Networking – how to solve the “find the utilization for a TCP sender that is continuously sending” 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.

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

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:

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) = 0.250514 time to send one segment

Step 3: Calculate the Round Trip Time (RTT)

RTT = propagation delay * 2 
33.3 * 2 = 66.6 round trip time

Step 4: Calculate the delay per packet

delay per packet = RTT + time to send one
66.6 + 0.250514 = 66.850514 delay per packet

Step 5: Calculate the utilization!

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

Networking – how to solve the “effective delay” 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.

Given an effective delay of 99ms when network usage is 76%, what is the effective delay when network usage is 53%?

Suppose they want the answer in milliseconds rounded to 1 decimal place.

You might’ve seen a variation of this problem where the delay is given in ms and the network usage is 0%. If you’re given a delay with a network that’s in-use, you have to first figure out what the the delay would be if the network congestion was 0%.

Let’s call that value x and solve for it.

We will use the same D = Do/(1-U) formula we would use if we already knew the delay for a zero-congestion network, but we will put our 99 ms delay where D goes and our unknown “x” value for Do.

Do / (1 - U) = D
x / (1- 0.76) = 99ms 
x / (0.24) = 99

Multiply both sides by 0.24 to turn that into a value for x:

x = 0.24 * 99
x = 23.76 <-- network delay in ms when congestion is at 0%

Now that we know the delay in ms when congestion is at 0%, we can plug it into the D = Do/(1-U) formula again.

D = 23.76 / (1 - .53) 
D = 50.6 ms

It’s not a difficult problem but my class’s materials only covered the “easy” version so I wanted to demonstrate the “missing steps” for anyone else who might be wondering what to do when they give you the delay for a partially-congested network instead.

Networking – how to calculate the size of MSS for a “mythical set of protocols”

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.

Imagine a mythical set of protocols with the following details:

  • Maximum Link-Layer data frame: 1,331 bytes
  • Network-Layer header size: 29 bytes
  • Transport-Layer header size: 34 bytes

What is the size, in bytes, of the MSS?


1331 - 29 - 34 = 1268

You just subtract the header sizes from the data frame. Easy.

You probably could’ve guessed that, but if you’re in CS372 and you’re reading this then it might be because you weren’t 100% sure wanted to check your work but the class materials didn’t really cover it. And that’s the thing that irked me about this problem: the class just threw it at us as a surprise on the weekly (graded but open-book) exercises in week 6. Then you go a-Googlin’ and end up somewhere like who wants $15 to sell you a solution! That’s just silly.

Now you know: calculating the maximum segment size is just a simple subtraction problem.

Networking – how to solve the “non-persistent vs persistent HTTP” 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.

A client’s browser sends an HTTP request to a website. The website responds with a handshake and sets up a TCP connection. The connection setup takes 2.5 ms, including the RTT. The browser then sends the request for the website’s index file. The index file references 2 additional images, which are to be requested/downloaded by the client’s browser. How much longer would non-persistent HTTP take than persistent HTTP?

Suppose they want the answer in milliseconds rounded to 2 decimal places.

You can get fancy with writing out each and every request (and this is what they’ll show in the worksheet), but there’s a faster way, especially if the number of referenced objects is more than a few.

Step 1: Count the total number things you’re getting, ie: 1 index file + 3 images = 3. That’s your N value.

index.html + image1.jpg + image2.jpg = 3 things total

Step 2: For non-persistent HTTP, your answer will be the result of N x 2.

3 x 2 = 6 connections needed for non-persistent HTTP

(That’s because for each and every item you request, you also need to request a TCP connection.)

  1. TCP request
  2. index.htm request
  3. TCP request
  4. image1.jpg request
  5. TCP request
  6. image2.jpg request

Step 3: For persistent HTTP, do N+1

3 + 1 = 4 connections needed for persistent HTTP

(That’s because you only need to connect once, and then request all the things you want.)

