# Safety Stock with Fill Rate Criterion

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?

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 $$\beta\to1^-$$.

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:

$$ss = \sigma \times z$$

Example:

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

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

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

then

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

$$z \approx 0.8$$

and

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

$$\approx4$$

• 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 $4.85$. Nov 22, 2021 at 10:27
• 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. Nov 22, 2021 at 10:57
• 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. Nov 22, 2021 at 10:59
• 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. Nov 22, 2021 at 12:28