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

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.

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– EhsanK
Nov 12, 2021 at 2:17

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.