Warehouse storage policies

A place for everything, and everything in its place Benjamin Franklin

🧠 Learning objectives

• Describe the main storage policies.
• Formulate mathematical optimization models for each storage policy.
• Discuss formulation trade-offs and caveats.

Storage policies

Main issue: Decide how to allocate the various storage locations of a uniform storage medium to a number of SKUs.

Policies:

• Random/Haphazard/Chaotic
• Dedicated
• Class-based
• Cube-per-order index

Example of a warehouse with 16 uniform storage locations:

1	2	3	4
5	6	7	8
9	10	11	12
13	14	15	16

The four-by-four grid is a course schematic showing sixteen uniform candidate locations. The source passages below define the policy choices summarized on the slide.

“There are five main policies that are used to store incoming items in the warehouse. The simpler policy, called random storage policy, is to store an incoming item in any available location. If more than one location is available for storage, theoretically, the incoming item has an equal probability of being assigned to any of the available locations. In practice, however, it gets assigned to the closest available space. As pointed out in Francis et al. (1992) and Tompkins et al. (2010), storage and retrieval under the random policy are not so ‘purely’ random in practice. Operators tend to store or retrieve items from the closest location.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 292).

“The dedicated policy, on the other hand, requires assignment of items to prespecified locations that depend on the item type being stored. Each of the two policies has its own advantages and disadvantages. For example, for the same volume and frequency of S/R operations, the random policy requires less storage space than the dedicated policy. This is primarily because under the latter policy, the space dedicated to each item must equal the maximum level of inventory of that item.”

— Facilities Design (Heragu 2022, sec. 8.6.7, pp. 292–294).

“Thus, we classify incoming items into one of three classes—A, B, and C—depending on the level of S/R activity it generates. If it is between 0% and 5% of the total S/R activity, the item is a class C item; if it is between 5% and 20%, the item is a class B item. Otherwise, it is a class A item. To minimize the time spent in storage and retrieval, class A items must be stored closest to the input/output point, class B to the next closest location, and so on.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

“The third policy—cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

Random storage policy

• Store an incoming item in any available location.
  • In practice not “purely” random.
  • Operators tend to store or retrieve items from the closest location.

“There are five main policies that are used to store incoming items in the warehouse. The simpler policy, called random storage policy, is to store an incoming item in any available location. If more than one location is available for storage, theoretically, the incoming item has an equal probability of being assigned to any of the available locations. In practice, however, it gets assigned to the closest available space. As pointed out in Francis et al. (1992) and Tompkins et al. (2010), storage and retrieval under the random policy are not so ‘purely’ random in practice. Operators tend to store or retrieve items from the closest location.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 292).

Random storage policy advantages

• Random policy requires less storage space:
  • The maximum aggregate inventory level is \(\leq\) the sum of the individual items’ maximum inventory levels.
  • Products are typically replenished at different times.
  • Appropriate for automated warehouses (item location/info stored electronically).

“The dedicated policy, on the other hand, requires assignment of items to prespecified locations that depend on the item type being stored. Each of the two policies has its own advantages and disadvantages. For example, for the same volume and frequency of S/R operations, the random policy requires less storage space than the dedicated policy. This is primarily because under the latter policy, the space dedicated to each item must equal the maximum level of inventory of that item. Of course, the maximum level is observed as soon as the item is replenished. With the random policy, however, the maximum aggregate inventory level is lower because the products are typically replenished at different times (see Exercise 33). In other words, because the products are replenished at different times, the maximum aggregate inventory level tends to be lower than the sum of the individual item’s maximum inventory levels.”

— Facilities Design (Heragu 2022, sec. 8.6.7, pp. 292–294).

“With a random storage policy, items are stored randomly at open and available storage locations. Such a policy is appropriate for automated warehouses because the storage location and item information are stored electronically and can be readily accessed by the automated devices, which can travel to the corresponding pick location without wasting any time looking for that item.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, sec. 15.4.2, p. 343).

Random storage policy disadvantages

• Retrieval may take longer due to searching time (more so for items that are frequently delivered).
• Low throughput rate (i.e., the number of S/R transactions per unit time).
  • S/R equipment not effectively utilized.
• Warehouse tends to be disorderly.

“Although the random policy requires less storage space, one of the problems is that if there are numerous items to be stored, the retrieval takes more time as some time is lost in searching the item’s location (more so for items that are frequently delivered). Thus, throughput rate (i.e., the number of S/R transactions per unit time) of the system goes down, the S/R equipment will not be effectively utilized, the warehouse tends be disorderly with items scattered all over (sometimes even put aside in aisles), and so on.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

Design model for random storage policy

• Given that \(n\) storage spaces are required:
  1. Determine the storage space layout that minimizes the total expected travel distance between each storage space \(j\) and \(p\) I/O points: \[ \sum_{k=1}^{p}{d_{kj}} \]
  2. Arrange the spaces in non-decreasing order of the sum of these distances, and pick the \(n\) closest storage spaces.

Note that \(n\) depends on the inventory levels of all the items.

