 # Pooling of inventory

When there are two or more demand points for a product, it may possible to save money if the separate inventories for these demand points can be combined or "pooled." There is a "square root economy of scale" from pooling for two of the simplest inventory models: a) the Economic Order Quantity (EOQ) model, and b) the Newsboy (NV) model.

For the EOQ case, where we want to minimize the fixed charge of ordering plus the holding cost of inventory carried, if we have two separate inventories for two demand points, each with constant demand rate $LaTeX: D,$ fixed charge of ordering $LaTeX: K,$ and holding cost/unit of inventory $LaTeX: h,$ then the minimum total cost is $LaTeX: \textstyle \sqrt{2\, D\, K\, h}$ at each demand point for a total cost of $LaTeX: \textstyle 2\, \sqrt{2\, D\, K\, h}.$ If we combine the two demand streams and carry a single inventory for them, then the minimum total cost is $LaTeX: \textstyle \sqrt{2\, (2\, D)\, K\, h} = 2\, \sqrt{D\, K\, h},$ which decreases the total cost by a factor of $LaTeX: \textstyle 1/\sqrt{2}.$

For the NV case, if we have two demands, each independently Normal distributed with mean $LaTeX: D$ and standard deviation $LaTeX: \sigma,$ a specified service level $LaTeX: \textstyle \alpha = \prob \left \{\mbox{Demand} < \mbox{inventory}\right\}$ at an inventory point, and $LaTeX: z_\alpha$ being the number of standard deviations corresponding to a left tail probability of $LaTeX: \alpha,$ then we must carry inventory $LaTeX: \textstyle D + z_\alpha \sigma$ at each inventory point for a total inventory of $LaTeX: \textstyle 2D + 2z_\alpha \sigma.$ If we can combine the two demands so we carry just one inventory, then the mean and standard deviation for the combined demands are $LaTeX: 2D$ and $LaTeX: \sqrt{2}\sigma.$ Thus, the total inventory we must carry is $LaTeX: \textstyle 2D + \sqrt{2}z_\alpha\sigma,$ and there is a square root economy of pooling in the amount of safety stock we must carry. If we say that the system service level should be $LaTeX: \alpha,$ then the benefits of pooling are even greater.

There is an interesting anomaly wherein pooling can increase inventory. Specifically, for the NV problem, if instead of there being a constraint on service level, the objective is simply to minimize the expected cost of unsatisfied demand and the cost of surplus inventory, and the demand distributions at each of two demand points (with associated inventory) are right skewed, e.g., the mean is greater than the median, then pooling the two inventories may in fact result in a higher total optimal inventory. For example, if the cost/unit short is 5, the cost/unit of inventory left over is 4, the demands at each of two demand points are independent Poisson distributed with mean 3.3, then the optimal amount to stock at each location is 3. If, however, the demands are pooled and a single inventory is carried so the demand is Poisson with mean 6.6, then the optimal amount to stock is 7. So pooling increased the optimal inventory. See Myths and Counter Examples for an additional example.

# Economic order quantity Abbreviated EOQ. This is the level of inventory to order that minimizes the sum of holding and ordering costs. The inventory function, $LaTeX: I(t)$, is the following periodic sawtooth function, where $LaTeX: T$ is the time between orders, and $LaTeX: Q$ is the ordering quantity: $LaTeX: I(t) = Q - dt \;$ for $LaTeX: \; 0 \le t \le T$,

where $LaTeX: d$ is the rate of demand (inventory units per time units), and $LaTeX: I(t) = I(t-T)$ for $LaTeX: t > T$. The inventory becomes zero at $LaTeX: T = Q/d$, which requires a new order of $LaTeX: Q$ units. The model is thus: $LaTeX: \min \; (1/2) hdT + K/T$,

where $LaTeX: h$ is the holding cost (currency per time units), so $LaTeX: (1/2) hdT$ is the average holding cost, and $LaTeX: K$ is the fixed cost of ordering, so $LaTeX: K/T$ is the average ordering cost. The solution is $LaTeX: T^* = (2K/hd)^{1/2}$, which yields the Economic Order Quantity (EOQ): $LaTeX: Q^* = (2Kd/h)^{1/2}$. See the more general production scheduling problem.