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Full exam for Internet of Things in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Internet of Things, Exam 20-2-2017 Available time: 1 hour, 30 minutes 1 – Exercise (6 points) A personal area network (PAN) is operated according to the IEEE 802.15.4 beacon enabled mode. The Collision Access Part (CAP) is composed of 20 slots, the inactive part is composed of

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Full exam for Internet of Things in the Computer Engineering degree programme at Politecnico di Milano. The document covers: Internet of Things, Exam 20-2-2017 Available time: 1 hour, 30 minutes 1 – Exercise (6 points) A personal area network (PAN) is operated according to the IEEE 802.15.4 beacon enabled mode. The Collision Access Part (CAP) is composed of 20 slots, the inactive part is composed of

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Internet of Things, Exam 20-2-2017 Available time: 1 hour, 30 minutes 1 – Exercise (6 points) A personal area network (PAN) is operated according to the IEEE 802.15.4 beacon enabled mode. The Collision Access Part (CAP) is composed of 20 slots, the inactive part is composed of 2000 slots and the duty cycle is 2%; each slot (of CAP and CFP) carries 127 [byte] packets and the nominal data rate is 250 [kb/s]. Find (i) the total number of available slots in the CFP, (ii) the equivalent rate of the channel defined as “one slot per beacon interval”. Using the numbers found in (i) and (ii), suppose now that two motes are active in the PAN with a required uplink rate, r, defined as follows P(r=1200 [bit/s])= 0.3, P(r=600 [bit/s])= 0.3, P(r=200 [bit/s])= 0.4. (iii) Define a consistent slot assignment in the CFP for the two motes (the motes do not use the CAP). (iv) how many additional motes requiring a channel of 120 [bit/s] could be added to the network? Solution The slot duration is Ts= 127[byte]/250[kb/s]=4.064[ms] Knowing that the duty cycle is 2%, we can write: 0.02=Ntot - Ninactive / Ntot, being Ntot and Ninactive the number of slots in the entire beacon interval and in the inactive part, respectively. Ninactive = 2000, (i) thus we can find Ntot = 2041 (approximating to the closest larger integer). It follows Ncap=20, Ncfp=1 and BI = Ts * Ntot = 8,29[s]. (ii) The equivalent rate of “one slot per beacon interval” is: r=127[byte]/8.29[s] = 122[bit/s]. (iii) each one of the two motes requires in the worst case a rate r=1200[bit/s], thus 10 slots must be assigned to each one of the motes. (iv) there’s no additional space in the CFP for other motes. 2 – Exercise (5 points) A personal area network (PAN) is composed of 10 motes and a PAN coordinator. The Beacon Interval duration…

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