“Consider the following random storage policy. Items coming into a warehouse are stored randomly in one of several storage spaces that are empty and available. Each empty storage space has an equal probability of being selected for storage. As pointed out previously, the storage or retrieval may not be purely random. However, we will assume a ‘pure’ random policy to make the modeling easier. Because each storage location has an equal likelihood of being accessed for storage or retrieval, the problem of interest here can be defined as follows. Given that \(n\) storage spaces are required, determine the storage space layout so that the total expected travel distance between each storage space and \(p\) I/O points is minimized. Following the algorithm in the previous section, we can determine the sum of the distances of each storage space with each I/O point as \(\sum_{k=1}^{p} d_{kj}\), arrange the spaces in nondecreasing order of the sum of these distances, and pick the \(n\) closest storage spaces. Note that \(n\) depends on the inventory levels of all the items, and as demonstrated in Exercise 33, the total number of spaces \(n\) would be less than that required under the dedicated policy.”

— Facilities Design (Heragu 2022, sec. 8.6.7.3, p. 301).

Example

It is desired to determine the storage space layout for 20 storage spaces in a rectangular \(100\ \text{m} \times 50\ \text{m}\) warehouse (50 storage spaces). A random policy will be used, and the designer wants to minimize the total distance traveled. Each storage space is \(10\ \text{m}^2,\) and the I/O point is located in the middle of the south wall.

Distances from I/O in the middle of the south wall

9	8	7	6	5	5	6	7	8	9
8	7	6	5	4	4	5	6	7	8
7	6	5	4	3	3	4	5	6	7
6	5	4	3	2	2	3	4	5	6
5	4	3	2	1	1	2	3	4	5

Solution: Max. distance = 40m, Avg. distance = \(\frac{60\text{m}}{20}\) = 3m

				4	4
			4	3	3	4
		4	3	2	2	3	4
	4	3	2	1	1	2	3	4

“Design Model for Random Storage Policy Consider the following random storage policy. Items coming into a warehouse are stored randomly in one of several storage spaces that are empty and available. Each empty storage space has an equal probability of being selected for storage. As pointed out previously, the storage or retrieval may not be purely random. However, we will assume a”pure” random policy to make the modeling easier. Because each storage location has an equal likelihood of being accessed for storage or retrieval, the problem of interest here can be defined as follows. Given that n storage spaces are required, determine the storage space layout so that the total expected travel distance between each storage space and p I/O points is minimized. Following the algorithm in the previous section, we can determine the sum of the distances of each storage space with each I/O point as dkj k p = å 1 , arrange the spaces in nondecreasing order of the sum of these distances, and pick the n closest storage spaces. Note that n depends on the inventory levels of all the items, and as demonstrated in Exercise 33, the total number of spaces n would be less than that required under the dedicated policy.”

— Facilities Design (Heragu 2022, 301).

Dedicated policy

• Items to prespecified locations that depend on the item type being stored.
• The space dedicated to each item must equal the maximum level of inventory of that item.
  • Maximum level is observed as soon as the item is replenished.
• Locations of SKUs do not change:
  • Popular items (e.g., based on turnover) can be stored in more convenient locations.
  • Workers can learn the topology of the warehouse.

For the same volume and frequency of S/R operations, the random policy requires less storage space than the dedicated policy.

“The dedicated policy, on the other hand, requires assignment of items to prespecified locations that depend on the item type being stored. Each of the two policies has its own advantages and disadvantages. For example, for the same volume and frequency of S/R operations, the random policy requires less storage space than the dedicated policy. This is primarily because under the latter policy, the space dedicated to each item must equal the maximum level of inventory of that item. Of course, the maximum level is observed as soon as the item is replenished. With the random policy, however, the maximum aggregate inventory level is lower because the products are typically replenished at different times (see Exercise 33). In other words, because the products are replenished at different times, the maximum aggregate inventory level tends to be lower than the sum of the individual item’s maximum inventory levels.”

— Facilities Design (Heragu 2022, sec. 8.6.7, pp. 292–294).

“In a manual system, the operator would have to search for that item, hence the dedicated storage policy is preferred. In this policy, a set of storage locations is designated for each SKU, based on its turnover. With a dedicated policy, each location can only contain the item designated to be stored in that space. Thus, although an arriving item may not have an available space in its designated location, it cannot be stored in another SKU’s location, even if that location is open and available. The dedicated policy, therefore, requires more storage spaces for the same types and volumes of SKU handled.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, sec. 15.4.2, p. 343).

“Because the locations of SKUs do not change, popular items can be stored in more convenient locations and workers can learn the topology of the warehouse. On the negative side, dedicated storage does not use space efficiently, since storage space has to be dimensioned according to the maximum amount of stored items per SKU.”

— Scattered Storage: How to Distribute Stock Keeping Units All Around a Mixed-Shelves Warehouse (Weidinger and Boysen 2018, 1).

Location assignment under dedicated storage policy

Goal: Enhance the throughput of storage and retrieval operations by reducing travel time/distance.

