Derivation of safety stock formula of an inventory model with probabilistic demand

Question

I am applying the "Probabilistic inventory model with safety stock". There are some assumptions in this model: normally distributed demand, fixed lead time (time between placing the order of new items and their arrival). The next figure shows it clearly:

And the formula to calculate the number of items to order is the next:

Where $d$ is the average demand, $T$ is the time between orders, $L$ is the time that takes the new order to arrive, $z$ is the number of standard deviation for a certain service level, $\sigma_{T+L}$ is the standard deviation of the period $T+L$, and $I$ is the current stock.

My concern is that the multiplication of $z$ and $\sigma_{T+L}$ is called "Safety Stock", but I don't know why you should multiply these two values. I was thinking that it is just a rearrangement of values of the formula to calculate the probability of a certain value of a normally distributed random variable. In the next image I wrote my thoughts:

Am I correct or am I misunderstanding the topic? Thanks a lot in advance and sorry if I just pasted images since I don't know how to type formulas directly.

Answer 1

You are thinking along the right lines. There are various ways to set the safety stock, but in this case the logic is something like what you said.

When you place an order at time $t$ (say, at the first "Place order" point in your figure), that order has to last you until the end of the lead time for your next order. The total demand during those $T+L$ periods is a random variable (call it $D$) that has mean $\bar{d}(T+L)$ and standard deviation $\sigma_{T+L}$.

If you set $Q$ as in your formula, then when the next lead time ends, the inventory level will be $Q+I-D$, since the inventory after we order is $Q+I$. (I'm skipping some details of the inventory dynamics here.) The inventory level set by your formula will be sufficient to meet the demand iff $D \le Q+I$, and the probability that this happens is $$P(D \le Q+I) = \Phi\left(\frac{Q+I - \bar{d}(T+L)}{\sigma_{T+L}}\right) = \Phi(z).$$ Suppose we want the probability of not having a stockout during the lead time to be $\alpha$. Then we set $z = z_\alpha \equiv \Phi^{-1}(\alpha)$. So the probability of not having a stockout is $$\Phi(z_\alpha) = \alpha,$$ as desired.