  1. TCP request
  2. index.htm request
  3. image1.jpg request
  4. image2.jpg request

Step 4: Now you have:

6 connection requests for non-persistent HTTP
4 connection requests for persistent HTTP

Step 5: Find the difference.

6 - 4 = 2

Non-persistent HTTP needs 2 more connections than persistent.

Step 6: Multiply that difference by how long each connection takes to set up.

2 x 2.5 milliseconds = 5 milliseconds. 

Double-check your rounding and units! You might be asked to give this in seconds instead of milliseconds.

In summary

  • Sum the things (index file and images) to get N
  • Non-persistent HTTP = N x 2
  • Persistent HTTP = N + 1
  • Find the difference
  • Multiply the difference by how long a round trip takes

Networking – how to solve the “fileX and fileY” 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. 

You’re given a link with a maximum transmission rate of 70.5 Mbps. Two computers, X and Y, want to transmit starting at the same time. Computer X sends File X (18 MiB) and computer Y sends File Y (130 KiB), both starting at time t = 0.

  • Statistical multiplexing is used, with details as follows
    • Packet Payload Size = 1000 Bytes
    • Packet Header Size = 24 Bytes (overhead)
  • Ignore Processing and Queueing delays
  • Assume partial packets (packets consisting of less than 1000 Bytes of data) are padded so that they are the same size as full packets.
  • Assume continuous alternating-packet transmission.
  • Computer X gets the transmission medium first.

At what time (t = ?) would File X finish transmitting?

Suppose they want the answer in seconds (hah – not milliseconds like usual!) and they want it rounded to two decimal places.

The “twist” of this problem is that File X and File Y take turns being sent in little chunks of 1000 bytes (plus 24 bytes of overhead).

First a piece of X goes, then a piece of Y, then back to X, etc.

Y is smaller so you run out of Y before you run out of X, but fortunately for us, this problem wants to know when X is done being sent. So really what we’re looking for is how long it takes to send both X and Y.

Here’s a drawing I made to visualize the problem (obviously, X and Y are both way more than just 10 and 5 packets in our real problem).

Diagram depicting X1, Y1, X2, Y2, X3, Y3, X4, Y4, X5, Y5, X6, X7, X8, X9, X10 interwoven with each other. Each block represents a packet.

Step 1: Figure out how many packets have to be sent, total

Our packet size (1000 bytes) and our overhead size (24 Bytes) are given in bytes, so I’m going to turn the two filesizes into Bytes instead of bits for this one. (We can always turn it into bits later if needed.)

File X = 18 MiB // turn to bytes by multiplying by 1024 twice
= 18 * 1024 * 1024 
= 18874368 bytes
File Y = 130 KiB // turn to bytes by multiplying by 1024 once 
= 130 * 1024
= 133120 bytes

Now X and Y are in the same units (bytes) and we can use that to figure out how many packets of 1000 bytes each one would be divided into.

File X = 18874368 Bytes / 1000 = 18874.368 packets

Our result isn’t a whole number. What should we do with the “partial packet”? Well, the problem tells us that any partial packets should be padded up to a full size. In practice, this just means we round up to the next nearest whole number. 18874.368 becomes 18875 packets for File X.

Now do the same for File Y:

File Y = 133120 Bytes / 1000 = 133.12 
Round up to get 134 packets for File Y

(Save these numbers for later, we’ll come back to them.)

Step 2: Calculate the time needed to send one packet plus its overhead

Add the packet size and the overhead size (both must be in the same units; here, that’s bytes but your variation of this problem may differ). Since transmission rate is in bits, let’s also multiply the packet size + overhead size by 8 to put it in bits, and multiply the transmission rate by 10^6 to put it in bits also. That looks like…

L/R, where R's units is what the outcome of this calculation will be 
= (1000 Bytes + 24 Bytes * 8) / (70.5 Mbps x 10^6)
= (8192) / (70.5 x 10^6)
= 0.0001161985816 seconds // same units as R (bits per second) 
= 0.11619858156 msec // multiply by 1000 to get milliseconds

We now know that it takes 0.11619858156 milliseconds to send one packet (and its overhead). Save this number for later.

