Extensions of a DRFLP model

In the classic models for the DRFLP, see, e. g., [8, 33, 42], the following three assumptions are made

1. the total size of the area needed for the arrangement is not limited, 2. vertical distances between the departments are neglected,

3. each department can be assigned to any of the two rows.

The approach of [42] enumerates over all distinguishable row assignments of the departments.

Then, the dummy department n+ 1 (n+ 2) is arranged at the left (right) border with `_n+1 =

`_n+2 = 0, w_ij = w_ji = 0, i ∈ [n+ 2], j ∈ {n+ 1, n+ 2}, i < j, and an MILP model is used to solve several DRFLP with fixed-row assignment (FR-DRFLP). In order to consider a fixed row assignments, let R₁ ⊆[n] (R2 ⊆[n]) denote the set of departments assigned to row 1 (row 2).

The dummy departments n+ 1 and n+ 2 are assigned to both rows. The MILP model of [42]

makes use of betweenness variables x_ikj =x_jki =

(1, k lies betweeni andj in the same row, 0, otherwise,

for l ∈ {1,2}, i, j, k ∈ R_l∪ {n+ 1, n+ 2},|{i, j, k}| = 3, i < j. Further, let d_ij = d_ji, i, j ∈ [n+ 2], i < j,measure the horizontal center-to-center distance betweeniandj. The betweenness variables containing a dummy departmentn+ 1 orn+ 2 are combined with the distance variables via big-M-inequalities. We refer to [42] for a description of theMILP model and to Section C for a summary of the MILP model.

Usually, in factory planning the incoming warehouse and the shipping warehouse of a factory are arranged at the left and at the right border, respectively. If this is the case, the dummy departments n+ 1 andn+ 2 can be interpreted as these warehouses, see Figure 3.3.1. Of course, we might obtain a better overall solution value if we drop the restriction on the position of both warehouses. In this case they are treated as ordinary departments that have transport weights to other departments and need a certain space.

In this section we extend the approach of [42] in order to overcome these three assumptions.

Let the departments be given as two-dimensional objects such that each department has a length and a height. Then, the area of a layout is defined in the following way:

Definition 3.3.1. The area of a given layout is defined as the area of the minimum boundary rectangle containing all departments.

By definition the area of a layout is equal tod_(n+1)(n+2)·h, whereh is the height of the layout.

Let h_i denote the height of departmenti∈[n]. In the FR-DRFLPwe compute the height of the

incomingwarehouse shippingwarehouse

Figure 3.3.1: Visualization of an extended DRFLP layout where we fixed the incoming and the shipping warehouse to the border of the layout. One motivation for this arrangement is that one hopes to receive rather linear transport flows between the departments.

layout by summing up the height of a department with the largest height in each row plus the width of the pathw^D_path between the two rows, i. e.,h= maxi∈R1h_i+ maxi∈R2h_i+w_path^D . So for a fixed row assignment the height of a layout is constant. Assume that the area of the layout may be at most F ∈R≥0. Then, the linear inequality d_(n+1)(n+2)≤ ^F_h ensures that the area of the layout is bounded by F. Further, we can neglect all row assignments where the sum of the lengths of the departments in the same row exceeds ^F_h.

Apart from a restriction of the used area, there might appear so called blocked areas in real-world factory planning problems. It is not allowed to place departments in these areas. This might be due to already existing departments or due to safety restrictions. Let B₁ ={[b1, b₁+g₁], . . . ,[bu, bu+ g_u]}be the blocked areas in row 1 andB₂ ={[bu+1, b_u+1+g_u+1], . . . ,[bv, b_v+g_v]}be the blocked areas in row 2 for given b_k, g_k ≥0, k∈[v]. For each blocked area we introduce a new dummy department, which we will call blocked department, with length equal to the length of the blocked area. We place the center of the blocked department in the middle of the blocked area. So we get the blocked departments n+ 3, n+ 4, . . . ,(n+ 2 +|B₁|+|B₂|) with length`_n+2+k=g_k fork∈[v].

The row assignment of the blocked departments is fixed, namelyR¹={(n+3), . . . ,(n+2+|B₁|)}

are assigned to row 1 and R² = {(n+ 2 +|B₁|+ 1), . . . ,(n+ 2 +|B₁|+|B₂|)} to row 2. To ensure that the blocked department n+ 2 +klies exactly on the interval [bk, b_k+g_k], we set the distance variable to

d(n+1)(n+2+k)=b_k+g_k

2 , k∈[v].

Next we include vertical distances between the departments in order to overcome the second assumption stated in the beginning of this section. The departments are given as 2-dimensional objects. The vertical center-to-center distance between two departments in the same row equals the sum of the heights of these two departments. If two departments are in distinct rows, one additionally adds w_path^D . An example is illustrated in Figure 3.3.2. In conclusion, we only need to add the following constant value to the objective value of some FR-DRFLPto include vertical distances

X

i,j∈[n]

i<j

h_i+h_j

2 (wij+w_ji) + ^X

i∈R1

j∈R₂

w_path^D (wij+w_ji).

Naturally, the third assumption mentioned at the beginning of this section can easily be dropped. If the row assignment of some departments is fixed in advance, this only helps us because the number of possible row assignments decreases.

In this exact approach, we enumerate over all row assignments of the departments and solve some MILP model in each step. One can fix one department to row 1, so there are in general

12 ·2ⁿ distinguishable row assignments that have to be considered. In realistic instances there

n+ 1 n+ 2

Figure 3.3.2: Visualization of the vertical and horizontal center-to-center distances between three departments.

appear departments of the same type, i. e., departments that have the same length and the same transport weightw to all other departments. In our test case in Section 3.3.3 we consider such a realistic instance. We use this additional information to reduce the number of distinguishable row assignments significantly.

Theorem 3.3.2. Letk denote the number of different department types and leta_i be the number of departments of type i∈[k]. Then there are at most





 1 2

Y

i∈[k]

(ai+ 1)





 distinguishable row assignments.

This formula is also correct if all departments have different types, because thenai= 1 for all i∈[k] andk=n. We illustrate the advantages of Theorem 3.3.2 by a realistic example, see [95]

and Section 3.3.3.

Example 3.3.3. We are given n= 21departments, where two departments appear four times, three departments twice and seven departments just once. Without reduction, we have to test 2²⁰= 1048576 row assignments. By Theorem 3.3.2 we obtain at most ¹₂·5·5·3·3·3·2⁷ = 43200 distinguishable row assignments.

Considering departments of the same type, we can strengthen our MILP model. We break symmetries by fixing the ordering of departments of the same type in the same row. This symmetry breaking is done in such a way that at least one optimal solution is preserved. Let a_i1 departments of the same type i, i∈[k],be in row 1. We denote these departments, w. l. o. g., by 1, . . . , ai1. Then, we fix the ordering of these departments by additional constraints, w. l. o. g., we use an ascending order. Since these departments are of the same type, they have the same length and we can add

d_(n+1)1≤d_(n+1)2+`<sub>1</sub>≤. . .≤d<sub>(n+1)(ai1)</sub>+ (ai<sub>1</sub>−1)·`₁

to our model. It follows immediately that we can set the betweenness variables which contain dummy departments to

x_(n+1)kl=

(1, k, l∈[ai₁], k < l, 0, k, l∈[ai1], k > l.

Similar equations can be added for department n+ 2. Furthermore, we fix the associated betweenness variables

x_kuv =

(1, k, u, v∈[ai1], k < u < v,

0, k, u, v∈[ai1], k < v and (u < kor u > v).