# Series systems

Russian: Линейная структура

This section analyzes a series system. There are J items numbered $\,j=1,...,J$ from first to last. The items represent the outputs of successive production stages, or stocking points along a supply chain. Demand occurs, at rate $\,\lambda$, only for item J, and an external source supplies item 1. All other supply links are internal; item 1 supplies item 2, item 2 feeds item 3, and so on. See Figure 1. Another word for item in this context is stage. A series system is sometimes called a multistage system.

Stock moves in discrete batches, as in the EOQ model. An order is a decision to move a batch to any stage, whether the batch comes from the supplier or a prior stage. The stages do not make their own order decisions, however; information and control are fully centralized. The order decisions must be coordinated; it makes no sense to order a batch to be sent to one stage, when the prior stage has insufficient inventory. (Nevertheless, the system can be operated effectively in a decentralized manner.) The external supplier always has ample stock available.

All demand must be met as it occurs; stockouts are forbidden. There are economies of scale in the form of fixed costs for all orders. (If not, the entire system could operate as a perfect flow system.) So, the basic issue here, as in the EOQ model, is to find a good balance between these fixed costs and inventory holding costs. Denote $\,k_{j}$= fixed cost for orders of item $\,j$ $\,h'_{j}$ = inventory holding-cost rate for item $\,j$ $\,I'_{j} (t)$= inventory of item $\,j$ at time t

Each stage requires precisely one unit from its predecessor to produce a unit of its own item. This is no real restriction: If a stage requires more or less than one unit, just redefine the quantity units of items $\,i to reflect their usages in the end product. For example, consider a two-stage system, where both items are originally measured in tons, and it requires two tons of item 1 to make one ton of item 2. Measure item 1 in two-tons instead; just one of these new units is needed to produce a unit of item 2. Accordingly, revise $\,h'_{j}$ to be twice its original value.

For simplicity, then, assume each leadtime is equal to 0. So, it is possible to order stock from the supplier and pass it through the successive stages, even all the way to stage J, instantaneously. Assume the system starts empty, i.e., $\,I'_{j} (0^{-} )=0$ for all $\,j$.

## Echelons and Echelon Inventories

Here is a fundamental concept in multiitem systems: The echelon of stagey (or echelon j for short) comprises stagey itself and all downstream stages, i.e., all stages $\,i\ge j$. The echelons of a four-stage system are indicated by rectangles in Figure 2. This notion captures the supply-demand relationships in a useful manner: First, stage $\,J$ is its own echelon. The external supplier and all the prior stages can be viewed as stage J 's supply process. Likewise, consider echelon $\,J-1$, i.e., the last two stages. This is another subsystem, whose supply process includes the earlier stages $\,i. Continuing in this manner, the entire system can be viewed as a hierarchy of nested subsystems, the echelons, each with a clearly defined supply process.

Imagine a multistage production process. Item 1 is a raw material. At each stage the material is transformed somehow, or various enhancements are added, until the final product emerges as item J. So, in a sense, a unit of item $\,i>j$ includes one of item $\,j$. The total system inventory of item $\,j$ thus comprises, not just $\,I'_{j} (t)$, but also the inventories downstream. To express this idea, define $\,I_{j} (t)=\sum _{i\ge j}I'_{i} (t)$

The original $\,I'_{j} (t)$ is sometimes called the local or installation inventory of item $\,j$. The prime indicates that it is a local quantity. Also, let $\,h_{j} =h'_{j} -h'_{j-1}$

where $\,h_{0} =0$. Assume that each $\,h_{j} >0$. (This is usually the case: Suppose $\,c_{j}$ is the variable order-cost rate for item $\,j$, and $\,h'_{j}$ includes financing costs only. To create a unit of item $\,j$ incurs all the costs $\,c_{i}$, $\,i, so $\,h'_{j} =\alpha \sum _{i\le j}c_{i}$ where $\,\alpha$ is the interest rate. Thus, the echelon holding cost is just $\,h_{j} =\alpha c_{j}$ that is, $\,h_{j}$ reflects the value added at stage $\,j$. And $\,c_{i} >0$ immediately implies $\,h_{i} >0$. If $\,h_{j}$\textit{ }also includes a physical handling cost $\,\underline{h'}_{j}$, the $\, \underline{h'}_{j}$ must also increase, or at least not decrease too fast. This is usually true also; physical handling tends to be more expensive downstream.)

With these definitions the systemwide inventory cost rate becomes $\,\sum _{j}h'_{j} I'_{j} (t) =\sum _{j}h_{j} I_{j} (t)$ for all t

Thus, the echelon inventories track stocks and their costs throughout the system just as well as the local inventories.

## Policy Characteristics

For practical purposes, and analytical purposes too, we would like to focus on relatively simple policies. Fortunately, good policies share certain simplifying qualitative characteristics.

A policy is nested if, for all $\,j$, whenever stage $\,j$ orders, so does stage $\,j+1$. Here, an order at one stage triggers orders at all downstream stages; in other words, all stages in an echelon order together. Consequently, stage J orders most frequently among all the stages, and stage 1 least. It turns out that nested policies are the only ones we need consider:

Every non-nested policy is dominated by a nested policy. (The nested policy has lower inventories and no more orders.)

A stationary-interval policy means just that: For each item the time intervals between orders are equal. Such policies are generally easier to implement than others, and they are certainly easier to analyze.

Consider a policy with all these properties. Let $\,u_{j}$= order interval for item $\,j$ $\,U=(u_{1} ,...,u_{J} )$ $\,g_{j} =h_{j} \lambda$ $\,C(U)$= average cost of the policy specified by $\,U$

Each $\,I_{j} (t)$ describes the periodic pattern familiar from the EOQ model. Therefore, applying the EOQ model's calculations to each item, $\,C(U)=\sum _{j}\left(\frac{k_{j} }{u_{j} } +\frac{1}{2} g_{j} u_{j} \right)$

Each $\,u_{j}$ is a positive-integer multiple of $\,u_{j+1}$. The problem of selecting the best such policy can be stated as mixed-integer problem: $\,\min C(U)=\min \sum _{j}\left(\frac{k_{j} }{u_{j} } +\frac{1}{2} g_{j} u_{j} \right)$ $\,u_{j} =\xi _{j} u_{j+1}$ $\,\xi _{j} >0$ $\,\xi _{j}$-- целое, $\,j