Step 3: Calculate the time it takes to send all packets

This step is easy: we know how many packets we have between File X and File Y, and we know how long it takes to send one of them, so let’s figure out how long it takes to send all of them.

18875 packets + 134 packets = 19009 packets to send
19009 packets x 0.11619868156 msec = 2208.820738 milliseconds to send all 

Step 4: Convert to seconds and round to 2 decimal places to get the final answer

This variation of the problem wants the solution in seconds, so divide our result from Step 3 by 1000 to get 2.208820738 seconds and round it to two decimal places to get 2.21 seconds, which is when time at which File X will be done sending.

What if it wanted to know when the smaller file was done sending?

Well, let’s look at our diagram again. Here we can see that for every packet of Y sent, a packet of X has already been sent.

For every packet of Y that is sent, one from X precedes it. In other words, to send all 5 packets of Y we must also send 5 packets of X.

We know Y is made up of 134 packets, so that many packets of X will also be sent. That means for our final answer all we have to do is figure out how long it’ll take to send 134*2 packets.

134 * 2 = 268 // all of Y + same amount of X
268 packets x 0.11619868156 msec = 31.14124666 msec to send File Y

Convert that answer to seconds and there we have it: 0.03 seconds have elapsed when File Y (the smaller file) is done sending.

Networking – how to solve the “routers in a sequence” 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. Yeah, I’m still cranky that this class makes no effort to teach this stuff. They just send you a-Googlin’. Welcome, Googlers…

Suppose there are 3 routers in sequence between Host A and Host B, all of which use store-and-forward routing. What is the total end-to-end delay for a packet originating from Host A with destination Host B, under the following conditions.

  • Each of the link transmission rates are 5.1 Mbps
  • The total distance from Host A to Host B along its path of transmission is 175.1 km
  • The speed of propagation through the transmission medium is is 2.7 x 108 m/s
  • The packet size is 2 KiB

Remember that you must also uplink from Host A to the first router. Suppose this one wants the answer in milliseconds rounded to 1 decimal place.

Step 1: Figure out the transmission delay for one link

The “twist” of this problem is that there are multiple transmission delays to account for. Every “hop” incurs a transmission delay.

We can use the L/R formula with the known packet size (2 KiB and known transmission rate (5.1 Mbps) to calculate the transmission delay for one node to another. Node here means host or router. Be sure to convert both into bits.

(2 KiB * 8192) / (5.1 Mbps * 10^6) 
= 16,384/5,100,000 = 0.00321254... seconds

Remember that the units R is in determines the outcome, so this result is in seconds. Multiply it by 1000 to turn it into milliseconds and save this result for later.

0.00321254 x 1000 = 3.21254901... milliseconds

Step 2: Figure out the entire path’s transmission delay

Recall that our three routers and two hosts basically look like this:


We know the transmission delay (the time it takes for a bit to be placed on the transmission medium) at one host or router, and we know how many routers there are total, so we can figure out how many “transmission delays” our bits will go through by multiplying the total number of routers plus one (to account for being placed on the transmission medium by A) by the transmission delay we calculated in step 1.

(R + 1) * 3.21254901  // R is the number of routers
= (3 + 1) * 3.21254901 // add 1 to account for initial link
= 4 * 3.21254901 
= 12.85019604 milliseconds

This number represents the total amount of transmission delay experienced by bits traversing this multi-router path.

Step 3: Figure out the propagation delay

Recall that our distance is 175.1 km and our speed of propagation is 2.7 x 108 m/s and the propagation delay formula is Tprop = d/s. Note that km has to be turned into meters and the result will be in seconds.

= (175.1 * 1000) / (2.7 x 108 m/s) = 0.0006485185... seconds 
0.0006485185... seconds x 1000 = 0.64851 milliseconds

Step 4: Add it all up

 3.21254901 + 12.85019604 + 0.64851 = 16.71125505 milliseconds

And that’s it, you’re done! Hope you found this helpful!