How?: By allocating the most active units to the most convenient locations.

The assignment rule links activity to location convenience: high-trip items receive lower-distance spaces so that the movement term in the objective falls. Stable dedicated locations also let workers learn the warehouse topology.

“To maximize throughput when using dedicated storage, SKUs should be assigned to storage locations based on the ratio of their activity to the number of openings or slots assigned to the SKU. The SKU having the highest ranking is assigned to the preferred openings and so on, with the lowest-ranking SKU being assigned to the least-preferred openings. Because ‘fast movers’ are up front and ‘slow movers’ are in back, throughput is maximized.

In ranking SKUs, it is important to define activity as the number of storages/retrievals per unit time, not the quantity of materials moved. Also, it is important to think of ‘part families’ as well. ‘Items that are ordered together should be stored together’ is a maxim of activity-based storage.”

— Facilities Planning (Tompkins 2010, 419–20).

“Consider the following warehouse storage problem. A warehouse has \(p\) I/O points through which \(m\) items enter and leave the warehouse. The items are stored in one of \(n\) storage spaces or locations. Each location requires the same storage space. It is known that item \(i\) requires \(S_i\) storage spaces. Ideally, we would like \(\sum_{i=1}^{m}S_i=n\). However, if the left-hand side is less than that of the right-hand side, we can always add a dummy product \((m+1)\) to take up the remaining \(n-\sum_{i=1}^{m}S_i\) spaces. We will assume throughout our discussion that the aforementioned equality holds. (Of course, if the right-hand side is less than the left-hand side, there is no feasible solution to the problem as there are not enough storage spaces to accommodate all the items.)

There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1, p. 295).

“Because the locations of SKUs do not change, popular items can be stored in more convenient locations and workers can learn the topology of the warehouse. On the negative side, dedicated storage does not use space efficiently, since storage space has to be dimensioned according to the maximum amount of stored items per SKU.”

— Scattered Storage: How to Distribute Stock Keeping Units All Around a Mixed-Shelves Warehouse (Weidinger and Boysen 2018, 1).

Convenient locations

• Locations \(j\) with the smallest distance \(d_j\) to the I/O point.
• If the material transfer is performed through a forklift truck (or a similar type of MHE¹), a proper distance metric is the rectilinear or Manhattan metric (or \(L_1\) norm): \[ d_j = |x(j)-x(I/O)| + |y(j)-y(I/O)| \]
• For an AS/RS type of storage mode, where the S/R unit can move simultaneously in both axes, with uniform speed, the most appropriate distance metric is the Tchebychev metric (or \(L_\infty\) norm): \[ d_j = max (|x(j)-x(I/O)|,|y(j)-y(I/O)|) \]

1. MHE: material handling equipment

Active SKUs

• SKUs that cause a lot of traffic (S/R transactions).

• Every SKU \(i\) gets a number of storage locations, \(S_i\), exclusively allocated to it.

• \(S_i\) reflects its max. storage needs and it must be determined through inventory activity profiling.

• Ideally, given \(n^{\mathrm{SKUs}}\) SKUs, the total n. of products is equal to the n. of storage spaces \(n^{\mathrm{storage}}\): \[ \sum_{i=1}^{n^{\mathrm{SKUs}}}{S_i} = n^{\mathrm{storage}} \]

• If LHS < RHS, add a dummy SKU \((n^{\mathrm{SKUs}} +1)\) to take up the remaining space \((n^{\mathrm{storage}} - \sum_{i=1}^{n^{\mathrm{SKUs}}}{N_i})\).

• If RHS < LHS, no feasible solution.

“Consider the following warehouse storage problem. A warehouse has \(p\) I/O points through which \(m\) items enter and leave the warehouse. The items are stored in one of \(n\) storage spaces or locations. Each location requires the same storage space. It is known that item \(i\) requires \(S_i\) storage spaces. Ideally, we would like \(\sum_{i=1}^{m} S_i = n\). However, if the left-hand side is less than the right-hand side, we can always add a dummy product \((m + 1)\) to take up the remaining \(n - \sum_{i=1}^{m} S_i\) spaces. We will assume throughout our discussion that the aforementioned equality holds. Of course, if the right-hand side is less than the left-hand side, there is no feasible solution to the problem because there are not enough storage spaces to accommodate all the items.

There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\), and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in Model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.

The preceding model appears similar to the quadratic assignment problem, except that the objective function is linear and the first assignment constraint is more general—the right-hand side is not 1 but \(S_i\). If the right-hand side were 1 for constraint (8.13), then the preceding model would be the well-known assignment problem. It is called the generalized assignment problem for obvious reasons. Like the transportation and assignment problems, the optimal solution will always be such that the \(x_{ij}\) variables are integers. Hence, integer constraint (8.15) can be replaced by a nonnegativity constraint. Because the generalized assignment problem is a special case of the transportation problem, it can be solved using the transportation algorithm. The latter is computationally efficient.”

— Facilities Design (Heragu 2022, 295–96).

Dedicated assignment model notation

Table 1: Parameters and variables (Heragu 2022).

