WH - Storage Configuration

7/29/2019 WH - Storage Configuration

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Warehouse Storage Configurationand Storage Policies

Bibliography

Bartholdi & Hackman: Chapter 6

Francis, McGinnis, White: Chapter 5

Askin and Standridge: Sections 10.3 and 10.4


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Storage Policies

Main Issue: Decide how to allocate the various storagelocations of a uniform storage medium to a number of

SKUs.

I/O


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Types of Storage Policies

Dedicated storage: Every SKU i gets a number of storagelocations, N_i, exclusively allocated to it. The number ofstorage locations allocated to it, N_i, reflects its maximumstorage needs and it must be determined through inventory

activity profiling. Randomized storage: Each unit from any SKU can by

stored in any available location

Class-based storage:SKUs are grouped into classes. Eachclass is assigned a dedicated storage area, but SKUswithin a class are stored according to randomized storagelogic.


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Location Assignment underdedicated storage policy

Major Criterion driving the decision-making process:

Enhance the throughput of your storage and retrieval

operations by reducing the travel time reducing the

travel distance How? By allocating the most active units to the most

convenient locations...


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Convenient Locations

Locations with the smallest distanced_j to the I/O point! In case that the material transfer is performed through a

forklift truck (or a similar type of material handlingequipment), a proper distance metric is the, so-called,rectilinearorManhattan metric (orL1 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 unitcan move simultaneously in both axes, with uniform speed,the most appropriate distance metric is the, so-called

Tchebychev metric (orL norm):d_j = max (|x(j)-x(I/O)|,|y(j)-y(I/O)|)


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Active SKUs

SKUs that cause a lot of traffic!

In steady state, the appropriate activity measure for a

given SKU i:

Average visits per storage location per unit time =(number of units handled per unit of time) /

(number of allocated storage locations) =

TH_i / N_i


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A fast solution algorithm

Rank all the available storage locations in increasing

distance from the I/O point, d_j.

Rank all SKUs in decreasing turns, TH_i/N_i.

Move down the two lists, assigning to the next most highly

ranked SKU i, the next N_i locations.


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Example

I/O

I/O

0 11

2

2

2 2

2

3

3

3

3

4

3

3

34

4

4

4

5

4

4

4

45

5

5

5

5 5

5

5

5

5

6

6

6

6 6

66

6

7

7

7 7

77

8

8 8

8

9 9A: 20/10=2

B: 15/5 = 3

C: 10/2 = 5

D: 20/5 = 4

CC

D

D

D

D

B

D

BB

B

B

A

A

AA

A

A

A

A AA


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Problem Formulation

Decision variables: x_ij = 1 if location j is allocated toSKU i; 0 otherwise.

Formulation:

min

S_i

S_j [(TH_i/N_i) * d_j] * x_ij

s.t.

i, S_j x_ij = N_i

j, S_i x_ij = 1 i, j, x_ij {0,1} => x_ij 0


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Problem Representation

SKU Location

N_1

N_i

N_S

1

1

1

c_ij = (TH_i/N_i)*d_j

1

i

S

1

j

L


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Remarks

The previous problem representation corresponds to abalanced transportation problem: Implicitly it has been

assumed that: L =S_iN_i For the problem to be feasible, in general, it must hold that:

L S_iN_i If L -S_iN_i > 0, the previous balanced formulation is

obtained by introducing a fictitious SKU 0, with

N_0 = L -S_iN_i and TH_0 = 0


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Locating the I/O point

In many cases, this location is already predetermined bythe building characteristics, its location/orientation with

respect to the neighboring area/roads/railway tracks, etc.

Also, in the case of an AS/RS, this location is specified by

the AS/RS technical/operational characteristics. In case that the I/O point can be placed at will, the ultimate

choice should seek to enhance its proximity to the

storage locations.


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Locating the I/O point: Example 1Option A

I/O0 11

2

2

2 2

2

3

3

3

3

4

33

34

44

4

5

44

4

45

5

5

5

5 5

5

5

5

5

66

66 6

6

66

7

7

7 7

7

7

8

8 8

8

9 9

Option B

I/O

0 1

2 2

2

3 3 4

3

3

4

4 5

5

5

44 5

55 6

6

6

6

6

7

7

7

7

7

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8

8

8

8

9

9

9

9

9

10

10

10

10

10

11

11

11

11

12

12

12

13

13

14


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Locating the I/O point: Example 2

I/O

0 22

4

4

4 4

4

6

6

6

6

8

6

6

68

8

8

810

88

8

810

10

10

1010 10

10

10

10

10

12

12

1212 12

12

12

12

14

1414 14

14

14

16 16 16 1618 18

Option C

15 13

15 13

11

15 13 11

11

9

9

7 6

7

9

1315 11

1315 11

9

7

7

9

7

7

7

9

7

7

79

11

6

7

9

11

13

7

9

1113

15

9

11

13

15

11

13

15

13

15

15

I

O

Option A

7


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Example 2 (cont.)