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.

OSU eCampus CS475 Parallel Programming review & recap

This post is part of an ongoing series recapping my experience in Oregon State University’s eCampus (online) post-baccalaureate Computer Science degree program. You can learn more about the program here.

Six-word summary: A well-designed and satisfying course

I feel like I sometimes write these recap articles to complain and warn other students of landmines, but CS475 was what every OSU online CS course should be: organized, succinct, and educational. I learned so much and enjoyed the course from start to finish.

CS475 Review

It only runs in the Spring quarters, but it’s worth holding out for! You’ll learn a lot, the material is interesting, the course is well-paced, and the exams felt fair.

I picked this class because parallel programming was something I thought sounded intimidating and was unlikely to be something I would attempt on my own. I’m very glad I chose it! It was way less intimidating than I feared, and now I feel entry-level competent in a subject I formerly knew nothing about.

Class structure

  • 10 weeks
  • 8 programming assignments where you code, run the code, and then write an analysis of it
  • I spent about 7-10 hours on this class each week, making it one of the lighter workloads in the program
  • Proctored midterm, proctored final
  • Weekly quizzes in Canvas (usually 5 questions with a full hour to answer them)
  • About 1-1.5 hours of video lectures each week (sadly they are without captions, but they are very high quality otherwise)
  • Some code is provided for you to use, saving time for focusing on what the class is really about
  • Instructor attends office hours! He’s really approachable, you should go even if you don’t have a specific question!

The instructor is SUPER active! You can get on a video chat with him – he’s just sitting there at pre-announced times, waiting for people to show up to his video chat room! He’ll look at your code! He’ll talk to you 1:1 about topics! I’ve never had such a productive “office hours” experience as I did in this class, and I didn’t even show up with any questions.

Contrast this with how most classes in this program hand you a problem and set you loose on figuring it yourself (via Google, Stack Overflow, old projects posted to GitHub, etc). There’s a time and a place for that, sure, but for $2000 in tuition I’m happiest when the class makes me feel like I learned something from the person teaching it. I want an experience I can’t get from Google and an independent project. I want an intelligent, skilled person to say “here’s what I know, and here’s how I think about it” and that’s how Professor Bailey ran this class.

Prof. Bailey also sends out an anonymous survey at the end of the course asking for feedback on each and every assignment as well as time spent. He genuinely cares about the class.

CS475 Project Structure

Unlike most classes, CS475 hands you some “starter code” for every assignment. You still have to put it in the right place, call it at the right time with the right data, and format and analyze the output yourself, but being provided this “starter code” was great for a few reasons:

  • Less time spent on things that aren’t the point of the assignment
  • No super fancy math skills needed (the bezier volume calculation formula is given to you, for example)
  • Less likely that you’ll go off into the weeds and do something ridiculously wrong and miss the point of the assignment
  • Less time needed overall for each assignment

This class has you throw nearly 100% of your efforts towards parallel programming related coding and analysis. For those of us with a finite number of hours in a week it was so good to just focus on the subject of the class and skip some of the boilerplate project setup. This class would pair well with another less-time consuming class, if you need (or prefer) to double up.

Most of this class’s projects go something like this: start a new C project, bring in the starter code, refactor/add to it to make it do what you need, write a bash script (or whatever you prefer) to run a loop and pass different parameters to the program each run, get all of the output in your Terminal (some people write it to a file), dump the output into a spreadsheet (I used Google Sheets), create graphs and write your analysis (in response to prompts).

Typical project output – here’s my output from Project 2

Bring these into your favorite spreadsheet software and create a pivot table. Then, generate the needed graphs for your analysis.

Somehow I made it this far in life without ever needing a pivot table. In this class, pivot tables made graphs a cinch.

And that’s pretty much how the assignments go.

I was a bit intimidated by this class before I took it. People say it’s dry and the subject has a reputation for being challenging, so I wanted to show other prospective students what’s actually involved and how approachable it really is.

