In this chapter we evaluate the performance of DFFR by simulation. We consider two tier interference sources, that is, for a cell there are two circles of neighboring cells around itself. So the system layout is set with 19 hexagonal cells and each cell equips with omni-antenna. These cells consist of three clusters which are 7 cluster-A cells, 6 cluster-B cells and 6 cluster-C cells respectively. The cell arrangement is illustrated in Figure 14.
Figure 12 - Simu lation system layout with 19 he xagonal cell which consist of 7 cluster-A ce lls, 6 c luster-B cells and 6 cluster-C cells
Because DFFR is proposed to not only to reduce the inter-cell interference from neighboring cells but also to overcome the user distribution problem, we will compare the
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effect with EFFR with different proportion of high power sub-band to normal power sub-band to evaluate the performance. The performance is evaluated by outage network probability, resource utilization efficiency and average CEU throughput. Outage network probability and Resource utilization efficiency are defined by (11) and (12) respectively. is the number of users cannot access the radio system and is the number of all users in the system.
and is the number of allocated RBs and number of all the RBs respectively.
(11)
(12)
We mainly focus on these two indices. Because these two indices are used to indicate the situation we mentioned in chapter 2.3 (Figure 9). In this situation, there are a lot of CEUs cannot access the system so that the network outage probability is very high. However, the high network outage probability does not cause by too many users in the system. It is because the CEU cannot access the normal power RBs due to the poor channel quality. Hence, lots of idle normal power RBs cut down the resource utilization efficiency. In this case, we expect the proposed DFFR to increase resource utilization efficiency and cut down network outage probability. That is, network outage probability only rises until resources are exhausted.
As mentioned in chapter 3, we use the SINR value as the criterion for grouping UEs.
Based on the equation (13) we can derive a SINR to data rate mapping table [19].
(13)
In equation (13), are the carried bits per symbol if we adopt (e.g.
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QPSK (1/2) or 64-QAM (3/4)) and is the number of symbols per slot can carry on.
is the number of slots per TTI and is the number of subcarriers per RB. Recall the resource block structure in chapter 1.2, RB size is determined by number of slots per TTI.
In our simulation, there are two slots in a TTI. Thus we can derive Table 1. value that UE can apply QPSK (1/2) which is 1.7dB according to Table 1. It means that if the SINR value of an UE is less than 1.7dB, then it belongs to CEU. Otherwise, it belongs to CCU. For comparison with EFFR, we apply as 3 and the other system settings of EFFR are same as DFFR. Other simulation parameters are listed in Table 2.
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Table 2 - Simulation parameter list
Parameter Value
Cellular layout Hexagonal grid, 19 cells
Cell range 1000m
Resource block size 180KHz, total: 50 RBs
Thermal noise -174dBm/Hz
Notice that the CEU RBs has to further divide into each cluster, each cell got only one third of CEU RBs due to we use 3 clusters in the simulation. For example, EFFR[9.6:0.4] in Figure 15
denotes that each cell has CEU RBs and CCU RBs.
Because the user distribution problem is the mainly problem we focus on, so we consider the situation similar with Figure 9 first.
In Figure 15, the UE distribution is CEU:CCU=8:2 in each cell. We observe that the outage network probability of EFFR[9.6:0.4] is close to DFFR. It is because DFFR can adjust CEU RBs and CCU RBs according to the UE distribution during the frequency and power allocation part, the simulation results may close to EFFR with appropriate proportion of CEU RBs to CCU RBs. With inappropriate proportion, see the curve of EFFR[4.2:5.8], the outage network probability increases dramatically when number of UEs is less than 50. The reason is that there are only 7 CEU RBs to allocate to CEU. Once there are more CEUs, they cannot access the system because CEU RBs exhausted. Although there are a lot of available CCU
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RBs, they cannot access due to the poor channel quality. To put Figure 15 and F igure 16 together, we can see that EFFR with wrong RB proportion will cause high outage network probability and low resource utilization efficiency. This result indicates that lots of RBs are wasted. For example, notice that the case of 20 UEs which is circled in red in Figure 15 and Figure 16, the outage network probability of EFFR[4.2:5.8] rises to about 46% and the resource utilization efficiency is only 31%. Compare with EFFR[9.6:0.4] and DFFR, outage network probability is less than 10% and resource utilization efficiency is about to 100%.
Notice that the outage network probability of EFFR[4.2:5.8] is less than DFFR and EFFR[9.6:0.4] when number of UE is over 60, it does not means the former got better performance. The reason is that the number of total available RBs in these three schemes is different. DFFR is similar with EFFR[9.6:0.4] which has 16 CEU RBs and 2 CCU RBs per cell and EFFR[4.2:5.8] has 7 CEU RBs and 29 CCU RBs per cell. In the case of 20 UEs, there are about 4 CCUs and 16 CEUs. In this situation DFFR and EFFR[9.6:0.4] are out of resources because all the CCU RBs and CEU RBs are allocated to UEs. Hence, if there are more UEs coming, no more resources can be allocated, outage network probability rises.
EFFR[4.2:5.8] has much more available RBs than DFFR and EFFR[9.6:0.4]. So when number of UEs comes to 60, it still has CCU RBs to allocate to CCU. However, at the CEU aspect, see Figure 17, average CEU throughput of EFFR[9.6:0.4] and DFFR is much higher than EFFR[4.2:5.8]. From these results we can observe that when number of UEs getting more, it is a tradeoff between serving more CEUs and reducing the outage network probability.
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Figure 13 - Co mparison of outage network probability between DFFR and EFFR with different proportion in case of CEU:CCU=8:2 UE distribution.
Figure 146 - Co mparison of resource utilization effic iency between DFFR and EFFR with different proportion in case of CEU:CCU=8:2 UE distribution.
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Figure 15 - Comparison of CEU average throughput between DFFR and EFFR with different proportion in case of CEU:CCU=8:2 UE distribution.
To further confirm that DFFR has the similar performance with EFFR with appropriate proportion of CEU RBs to CCU RBs, we choose another case to test again. Considering the UE distribution of half CEU and half CCU, we compare with EFFR with different proportion of CEU RB to CCU RB to see outage network probability and resource utilization efficiency.
In EFFR[7.2:2.8], each cell has 12 CEU RBs and 14 CCU RBs, it is close to the proportion of half CEU to half CCU. As we expected, Figure 18 shows that DFFR has nearly the same curve as EFFR[7.2:2.8]. Both of them have better performance in outage network probability.
Especially in the case of 30 UEs, the difference DFFR and EFFR[7.2:2.8] is up to 25%. From Figure 19 we can see that although resource utilization efficiency of EFFR[9.6:0.4] is higher than both DFFR and EFFR[7.2:2.8] when UE less than 30, the outa ge network probability of the former is higher than the latter two. It is because the total available RBs of EFFR[9.6:0.4]
is less than EFFR[7.2:2.8] and DFFR. Once any RB being allocated, resource utilization
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efficiency raise up rapidly. This is also the reason that EFFR[9.6:0.4] has higher outage network probability, because all the resources exhausted when number of UEs is over than 20.
Figure 168 - Comparison of outage network proportion between DFFR and EFFR with different proportion in case of CEU:CCU=5:5 UE distribution.
Figure 179 - Comparison of resource utilization efficiency between DFFR and EFFR with different pro portion in case of CEU:CCU=5:5 UE distribution.
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