Option A: U-shaped or cross-docking configuration

amplifies the convenience/inconvenience of close/distant locations appropriate for product movement with strong ABC skew

provides flexibility for interchanging between shipping and receivingdocking capacity

allows for dual command operation of forklifts, reducing, thus, the

deadhead traveling minimizes truck apron and roadway

Option C: Flow-through configuration

attenuates the convenience difference among storage locations

conservative design: more reasonably convenient storage locations

but fewer very convenient more appropriate for extremely high volume

preferable when the building is long and narrow

limits the opportunity for efficiencies for dual command operations


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Storage Sizing

Randomized Storage:

How many storage locations, N, should be employed for thestorage of the entire SKU set?

Dedicated storage:

How many storage locations, N_i, should be dedicated to each

SKU i? Given a fixed number of available locations, L, how should these

locations be distributed among the various SKUs?

Class-based storage:

How should SKUs be organized into classes? How many storage locations, N_k, should be dedicated to each

SKU class k?


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Possible Approaches toStorage Sizing

Quite often, this issue is resolved/predetermined from the overalloperational context: e.g., replenishment policies, contractualagreements, etc., which impose some structure on the manner in whichrequests for storage locations are posed by the various SKUs

Service-level type of analysis:

Determine the number of storage locations, N_i to be assigned toeach SKU i so that the probability that there will be no shortage ofstorage space in any operational period (e.g., day) is equal to orgreater than a pre-specified value s.

Cost-based Analysis

Select N_is in a way that minimizes the total operational cost overa given horizon, taking into consideration the cost of owning andoperating the storage space and equipment, and also any additionalcosts resulting from space shortage and/or the need to contractadditional storage space.


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Sizing randomized storage based onservice level requirements

Q = max number of storage locations requested at any single operational period(a random variable)

p_k = Prob(Q=k), k=0,1,2, (probability mass function for Q)

F(k) = Prob(Qk) =S_{j=0,,k}p_j (cumulative distribution function for Q)

Problem Formulation

Find the smallest number of locations N, that will satisfy a requested service

level s for storage availability, i.e.,

minN

s.t.

F(N) sN 0

Solution:

N = min{k: S_{j=0,,k}p_j s}


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Sizing dedicated storage based onservice level requirements

Q_i = max number of storage locations requested at any singleoperational period for the storage of SKU i (random variable)

F_i(k) = Prob{Q_i k} (cumulative distribution function of Q_i) If a distinct service level s_i is defined for each SKU i, then the

determination of N_i is addressed independently for each SKU, accordingto the logic presented for the randomized storage policy.

Next we address the problem of satisfying a single service levelrequirement, s, defined for the operation of the entire system, i.e.,

Prob{no storage shortages in a single day} sunder the additional assumption that the storage requirements posed byvarious SKUs are independent from each other.

Then, for an assignment of N_i locations to each SKU i,

Prob{no storage shortages in a single day} = _i F_i(N_i)and

Prob{1 or more storage shortages} = 1 - _i F_i(N_i)


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Sizing dedicated storage based onservice level requirements (cont.)

Formulation I: Fixed service level, s

min S_i N_is.t.

_i F_i(N_i) sN_i 0 i

Formulation II: Fixed number of locations, L

max _i F_i(N_i)

s.t.S_i N_i L

N_i 0 i


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Class-Based Storage Sizing andLocation Assignment

Divide SKUs into classes, using ABC (Pareto) analysis, based ontheir number of turns TH_i/N_i.

Determine the required number of storage locations for each classC_k

ad-hoc adjustmentof the total storage requirement of the class SKUs

N_k = p * S_{iC_k } N_i, where 0 < p < 1 Class-based service-level type of analysis:

Q_k = storage space requirements per period for class k = S_{iC_k} Q_iFor independent Q_i:

p_k(m) = Prob(Q_k=m) =S_{m_i: S_i m_i = m}[_ip_i(m_i)]where p_i( ) : probability mass function for Q_i.

Assign to each class the requested storage locations, prioritizingthem according to their number of turns,

TH_k/N_k where TH_k = S_{iC_k } TH_i


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A simple cost-based modelfor (dedicated) storage sizing

Model-defining logic:Assuming that you know your

storage needs d_ti, for each SKU i, over a planning horizon

T, determine the optimal storage locations N_i for each

SKU i, by establishing a trade-off between the

fixed and variable costs for developing this set of locations, andoperating them over the planning horizon T, and

the costs resulting from any experienced storage shortage.


