Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and with a link transmission rate 26 Mbps an end-to-end propagation delay of 22.1 ms, what is the utilization

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%

answer; i believe its compare (b3: b4) ;

answers

a. nitrogen (n)

d. oxygen (o)

hope it