; calculate Q* and R?
Answer
a) Q0 = EOQ =
b) 1 – F (R0) =
Or,
P
Now,
P
The plot shows the
"bell" curve of the standard normal pdf, with µ = 0 and
Area = 0.04
We
get
and, n(R0) = L (
= 25 L (1.75) = 25 X 0.0162 = 0.405
h=
20% of Rs 10/- = 2/-
Q1 =
Continuing
the iterations we get R1 =
143, and Q2 =110.8
and , R2 =143 . We can see here that the convergence is
reached, as values of R1 and
R2 are same (= 143).
Therefore,
Q* = 110.8, and R = 143.
Probability
that no stock loss occurs in a cycle = P (D
Proportion
of demand that is stock out =
Even
though there is more than 4% chance of stock loss but the proportion of demand
that is stock out is just 0.4%
TIME TO THINK One might think
intuitively that a basic flaw in this system is that a reorder quantity Q is
placed at every reorder point, and the inventory stock at the end of the lead
time increases by Q irrespective of the inventory level at that time. Another
sensible way of managing the system would be to forecast at the reorder point
p, the inventory the system will have at the end of lead time t, and place an order Q’ such that maximum inventory
in the system remains more or less constant. In other words Si’ + Qi’ = k
(constant). Where Si’ is the forecasted demand at the reorder point for the end
of lead time, and Qi’ is the amount of stock ordered at reorder point of period
i. However, whether the latter case is better than the former or not, is still
to be proved.