Consider the problem of Inventory Management for a new retail refrigerator store. The store manager needs to consider at least two things: the maximum size of inventory that can be stored on site, say I, and the number of shipments of new refrigerators per year, say N, as required to keep inventory on hand. One usually assumes that each shipment of refrigerators will arrive just as the existing inventory runs out, so each shipment would have exactly 'I' refrigerators. There are costs associated to how the inventory is managed, apart from the costs of the actual refrigerators. Let's suppose that the shipping company charges a flat fee of $ 2000 per shipment (remember, we're assuming all the shipments contain 'I' refrigerators.) Another cost is storage: let's assume storage costs are $ 20 per square foot per year and that each refrigerator stored takes 5 square feet of space (remember, there are never more than 'I' refrigerators on site). Finally, let's assume that the store expects to sell 500 refrigerators each year. (This tells you that the two variable 'I' and 'N' are RELATED to one another.) The sum of all the costs associated to managing inventory is often called the Inventory Control Cost . It is usually more complicated than this example.

Question

a) Give a formula for the annual inventory control costs, C, as a function of the maximum storage size, I

b) Give a formula for the annual inventory control costs, C, as a function of the number of shipments, N

c) Suppose the store manager chooses to have at most 25 refrigerators on site at any one time. Calculate the annual inventory control costs

d) If you look at a graph, you should be able to figure out the OPTIMUM number of shipments, the OPTIMUM storage capacity for the refrigerator store and the minimum possible annual inventory control costs.

d-a) Optimum number of shipments
d-b) Optimum storage capacity
d-c) Minimum possible inventory control costs

Answer 1

I'll use x for the size of inventory, so the upper-case I does not get confusing to read. Then we have
C(x,n) = 2000n + 20*5x
But, since n=500/x, that makes
C(x) = 1,000,000/x + 100x

Now suppose you try the other parts. Show what you tried if you get stuck.