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 $y_2^0$.


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

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    $\begingroup$ 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. $\endgroup$ Jun 7, 2019 at 1:55
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    $\begingroup$ (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 $y_1^0$? $\endgroup$ Jun 7, 2019 at 1:55
  • $\begingroup$ 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) $\endgroup$
    – user441848
    Jun 7, 2019 at 2:02
  • $\begingroup$ @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 $\endgroup$
    – user441848
    Jun 7, 2019 at 2:05
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    $\begingroup$ It would be good to add these definitions, explanations, and citations to the question itself. $\endgroup$ Jun 7, 2019 at 2:43

1 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.

  • $\begingroup$ 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? $\endgroup$
    – user441848
    Jun 8, 2019 at 17:44
  • $\begingroup$ 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 ? $\endgroup$
    – user441848
    Jun 8, 2019 at 18:00
  • $\begingroup$ That depends on whether you need integer solutions or not, which in turn depends on the problem statement and/or your instructor. $\endgroup$ Jun 9, 2019 at 3:56
  • $\begingroup$ Yes, integer solutions. $\endgroup$
    – user441848
    Jun 9, 2019 at 17:52
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    $\begingroup$ That sounds frustrating, but SE is not a site for open-ended homework help, so at this point you'll have to rely on your instructor and other resources, I'm afraid. $\endgroup$ Jun 11, 2019 at 0:02

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