Synthetic Test Problems for the Size Assortment Problem

 

 

Generation Procedure

 

Six different beta distributions, with parameters α, β listed below, were used as base distributions:

 

α = 0.5,  β = 0.5

α = 1.0,  β = 2.0

α = 2.0,  β = 3.0

α = 2.0,  β = 2.0

α = 3.0,  β = 1.0

α = 3.0,  β = 2.0

 

Each such distribution was discretized at s equally spaced points, where s is the number of sizes. These base probabilities were then each independently multiplied by a random factor uniformly distributed over the interval [1 – ε, 1 + ε]. Since the resulting values do not in general sum to 1, we renormalized the values to produce genuine probabilities for the s sizes in this generated profile for this store. The store volume, randomly generated from [5, 180], was then multiplied by the profile values to obtain the demand distributions tabulated later.

 

Experimental Design

 

Data sets were produced for n = 30, 90, 150 stores with equal numbers of profiles generated from each of the m = 6 base distributions. In the problems below, the number of sizes is s = 4 and the perturbation parameter ε was selected from {0.15, 0.20, 0.25}. For each pair (n, ε) five replications were randomly generated.

 

Data Sets and Solutions

 

The three tables below correspond to  n = 30, 90, 150 stores. The three columns of each table correspond to the values 0.15, 0.20, 0.25 of the perturbation parameter ε. The five rows correspond to the five replications for each pair (n, ε).

 

Each data set in the table has a header record containing the value n, s, m. The succeeding records contain a store number (repeated in the first and second  columns) followed by the s demand values generated for that store.

 

In the majority of the cases, the best solution found was the “natural” solution with bins containing all the n/m profiles generated from the same base distribution. Specifically, the natural binning has the first n/m stores from {1, 2, ..., n} in the first bin, the next n/m stores from {1, 2, ..., n} in the second bin, and so forth. In 8 of the 45 test problems, however, the natural solution was not the best found, and the best solutions known in these cases are given in the files soln(1), soln(2), ..., soln(8).

 

 

n = 30

 

 

 

0.15

0.20

0.25

replication 1

30e15a

30e20a

30e25a, soln(1)

replication 2

30e15b

30e20b

30e25b

replication 3

30e15c

30e20c

30e25c, soln(2)

replication 4

30e15d

30e20d

30e25d

replication 5

30e15e

30e20e

30e25e

 

 

n = 90

 

 

 

0.15

0.20

0.25

replication 1

90e15a

90e20a

90e25a

replication 2

90e15b

90e20b

90e25b

replication 3

90e15c

90e20c, soln(3)

90e25c, soln(4)

replication 4

90e15d

90e20d

90e25d

replication 5

90e15e

90e20e

90e25e

 

n = 150

 

 

 

0.15

0.20

0.25

replication 1

150e15a

150e20a

150e25a, soln(5)

replication 2

150e15b

150e20b, soln(6)

150e25b, soln(7)

replication 3

150e15c

150e20c

150e25c

replication 4

150e15d

150e20d

150e25d, soln(8)

replication 5

150e15e

150e20e

150e25e