The single facility location problem with time-dependent weights and relocation cost over a continuous time horizon

Abstract

In this study, we investigate the problem of locating a facility in continuous space when the weight of each existing facility is a known linear function of time. The location of the new facility can be changed once over a continuous finite time horizon. Rectilinear distance and time- and location-dependent relocation costs are considered. The objective is to determine the optimal relocation time and locations of the new facility before and after relocation to minimize the total location and relocation costs. We also propose an exact algorithm to solve the problem in a polynomial time according to our computational results.

Appendices

Appendix A

The algorithm by Drezner and Wesolowsky (1991)

Step 1::

Calculate X ⁽¹⁾=(x ⁽¹⁾, y ⁽¹⁾) and X ⁽²⁾=(x ⁽²⁾, y ⁽²⁾) when B=0. Set B=0, K ^*=K(0, X ⁽¹⁾, X ⁽²⁾) and B ^*=0.

Step 2::

Calculate the eight break points B _n ′ for moving the location of the new facility to its adjacent location (note that there are two break points for each movement) by B′ _n =2(∑ _i=1 ^j u _i −∑ _i=j+1 ^m u _i )/(∑ _i=j+1 ^m v _i −∑ _i=1 ^j v _i ), (for moving from j to j+1 or vice versa), for n=1, …, 8.

Step 3::

Find the minimum of these break points B″=min _{n=1, …, 8}{B′ _n |B′ _n >B}. If B″>T or B″ is not defined, set B″=T. The next segment is [B, B″] Find B _min by B _min =−∑ _i=1 ^m[u _i (d _i ⁽¹⁾−d _i ⁽²⁾)]/∑ _i=1 ^m[v _i (d _i ⁽¹⁾−d _i ⁽²⁾)] where d _i ⁽¹⁾=∣x ⁽¹⁾−x _i ∣+∣y ⁽¹⁾−y _i ∣ and d _i ⁽²⁾=∣x ⁽²⁾−x _i ∣+∣y ⁽²⁾−y _i ∣. If B⩽B _min ⩽B″ and the denominator of above relation for B _min is positive, set B ⁰=B _min ; otherwise, set B ⁰=B″. Calculate K ⁰=K(B ⁰).

Step 4::

If K ⁰<K ^*, update K ^*=K ⁰ and B ^*=B ⁰.

Step 5::

If B″=T, then B ^* is the optimal solution and K ^* is the optimal objective value. Otherwise, go to Step 6.

Step 6::

Change the value of x ⁽¹⁾, y ⁽¹⁾, x ⁽²⁾ or y ⁽²⁾ to its new value based on the value of n for which B″ is obtained in Step 3. Calculate the next B′ _n for this particular n. Set B=B″ and go to Step 3.

Appendix B

Proof of Lemma 2 Let A ^*=(B ^*, X ^(1)*, X ^(2)*)=(B ^*, x ^(1)*, y ^(1)*, x ^(2)*, y ^(2)*) and W _i ⁽¹⁾=∫₀ ^B* w _i (t)dt, W _i ⁽²⁾=∫ _B* ^T w _i (t)dt; i=1, …, m.

Therefore, we have

and

Now, we aim to prove that (8) and (9) will hold at optimality by using contradiction.

Suppose that x _l <x ^(1)*<x ^(2)*<x _l=1, x ^(1)*−x _l =d>0, x ^(2)*−x ^(1)*=a>0 and x _l+1−x ^(2)*=b>0. Then, we obtain

Now, suppose x ^(1)* and x ^(2)* are both decreased to obtain two new x-coordinates x′⁽¹⁾ and x′⁽²⁾ such that x′⁽¹⁾=x ^(1)*−d=x _l and x′⁽²⁾=x ^(2)*−d=x _l +a, where both a and d are positive. (Note that we do not decrease either y ^(1)* or y ^(2)*; ie y′⁽¹⁾=y ^(1)* and y′⁽²⁾=y ^(2)*.)

Then, we get

Noting that W _i ⁽¹⁾+W _i ⁽²⁾=∫₀ ^B* w _i (t)dt+∫ _B* ^T w _i (t)dt=∫₀ ^T w _i (t)=V _i (T) and the above equations, we obtain

Now, suppose that both y ^(1)* and y ^(2)* are increased to obtain two new x-coordinates x″⁽¹⁾ and x″⁽²⁾ such that x″⁽¹⁾=x ^(1)*+b=x _l+1−a and x″⁽²⁾=x ^(2)*+b=x _l+1. Then, we have

However, without loss of generality, the expression ∑ _i=1 ^l V _i (T)−∑ _i=l+1 ^m V _i (T) can be equal to zero, positive, or negative. (If the expression is equal to zero, ie ∑ _i=1 ^l V _i (T)−∑ _i=l+1 ^m V _i (T), then decreasing or increasing x ^(1)* and x ^(2)*simultaneously will not change the optimal cost and hence we have multiple optimal solutions.) If it is positive, then F(B ^*, X′⁽¹⁾, X′⁽²⁾) is less than F(B ^*, X ^(1)*, X ^(2)*). If the expression is negative, then F(B ^*, X″⁽¹⁾, X″⁽²⁾) is less than F(B ^*, X ^(1)*, X ^(2)*). Therefore, there are situations in which F(B ^*, X′⁽¹⁾, X′⁽²⁾) or F(B ^*, X″⁽¹⁾, X″⁽²⁾) is less than F(B ^*, X ^(1)*, X ^(2)*), which implies that there is a solution that is better than the optimal solution! This is a contradiction. Therefore, an optimal solution cannot satisfy x _l <x ^(1)*<x ^(2)*<x _l+1 and x _l <x ^(2)*<x ^(1)*<x _l+1. Similarly, the optimal solution cannot satisfy y _k <y ^(1)*<y ^(2)*<y _k+1 and y _k <y ^(2)*<y ^(1)*<y _k+1. Therefore, (8) and (9) are held at an optimal solution of F. Because (8) and (9) are held at optimality, the triangle inequalities (4) and (5) will become the equalities and hence, F(A ^*)=H(A ^*). □

Appendix C

illustration

Note that the performance of the above algorithm can be accelerated by checking if a certain BASELP has already been solved.