# Planned backorders

Russian: Задолженный заказ

## The Setting

Consider the Basic economic order quantity (EOQ) system, but relax the requirement that all demand be met from stock on hand. All demand is ultimately filled, though perhaps after a delay. That is, demand not filled immediately is backordered. Any inventory on hand is used to fill demand; backorders accumulate only when a warehouse runs out of stock entirely.

## Reorder-Point/Order-Quantity Policies

Define some functions and redefine some of those used in the EOQ model: $\,I(t)$ -- inventory at time $\,t$; $\,B(t)$ -- backorders at time $\,t$; $\,IN(t)$ -- net inventory at time $\,t$, $\,IN(t)=I(t)-B(t)$; $\,IO(t)$ -- stock on order at time $\,t$; $\,IP(t)$ -- inventory position at time $\,t$, $\,IP(t)=IN(t)+IO(t)$.

The new functions are $\,B(t)$ and $\,IN(t)$, and $\,IP(t)$ has a new meaning.

The net inventory $\,IN(t)$ captures the information in both $\,I(t)$ and $\,B(t)$: At any given time, at least one of those two functions is zero, since we use any available stock to fill demand. Therefore, $\,IN(t)=\left\{\begin{array}{ll} {I(t)} & {when\, IN(t)\ge 0} \\ {-B(t)} & {when\, IN(t)<0} \end{array}\right.$

The definition of $\,IN(t)$ treats backorders as negative inventories, and indeed they function in this way: Between receipts of orders, $\,IN(t)$ decreases at the constant rate $\,\lambda$, regardless of whether $\,IN(t)$ is positive or negative.

When an order arrives, $\,IN(t)$ jumps up by precisely $\,q$ in all cases; some of the batch may be used to fill backorders, and the rest is added to inventory.

Thus, $\,IN(t)$ behaves much like $\,I(t)$ did before, when backorders were forbidden, while now $\,I(t)$ itself is more complex ( $\,IN(t)$ is sometimes called the inventory level.), fig. 1.

Consider, $\,IN(t+L)=IN(t)+IO(t)-D=IP(t)-D \, \, \, \, \, \, \, (1)$

It is the conservation-of-flow law for this system: Between $\,t$ and $\,t + L$, $\,IO(t)$ gets added to the net inventory, and $\,D$ gets subtracted.

Thus, $\,IP(t)$ summarizes all the information needed to predict the net inventory a leadtime into the future.

As in the EOQ model, assume that all orders are of the same size $\,q>0$. The issue of when to order is now more complex. We need a second policy variable in addition to $\,q$: $\,r$ - reorder point (quantity-units)

This variable can take on any real value, positive or negative. Consider the following policy:

Monitor the inventory position $\,IP(t)$ constantly. When $\,IP(t^{-} )=r$, place a new order of size $\,q$ at time $\,t$.

(The EOQ model's policies are special cases with $\,r = D$.)

In honor of the two variables, a policy of this kind is called a reorder-point/order-quantity or $\,(r,q)$ policy.

Figure 2 illustrates the behavior of $\,IN(t)$ and $\,IP(t)$ under such a policy.

As in the EOQ model the graph retains the same sawtooth pattern, but now the pattern can shift vertically, depending on the choice of $\,r$.

## Performance Criteria

The relevant criteria include average inventory $\,\overline{I}$ and average frequency $\,\overline{OF}$.

The primary backorder-related performance measure is the following: $\,\overline{B}$ - long-term average outstanding backorders, $\,\overline{B}=\mathop{\lim }\limits_{T\to \infty } \frac{1}{T} \int _{0}^{T}B(t)dt$

(The limit here is analogous to the one defining $\,\overline{I}$). $\,v$ -- safety stock, $\,v=r-D,$ $\,y$ -- safety time, $\,y=\frac{v}{\lambda } .$

Like $\,v$, $\,y$ can be negative.

For any given $\,q$, only certain values of $v$ make sense: Equation (1) implies that the net inventory $\,IN(t^{-}$ ) just at the end of a cycle is precisely $\,v$, so $\,IN(t)=v+q$ at the beginning of a cycle. Therefore,

If $\,v>0$, then for all $\,t$ $\,I(t)>v>0$ and $\,B(t)=0$

If $\,v<-q$, then for all $\,t$ $\,I(t)=0$ and $\,B(t)>-(v+q)>0$

Neither conclusion is appealing: In case 1, we have more inventory than we actually need, and in case 2 we never fill all backorders. So, we can and do restrict attention to the range $\,-q\le v\le 0$. So, $\,v$ is negative (more precisely, nonpositive), as is $\,y$, and each arriving order fills all current backorders.

Thus, a cycle consists of two parts, one of length $\,u+y=(q+v)/\lambda$, during which inventory is held, and a second part of length $\,-y=-v/\lambda$, when backorders accumulate (see fig. 3). These intervals correspond to the fractions $\,(q+v)/q$ and $\,-v/q$, respectively, of the full cycle.

The average inventory is simply $\,\frac{1}{2} (q+v)$ during the first part and zero during the second. The average over a full cycle is a weighted average of these quantities: $\,\overline{I}=\left(\frac{q+v}{q} \right)\left(\frac{1}{2} (q+v)\right)+\left(\frac{-v}{q} \right)(0)=\frac{1}{2} \frac{(q+v)^{2} }{q} .$

Likewise, the average backorders in the first part of the cycle is zero, and $\,\frac{1}{2} (-v)$ in the second, so $\,\overline{B}=\left(\frac{q+v}{q} \right)(0)+(-v/q)\left(\frac{1}{2} (-v)\right)=\frac{1}{2} \frac{v^{2} }{q} .$

Finally, the cycle length is $\,u=\frac{q}{\lambda }$, so $\,\overline{OF}=\frac{\lambda }{q} .$

Clearly these criteria are in direct conflict. Let us translate them into monetary terms. We continue to use the cost factors $\,k$, $\,c$ and $\,h$ defined earlier. Suppose we can also estimate a factor for backorders analogous to $\,h$: $\,b$ - penalty cost for one unit backordered during one time-unit (moneys/[quantity-unit time-unit])

This parameter summarizes all the drawbacks of backorders mentioned above. The total average cost then becomes $\,C(v,q)=(k+cq)\overline{OF}+h\overline{I}+b\overline{B}=$ $\,=c\lambda +\frac{k\lambda }{q} +\frac{1}{2} \frac{h(q+v)^{2} }{q} +\frac{1}{2} \frac{bv^{2} }{q} .$

## The Optimal Policy and Sensitivity Analysis

The cost $\,C(v,q)$ is now a function of two variables. To minimize it, we equate its partial derivatives to zero. ( $\,C$ is continuously differentiable and strictly convex on its domain, so this approach works.) That is, $\,\frac{\partial C}{\partial v} =\frac{h(q+v)}{q} +\frac{bv}{q} =0$ $\,\frac{\partial C}{\partial q} =\frac{-k\lambda }{q^{2} } +\frac{1}{2} \frac{h(q^{2} -v^{2} )}{q^{2} } -\frac{1}{2} \frac{bv^{2} }{q^{2} } =0$

Defining the cost ratio $\,\omega =\frac{b}{b+h}$

gives the unique solution to these equations as $\, q^{*} =\sqrt{\frac{2k\lambda }{h} } \sqrt{\frac{1}{\omega } }$ $\, v^{*} =-(1-\omega )q^{*}$ $\, C^{*} =C(v^{*} ,q^{*} )=c\lambda +\sqrt{2k\lambda h\omega }$