Symbol	Description
\(i\)	SKU index
\(j\)	Storage location index
\(k\)	I/O point index
\(n^{\mathrm{IO}}\)	Number of I/O points
\(n^{\mathrm{SKUs}}\)	Number of SKUs (supply points)
\(n^{\mathrm{storage}}\)	Number of storage locations
\(S_i\)	Number of product locations required by SKU \(i\)
\(d_{kj}\)	Distance of storage space \(j\) from I/O point \(k\)
\(c_{ik}\)	Cost of moving a unit load of item \(i\) to and from I/O point \(k\)
\(f_{ik}\)	Number of trips of item \(i\) through I/O point \(k\)
\(w_{ij}\)	=\(\frac{\sum_{k=1}^{n^{\mathrm{IO}}} c_{ik} f_{ik} d_{kj}}{S_i}\)
\(x_{ij}\)	1 if location \(j\) is allocated to SKU \(i\); 0 otherwise

Heragu’s source model defines the dimensions, parameters, and binary assignment variable used on this slide. Calling an SKU a “supply point” is the course’s transportation-model analogy; in the source it is an item that requires \(S_i\) storage spaces.

“Consider the following warehouse storage problem. A warehouse has \(p\) I/O points through which \(m\) items enter and leave the warehouse. The items are stored in one of \(n\) storage spaces or locations. Each location requires the same storage space. It is known that item \(i\) requires \(S_i\) storage spaces. Ideally, we would like \(\sum_{i=1}^{m}S_i=n\). However, if the left-hand side is less than that of the right-hand side, we can always add a dummy product \((m+1)\) to take up the remaining \(n-\sum_{i=1}^{m}S_i\) spaces. We will assume throughout our discussion that the aforementioned equality holds. (Of course, if the right-hand side is less than the left-hand side, there is no feasible solution to the problem as there are not enough storage spaces to accommodate all the items.)”

— Facilities Design (Heragu 2022, sec. 8.6.7.1, p. 295).

“There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1 and Model 2, pp. 295–296).

Dedicated assignment model formulation

Objective Function

\[ \begin{align*} \min \quad Z & = \sum_{i=1}^{n^{\mathrm{SKUs}}} \sum_{j=1}^{n^{\mathrm{storage}}} w_{ij} x_{ij} \end{align*} \]

The slide abbreviates the source’s item–location cost coefficient as \(w_{ij}\). In the source, that coefficient is \(\left(\sum_k c_{ik}f_{ik}d_{kj}\right)/S_i\): trip frequency, movement cost, and I/O-to-location distance, spread over the item’s required spaces.

“There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1 and Model 2, pp. 295–296).

Dedicated assignment model constraints

All products \(S_i\) of a SKU \(i\) are allocated: \[ \begin{align} \sum_{j=1}^{n^{\mathrm{storage}}} x_{ij} = S_i, \quad i = 1,2,\ldots,n^{\mathrm{SKUs}} \\ \end{align} \]

Every location stores a single product: \[ \begin{align} \sum_{i=1}^{n^{\mathrm{SKUs}}} x_{ij} = 1, \quad j = 1,2,\ldots,n^{\mathrm{storage}} \\ \end{align} \]

Binary domain: \[ \begin{align} \quad & x_{ij} \in \{0,1\}, \quad i = 1,2,\ldots,n^{\mathrm{SKUs}}, \quad j = 1,2,\ldots,n^{\mathrm{storage}} \end{align} \]

The first equality gives item \(i\) exactly its required \(S_i\) locations. The second assigns every storage location to one item. The binary domain records whether the item–location pairing is selected.

“There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1, p. 295).

\[ \sum_{j=1}^{n}x_{ij}=S_i \qquad i=1,2,\ldots,m \tag{8.13} \]

\[ \sum_{i=1}^{m}x_{ij}=1 \qquad j=1,2,\ldots,n \tag{8.14} \]

\[ x_{ij}=0\text{ or }1 \qquad i=1,2,\ldots,m;\ j=1,2,\ldots,n \tag{8.15} \]

“The preceding model appears similar to the quadratic assignment problem (QAP) in Chapter 10, except that the objective function is linear and the first ‘assignment’ constraint is more general than that seen in the QAP—the right-hand side is not 1 but \(S_i\). If the right-hand side were 1 for constraint (8.13), then the preceding model would be the well-known assignment problem. It is called the generalized assignment problem for obvious reasons. Like the transportation and assignment problems, the optimal solution will always be such that \(x_{ij}\)’s are integers. (The reasons have been discussed elsewhere in the book, but also see Exercise 35.) Hence, integer constraint (8.15) can be replaced by a nonnegativity constraint. Because the generalized assignment problem is a special case of the transportation problem (notice that the transportation problem has \(B_j\) instead of 1 in the right-hand side of constraint (8.14)), it can be solved using the transportation algorithm. The latter is computationally efficient. It can be verified in Chapter 11 that the transportation algorithm examines the rows and columns of the cost (\(w_{ij}\)) matrix and only does addition or subtraction in each iteration. Even if we have several thousand storage spaces, the problem can be solved quickly on a computer.”

