Example 1 (Algorithms for star networks). Let n = 3, c 1 = a = 1. Suppose dataset D 1 is released at time r 1 = 0, has size α 1 = 4 and due date d 1 = 8, dataset D 2 has release time r 2 = 1, size α 2 = 2 and due date d 2 = 9, and the parameres ofAlgorithm FIFO sends datasets in the order (D 1 , D 2 , D 3 ). Hence, D 1 is transferred in the interval [0, 4], D 2 in the interval [4, 6], and D 3 in the interval [6, 8]. The order in which datasets are processed by the base station is (D 1 , D 3 , D 2 ), since d 3 < d 2 < d 1 , and dataset D 3 does not arrive until the processing of D 1 finishes. Thus, D 1 is processed in the interval [4, 8] and has lateness L 1 = 0, dataset D 3 is processed in the interval [8, 10], which results in L 2 = 3, and finally, D 2 is processed in the interval [10, 12] and has lateness L 3 = 3. Hence, the maximum lateness obtained by the FIFO rule is L max = 3.Algorithm SRT starts with sending dataset D 1 in the interval [0, 1]. At time r 2 = 1, dataset D 2 becomes ready, and its size α 2 = 2 is smaller than the remaining size of dataset D 1 , which is β 1 = 3. Hence, a preemption takes place, and the transfer of D 2 starts at time 1. At time r 3 = 2, dataset D 3 becomes ready, but its size is larger than the remaining size of D 2 , which is β 2 = 1. Thus, D 2 is sent without preemptions in the interval [1, 3]. Afterwards, D 3 is sent in the interval [3, 5], followed by the remaining part of D 1 in the interval [5, 8]. The intervals in which datasets are processed are: [3, 5] for D 2 , then [5, 7] for D 3 , and finally, [8, 12] for D 1 . The resulting lateness values are L 1 = 4, L 2 = −4 and L 3 = 0. Thus, SRT achieves L max = 4. Since d 2 > d 1 and d 3 < d 1 , algorithm EDD does not preempt the transfer of D 1 at time r 2 = 1, but a preemption takes place at time r 3 = 2. Hence, dataset D 1 is sent in the intervals [0, 2] and [4, 6], D 3 is transferred in the interval [2, 4], and finally, D 2 is sent in the interval [6, 8]. The datasets are processed in the order (D 3 , D 1 , D 2 ), in the intervals [4, 6], [6, 10] and [10, 12], correspondingly. Their lateness values are L 1 = 2, L 2 = 3 and L 3 = −1, which gives L max = 3. Algorithm SD does not preempt the transfer of D 1 at time r 2 = 1 because d 2 > d 1 . At time r 2 = 2, when dataset D 2 with due date d 2 < d 1 becomes ready, the remaining part of dataset D 1 has size β 1 = 2. Hence,
