# One and two period policy for inventory situation

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 $$y^0_1= -41.2896\text{ and }y^0_1=25.064.$$

• Hi @user441848, and welcome to OR.SE. A few things about this question: (1) Please define the notation you use. I assume $p$ and $h$ are the stockout and holding cost parameters. What is $\alpha$? What are $y_1^0$ and $y_2^0$? You say the optimal value, but the optimal value of what? the order quantity? the order up to level? something else? And is period 1 the first period or the second? (Different inventory models number the periods in different ways.) (2) Please cite the book that you are copying and pasting from. Jun 7, 2019 at 1:55
• (3) It's really not clear what you are asking. The only question I see is "How will I interpret this value?" Are you also asking something about the calculation of $y_1^0$? Jun 7, 2019 at 1:55
• Hi @LarrySnyder610 thanks. The book I am using is Operational Research by Hillier 7th edition. 𝑝 and ℎ are indeed the stockout and holding cost parameters. $y^0_i$ is the optimal value for the order up to level. If I understand correctly, period 1 is the second period. (This is on page 978) Jun 7, 2019 at 2:02
• @LarrySnyder610 Well yes there is only one question because I got the same problem for $y^0_1$ so solving for $y^0_2$ will help me with the other optimal value Jun 7, 2019 at 2:05
• It would be good to add these definitions, explanations, and citations to the question itself. Jun 7, 2019 at 2:43

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.
• Thank you. I use Mathematica and now the results were ${{x -> -37.3142}, {x -> 23.2932}}$ but there was a note that stated Inverse functions are being used by Solve, so some solutions may not \ be found; use Reduce for complete solution information. I think I should choose $y^0_2= 23.2932$. What do you think? Jun 8, 2019 at 17:44
• Also Larry, should I take $y^0_2=7$ and $y^0_1=23$ so that I can write appropiately the optimal policy? In the book there was an example related to this. Where analyze each case as what happens with value $y^0_2=6$ and then what happens with value $y^0_2=7$. Is it really necessary ? Jun 8, 2019 at 18:00