*In this post:** A detailed, step-by-step guide demonstrating the steps I use to solve a problem similar to one encountered in my CS 372 Intro to Networking class.*

You have a TCP sender that is continuously sending a 1,096-byte segment. If a TCP receiver advertises a window size of 8,551 bytes, and with a link transmission rate 35 Mbps an end-to-end propagation delay of 33.3 ms, what is the utilization?

(Assume no errors, no processing or queueing delay, and ACKs transmit instantly. Also assume the sender will not transmit a non-full segment.)

Suppose they want the answer as a percentage rounded to one decimal place.

This one breaks down into a series of simple steps:

## Step 1: Calculate how many segments the pipeline holds

formula:segments = floor(pipeline size / segment size)

8551 / 1096 = 7.802 =7 segments

Remember to round down! There are no partial segments in this pipeline.

## Step 2: Calculate how long it takes to send one segment

This uses the L/R formula, which is:

formula:time to send one segment = length of segment / transmission rate

(Don’t forget to use the same units for both. I convert everything to bits for consistency.)

(1096 * 8) / (35 x 10^6) =0.250514time to send one segment

## Step 3: Calculate the Round Trip Time (RTT)

formula:RTT = propagation delay * 2

33.3 * 2 =66.6round trip time

## Step 4: Calculate the delay per packet

formula:delay per packet = RTT + time to send one

66.6 + 0.250514 =66.850514delay per packet

## Step 5: Calculate the utilization!

formula:utilization = (segments * time to send one) / delay per packet

(7 * 0.250514) / 66.850514 = 0.026316

Move the decimal to places to the right to get the percentage.

= 2.6%utilization

I cannot thank you enough for taking the time to put these out here! Your problem work-throughs have helped me so incredibly much in this class!

Thanks for taking the time to post this one up . I am taking cs372 right now and missed this problem because of the ambiguous wording , specifically:

(Assume no errors, no processing or queueing delay, and ACKs transmit instantly. Also assume the sender will not transmit a non-full segment.)

The phrase “ACKs transmit instantly” might be construed to mean that they arrive instantaneously, but that is not the case here as we still count the propagation time (magicically fast router to transmit, but still limited by the near-speed-of-light of twisted pair links)

You have a lovely website !