A company has determined from its analysis of production and accounting data that, for a part number KC-438, the annual demand is equal to 10,000 units, the cost to purchase the item is Rs 36 per...

A company has determined from its analysis of production and accounting data that, for a part number KC-438, the annual demand is equal to 10,000 units, the cost to purchase the item is Rs 36 per order, and the holding cost is Rs 2/unit/year.

a) What is the economically optimal (profit-maximizing) size of each order the company should place?

b) How many days of supply should the company hold as inventory? (assume 300 working days in a year) 

Expert Answers
pnrjulius eNotes educator| Certified Educator

We know that we want to sell 10,000 units over the course of the year, but we don't know whether to order them all once, or split up our orders into smaller quantities to save on the cost of holding inventory. (Apparently the unit cost of buying each unit is the same either way, so we're ignoring it.)

Since our revenue won't change, we maximize profit by minimizing cost.

If we place n orders over the course of a year, we know that the size of each order must be 10,000/n.

The time that this order must be held in inventory is 1/n years; but since we're depleting it at a constant rate, the average item in each batch will only be held for half that, 1/(2n) years.

(If we were to draw a graph of how many units we hold in inventory at any given time, it would be a line slanted downward; the area under that line would give us the number of unit-days of inventory held. That area is a triangle, 1/2*base*height; that's how we get the 1/2 number.)

So, each order will cost Rs 36 to buy, and then we will need to hold 10,000/n units for an average of 1/(2n) years each, paying Rs 2 for each unit-year.

Thus, the cost of each order is:

c = 36 + (10,000/n)*(1/(2n))*2
c = 36 + 10,000/n^2

The total cost of n orders is therefore:

C = n c = 36 n + 10,000/n

We want to minimize this, so we take the derivative; the first-order condition is that the derivative is zero:

dC/dn = 0 = 36 - 10,000/n^2
10,000/n^2 = 36
10,000/36 = n^2
n = 16.666...

(It's a good idea to check the second-order condition and make sure this is really a minimum: d2C/dn2 = 20,000/n^3 > 0, so yes, it is.)

The size of each order is:
10,000/n = 600

So we want to place orders of 600 units.
That's answer (a) 600 units per order.

We can't actually place 16.666... orders; so each year we'll place 16 orders of 600 units, and then 1 final order of 400 units.

If we assume there are 300 business days in a year, and we sell 10,000 units in a year, then one day of units is 33.333...

Since each order is 600 units, the maximum inventory we need to be able to support is 600, which is 18*(33.333...); that is, we need enough inventory space to store 18 days worth of sales. On average, we will only have half that (9 days worth), but we need the space for 18. That's answer (b), 18 days of inventory.

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