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Diffstat (limited to '3446/CH5/EX5.1/Ex5_1.sce')
| -rw-r--r-- | 3446/CH5/EX5.1/Ex5_1.sce | 29 |
1 files changed, 29 insertions, 0 deletions
diff --git a/3446/CH5/EX5.1/Ex5_1.sce b/3446/CH5/EX5.1/Ex5_1.sce new file mode 100644 index 000000000..7053a8d1b --- /dev/null +++ b/3446/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,29 @@ +// Exa 5.1
+// To Calculate
+// A) The system capacity if the cluster size, N (reuse factor), is 4 and
+// B) The system capacity if the cluster size is 7.
+// C) How many times would a cluster of size 4 have to be replicated to cover the entire cellular area?
+// D) Does decreasing the reuse factor N increase the system capacity?
+
+clc;
+clear all;
+
+ToCH=960;// Total available channels
+Cellarea=6; //in km^2
+Covarea=2000;//in km^2
+N1=4; // Cluster Size
+N2=7; //Cluster Size
+
+//solution
+Area1=N1*Cellarea;//for N=4
+Area2=N2*Cellarea;//For N=7
+No_of_clusters1=round(Covarea/Area1);
+No_of_clusters2=round(Covarea/Area2);
+No_of_CH1=ToCH/N1; // No of channels with cluster size 4
+No_of_CH2=ToCH/N2; // No of channels with cluster size 7
+SysCap1=No_of_clusters1*ToCH;
+SysCap2=No_of_clusters2*ToCH;
+printf(' System Capacity with cluster size 4 is %d channels \n ',SysCap1);
+printf(' Number of clusters for covering total area with N equals 4 are %d \n ',No_of_clusters1);
+printf(' System Capacity with cluster size 7 is %d channels \n',SysCap2);
+disp(" It is evident when we decrease the value of N from 7 to 4, we increase the system capacity from 46080 to 79680 channels. Thus, decreasing the reuse factor (N) increases the system capacity.")
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