— Facilities Design (Heragu 2022, Model 2, eqs. 8.13–8.15, pp. 295–296).

Example: Problem statement

Consider two SKUs (\(a\), \(b\)) and three storage locations (1, 2, 3). SKU \(a\) demands 1 storage location (\(S_a = 1\)), and SKU \(b\) demands 2 storage locations (\(S_b = 2\)). The I/O point \(p\) is at the leftmost side of the warehouse, such that \(d_{p1} = 1\), \(d_{p2} = 2\), and \(d_{p3} = 3\). How should the items be allocated to minimize travel costs between the I/O point and the locations under a dedicated policy?

The two SKUs, three locations, one I/O point, capacities \(S_a=1\) and \(S_b=2\), and distances 1–3 form an instructor-created numerical instance of Heragu’s general dedicated-storage model.

“Consider the following warehouse storage problem. A warehouse has \(p\) I/O points through which \(m\) items enter and leave the warehouse. The items are stored in one of \(n\) storage spaces or locations. Each location requires the same storage space. It is known that item \(i\) requires \(S_i\) storage spaces. Ideally, we would like \(\sum_{i=1}^{m}S_i=n\). However, if the left-hand side is less than that of the right-hand side, we can always add a dummy product \((m+1)\) to take up the remaining \(n-\sum_{i=1}^{m}S_i\) spaces. We will assume throughout our discussion that the aforementioned equality holds. (Of course, if the right-hand side is less than the left-hand side, there is no feasible solution to the problem as there are not enough storage spaces to accommodate all the items.)”

— Facilities Design (Heragu 2022, sec. 8.6.7.1, p. 295).

“There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1 and Model 2, pp. 295–296).

Example: Assignment inputs

Locations

I/O

1

2

3

Distances (\(d_{pj}\))

	1	2	3
I/O	1	2	3

N. of trips/frequency

	a	b
I/O	20	10

Costs (\(c_{pj}\))

	a	b
I/O	1	1

Example: Weights and solution

Weights (\(w_{ij} = \frac{\sum_{k=1}^{n^{\mathrm{IO}}} c_{ik} f_{ik} d_{kj}}{S_i}\))

	1	2	3
a	\(\frac{1 \times 20 \times 1 }{1}=20\)	\(\frac{1 \times 20 \times 2 }{1}=40\)	\(\frac{1 \times 20 \times 3 }{1}=60\)
b	\(\frac{1 \times 10 \times 1 }{2}=5\)	\(\frac{1 \times 10 \times 2 }{2}=10\)	\(\frac{1 \times 10 \times 3 }{2}=15\)

Possible solutions

Solution	1	2	3	Travel cost
1	a	b	b	20 + 10 + 15 = 45
2	b	a	b	5 + 40 + 15 = 60
3	b	b	a	5 + 10 + 60 = 75

For each item–location pair, the slide evaluates Heragu’s three-factor cost. With one I/O point and unit movement cost, the weights are trip frequency times distance divided by required spaces. The three feasible assignments total 45, 60, and 75; therefore assigning \(a\) to location 1 and \(b\) to locations 2 and 3 gives the minimum of 45. These numbers are the course’s worked instance, not a reproduced source example.

“There are \(f_{ik}\) trips of item \(i\) through I/O point \(k\). The cost of moving a unit load of item \(i\) to and from I/O point \(k\) is \(c_{ik}\) and the distance of storage space \(j\) from I/O point \(k\) is \(d_{kj}\). Given the aforementioned and a binary decision variable \(x_{ij}\) that specifies whether or not item \(i\) is assigned to storage space \(j\), we are ready to formulate a model to assign the items to storage spaces in such a way that the cost of moving the items in and out of the I/O points is minimized. This is shown next in model 2. Note that the objective spreads the cost of moving item \(i\) in and out of storage evenly to all the \(S_i\) spaces occupied by the item. The cost for each item–space pair is calculated as a product of three factors: (1) frequency of trips made to each I/O point, (2) travel distance to each I/O point, and (3) cost per unit distance. It assumes that the cost per unit distance may vary for each I/O point–storage space combination.”

— Facilities Design (Heragu 2022, sec. 8.6.7.1 and Model 2, pp. 295–296).

Cube-per-order index (COI) policy

• Cube (size) per order (frequency—annual demand, S/R transactions).
• The COI policy places items near the I/O point when they
  1. have a large number of S/R requests and
  2. require less storage.
• COI for an item \(i\) in the period of interest \(T\) is defined as: \[ \text{COI} = \frac{\text{item storage space requirement (its cube)}}{\text{number of S/R transactions for the item}} \]
• Formally (see Manzini 2012): \[ \text{COI}_{i,T} = \frac{v_{i,T}}{\sum_{\text{order} \; j \; \in \; T}{x_{ij}}} \]
  • \(v_{i,T}\) = average storage level of SKU \(i\) in period \(T\).
  • \(x_{ij}\) = 1 if item \(i\) occurs in order \(j\); 0 otherwise.