Tips for CS475

NVIDIA CUDA support is limited on some versions of Mac computers. (For example, Apple just doesn’t support it for OS 10.14 at the time of this writing). If you’re on a Mac, I suggest using the school-provided rabbit server for your CUDA assignments and not wasting hours like I did trying to get it all working on a Mac.

Go to the office hours! Seriously, I went on a whim near the end of the quarter and was surprised to find myself alone in a video chat with the professor. I was just going to lurk but it ended up being a great chance to discuss a bunch of class topics one on one. Another student showed up and he stepped through her previous assignment code with her. You can’t beat the value of having an expert review your code with you!

The exams are straightforward. Some classes in this program use exams as an opportunity to test your speediness and your ability to keep track of registers through multiple iterations of multiple loops, but this class kept it fair and simple: watch the lectures, take notes, study those notes and you’ll do fine! He even gives a generous amount of time for the exams.

Overall, it’s just a high quality course. The lectures were freshly recorded, well-paced, on-topic, and relevant to the assignments and tests. The weekly assignments always looked challenging at first but were ultimately quite doable and enjoyable.

I’ve got just 4 classes left to complete my degree. I’ll be back in the fall quarter for Intro to Networks! See you all then!

OSU eCampus CS344 Operating Systems review & recap

This post is part of an ongoing series recapping my experience in Oregon State University’s eCampus (online) post-baccalaureate Computer Science degree program. You can learn more about the program here.

Six-word summary: Four final projects in a row

CS344 is your chance to work on four large, back-to-back projects that cover a breadth of topics. They feel a bit contrived (ie: more like schoolwork than real-world problems) at times and their application to real problems might be hard to see at first, but there’s no denying it: doing four big projects in a row will level up your skills a whole bunch.

CS344 Review

This course picks up where other C courses left off and introduces some new and exciting features, such as bash scripting and mutual exclusion locks.

The projects are utterly massive – any one of them could’ve been CS162’s notoriously huge final project – and the course comes off as feeling more like an advanced C course than an operating systems course because of it. You do a lot of project setup and boilerplate in this class (but you do touch on a long list of OS topics as well). Sometimes the new thing (whatever the actual point of the assignment was) only takes minutes to add, like a little cherry on top of a 5-tier wedding cake of a C project.

Class structure

  • 10 weeks
  • 4 programming projects with about 2 weeks each
  • Bash and C programming language
  • No midterm, open-book (but timed) final
  • Due dates are kinda all over the place, best to triple-check them (they don’t follow a pattern)

Grading script FTW

CS344’s best feature is the clarity of its expectations. Unlike many of the courses that came before it in this program, this one is VERY CLEAR about what’s expected and how you’ll be graded on it. They hand you a grading script that runs your program code and spits out a list of results (pass/fail) or something for you to interpret. When it all passes, so do you. It’s that simple.

Overall, this class is fair and straightforward. It’s one of those classes you just gotta be disciplined and hardworking in, and it’ll reward you by introducing you (in a fairly friendly way) to topics such as sockets, locking threads, writing bash script, making your own shell, and more. I’m not sure I would have pushed myself through some of this stuff on my own – the project where we wrote our own shell was definitely intimidating at first – but I’m glad the program included this class and these opportunities. I think I learned a lot in CS344!

Tips for CS344

You don’t have to code on the school server. The class will tell you up and down and over and over to code in VIM on the school server. You don’t have to. I did the whole class on my MacBook in SublimeText and I transferred the files to the school server for testing using FileZilla. It’s just bash and C code, there’s no reason to isolate it the way the class tells you to.

I think this whole “only work on the school server” stuff is bad advice. The programs in this class are large enough to justify a text editor and working locally makes it easy to use version control (which you should absolutely be using – at least one person in my class lost their work by compiling over their .c files on accident :( ).

