Answer:
for the 5 segments, the utilization is 3.8%
Explanation:
Given the data in the question;
segment size = 1090 bytes
Receiver window size = 5,718 bytes
Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps
propagation delay = 22.1 ms
so,
Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms
we determine the total segments;
Total segments = Receiver window size / sender segment or segment size
we substitute
Total segments = 5718 bytes / 1090 bytes
Total segments = 5.24587 ≈ 5
Next is the throughput
Throughput = Segment / Round trip
Throughput = 1090 bytes / 44.2 ms
1byte = 8 bits and 1ms = 10⁻³ s
Throughput = ( 1090 × 8 )bits / ( 44.2 × 10⁻³ )s
Throughput = 8720 bits / ( 44.2 × 10⁻³ s )
Throughput = 197.285 × 10³ bps
Now Utilization will be;
Utilization = Throughput / Bandwidth
we substitute
Utilization = ( 197.285 × 10³ bps ) / ( 26 × 10⁶ bps )
Utilization = 0.0076
Utilization is percentage will be ( 0.0076 × 100)% = 0.76%
∴ Over all utilization for the 5 segments will be;
⇒ 5 × 0.76% = 3.8%
Therefore, for the 5 segments, the utilization is 3.8%
