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Is it possible to modify the EOQ model to work in a purchasing environment when ordering costs are effectively $0$?

The classic EOQ model is: $$ Q=\sqrt{2aK/h} $$

with $a$ being demand, $K$ ordering costs, and $h$ the holding costs. If $K$ is $0$ it yields an order quantity of $0$. Is there any optimal way of determining the order quantity when there are no order costs?

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  • $\begingroup$ Usually in these cases where formulas fail (happens mostly when something is 0 or infinity) you have to think outside the formula. What you get then is a "corner solution", in which you have to think logically about the problem and hit some contraint (again often involving 0 or infinity if there is no explicit constraint (like a budget constraint)). I'm not familiar with this model, but I would expect here if there is no ordering cost the ordering quantity would be infinity. Hope this helps at all! :) $\endgroup$ – BB King Dec 29 '15 at 14:27
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With ordering costs equal to zero, one has only to optimize holding costs.

Since total holding costs are

$$TC_h=\frac {hQ}{2}$$

it is monotonically decreasing in $Q$. $Q$ may be constrained, say,

$$Q \geq Q_m >0$$

where $Q_m$ express any constraints that may be imposed by the suppliers as regards "minimum order quantity".

So the optimal Economic Order Quantity is "as little quantity as possible per order".

Formally the Lagrangean of the problem can be written as

$$\Lambda = \frac {hQ}{2} + \lambda (Q_m-Q) $$

the first order conditions are

$$\frac {\partial \Lambda}{\partial Q} \leq 0 \implies \frac {h}{2} \leq \lambda$$

$$\frac {\partial \Lambda}{\partial \lambda} \leq 0 \implies Q_m-Q^* \leq 0$$

Since $h>0 \implies \lambda >0$. If the multiplier at the optimum is not zero, then the constraint is binding and we get

$$Q^* = Q_m$$

This would again become a more complex and interesting problem, if the unit price of the product became dependent (and decreasing) on quantity per order (because, say, fewer deliveries by the suppliers reduce their transportation costs and so they give you a discount if you order more per order).

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The total ordering cost are

$TC(Q)=P\cdot a+\frac{a\cdot K}{Q}+\frac{h\cdot Q}{2}$

If K=0 TC becomes

$TC(Q)=P\cdot a+\frac{h\cdot Q}{2}$

P\cdot a is a constant. Let denote it as c.

$TC(Q)=c+\frac{h}{2}Q$

You can see that TC is a straight line with a slope of $\frac{h}{2}$ and a domain of $Q \in \mathbb R^+$. The marginal cost is

$\frac{dTC}{dQ}=\frac{h}{2}$

It is a constant. Therefore there is no single optimum order quantity.

The store has a capacity. This is the upper limit of the order. If you have still some goods in the store the maximum order is the capacity (c) minus the goods in the store (x).

Therfore you can order bewtween $0$ (exclusive) and $c-x$ (inclusive) goods. The total cost of inventory will be always the same for a given demand in a period.

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  • $\begingroup$ Are you sure about your answer? Total holding costs do not depend on total quantity demanded/purchased $a$ but on quantity $Q$ per order. It looks like the lower $Q$ is, the lower Total holding costs will be. $\endgroup$ – Alecos Papadopoulos Dec 30 '15 at 21:13
  • $\begingroup$ I just took the eoq formula: en.wikipedia.org/wiki/Economic_order_quantity And if $K=0$, then the total costs becomes higher the higher Q is. Inside the limits you can order an arbitrary quantity. $\endgroup$ – callculus Dec 31 '15 at 15:02
  • $\begingroup$ Why order "an arbitrary quantity" (per order) if it is optimal (cost-minimizing) to order the minimum possible? $\endgroup$ – Alecos Papadopoulos Dec 31 '15 at 17:15

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