Warehouse Storage Configuration and Storage Policies. Bibliography Bartholdi & Hackman: Chapter 6 Francis, McGinnis, White: Chapter 5 Askin and Standridge: Sections 10.3 and 10.4. I/O. Storage Policies.
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Warehouse Storage Configuration and Storage Policies
Bibliography
Bartholdi & Hackman: Chapter 6
Francis, McGinnis, White: Chapter 5
Askin and Standridge: Sections 10.3 and 10.4
I/O
d_j = max (|x(j)-x(I/O)|,|y(j)-y(I/O)|)
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
5
9
8
7
5
6
7
8
9
5
6
7
6
5
4
4
4
5
8
8
6
7
4
5
4
3
3
3
6
7
6
7
5
6
5
4
2
2
2
3
6
3
4
5
3
5
4
3
2
1
0
1
2
4
5
I/O
I/O
A: 20/10=2
B: 15/5 = 3
C: 10/2 = 5
D: 20/5 = 4
A
A
A
A
A
B
B
A
D
D
D
A
A
B
B
A
A
C
C
D
D
B
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
Location
SKU
N_1
1
1
1
c_ij = (TH_i/N_i)*d_j
N_i
i
1
j
N_S
S
L
1
L S_iN_i
N_0 = L - S_iN_i and TH_0 = 0
5
9
8
7
5
6
7
8
9
5
6
7
6
5
4
4
4
5
8
8
6
7
4
5
4
3
3
3
6
7
6
7
5
6
5
4
2
2
2
3
6
3
4
5
3
5
4
3
2
1
0
1
2
4
5
Option B
6
7
9
10
12
13
14
11
5
5
8
4
8
11
4
5
6
7
9
10
12
13
3
3
4
5
6
8
9
10
11
7
12
4
6
8
10
2
11
2
3
5
7
9
0
1
2
3
4
5
6
7
8
9
10
I/O
I/O
Option A
I/O
Option A
12
10
10
10
18
16
14
12
14
16
18
16
14
12
10
8
8
8
10
16
12
14
8
14
12
6
10
10
8
6
6
12
14
12
6
4
8
10
10
8
4
4
6
12
10
6
8
6
4
2
0
2
4
8
10
O
Option C
11
9
6
7
11
13
15
15
9
13
7
15
7
11
11
9
7
7
9
13
15
13
15
13
11
9
7
9
11
13
7
7
15
9
7
9
13
15
15
13
11
7
7
11
15
13
11
9
7
6
7
9
11
13
15
I
Find the smallest number of locations N, that will satisfy a requested service
level s for storage availability, i.e.,
min N
s.t.
F(N) s
N 0
N = min{k: _{j=0,…,k} p_j s}
Prob{no storage shortages in a single day} s
under the additional assumption that the storage requirements posed by various SKU’s are independent from each other.
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)
min _i N_i
s.t.
_i F_i(N_i) s
N_i 0 i
max _i F_i(N_i)
s.t.
_i N_i L
N_i 0 i
N_k = p * _{iC_k } N_i, where 0 < p < 1
Q_k = storage space requirements per period for class k = _{iC_k} Q_i
For independent Q_i:
p_k(m) = Prob(Q_k=m) = _{m_i: _i m_i = m}[_ip_i(m_i)]
where p_i( ) : probability mass function for Q_i.
TH_k/N_k where TH_k = _{iC_k } TH_i
S_i [C_0 N_i + S_t {C_1 [min(d_ti, N_i)] + C_2 [max(d_ti - N_i, 0)]}]
C’ = C_0/ (C_2-C_1)
stop; the optimum capacity for SKU i, N_i, equals the corresponding demand level.
N=1; T=6;d = < 2, 3, 2, 3, 3, 4,>;C_0 = 10, C_1 = 3, C_2 = 5
C’ = C_0/(C_2-C-1) = 10/(5-3) = 5
=> N = 2
Criterion: Maximize productivity by reducing the traveling effort / cost
Criterion: Maximize productivity by reducing the traveling effort / cost
Criterion:
(based on Bartholdi & Hackman, Section 6.3)
Lane Height
Lane Depth
(3-deep)
Lanes
Aisle
x_opt = [(a/2dn)*_i (N_i /z_i)]
x_opt = [(a/2d)*(N_i /z_i)]
x_opt = [(a/2dn)*_i N_i ]