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 locationdependent 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.
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Acknowledgements
We are grateful for the constructive comments of the associate editor and two reviewers. The second author acknowledges that his research was supported by a grant from the National Natural Science Foundation of China (71271183) and two grants (201011159026, 201111159056) from the University Research Committee of the University of Hong Kong.
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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 xcoordinates 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 xcoordinates 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
Note that the performance of the above algorithm can be accelerated by checking if a certain BASELP has already been solved.
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Zanjirani Farahani, R., Szeto, W. & Ghadimi, S. The single facility location problem with timedependent weights and relocation cost over a continuous time horizon. J Oper Res Soc 66, 265–277 (2015). https://doi.org/10.1057/jors.2013.169
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DOI: https://doi.org/10.1057/jors.2013.169