In this article, we will perform the derivation of several useful formula for EOQ (Economic Order Quantity) Model with and without planned shortages.
EOQ Model: No Planned Shortages
Before formulating the model, we define some variables that is required. Given that
- $Q$ = order quantity, in unit
- $K$ = setup cost per batch, in $
- $c$ = unit cost, in $ per unit
- $h$ = holding cost, in $ per unit per unit time
- $d$ = ordering rate, in unit per unit time
The cycle length is $\frac{Q}{d}$ unit time.
Therefore, the production or ordering cost per cycle is $K + cQ$ in $.
Assume that the average inventory level during a cycle is $\frac{Q+0}{2} = \frac{Q}{2}$ units, and its corresponding holding cost is $\frac{hQ}{2}$ in $ per unit time.
Therefore, the holding cost per cycle is, in $,
\[\frac{hQ}{2} \times \frac{Q}{d} = \frac{hQ^2}{2d}\]Then, the total cost per cycle is, $K + cQ + \frac{hQ^2}{2d}$, in $.
We denote by $T$, the total cost (in $) per unit time, as follows:
\[T = \left( K + cQ + \frac{hQ^2}{2d} \right) \times \frac{1}{Q/d} = \frac{dK}{Q} + cd + \frac{hQ}{2}\]We denote by $Q^{\ast}$, the value of $Q$ that minimizes $T$ as follows:
\[Q^{\ast} = \sqrt{\frac{2dK}{h}}\]$Q^{\ast}$: The Derivation
Given that
\[T = \frac{dK}{Q} + cd + \frac{hQ}{2} \notag\]we differentiate $T$ with respect to $Q$ as follows:
\[\frac{dT}{dQ} = -\frac{dK}{Q^2} + \frac{h}{2}\]To find $Q$ that minimizes $T$, we want to find the value of $Q^{\ast}$ such that $\frac{dT}{dQ} = 0$. Hence,
\[\begin{align} -\frac{dK}{(Q^{\ast})^2} + \frac{h}{2} &= 0 \\ (Q^{\ast})^2 &= \frac{2dK}{h} \notag \\ Q^{\ast} &= \sqrt{\frac{2dK}{h} \notag} \end{align}\]The value of $Q^{\ast}$ answers the following question:
How many products to produce in each batch of order?
Next, we will look at the derivation of $t^{\ast}$, the optimal cycle length.
$t^{\ast}$: The Derivation
From earlier section, we know that the cycle length is $t = \frac{Q}{d}$, therefore
\[t^{\ast} = \frac{Q^{\ast}}{d} = \sqrt{\frac{2K}{dh}}\]The value of $t^{\ast}$ answers the following question:
When to make the order?
The formula for $Q^{\ast}$ and $t^{\ast}$ above enables us to calculate the optimal unit of product to order at an optimal time. However, this is only the special case where planned shortage is not allowed.
So, what will happen if we change the assumption: to allow planned shortage?
EOQ Model: With Planned Shortages
Similar to previous section, we define some variables that is required. Given that
- $Q$ = order quantity, in unit
- $K$ = setup cost per batch, in $
- $c$ = unit cost, in $ per unit
- $h$ = holding cost, in $ per unit per unit time
- $d$ = ordering rate, in unit per unit time
- $p$ = (new) shortage cost, in $ per unit per unit time
- $S$ = (new) inventory level (in unit) just after a batch of $Q$ units are added to inventory
- $Q - S$ = (new) shortage in inventory (in unit) just before a batch of $Q$ units are added to inventory.
Again, the cycle length is $\frac{Q}{d}$ unit of time.
Therefore, the production or ordering cost per cycle is $K + cQ$, in $.
Assume that the average inventory level during a cycle is $\frac{S+0}{2} = \frac{S}{2}$ units (we use $S$ instead of $Q$ because $S$ is the quantity of product in the inventory), and its corresponding holding cost is $\frac{hS}{2}$ in $ per unit time.
Therefore, the holding cost per cycle is, in $,
\[\frac{hS}{2} \times \frac{S}{d} = \frac{hS^2}{2d}\]Similarly, assume that the average shortage during a cycle is $\frac{Q-S}{2} $ units, and its corresponding shortage cost is $\frac{p(Q-S)}{2}$ in $ per unit time.
Therefore, the shortage cost per cycle is, in $,
\[\frac{p(Q-S)}{2} \times \frac{Q-S}{d} = \frac{p(Q-S)^2}{2d}\]Here, the total cost per cycle is, $K + cQ + \frac{hS^2}{2d} + \frac{p(Q-S)^2}{2d}$, in $.