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A simple cost-based model

for (dedicated) storage sizing (cont.) Model Parameters:

T = length of planning horizon in time periods

d_ti = storage space required for SKU i during period t

C_0 = discounted present worth cost per unit storage capacityowned during the planning horizon T

C_1 = discounted present worth cost per unit stored in ownedspace per period

C_2 = discounted present worth cost per unit of space shortage (e.g., per unitstored in leased space) per period

Model Decision Variables:

N_i = owned storage capacity for SKU i

Model Objective: min TC (N_1,N_2,,N_n) =

S_i [C_0 N_i + S_t {C_1 [min(d_ti, N_i)] + C_2 [max(d_ti - N_i, 0)]}]

A f l i l i h f h f


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A fast solution algorithm for the case oftime-invariant costs

For each SKU i: Sequence the storage demands appearing in the d_ti, t=1,T,

sequence in decreasing order.

Determine the frequency of the various values in the ordered

sequence obtained in Step 1.

Sum the demand frequencies over the sequence. When the obtained partial sum is first equal to or greater than

C = C_0/ (C_2-C_1)

stop; the optimum capacity for SKU i, N_i, equals the

corresponding demand level.


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Example

Problem Data:

N=1; T=6; d = < 2, 3, 2, 3, 3, 4,>; C_0 = 10, C_1 = 3, C_2 = 5

Solution:

Stor. Demand Frequency Partial Su

4 1 1

3 3 4

2 2 6

C = C_0/(C_2-C-1) = 10/(5-3) = 5

=> N = 2

St C fi ti d P li i


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Storage Configuration and Policiesfor Unit Load warehouses:

Topics covered

Storage Policies: Assigning storage locations of a uniform

storage medium to the various SKUs stored in that

medium

Dedicated Randomized

Class-based

Criterion: Maximize productivity by reducing the traveling effort /

cost

The placement of the I/O point(s)

Criterion: Maximize productivity by reducing the traveling effort /

cost

St C fi ti d P li i


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Storage Configuration and Policiesfor Unit Load warehouses:

Topics covered (cont.)

Storage sizing for various SKUs: Determine the numberof storage locations to be assigned to each SKU / group ofSKUs.

Criterion:

provide a certain (or a maximal) service level minimize the total (space+equipment+labor+shortage) cost over a

planning horizon

Next major theme: Storage Configuration for better spaceexploitation

floor versus rack-based storage for pallet-handling warehouses determining the lane depth (mainly forrandomized storage)

(based on Bartholdi & Hackman, Section 6.3)

D t i i th E l t ( d


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Determining the Employment (andConfiguration) of Rack-based storage

Basic Logic:

For each SKU,

compute how many pallet locations would be created by moving itinto rack of a given configuration;

compute the value of the created pallet locations;

move the sku into rack if the value it creates is sufficient to justify the

rack. Remark: In general, space utilization will be only one of

the factors affecting the final decision on whether to movean SKU into rack or not. Other important factors can be

the protection that the rack might provide for the pallets of the

considered SKU; the ability to support certain operational schemes, e.g., FIFO

retrieval;

etc.


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Examples on evaluating the efficienciesfrom moving to rack-based storage

Case I: Utilizing 3-high pallet rack for an SKU of N=4(pallets), which is not stackable at all. Current footprint: 4 pallet positions

Introducing a 3-high rack in the same area creates 3x4=12 position,8 of which are available to store other SKUs. What are the gains

of exploiting these new locations vs the cost of purchasing andinstalling the rack?

Case II: Utilizing a 3-high pallet rack for an SKU withN=30 (pallets), which are currently floor-stacked 3-high, tocome within 4 ft from the ceiling.

Current footprint: 10 pallet positions

Introducing a 3-high rack does not create any new positions, and itwill actually require more space in order to accommodate the rackstructure (cross-beams and the space above the pallets, required for

pallet handling)

D t i i ffi i t l d th


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Determining an efficient lane depth(in case of randomized storage)

A conceptual characterization of the problem:

More shallow lanes imply more of them, and therefore, more spaceis lost in aisles (the size of which is typically determined by the

maneuvering requirements of the warehouse vehicles)

On the other hand, assuming that a lane can be occupied only by

loads of the same SKU, a deeper lane will have many of its

locations utilized over a smaller fraction of time(honeycombing).

So, we need to compute an optimal lane depth, that balances out

the two opposite effects identified above, and minimizes the

averagefloor space required for storing all SKUs.

LanesLane Depth

(3-deep)

Lane Height

Aisle


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Notation w = pallet width

d = pallet depth

g = gap between adjacent lanes

a = aisle width

x = lane depth

n = number of SKUs

N_i = max storage demand by SKU i

z_i = column height for SKU I

lane footprint = (g+w)(d*x+a/2)


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Key results

Assuming that the same lane depth is employed across all n SKUs,under floor storage, the average space consumed per pallet isminimized by a lane depth computed approximately through thefollowing formula:

x_opt =[(a/2dn)*_i (N_i /z_i)] The optimal lane depth for any single SKU i, which is stackable z_i

pallets high, is

x_opt =[(a/2d)*(N_i /z_i)] Assuming that the same lane depth is employed across all n SKUs,

under rack storage, the average space consumed per pallet isminimized by a lane depth computed approximately through the

following formula:

x_opt =[(a/2dn)*_i N_i ]

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