8.22.Mini-Case:You have recently acquired a franchise forDunkin'Donuts. You have reviewed the operations of the business carefully, from marketing to management, from accounting to finance. You have noticed that the store manager does not have a proper inventory policy. You want to optimize the amount of coffee that you should order at one time. The price of coffee is $4.50 per lb and it comes in 25-lb bags. The price per bag is 4.5*25 = $112.50. The store uses 3000 lb of coffee every year, that is, 3000/25 = 120 bags. The coffee is stored in a back room, where other supplies are also stored. You have found that the storage cost of coffee per year, based on average inventory is $1.25 per lb per year. Or, it is 1.25*25 = $31.25 per bag per year. The cost of capital for this business is 15% per year. The ordering cost of coffee is $35 per order.

From finance, you have learned that you should find the present value of all the payments you have made to (1) order, (2) purchase, and to (3) store the coffee. Of course, you need not include the financing cost because it is included in the present value calculations. You can find the optimal solution by minimizing the PV of the total cost.

You realize that you are working on a calculus problem. You set out to develop a new function,P, which is the present value of all payments for ordering, purchasing, and storing coffee. You assumen, the number of orders placed in a year, as the independent variable. Since you are going to use 120 bags of coffee per year, you will order 120/nbags at one time, at the cost of $112.50 per bag.

You place your first order right now.

The ordering cost is $35.

The purchasing cost is ($112.50/bag) * (120/nbags) = $13,500/n.

Since you buy 120/nbags at one time, the average inventory is 60/nbags. The storage cost is proportional to the average inventory. You will store the first order for 1/nyear and then you will get another order. Recall that the storage cost per bag per year is $31.25. Therefore, the storage cost of the first order is

($31.25/bag/year) * (60/nbags) * (1/nyear) = 1875/n²

PV of the first order = 35 + 13,500/n+ 1875/n²

You place another order after 1/nyear. The cost of the second order is the same as the first one except that this cost occurs after 1/nyear. To find its present value, use the discount rate of 15% and a time of 1/nyear. This gives

PV of the second order =35 + 13500/n+ 1875/n²

1.15^1/n

You will place another order after 2/nyears. Likewise, its present value is

PV of the third order =35 + 13500/n+ 1875/n²

1.15^2/n

Comparing the terms, you discover that they form an infinite series, witha= 35 + 13,500/n+ 1875/n²and multiplicative factorx= 1/1.15^1/n. Mathematically, it is easier to find the sum of an infinite geometric series. Using equation (1.5), you get

The present value of infinite orders is =35 + 13500/n+ 1875/n²

1 1/1.15^1/n

To minimize the cost, you can differentiate it with respect tonand set the derivative

equal to zero. This is

d (35 + 13500/n+ 1875/n²)= 0

dn 1 1/1.15^1/n

You can use Maple to solve the above equation by using these instructions,

diff((35+13500/n+1875/n^2)/(1-1/1.15^(1/n)),n)=0; fsolve(%); 

The result isn= 9.027667691. The number of bags that you should buy at one time is 120/9.027667691 = 13.29246978 = 13 bags.

You expect that the price of coffee will rise by 15% during the next year due to flooding in the coffee-producing areas. You want to overcome the possibility of rising prices by ordering a larger quantity of coffee for each order. Assuming that you are right, what is your optimal order quantity?