# Help ; Probabilistic inventory models

• Mar 26th 2013, 11:16 PM
Trish
Help ; Probabilistic inventory models
The demand for printed books is a random variable D with distribution approximated by a continuous uniform distribution on the interval[20,60]. Each book printed costs \$15 to produce and sells for \$33. Any books left unsold will be disposed of at a cost of \$10 per book plus a fixed disposal fee of \$80. If a copy of the book is unavailable then a copy will be fast printed but this will cost the bookshop \$50 instead of the usual \$15. How many books should be printed at the start to minimise expected costs?
• Mar 27th 2013, 06:31 PM
chiro
Re: Help ; Probabilistic inventory models
Hey Trish.

Can you show us what you have tried? (Hint: Can you get random variables for each of your quantities including unsold books, fast printed books, and sold books with an equation for expected costs as a function of these random variables?)
• Mar 27th 2013, 08:30 PM
Trish
Re: Help ; Probabilistic inventory models
Hi Chiro.

This is what I have.

Let q = quantity printed d = demand
q - d = unsold books
10(q - d) + 80 = cost of disposing of unsold books
d - q = fast printed books needed
50(d-q) = cost of fast printed books
Expected total cost = 15q + 50(d-q) + 10(q-d) + 80
Now, my understanding is that I have to integrate this function between 20 and 60 but I am not really sure.
Thanks
• Mar 27th 2013, 10:21 PM
chiro
Re: Help ; Probabilistic inventory models
The thing is that you have to quantify what attribute you are trying to find.

Is the expected value (mean)? Perhaps you want a confidence interval that corresponds to some probability (85% of all values two sides of the mean?)
• Mar 28th 2013, 02:35 AM
Trish
Re: Help ; Probabilistic inventory models
I'm sorry but you have lost me. I know that the mean is 20 and the standard deviation is 60. Also the expected demand is the mean, therefore 20. I don't know what a confidence interval is.
• Mar 28th 2013, 05:13 PM
chiro
Re: Help ; Probabilistic inventory models
Because you have a random variable, it means you can have a range of possible solutions.

You can construct a confidence interval that corresponds to a range of values that satisfies some probability.

A 90% interval would for example capture the mean and 45% of the values either side of the mean and you would get an interval (a,b) for that probability.

Alternatively you could just get the average (expectation) and use that.

So you have two choices: get a single value (mean) or a interval value (probability).