Lineamiento para evaluar la propuesta

We next present a new formulation which is obtained by aggregating some of variables in the time-indexed formulation presented above. This formulation, as we describe below, is a relaxation for ATSPTW and some feasible solutions to this formulation may not correspond to feasible tours. As we discuss in the next section, we add the so-called infeasible path inequalities to this formulation to make it an exact formulation for ATSPTW.

We call this formulation the time bucket formulation (TBF) where, unlike the time- indexed formulation, we do not define variables associated with each time index. We instead, partition the time window of a node into a collection of non-overlapping in- tervals and define variables associated with these intervals. More precisely, the time

window Wi for a node i ∈ V is divided in a set of buckets (intervals) Bi = {bi1, . . . , biL}, with b = [rb, db], rbi

1 = Ri, dbiL = Di, and the start time of b i

j+1 is greater than the end time of bi

j. We allow Wi 6= S

b∈Bi[rb, db] as long as each missing time instant t ∈ Wi cannot be a valid starting time of the node i associated with a feasible tour. Using earlier notation, a sufficient condition for such a time instant is S_k∈V−_(i)I_k(i, t) = ∅.

As in the previous formulations, binary variables xij indicate whether or not arc (i, j) ∈ A is a part of the tour. We define binary variables zb

i associated with each bucket b ∈ B_i of a node i ∈ V such that zb

i = 1 if and only if bucket b is selected for node i (b is then called the starting bucket at i). We also define binary variables yb

ij associated with each arc (i, j) ∈ A and each bucket b ∈ B_i such that yb

ij = 1 if and only if xij = 1 and zib= 1.

We also use I_k(i, b) to denote the collection of possible starting buckets at node k assuming that arc (k, i) is selected and the starting bucket at node i is b. More precisely, for a bucket b`∈ Bi

I_k(i, b_{`</sub>) = {b ∈ B<sub>k</sub> : d<sub>b`−1} < r_b+ θ_ki ≤ d_b`}

where we assume d_b0 = −∞. Conversely, if the starting bucket at node k is b and arc (k, i) is selected then the starting bucket at node i is denoted by Ni(k, b) and is defined as the bucket β ∈ Bi such that b ∈ Ik(i, β). If rb + θij > Dj, we define Ni(k, b) to be null. The following is the time bucket relaxation (TBR) of ATSPTW:

min X (i,j)∈A cijxij X b∈Bi z_ib = 1 ∀ i ∈ V (7.14) X j∈V+_(i) y_ijb = z_ib ∀ i ∈ V \ {q}, ∀ b ∈ Bi (7.15) X k∈V−_(i) X β∈Ik(i,b) yβ_ki= z_ib ∀ i ∈ V \ {p}, ∀ b ∈ B_i (7.16) X b∈Bi y_ijb = xij ∀ (i, j) ∈ A (7.17) z_ib ∈ {0, 1} ∀ i ∈ V, ∀ b ∈ Bi y_ijb ∈ {0, 1} ∀ (i, j) ∈ A, ∀ b ∈ B_i xij ∈ {0, 1} ∀ (i, j) ∈ A

y_ijb = 0 f or all (i, j) ∈ A, b ∈ Bi such that rb+ θij > Dj. Note that equation (7.14) implies that for each node i ∈ V , there is precisely one b ∈ B_i for which zb

i = 1. Therefore, if i 6= q, equation (7.15) implies that precisely one yb

ij = 1 for some j ∈ V+(i). Similarly, if i 6= p, equation (7.16), implies that precisely one y_kiβ = 1 for some k ∈ V−_{(i). Therefore, if i 6= p, there is precisely one node}

j ∈ V+_{(i) for which x}

xki = 1. Consequently, a feasible solution to the TBR does not necessarily give a tour (a permutation of the nodes) but instead gives a directed path from p to q that does not necessarily visit all other nodes on the way. If there are nodes that do not appear on this path, they are partitioned into cycles. Therefore, as presented above, the TBR has feasible solutions that do not correspond to feasible tours for ATSPTW.

We next show that all feasible tours for the ATSPTW are feasible for the this formulation and therefore TBR is a relaxation for ASTPTW.

Proposition 7.2 Any feasible tour for ATSPTW is also feasible for the TBR and the- refore TBR is a relaxation for ATSPTW.

Proof. Let V = {1 . . . n} and without loss of generality consider a feasible tour (1 → 2 → . . . → n) for which there are feasible starting times si ∈ Wi for each i ∈ V such that s1 = R1 and si = max{Ri, si−1+ θi−1,i} for i ≥ 2. For i ∈ V , remember that

Bi = {b1, . . . , bL}, with b = [rb, db], rb1 = Ri, dbL = Di, and rbj+1 > dbj.

For all i ∈ V , we next identify a time instant τi and the associated bucket β(i) and then construct a feasible y vector using these buckets. Let τ₁ = R₁ and β(1) be the first bucket of node 1 and notice that τ1 = rβ(1) = s1. For i > 1, we iteratively define

τi = max{Ri, rβ(i−1)+ θi−1,i} and we define β(i) to be the bucket b` ∈ Bi such that

d_{b`−1</sub> < τ<sub>i</sub> ≤ d<sub>b`} where we let d_b0 = −∞. Notice that as R_i≤ τ_i, if τ_i ≤ s_i, then τ_i∈ W_i and it is possible to identify the time window β(i) ∈ Bi.

We next inductively argue that τi ≤ si for all i ∈ V . First note that τ1 = s1, and assume that the claim is true for i ∈ V . As τ_i ≤ s_i, τ_i ∈ W_i and there exists a time window β(i) ∈ Bi. If τi ≥ rβ(i), then using the definition of si and τi, we clearly have

τi+1≤ si+1. If, on the other hand, τi < rβ(i) then τi ∈ Wi is a time instant that cannot be a valid starting time of the node i associated with a feasible tour. Let ˆτ > τ_i be the earliest possible starting time for a feasible tour. Clearly ˆτ is contained in some bucket for node i and ˆτ ≥ r_β(i). Furthermore, as si ≥ τi and as si is a valid starting time at node i, we have s_i ≥ ˆτ implying s_i ≥ r_β(i) and therefore τ_i+1≤ s_i+1as desired. Therefore, time windows β(i) ∈ Bi can be constructed for all i ∈ V as claimed.

The last step is to set xi,i+1 = ziβ(i)= yi,i+1β(i) = 1 for i ∈ {1, . . . , n − 1} and znβ(n) = 1 to obtain a feasible solution to TBR. As this solution has the same cost as the starting

solution to ATSPTW, the proof is complete. 2

Therefore, all feasible solutions and in particular the optimal solution to the AT- SPTW corresponds to a feasible solution for the TBR. However, the converse is not true as feasible solutions to the TBR do not have to give feasible solutions to ATSPTW.