The literature uses two closely related numerator conventions. Heragu and Zijm describe the cube or required storage space per transaction; Manzini’s displayed formula uses average storage level \(v_{i,T}\) per order incidence. The slide shows both the conceptual rule and Manzini’s period-specific formalization.

“The third policy—cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI. Allocate the first item in this list to the required number of storage spaces that are nearest to the input/output (I/O) point. Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point and so on until all the items are allocated. Thus, the COI policy puts items having a large number of S/R requests and requiring less storage space near the I/O point.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

“The cube-per-order index (COI) policy is between the two policies. While each item has its set of designated spaces, these spaces are located near the warehouse entry/exit points based on their size and frequency of usage—hence the name cube (size) per order (frequency). This policy essentially calculates the ratio of each SKU’s size to the frequency with which it is demanded (e.g., annual demand) and ranks the SKU’s in non-decreasing order of their COIs. The first SKU in the list is then assigned to the set of closest spaces required by that SKU, then the next SKU in the rank-ordered list is assigned to the next set of closest spaces required by that SKU, and so on, until all the SKUs are assigned. Next, a mathematical model (Model 3) for the COI storage policy is presented. The model determines the spaces to which each SKU must be assigned. It uses the following notation.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, sec. 15.4.2, pp. 343–344).

“Popularity (P). This is defined as the number of times an item belongs to an order in a given set of picking orders which refer to a period of time \(T\).

Cube per order index (COI). This can be defined as the ratio of volume storage to inventory for the generic SKU to the average number of occurrences of SKU in the order picking list for a given period of time (Haskett 1963). Given an SKU \(i\), COI is defined as the ratio of the volume of the stocks to the value of its popularity in the period of interest \(T\). Formally: \(COI_{i,T}=v_{i,T}/\sum_{\text{order }j\text{ in period }T}x_{ij}\), where \(v_{i,T}\) [is the] average storage level of SKU \(i\) in time period \(T\).”

— Warehousing in the Global Supply Chain (Manzini 2012, chap. 14, p. 376).

COI assignment model notation

Table 2: Parameters and variables (Zijm et al. 2019).

Symbol	Description
\(i\)	SKU index
\(j\)	Storage location index
\(k\)	I/O point index
\(n^{\mathrm{IO}}\)	Number of I/O points
\(n^{\mathrm{SKUs}}\)	Number of SKUs (supply points)
\(n^{\mathrm{storage}}\)	Number of storage locations
\(S_i\)	Number of product locations required by SKU \(i\)
\(f_{i}\)	Frequency of usage of SKU \(i\)
\(w_{j}\)	Distance of storage location \(j\) from the I/O point
\(x_{ij}\)	1 if location \(j\) is allocated to SKU \(i\); 0 otherwise

“cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI. Allocate the first item in this list to the required number of storage spaces that are nearest to the input/output (I/O) point. Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point and so on until all the items are allocated. Thus, the COI policy puts items having a large number of S/R requests and requiring less storage space near the I/O point. The COI policy has been investigated by several researchers. For references and results, see Malmborg and Bhaskaran (1990).”

— Facilities Design (Heragu 2022, 294).

“The cube-per-order index (COI) policy is between the two policies. While each item has its set of designated spaces, these spaces are located near the warehouse entry/exit points based on their size and frequency of usage—hence the name cube (size) per order (frequency). This policy essentially calculates the ratio of each SKU’s size to the frequency with which it is demanded (e.g., annual demand) and ranks the SKU’s in non-decreasing order of their COIs. The first SKU in the list is then assigned to the set of closest spaces required by that SKU, then the next SKU in the rankordered list is assigned to the next set of closest spaces required by that SKU, and so on, until all the SKUs are assigned. Next, a mathematical model (Model 3) for the COI storage policy is presented. The model determines the spaces to which each SKU must be assigned. It uses the following notation.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, 343).

“Cube per order index (COI). This can be defined as the ratio of volume storage to inventory for the generic SKU to the average number of occurrences of SKU in the order picking list for a given period of time (Haskett 1963). Given an SKU i, COI is defined as the ratio of the volume of the stocks to the value of its popularity in the period of interest T. Formally: COIi; T ¼ vi; T P order j in period T xij ð14:3Þ where vi,T average storage level of SKU i in time period T.”

— Warehousing in the Global Supply Chain (Manzini 2012, 367).

COI assignment model formulation

Objective Function

Minimizes the total distance of moving the SKUs into and out of their respective areas:

\[ \begin{align*} \min \quad Z & = \sum_{i=1}^{n^{\mathrm{SKUs}}} \sum_{j=1}^{n^{\mathrm{storage}}} \frac{f_{i}}{S_{i}}w_{j} x_{ij} \end{align*} \]

In Zijm’s one-I/O formulation, \(f_i/S_i\) measures activity per required location, \(w_j\) is location distance, and \(x_{ij}\) selects the assignment. Minimization therefore pairs high-activity-per-space SKUs with short-distance locations.