We denote by $T$, the total cost (in $) per unit time, as follows:
\[\begin{align} T &= \left( K + cQ + \frac{hS^2}{2d} + \frac{p(Q-S)^2}{2d} \right) \times \frac{1}{Q/d} \notag \\ &= \frac{dK}{Q} + cd + \frac{hS^2}{2Q} + \frac{p(Q-S)^2}{2Q} \notag \\ &= \frac{dK}{Q} + cd + \frac{hS^2}{2Q} + \frac{pQ^2 - 2pQS + p^2}{2Q} \notag \\ &= \frac{dK}{Q} + cd + \frac{(h+p)S^2}{2Q} + \frac{pQ}{2} - pS \end{align}\]$Q^{\ast}$ and $S^{\ast}$: The Derivation
Given that
\[T = \frac{dK}{Q} + cd + \frac{(h+p)S^2}{2Q} + \frac{pQ}{2} - pS \notag\]First, we differentiate $T$ with respect to $Q$:
\[\begin{align} \frac{\partial{T}}{\partial{Q}} &= -\frac{dK}{Q^2} - \frac{(h+p)S^2}{2Q^2} + \frac{p}{2} \notag \\ &= -\frac{2dK + (h+p)S^2}{2Q^2} + \frac{p}{2} \end{align}\]Next, we differentiate $T$ with respect to $S$:
\[\frac{\partial{T}}{\partial{S}} = -p + \frac{(h+p)S}{Q}\]To find $Q$ and $S$ that minimize $T$, we want to find the value of $Q^{\ast}$ and $S^{\ast}$ such that $\frac{\partial{T}}{\partial{Q}} = 0$ and $\frac{\partial{T}}{\partial{S}} = 0$ respectively. Hence,
\[\begin{align} -\frac{2dK + (h+p)(S^{\ast})^2}{2(Q^{\ast})^2} + \frac{p}{2} &= 0 \\ \frac{2dK + (h+p)(S^{\ast})^2}{(Q^{\ast})^2} &= p \notag \\ \frac{2dK + (h+p)(S^{\ast})^2}{p} &= (Q^{\ast})^2 \label{eq:eqone} \end{align}\]and
\[\begin{align} -p + \frac{(h+p)S^{\ast}}{Q^{\ast}} &= 0 \\ \frac{(h+p)S^{\ast}}{Q^{\ast}} &= p \notag \\ Q^{\ast} &= \frac{h+p}{p}S^{\ast} \label{eq:eqtwo} \end{align}\]Next, we substitute $\eqref{eq:eqtwo}$ into $\eqref{eq:eqone}$:
\[\begin{align} \frac{2dK + (h+p)(S^{\ast})^2}{p} &= \left( \frac{h+p}{p}S^{\ast} \right)^2 \notag \\ \frac{2dK}{p} + \frac{h+p}{p}(S^{\ast})^2 &= \left( \frac{h+p}{p} \right)^2 (S^{\ast})^2 \notag \\ \left[ \left( \frac{h+p}{p} \right)^2 - \frac{h+p}{p} \right] (S^{\ast})^2 &= \frac{2dK}{p} \notag \\ \frac{h}{p}(h+p)(S^{\ast})^2 &= 2dK \notag \\ (S^{\ast})^2 &= \frac{2dK}{h} \cdot \frac{p}{p+h} \notag \\ S^{\ast} &= \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p}{p+h}} \label{eq:eqthree} \end{align}\]Finally, we substitute $\eqref{eq:eqthree}$ into $\eqref{eq:eqtwo}$:
\[\begin{align} Q^{\ast} &= \frac{(h+p)}{p} \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p}{p+h}} \notag \\ &= \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p+h}{p}} \end{align}\]In summary, the values of $Q^{\ast}$ and $S^{\ast}$ can be calculated using the formula as follows:
\[\begin{align} Q^{\ast} &= \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p+h}{p}} \notag \\ S^{\ast} &= \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p}{p+h}} \notag \end{align}\]We can also deduce $Q^{\ast} - S^{\ast}$ as follows:
\[\begin{align} Q^{\ast} - S^{\ast} &= \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p+h}{p}} - \sqrt{\frac{2dK}{h}} \cdot \sqrt{\frac{p}{p+h}} \notag \\ &= \sqrt{\frac{2dK}{h}} \cdot \left( \sqrt{\frac{p+h}{p}} - \sqrt{\frac{p}{p+h}} \right) \notag \\ &= \sqrt{\frac{2dK}{h}} \cdot \left( \sqrt{\frac{(p+h)^2}{p(p+h)}} - \sqrt{\frac{p^2}{p(p+h)}} \right) \notag \\ &= \sqrt{\frac{2dK}{h}} \cdot \left( \frac{p+h}{\sqrt{p(p+h)}} - \frac{p}{\sqrt{p(p+h)}} \right) \notag \\ &= \sqrt{\frac{2dK}{h}} \cdot \frac{h}{\sqrt{p(p+h)}} \notag \\ &= \sqrt{\frac{2dK}{p}} \cdot \sqrt{\frac{h}{p(p+h)}} \end{align}\]$t^{\ast}$: The Derivation
We would also like to know when to make the order, therefore, the time $t^{\ast}$, which is the optimal cycle length, is given by
\[t^{\ast} = \frac{Q^{\ast}}{d} = \sqrt{\frac{2K}{dh}} \cdot \sqrt{\frac{p+h}{p}}\]References
- F. S. Hillier and G. J. Lieberman, Introduction to operations research. Boston: McGraw-Hill, 2010.