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DOWNLOAD MY Ph.D Thesis - UNAM

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Chapter 5Upstream channel capacity and characterisation pThe values selected for the different parameters are the defaults defined in thespecification where applicable. All references to packet sizes in the analysis that followsrefer to the size of the packet as it enters the system from a PC station. This would bethe packet that the NIU is going to submit for delivery over the cable network and doesnot include DVB MAC and PHY overheads. However, those include 18-byte Ethernet,20-byte IP, 20-byte TCP, 8-byte UDP and 12-byte RTP (for VoIP only) headers. Theseprotocol overheads should be taken into account when the results presented here areused to evaluate application performance. In addition, simulations will be run using bothcontention resolution algorithms adopted by the DVB/DAVIC protocol (exponentialbackoff algorithm and splitting tree algorithm) as defined by the scenarios.Finally, for the tests defined, the INA utilised a simple First In First Out (FIFO)scheduler. A more complex scheduler was not necessary since all the streams are treatedevenly.5.4 Performance characterisation of the DVB/DAVICprotocol5.4.1 Offered load scalabilityIn this section we analyse the system performance under increased offered load. Thesimulated network examined in this scenario consists of a small network size of 20active stations, each generating a variable amount of traffic load made up frommaximum Ethernet packets of 1518 bytes. The aim of this analysis is to find the highestachievable system performance in terms of system throughput and mean access delaysrather than the effect of packet size, which is addressed in Section 5.7.As mentioned in Section 4.4.3, a bound for the maximum upstream throughput is givenby Equation 4.42. In this analysis, we have reserved at least 2 slots (out of 18 slots) persignalling frame for contention access, CSs = 2 and RSs = 16. Therefore, the maximumtheoretical efficiency achieved is ≈ 67 %, which we may expect from our simulationanalysis (as shown below).Smax48⋅RSs==64 ⋅ ( RSs + CSs)48⋅16≈ 66.7%64 ⋅ (16 + 2)5-9

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