“The objective function (15.11) minimizes the total distance of moving the SKUs into and out of their respective areas. Constraint (15.12) ensures each SKU is assigned the required number of storage locations. Constraint (15.13) ensures that each storage location is assigned only one SKU. Constraint (15.14) is a binary constraint. The reader is referred to Heragu (2016) or Malmborg and Bhaskaran (1990) for additional details on the model.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, sec. 15.4.2, eqs. 15.11–15.14, p. 344).

COI assignment model constraints

Each SKU is assigned the required number of storage locations: \[ \begin{align} \sum_{j=1}^{n^{\mathrm{storage}}} x_{ij} = S_i, \quad i = 1,2,\ldots,n^{\mathrm{SKUs}} \\ \end{align} \]

Each storage location is assigned only one SKU: \[ \begin{align} \sum_{i=1}^{n^{\mathrm{SKUs}}} x_{ij} = 1, \quad j = 1,2,\ldots,n^{\mathrm{storage}} \\ \end{align} \]

Binary domain: \[ \begin{align} \quad & x_{ij} \in \{0,1\}, \quad i = 1,2,\ldots,n^{\mathrm{SKUs}}, \quad j = 1,2,\ldots,n^{\mathrm{storage}} \end{align} \]

The first constraint allocates exactly \(S_i\) locations to every SKU. The second permits one SKU in each location, and the binary domain makes each item–location decision yes/no.

“The objective function (15.11) minimizes the total distance of moving the SKUs into and out of their respective areas. Constraint (15.12) ensures each SKU is assigned the required number of storage locations. Constraint (15.13) ensures that each storage location is assigned only one SKU. Constraint (15.14) is a binary constraint. The reader is referred to Heragu (2016) or Malmborg and Bhaskaran (1990) for additional details on the model.”

— Operations, Logistics and Supply Chain Management (Zijm et al. 2019, sec. 15.4.2, eqs. 15.11–15.14, p. 344).

A fast solution algorithm

The COI policy can be summarized as follows:

1. Rank all the available storage locations in non-decreasing distance \(w_j\) from the I/O point.
2. List the items in a non-decreasing order of their COI.
3. Move down the two lists, assigning the next \(S_i\) locations to the next most highly ranked SKU \(i\).
   • Allocate the first item in this list to the required number of storage spaces nearest to the I/O point.
   • Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point.
   • … and so on until all the items are allocated.

The constructive rule coordinates two rankings: ascending location distance and ascending COI. After the SKUs are sorted, each receives its next \(S_i\) nearest available locations.

“The third policy—cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI. Allocate the first item in this list to the required number of storage spaces that are nearest to the input/output (I/O) point. Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point and so on until all the items are allocated. Thus, the COI policy puts items having a large number of S/R requests and requiring less storage space near the I/O point.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

Example: Problem statement

Consider two SKUs (\(a\), \(b\)) and three storage locations (1, 2, 3). SKU \(a\) demands 1 storage location (\(S_a = 1\)), and SKU \(b\) demands 2 storage locations (\(S_b = 2\)). The I/O point \(p\) is at the leftmost side of the warehouse, such that \(d_{p1} = 1\), \(d_{p2} = 2\), and \(d_{p3} = 3\). How should the items be allocated to minimize travel costs between the I/O point and the locations under a COI policy?

The two SKUs, three locations, capacities, trip frequencies, and location distances are an instructor-created numerical instance. The question applies the source rule: rank SKUs by COI and give the lower-COI SKU the nearest required locations.

“The third policy—cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI. Allocate the first item in this list to the required number of storage spaces that are nearest to the input/output (I/O) point. Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point and so on until all the items are allocated. Thus, the COI policy puts items having a large number of S/R requests and requiring less storage space near the I/O point.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

Example: COI inputs

Locations

I/O

1

2

3

Distances (\(d_{pj}\))

	1	2	3
I/O	1	2	3

N. of trips/frequency

	a	b
I/O	20	10

Costs (\(c_{pj}\))

	a	b
I/O	1	1

Example: COI ranking and assignment

• Location (distance):
  • 1(1)
  • 2(2)
  • 3(3)

• \(\frac{1}{COI}\) (\(\frac{f_{i}}{S_{i}}\)):
  • \(\frac{f_{a}}{S_{a}} = \frac{20}{1} = 20\)
  • \(\frac{f_{b}}{S_{b}} = \frac{10}{2} = 5\)

• Assignment:
  • 1=a
  • 2=b
  • 3=b

For the course instance, \(f_a/S_a=20/1=20\) and \(f_b/S_b=10/2=5\). Because \(f_i/S_i\) is the inverse of COI, sorting it in non-increasing order is equivalent to sorting COI in nondecreasing order. SKU \(a\) therefore receives nearest location 1; SKU \(b\) receives locations 2 and 3.

