# Networking – how to solve the “VOIP problem”

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.

### The “VOIP problem”

You’re given a diagram (you won’t need it) and some values in the form of…

• Host A converts analog to digital at a = 40 Kbps
• Link transmission rate R = 4.2 Mbps
• Host A groups data into packets of length L = 51 bytes
• Distance to travel d = 900 km
• Propagation speed s = 2.5 x 108 m/s
• Host A sends each packet to Host B as soon as it gathers a whole packet.
• Host B converts back from digital to analog as soon as it receives a whole packet.

The question asks, How much time elapses from when the first bit starts to be created until the conversion back to analog begins?

Let’s say they want the answer in milliseconds rounded to two decimal places (note: this varies – sometimes the answer has to be in seconds, sometimes they want it rounded to one decimal place – double check your instance of the problem!)

Step 1: Figure out how long it takes to make a single packet

“When the first bit starts to be created” means “when this imaginary machine starts building the very first packet”.

We know how big a packet is: 50 bytes.

We know the rate at which it produces packets: 40 Kbps.

This is the classic L/R (length over rate) you’ve hopefully seen if you’re keeping up with the course materials. We can use L/R here, but we have to turn 51 bytes into bits (multiply it by 8) and we have to turn 40 Kbps into bps (multiply it by 1,000).

`(51 * 8) / (40 * 1,000) = (408 / 40,000) = 0.0102 seconds`

We get 0.0102 seconds because the units that R is in determines the outcome. R is in bits per second (bps), so that’s how we know the units we’re looking at here is seconds.

We need this in milliseconds, because we’re going to put everything in milliseconds so that we’re working in the units the question expects. Multiply the seconds answer by 1000 to get milliseconds, like so:

`0.0102 x 1000 = 10.2 milliseconds`

That’s the time it takes to make one single, complete, ready-to-transmit packet. Save this value for later, we will come back to it in step 4.

Step 2: Figure out the transmission delay

Recall from the course materials that transmission delay is the time it takes to place every bit onto the transmission medium. Transmission delay is not the time it takes to actually send it!

If you like car analogies, consider transmission delay the time it takes to get your car onto the highway. Your car is being placed on the transmission medium, and the time it takes to do that is the transmission delay. (Again, this is not the time it takes to travel the length of the highway! It’s the time it takes to get your car onto the highway.)

We will use the L/R formula again for this step. This time L is the length of the packet (51 bytes) and R is the transmission rate (4.2Mbps). We need both of them in bits, so multiply 51 (bytes) by 8 to get bits, and multiply 3.8 by 1,000,000 to get bps.

`(51 * 8) / (3.8 x 1,000,000) = (408 / 3,800,000) = 0.00010737 seconds`

Again, the units of R determine what units the result is in, which in this case is seconds (because we had bits per second for our R).

To turn it into milliseconds, multiply by 1000:

`0.00010737 x 1000 = 0.10736842 milliseconds`

So, what do we have so far? We have the time it takes to make one packet and the time it takes to transmit that packet, both in milliseconds. There is one missing piece: propagation delay.

Step 3: Figure out the propagation delay

Propagation delay is the time it takes a bit to travel the given distance (900 km) at the given speed (2.5 x 108 m/s). The formula for propagation delay is d/s (distance over speed).

To go back to our car analogy, the propagation delay is the time it takes the car to zoom down the highway to its destination.

Like L/R, the d/s result will be in the same units as the bottom number (so, seconds). However, before we proceed, note that the distance we were given is in kilometers and the speed is given in meters per second. We need to convert kilometers to meters when we do our calculation, so multiply that 900 by 1,000.

`(900 km * 1,000 / 2.5 x 108) = 0.0036 seconds`

Remember to convert that answer to milliseconds:

`0.0036 x 1000 = 3.6 milliseconds`

Step 4: Add the previous three answers together

Time to generate packet + time to place it on the transmission medium + time for it to propagate = our answer

`10.2 msec + 0.10736842 msec + 3.6 msec = 13.90736842 milliseconds`

Step 5: Round to the requested number of decimal places

Since this problem wants the answer rounded to two decimal places, 13.90736842 becomes 13.91

I hope this breakdown of the “VOIP problem” helps someone else!

I have to admit I’m a bit salty this class had us sit through all those lectures, read all those book chapters, and do all those worksheets and couldn’t once be bothered to expose us to even a watered-down or partial version of it before throwing it as us on a graded, timed quiz.

Fortunately, once you’ve mastered solving this problem, a lot of the other “math problems” that CS 372 throws at you in week 1 and week 2 will feel like variations on it.

## One thought on “Networking – how to solve the “VOIP problem””

1. Kristen Orue says:

Oh my gosh! Thank you so much for posting these formulas and a step by step of why along with analogies. I am in this class currently and I felt the same frustration. I do have one question on figuring out the transmission delay (Step 2 of VOIP) problem: You list the Link transmission rate R = 4.2 Mbps but when you go to solve it you multiply “3.8” by 1000. If you could show me how you got to 3.8 that would be very helpful. Again thank you for taking the time to walk people through these and this website looks great!

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