Time Value of
Money
Σ ck
αk
K=1
If no discounting then
α=1
Average cost:
Choose the policy that
minimizes the average cost.
Figure 1:
No backlogging allowed
Making Decision: When to order and how
much to order.
Aim: to come up with the
best possible policy that minimizes average cost.
If backlogging
then:
Figure 2:
Backlogging Allowed
One thing is clear. We never
want to order at point A (refer Figure 1)
So lets consider the
following figure:
Figure 3:
Another Modification
In this if the λ rate line touches
the X-axis i.e. ordering when inventory becomes zero then it minimizes the cost
and hence savings.
N(t) + h ∫ I(t)
dt
No of times i
order
t
So, we narrow our search to
something like this:
Now, intuitively we can say
that the above triangles should be equal.
Lets take the following
example:
After two triangles, it
repeats. So for a fixed cycle time, if h1 and h2 are the heights of the
triangles, then for minimum area, each triangle should be of height
(h1+h2)/2
Alternatively, if A1 and A2
are fixed, then for the total area
be fixed, each should have an area of (A1+A2)/2 (here we have increased
the cycle time)
So, now we have:
Q
Q/λ
Cost/Cycle
K + hQ2
Fixed cost 2λ
We want cost/time:
K
+ hQ2/2λ
Q/λ
Q* =
√2kλ/h