“The third policy—cube-per-order index (COI) policy—is operationally very simple and a widely used policy for allocating storage space to items in a warehouse. It was first proposed by Heskett (1963). COI for an item is defined as the ratio of the item’s storage space requirement (its cube) to the number of S/R transactions for that item. The COI policy can be summarized as follows. List the items in a nondecreasing order of their COI. Allocate the first item in this list to the required number of storage spaces that are nearest to the input/output (I/O) point. Allocate the second item in the list to the required number of storage spaces that are next closest to the I/O point and so on until all the items are allocated. Thus, the COI policy puts items having a large number of S/R requests and requiring less storage space near the I/O point.”

— Facilities Design (Heragu 2022, sec. 8.6.7, p. 294).

Notice that \({COI}^{-1}\) can also be used, as long as the ratios are sorted in non-increasing order.

Class-based storage policy

• SKUs are grouped into classes.

• Typically, in a warehouse:

  • 20% of items generate 80% of S/R activity
    
  • 30% of items generate 15% of S/R activity
    
  • 50% of items generate 5% of S/R activity

• Each class is assigned a dedicated storage area.

  • Randomized storage within dedicated area.

• Item classification:

  • Classify incoming items as A, B, or C according to their activity level.
  • Store class A close to the I/O point, class B in the next closest locations, and class C farther away.
  • Class (Activity%)
    • C (0%–5%)
      
    • B (5%–20%)
      
    • A (20%–100%)

“In a warehouse, 80% of the S/R activity is due to 20% of the items, 15% due to 30% of the items, and the remaining 5% of the S/R activity is due to 50% of the items.”

— Facilities Design (Heragu 2022, 294).

“Taking into account that products need more space than their time average stock, our results show that the optimal number of classes is relatively small, and three classes give near-shortest travel time for all ABC demand curves between 20%/ 50% and 20%/90%.”

— Class-Based Storage with a Finite Number of Items: Using More Classes Is Not Always Better (Yu et al. 2015, 1246).

“Thus, we classify incoming items into one of three classes—A, B, and C—depending on the level of S/R activity it generates. If it is between 0% and 5% of the total S/R activity, the item is a class C item; if it is between 5% and 20%, the item is a class B item. Otherwise, it is a class A item. To minimize the time spent in storage and retrieval, class A items must be stored closest to the input/output point, class B to the next closest location, and so on.”

— Facilities Design (Heragu 2022, 294).

“Although each class of items has a dedicated storage space, an item can be stored randomly in any available storage space in the location assigned to its class, just as any item in a dedicated policy can be randomly stored in any available space dedicated to that item. It is perhaps apparent to the reader that the random and dedicated policies are two extremes, and the other two discussed so far fall somewhere in between. For example, if all the items were grouped into one class in the class-based storage policy, what we have is randomized storage. On the other hand, if we have as many classes as items in a class-based storage policy, we have dedicated storage policy.”

— Facilities Design (Heragu 2022, 294).

Yu et al. (2015) shows that three classes give near-shortest travel times (more classes do not significantly improve performance).

Relation between storage strategies

• Random and dedicated policies are two extremes.
• Class-based and COI policies fall somewhere between random and dedicated.
  • If items were grouped into one class in the class-based storage policy = randomized storage.
  • If we have as many classes as items in a class-based storage policy = dedicated storage policy.

“It is perhaps apparent to the reader that the random and dedicated policies are two extremes, and the other two discussed so far fall somewhere in between. For example, if all the items were grouped into one class in the class-based storage policy, what we have is randomized storage. On the other hand, if we have as many classes as items in a class-based storage policy, we have dedicated storage policy.”

— Facilities Design (Heragu 2022, 294).

References

“The storage assignment problem deals with the assignment of products to storage locations in order to identify which is the best location for the generic item (Cahn 1948), called SKU. The assignment problem has been formalised by Frazelle and Sharp (1989) and classified as a non-polynomial (NP) hard problem.”

— Warehousing in the Global Supply Chain (Manzini 2012, 366).

Manzini (2012) explains the storage assignment problem in detail.

Heragu, Sunderesh S. 2022. Facilities Design. Fifth edition. CRC Press.

Manzini, Riccardo, ed. 2012. Warehousing in the Global Supply Chain. Springer London. https://doi.org/10.1007/978-1-4471-2274-6.

Tompkins, James A., ed. 2010. Facilities Planning. 4th ed. J. Wiley.

Weidinger, Felix, and Nils Boysen. 2018. “Scattered Storage: How to Distribute Stock Keeping Units All Around a Mixed-Shelves Warehouse.” Transportation Science 52 (6): 1412–27. https://doi.org/10.1287/trsc.2017.0779.

Yu, Yugang, René B. M. De Koster, and Xiaolong Guo. 2015. “Class-Based Storage with a Finite Number of Items: Using More Classes Is Not Always Better.” Article. Production and Operations Management 24 (8): 1235–47. https://doi.org/10.1111/poms.12334.

Zijm, Henk, Matthias Klumpp, Alberto Regattieri, and Sunderesh Heragu, eds. 2019. Operations, Logistics and Supply Chain Management. Lecture Notes in Logistics. Springer International Publishing. https://doi.org/10.1007/978-3-319-92447-2.