When applying the base stock inventory policy, assuming the daily demands are normally distributed with parameter $(\mu, \sigma)$, we can find the optimal parameter $S$ (the base stock level) in several different ways: (say we have 0 lead time and review period is 1 unit period, and assume infinite horizon)

  1. From the holding cost/stockout cost criterion: if holding cost per item is $h$ and stockout cost per item is $p$, then $S = \mu + \sigma\Phi^{-1}(\frac{p}{p+h})$, where $\Phi$ is the cdf of the standard normal distribution.
  2. If the stockout cost $p$ is difficult to estimate for the firm, then a service-level-based approach is used, in particular, the two most basic types of service levels are Cycle Service Level (type 1 service rate) and Fill Rate (type 2 service rate): To achieve a type 1 service level of $\alpha$, we simply have $\alpha = \Phi((S-\mu)/\sigma)$, so the base stock level $S = \mu + \sigma\Phi^{-1}(\alpha)$.
  3. For a type 2 service level, the calculation is more complicated. The usual formula for approximating the fill rate is $\beta = 1-\frac{n(S)}{\mu}$, where $n(S) = \sigma \mathcal{L}(z), z=(S-\mu)/\sigma$, and $\mathcal{L}(z)$ is the standard normal loss function (see e.g. in the appendix of this book).

The first two approaches give the base stock level a nice structure: $\mu$ is the cycle stock to cover the average demand in lead time and review period, while $\sigma \Phi^{-1}(\alpha)$ is the safety stock to buffer the fluctuations in lead time demand, and we have a rather simple description of the relation between the safety stock and the service level $\alpha$. However, when using the fill rate approach, the base stock level $S$ is found by solving the nonlinear equation $\beta = 1-n(S)/\mu$, we can still compute the associated base safety stock level $ss = S-\mu$, but we no longer have a simple description on the relationship bewteen the safety stock level, the fill rate $\beta$ and the demand standard deviation $\sigma$.

So my question is: is there any approximate formula/asymptotic expression (as $\beta$ approaches $1$) that gives the rough relation between the safety stock level $ss = S-\mu$ and the fill rate $\beta$ when $S$ is found by solving the fill rate constraint?


2 Answers 2


We can work from scratch. Rearrange $\beta=1-n(S)/\mu$ to get $$-\frac{\mu\sqrt{2\pi}}\sigma(\beta-1)=e^{-z^2/2}-z\sqrt{2\pi}+z\int_{-\infty}^ze^{-t^2/2}\,dt$$ on using ${\cal L}(z)=\phi(z)-z(1-\Phi(z))$ and $\Phi'(z)=\phi(z)=e^{-z^2/2}/\sqrt{2\pi}$.

To determine the inverse asymptotic behaviour as $\beta\to1^-$, we now expand the integral term about infinity to obtain the Laurent series of $\Phi$, which is possible by expanding the reciprocal about zero. This gives \begin{align}y:=\frac{\mu\sqrt{2\pi}}\sigma(1-\beta)&=-e^{-z^2/2}\sum_{n=1}^\infty\frac{(-1)^n(2n-1)!!}{z^{2n}},\end{align} where for simplicity, we can cut the sum at the first term as $\mathcal O(1/z^4)$ decays much faster than $1/z^2$. Thus the RHS can be approximated as $f(z)=e^{-z^2/2}/z^2$ with inverse $$f^{-1}(z)=\sqrt{2W\left(\frac1{2z}\right)}$$ where $W$ is the Lambert $W$ function. This can be reduced to elementary functions through the asymptotic $W(z)\sim\log z-\log\log z+o(1)$, so the safety stock is asymptotically equal to $$\textsf{SS}=z\sigma\sim\sigma\sqrt{2\log\left(\frac\sigma{2\mu\sqrt{2\pi}(1-\beta)}\right)-2\log\log\left(\frac\sigma{2\mu\sqrt{2\pi}(1-\beta)}\right)}$$ which indeed tends to infinity as $\beta\to1^-$.

  • $\begingroup$ Quick question...it seems that this only works when $\beta$ gets very close to $1$, right? When $\mu$ is relatively big and $\beta$ goes significantly below $1$, like e.g. $0.95$, this seems to be not defined. Any comments on that? $\endgroup$ Oct 5, 2022 at 17:25
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    $\begingroup$ @Steven01123581321 Yes. If we wish to be more realistic and take $\beta\approx0.9$ we can use the same method illustrated by expanding about $0.9$ instead of $1$. $\endgroup$ Oct 9, 2022 at 9:32
  • $\begingroup$ okay, I get it. Would help the answer though if you would add that note, maybe with an example. $\endgroup$ Oct 10, 2022 at 6:40

You can solve:

$\beta = 1-\frac{n(S)}{\mu}$ wrt $\mathcal{L}(z) $

which gives you:

$$\mathcal{L}(z) = \frac{\mu\times(1-\beta)}{\sigma}$$

Then there is a a good approximation function (don't have the specific reference to it) that can convert $\mathcal{L}(z)$ to $z$:

$$z = 4.85-\mathcal{L}(z)^{1.3}\times0.3924-\mathcal{L}(z)^{0.135}\times 5.359 $$

Then the safety stock can be found as:

$$ss = \sigma \times z $$


$\beta$ or fill rate $= 0.98$, $\sigma = 5$ and $\mu=30$

$$\mathcal{L}(z) = \frac{30\times(1-0.98)}{5}$$

$$\mathcal{L}(z) \approx 0.12$$


$$z = 4.85-0.12^{1.3}\times0.3924-0.12^{0.135}\times 5.359$$

$$ z \approx 0.8 $$


$$safety\space stock = 5\times 0.8 $$


  • $\begingroup$ While this fits well for a range of small values of $1-\beta$ (I believe you used a numerical solver for this), it loses the asymptotic behaviour as $\beta\to1^-$. In particular, the safety stock should tend to infinity but your approximation tends to $4.85$. $\endgroup$ Nov 22, 2021 at 10:27
  • $\begingroup$ you're mathematically right, but in practice this asymptotic behavior is not desired. That's why this transformation is standard inventory control literature (assuming a normal distribution). Be aware that the approximation tends to 4.85 as being the z-value of a standard normal distribution, which is near to 1 for the CDF. $\endgroup$ Nov 22, 2021 at 10:57
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    $\begingroup$ I agree, I tested your function and it matches up to $\beta\le0.999$. I'd be interested to see a reference for the approximation equation. $\endgroup$ Nov 22, 2021 at 10:59
  • $\begingroup$ I know there is some reference, but I need to free up some time to dig in the books :) If I ever come across it, I'll make sure to edit the answer with the reference. $\endgroup$ Nov 22, 2021 at 12:28

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