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27 07 16sol

Esame completo di Internet of Things per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

Internet of ThingsEsame completo

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Esame completo di Internet of Things per il corso di Computer Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Exam July 27, 2016 July 7, 2016 21 PANC 3 4 5 6 7 8 Figure 1: Reference topology for Ex. 1. Exercise–1 ( July 27, 2016 ) The Personal Area Network in the figure is operated according to the IEEE 802.15.4 beacon enabled mode with the following parameters: (i) active part composed of CFP only (no CAP) with slots of 128 [byte] packets; (ii) nominal data rate is 250 [kbit/s]. The 8 motes in the figure are characterized by the following uplink traffic requirements: • motes 1, 3, 5 and 7 generate uplink traffic whose rate has the following distribution: P(r=25[bit/s])=0.4 P(r=50[bit/s])=0.6 • motes 2, 4, 6 and 8 generate uplink traffic whose rate has the following distribution: P(r=50[bit/s])=0.2 P(r=1[kbit/s])=0.8 Motes 1 and 2, besides sending up to the PANC their own traffic, have to relay in each BI the traffic generated by their siblings nodes. Define a consistent Beacon Interval structure including the number of slots in the CFP, the Beacon Interval duration, and the duty cycle. Define a consistent slot assignment in the CFP for all the devices in the network. Solution of Exercise–1 The lowest required rate is 25 [bit/s]. Thus, the BI can be set as: BI = 128[byte] 25[bit/s] = 40.96[s]. Motes 1, 3, 5 and 7 require 2 slots in the BI for their own traffic. Motes 2, 4, 6 and 8 require 40 slots in the BI for their own traffic. Moreover, Mote 1 1 and Mote 2 also require extra slots for receiving and delivering traffic from their sibling nodes. Namely, Mote 1 requires extra 88 slots and Mote 2 requires extra 164 slots. To sum up, we have: N3 =N5 =N7 = 40 N4 =N6 =N8 = 2 N1 = 90 N2 = 204 which leads to: NCF P = 40 × 3 + 2 × 3 + 90 + 204 = 422. The slot duration is Ts = 128[byte] 250[kbit/s]=4.096[ms]. The duration of the active and inactive parts and the duty cycle are, respectively: Tactive =…

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