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Algorithms for Routing Problems Involving UAVs  157



                              The basic idea is to construct a new graph G =(V ,E ) and the corres-
                           ponding cost function such that finding a single optimal tour on graph G
                           is equivalent to solving the SDTSP.Graph G =(V ,E ) is constructed as



                           follows:
                           •  Add additional m−1 vertices to V represented by V −1 ,V −2 ...V −(m−1) .The


                              new set of vertices V := V  {V −1 ,V −2 ...V −(m−1) }.

                           •  E contains
                               1. every edge present in E.
                               2. an edge (V −i ,V j )if(V 0 ,V j )ispresentin E, ∀i ∈{1, 2..(m − 1)} and
                                 ∀j ∈{1..n}.
                               3. an edge (V −i ,V −(i−1) ), ∀i ∈{1..(m − 1)}.


                           •  The new cost function c : E →  + is defined as follows:
                               1. c (V i ,V j )= c(V i ,V j ), ∀i = {1, 2..n}, ∀j = {1, 2..n} and edge

                                 (V i ,V j ) ∈ E.
                                                       1
                               2. c (V −i ,V j )= c(V 0 ,V j )+ C i , ∀i = {0, 1,..(m − 1)}, ∀j = {1, 2..n} and

                                                       2
                                 edge (V 0 ,V j ) ∈ E.
                                                1
                               3. c (V −i ,V −i+1 )= (C i−1 − C i ), ∀i ∈{1..(m − 1)}.

                                                2
                              An example of this transformation is shown in Fig. 10 and Fig. 11. The
                           main result in Rao [10] that helps us solve the SDTSP is stated in the fol-
                           lowing theorem.
                           Theorem 4. Solving the SDTSP on graph G is equivalent to solving a single

                           TSP on the transformed graph G .
                                                      V 5
                                                             c 56
                                                     c 45
                                                      V 4
                                                                       V 6
                                                 c 34   c 04
                                        V 3                      c 06

                                        c 23                 V 0
                                                                        depot
                                                         c 01
                                        V 2    c 12                     destination

                                                      V 1
                                Fig. 10. An example of a graph G with 3 vehicles present at the depot
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