0
$\begingroup$

This is very simple, however, I have the following setup

Suppose that the company ABC has a product that shows a constant annual demand rate of 3600 items. One item costs £3. Ordering cost is £20 per order and holding cost is 25% of the value of inventory.

What I want to do is calculate the EOQ

$$ EOQ = \sqrt{\frac{2DS}{H}} $$

Where

  • D = annual demand (here this is 3600)
  • S = setup cost (here that's £20)
  • H = holding cost
  • P = Cost per unit (which is £3 here)

I figured that I would have

$$ H = 0.25 \times 3 = 0.75 $$

I'm sceptical about this result though.

$\endgroup$
  • $\begingroup$ This seems to give $EOQ \approx 438$. Do you think this looks too large or too small? $\endgroup$ – Henry May 5 '18 at 20:33
  • $\begingroup$ Note that for the formula to be correct, $H$ must be holding cost per unit per year. $\endgroup$ – Adam Bailey May 6 '18 at 14:58
1
$\begingroup$

So your EOQ expression is suggesting that the optimal order size is for about $438$ items each time.

You can check the result if you wish. Suppose you order in batches of $Q$:

  • The average annual number of batches ordered is $\dfrac{3600}{Q}$ so the average annual cost of ordering is $£\dfrac{72000}{Q}$

  • The average number of items held in inventory is $\dfrac Q2$ worth $£\dfrac{3Q}{2}$ at a holding cost of $£\dfrac{3Q}{8}$

  • So the combined ordering and holding cost is $£\dfrac{72000}{Q}+£\dfrac{3Q}{8}$

  • For $Q=437$ this gives about $£328.6347$; for $Q=438$ this gives about $£328.6336$; for $Q=439$ this gives about $£328.6341$. This suggests that $438$ may indeed be the best order size

  • You can check the calculus: the derivative of $\dfrac{72000}{Q}+\dfrac{3Q}{8}$ is $\dfrac{3}{8} - \dfrac{72000}{Q^2}$ which is a increasing function of $Q$ and is zero when $Q^2=192000$ i.e. $Q \approx 438.178$, and this would minimise the combined cost

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.