One and two period policy for inventory situation

Question

The following exercise is in the book Operational Research by Hillier, 7th edition, page 978.

In this exercise $p$ and $p$ are the stockout and holding cost parameters, respectively. $𝑦^0_i$ is the optimal value for the order up to level. $\alpha$ is the parameter $\lambda$ of the exponential distribution. In this case $\lambda=\frac{1}{25}$. If I understand correctly, period $1$ is the second period.

Let $y^0_i$ be the optimal order-up-to level in period $i$.

19.7-1. Consider the following inventory situation. Demands in different periods are independent but with a common probability density function given by $$\varphi_D(\xi)=\begin{align}\begin{cases}\frac{e^{-\xi/25}}{25}&\quad\text{for}\,\,\xi\ge0\\0&\quad\text{otherwise}\end{cases}\end{align}$$ Orders may be placed at the start of each period without setup cost at a unit cost of $c=10$. There are a holding cost of $6$ per unit remaining in stock at the end of each period and a shortage cost of $15$ per unit of unsatisfied demand at the end of each period (with backlogging except for the final period).

(a) Find the optimal one-period policy.

(b) Find the optimal two-period policy.

The procedure for finding $y^0_1$ reduces to a simpler result for certain demand distributions. We summarize two such cases next. Suppose that the demand in each period has an exponential distribution.

Then $y_1^0$ satisfies the relationship $$(h+c)e^{-\alpha(y_1^0-y_2^0)}+(p+h)e^{-\alpha y_1^0}+\alpha(p+h)(y_1^0-y_2^0)e^{-\alpha y_1^0}=2h+c.$$ An alternative way of finding $y_1^0$ is to let $z^0$ denote $\alpha(y_1^0-y_2^0)$. Then $z^0$ satisfies the relation $$e^{-z^0}\left[(h+c)+(p+h)e^{-\alpha y_2^0}+z^0(p+h)e^{-\alpha y_2^0}\right]=2h+c,$$ and $$y_1^0=\frac1\alpha z^0+y_2^0.$$ When the demand has either a uniform or an exponential distribution, an Excel template is available in your OR Courseware for calculating $y_1^0$ and $y_2^0$.

Attempt.

I first found $y^0_2$ using $$\Phi(y^0_2)=\frac{p-c}{p+h}$$ and the definition of $$\Phi(x)=1-e^{-\frac{1}{25}x}.$$ Then I found $y^0_2=-6.8$.

How do I interpret this value (since it is negative)?

Then, as the book mentions, I tried to find $y^0_1$ using the relationship but the calculations started to get complicated, so I used Mathematica and the results were $$y^0_1= -41.2896\text{ and }y^0_1=25.064.$$

Answer 1

I think you simply made a mistake in the sign of one expression. You already figured out that: $$\Phi(y_2^0) = \frac{p-c}{p+h}.$$ So: $$\begin{align} &1 - e^{-\frac{1}{25}y_2^0} &&\hspace{-1cm}= \frac{5}{21} \\ \iff &e^{-\frac{1}{25}y_2^0} &&\hspace{-1cm}= \frac{16}{21} \\ \iff &-\frac{1}{25}y_2^0 &&\hspace{-1cm}= \ln\frac{16}{21} \\ \iff &y_2^0 &&\hspace{-1cm}= 6.7983. \end{align}$$ Now use Mathematica to determine $y_1^0$ as you did before.

By the way, it is possible to have a negative order-up-to level (i.e., if $y_2^0$ really did equal $-6.8$). It just means that we are operating in "backorder" mode, and we order enough to reduce the number of backorders to $6.8$. It's an unusual situation but it's mathematically possible and perfectly valid in a model like this.