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I have found a solved example of A Stochastic Two-Period Model with No Setup Cost in the book Operational Research by Hillier, 7th edition, that has a lot of complicated calculations to arrive to the solution.

In the following example, $c=$cost of ordering, $h=$cost of holding, $p=$shortage cost and $𝑦^0_𝑖$ is the optimal order-up-to level in period 𝑖.

enter image description here

The part where things start to complicate, is when making the substitution of $y_1^0$ into $C_1(x_1)$.

Clearly will not be an easy calculation by hand since $C_1$ has this function $L$, that it's evaluated with diferent arguments, which are two integrals.

I am not sure what form the argument of $\displaystyle \min_{y_1\ge x_1}\{\dots\}$ will have, nor what would be best to do next. My book mentions this OR Courseware but since I have the pdf file book I don't have the disk? is refering to.

And my question is what can I do here in order to find the optimal $y_1^0$ without having to deal with lots and lots of calculations by hand?

Any suggestions are very appreciated.

enter image description here

The $C_1(x_1):$ enter image description here

The $L(z):$ enter image description here

edit.

Following the comment of Larry, I substitute $y_1^0$ in the equation equal to zero (below the $C_1(x_1)$ definition in the image).

I assumed $y_1^0=5$ to be the optimal and this is what I got

$$-15+(15+10)\frac{5}{10}+(10-15)\Phi(5-2)+(15+10)\int_0^{3}\Phi(5-\xi)\phi_D(\xi)d\xi $$ $$=more\ calculations$$ $$=-\frac {11}{8}$$ This is different than $0$, which I don't understand why, the equation should have had the value $0$ because the book mentioned the optimal was $y_1^0=5$ and not $y_1^0=6$. Also the optimal should satisfy the equation equal to zero, but does not.

I really don't see what am I doing wrong? I did check twice the calculations of the expression and found no errors.

$\star$ Note: I asked this question on MSE https://math.stackexchange.com/questions/3246019/why-the-optimal-value-that-minimizes-a-function-does-not-satisfy-condition and currently has a bounty (grape period bounty to be exact) and I asked to migrate but it was not possible :(

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    $\begingroup$ This question is very hard to follow. You haven't defined the notation (e.g., $y_1^0$) or provided sufficient context. The actual questions you are trying to ask are sprinkled throughout the post and are not that clear. I would encourage you to edit the question to make the question more concise and easy to follow. Nevertheless, I will try to answer and see whether that helps. $\endgroup$ – LarrySnyder610 Jun 11 '19 at 2:47
  • $\begingroup$ @LarrySnyder610 Hi Larry, the context (the $y^0_1,y^0_2$) is the same as the other question of mine that you answered. I'll edit anyway. $\endgroup$ – user441848 Jun 11 '19 at 3:02
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    $\begingroup$ I know that. But other readers of this question will not know that. You didn't even link to the original question. Remember that SE sites are supposed to be building a body of knowledge that will potentially be of use to many people, not just the individual person asking the question. $\endgroup$ – LarrySnyder610 Jun 11 '19 at 3:12
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The text from the book tells you that the optimal value of $y_1$ is $y_1^0 = 5.42$. This comes from solving the optimality condition (the equation after "satisfies the equation") for $y_1^0$.

Presumably, somewhere in the example it says that the base-stock levels must be integers. Therefore, they plug the neighboring values, $y_1^0 = 5$ and $y_1^0 = 6$, into the objective function to see which one gives the better value. (This relies on the objective function being convex.) They found that $y_1^0=5$ is better. But it is not surprising that when you plug $y_1^0=5$ into the optimality condition, you don't get $0$; $y_1^0 = 5$ is only optimal because it has to be an integer.

(Similarly, if you want to minimize $(x-\frac14)^2$, the optimality condition is $2(x-\frac14)=0$; if $x$ must be an integer, you'd plug in $x=0$ and $x=1$ and pick the better value; but if you plug that value into the optimality condition, the LHS won't equal $0$.)

Now, to calculate the objective function for $y_1^0=5$ or $6$, I think they (or you) are skipping some logic here. The optimal policy is a base-stock policy, which means that you are looking for the optimal base-stock level, $y_1^0$. The cost of a given base-stock level $y_1$ is not given by calculating $C_1(y_1)$; it is given by setting $x_1 = $ something small (e.g., $-10$) and then calculating what's inside the $\min_{y_1\ge x_1}\{\cdots\}$ for that value of $y_1$. The idea is, assuming the inventory level $x_1$ is small enough that we will place an order in this period, what will be the cost if we place an order up to $y_1$?

The line in the book "substituting $y_1^0=5$ and $y_1^0=6$ into $C_1(x_1)$ seems misleading to me because it sounds like you're supposed to set $x_1=5$ and $6$, but that is wrong.

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