Understanding Hadley and Whitin's Reorder Point Formula for Logistics

[jacophile]

Hi, I am trying (in vain) to understand a common result in logistics due to Hadley and Whitin.

It is about the variance of the reorder point process in an inventory management system.
It is assumed that the demand X during lead time has a normal distribution with mean E(X) and variance Var(X). Then, the re-order point is set equal to:

R = E(X) + Z StdDev(X)

Where
X is a RV representing the demand for one lead time
R is the re-order point i.e. the inventory level at which a replenishment order should be placed
Z is the inverse CDF of the desired service level
The service level is the probability that there will be sufficient stock to meet demand

This makes sense to me because it transposes to

$$SL = F(\frac {R - x} {StdDev(X)})$$

so R is the mean of this process and Z.StdDev(X) is the safety stock

The lead time is the time it takes for a replenishment order to arrive and is also a RV independant of the demand

So,

X = d x LT

Where
d is the demand per unit time process (a RV) and
LT is a RV representing the lead time and
d an LT are independant

So my understanding is that

$$E(X) = E(d)E(LT)$$
$$Var(X) = Var(d)(E(LT))^2 + Var(LT)(E(d))^2 + Var(d) Var(LT)$$

The result that is normally given is (due to Hadley and Whitin 1963)

$$R = E(d)E(LT) + Z \sqrt {E(LT)Var(d) + E(d)^2Var(LT)}$$

But I can not see why.

can anyone help me out?

 

[jacophile]

Hmm, I guess my post was too confusing. Any way I think I worked it out..

Just to re-state the problem (hopefully more clearly),

What is the distribution of the sales demand for a stock item during the replenishment lead time (DLT), given that the daily demand D and the lead time N are independent random variables and the sequence of daily demand D, during the lead time period are iid.

The replenishment lead time N is the number of days between placing an order for more stock and actualy taking delivery of it. The general idea is to have enough stock buffer to cover the variance in D and N.

I was trying to imagine how to multiply D and N but I realized that the demand during lead time (DLT) is actually a sum of a random number of random variables:

$$DLT\;=\;\sum_{i=1}^{N}D_i$$

Then I discovered the characteristic function and moment generating functions (I had no idea about these things so it was a massive eye-opener...) by which I understand I can proceed as follows:

$$Define \;R \equiv e^{DLT \; t}$$

$$E(R \mid N=n)\;=\;E(e^{DLT \; t} \mid N=n) \; = \; E(e^{(D_1+D_2+...+D_n) \;t}) \; = \; (M_D(t))^n \; \;$$

where $$M_D(t)$$ is the moment generating function for D

Now let the condition be a rv so that the expectation above also becomes a rv defined as

$$E(R \mid N) \;= \; (M_D(t))^N$$

so the mean of this rv is

$$E(E(R \mid N)) \; = \; E(R) \; = \; E((M_D(t))^N)$$

Also,

$$E(R) \; = \; E(e^{DLT \; t}) \; = \; M_{DLT}(t)$$

by the definition of the mgf.
This leads to

$$M_{DLT}(t)\; = \; E((M_D(t))^N).$$

I had no trouble convincing myself that the zero'th, first and second derivatives of the mgf are the zero'th, first and second moments about zero, so now I can go ahead and differentiate

$$M_{DLT}'(t)\; = \; E(N M_D^{N-1}(t) M'_D(t))$$

$$M_{DLT}''(t)\; = \; E(N(N-1) M_D^{N-2}(t) M'_D(t) M'_D(t)+N M_D^{N-1}(t)M''_D(t))$$

It is then straight forward to show that

$$E(DLT) \;=\; M_{DLT}'(0)\; = \; E(D)E(N)$$

and that

$$Var(DLT) \;=\; E(N^2)Var(D)+E(N)Var(D)$$

which is exactly the result I was confused about.