determine optimal print-run size, given 500 units sold, and

[bjackson11]

OK....here is a homework problem that I have no idea how to get started on. Please help.

Inventory Cost

An author anticipates that he will sell 500 copies of his book annually. The set up cost for each print run is $300. The storage cost for each printed book is $1.20 per year. Assuming that the customer demand for the book is constant of the 250 business days of a year, determine the size of each print run that will minimize his inventory cost.

 

[galactus]

Re: Derivative Applications

This appears to be an Operations Research problem?.

They do not mention a leads time so I suppose we can assume it is 0.

First, find y*, which is the number of units for the optimum inventory policy.

\(\displaystyle y*=\sqrt{\frac{2KD}{h}}\)

Where K=300, D=500, h=1.20

Order y* units when the inventory drops to 0.

That is how I interpret it.

The total cost per unit time is \(\displaystyle TCU(y)=\frac{DK}{y}+\frac{hy}{2}\)

Order y* units every y/D time units.

 

[bjackson11]

Re: Derivative Applications

In class, we have just started learning about applications of maxima and minima. In this problem I think they are wanting to know what the minimum number of books for an order should be to save money on inventory costs. In my student solutions manual they have the following equations;

S(x)=1.20(x/2)

I don't understand where the 2 is coming from.

Also they have

C(x)=300(500/x)

I(x)=S(x)+C(x)

Finding the derivative from there and then finding the minimum I can do on my own. I just don't understand how or where the x/2 is in the S(x) equation. Any thoughts?

 

[galactus]

Re: Derivative Applications

That's just because it's an average.

Average Inventory level = y/2 units

If you notice, what I gave you should give the same result.

\(\displaystyle \frac{dTCU(y)}{dy}=\frac{-KD}{y^{2}}+\frac{h}{2}=0\)

Solve this for y and we get \(\displaystyle y=\sqrt{\frac{2KD}{h}}\)

 

[bjackson11]

Re: Derivative Applications

It does come out the same. I'm still a little confused on how to use 2 as an average. thanks for your help