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Diffstat (limited to '3446/CH17/EX17.11/Ex17_11.sce')
| -rw-r--r-- | 3446/CH17/EX17.11/Ex17_11.sce | 28 |
1 files changed, 28 insertions, 0 deletions
diff --git a/3446/CH17/EX17.11/Ex17_11.sce b/3446/CH17/EX17.11/Ex17_11.sce new file mode 100644 index 000000000..bd2c6334e --- /dev/null +++ b/3446/CH17/EX17.11/Ex17_11.sce @@ -0,0 +1,28 @@ +// Exa 17.11
+// To calculate average throughput and compare it with equal latency condition.
+
+clc;
+clear all;
+
+P1=1/2;//relative frequency of packets for user class1
+P2=1/3;//relative frequency of packets for user class2
+P3=1/6;//relative frequency of packets for user class3
+R1=16; //data rate in kbps for P1
+R2=64; //data rate in kbps for P2
+R3=1024; //data rate in kbps for P3
+S1=16;//number of slots assigned to the R1 user
+S2=8;//number of slots assigned to the R2 user
+S3=2;//number of slots assigned to the R3 user
+
+//solution
+//Using Equation 17.20 from page no 616
+Ravg=(P1*R1*S1+P2*R2*S2+P3*R3*S3)/(P1*S1+P2*S2+P3*S3);
+// For equal latency, using Eq 17.18
+Rav=1/(P1/R1+P2/R2+P3/R3);
+// For Latency ratio=4, using Eq 17.19 from page no 616
+PL=4;
+C=(P1+P2+PL*P3)/(P1/R1+P2/R2+P3/R3);
+printf('The average throughput for equal access condition is %.1f kbps \n ',Ravg);
+printf('The average throughput by considering equal latency is %.1f kbps \n ',Rav);
+printf('The average throughput by considering latency ratio as 4 is %.2f kkbps \n ',C)
+disp("Ii is observed that equal access provides the highest average output")
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