I kinda wish I’d taken it alone or with a less time-consuming class. I paired it with CS340 (Databases) and the two classes just took turns being hellishly time-consuming. It’s like right on the fence of being a “better off alone” class. By itself it might’ve felt light, but paired with anything I was constantly strapped for time. YMMV.

OSU eCampus CS340 Intro to Databases review & recap

This post is part of an ongoing series recapping my experience in Oregon State University’s eCampus (online) post-baccalaureate Computer Science degree program. You can learn more about the program here.

Six-word summary: Surprise, it’s a web development class!

You’ll glance at some database topics in CS340, but you’ll put most of your effort into building a big full-stack website.

CS340 Review

Get ready to sharpen your web skills! You’ll work with a partner the entire quarter (try to pick a good one!) to design and implement a website that interfaces with a database. This class is very heavy on web development. You’ll make tables to display data that comes in from the db, forms that can create, update, and delete data, and you’ll need it all up and running on the school’s servers by end of quarter. Course pacing is uneven and the numbering of assignments, projects steps, and assignments is borderline illogical.

I HIGHLY recommend finishing 290 before you take 340 so that you can use your last 290 project as starter code for 340’s. It’s criminal that people say you should take 290 and 340 simultaneously. Don’t do it. One of the last projects of 290 is a node.js website that interfaces with a database, and if you have this project complete, you have a fantastic head start on 340’s website project.

Class structure

  • 10 weeks
  • No midterm, no final
  • Pick your partner in the first week
  • Each week you make progress towards a completed website project with your partner
  • Some weeks require “peer review” in which people from outside your group look at what you turned in and critique it (and you, in turn, review others’ work)
  • Hits hard in the last few weeks of the quarter when you have to put the entire site together. Start early.

The first week is slow: you’ll pick a partner and you’ll import a database dump into the database tool of your choosing. You’ll sit on your hands after this ~30 minutes of work is complete.

This is a good time to figure out what you want your site to be. My partner and I did a museum ticketing system and found it easy to hit all the requirements. Don’t reinvent the wheel here.

Each week you’ll work on an ever-growing document outlining your plans. No one will read it, but you’ll submit it every week nonetheless. There is absolutely no TA feedback at any point in this class. I finished this class without so much as a “Looks good!” from a TA.

About midway through the quarter they’ll give you an assignment to build the HTML portion of your site. If this is all you do this week you’ll probably fall behind, because each week after this has a much higher workload. You should immediately look at your projects from 290 to get started with the routing and db queries once you’re happy with the HTML.

There are no exams and no quizzes.

CS340: Too light on database-specific stuff

For a database class, this class is (sadly) rather light on hands-on query writing. It tries to teach database design, which is appreciated, but I think it spends too long belaboring minutia relating to the design diagrams (we must’ve done three iterations of ours before the precise requirements finally became clear on Piazza). It also spends weeks on the diagrams – one would’ve been enough.

The two places CS340 has you get your hands dirty is on Mimir for a few query-writing assignments and in your quarter-long website project. This is where the class shines: where it’s actually teaching (or forcing you to figure out on your own) how to interact with a database to do something useful.

CS340’s Mimir assignments

Rather than write SQL queries in any kind of industry-standard tool, CS340’s query writing is done in a browser-based tool called Mimir. Deadlines are generous and the work takes maybe 2-5 hours to get through (per assignment). Mimir is slow and the feedback it gives is not as robust as what you might get in a better tool. CS340 does give you the db dump, so you can go play around in a better environment (I used MySQL Workbench) but sometimes syntax that works locally does not work in Mimir.

Nonetheless, the Mimir-based parts of the course are some of the best parts. Experience writing queries is a skill you can take to an interview and job. I just wish there had been more of it.

CS340 group work

“What one engineer can do in one week, two engineers can do in two weeks.” The adage holds up in CS340. My partner was great, but I think I would’ve moved faster through the project without having to coordinate with someone else. There’s a lot of overhead in keeping someone else in the loop, not duplicating work, waiting for input before continuing, etc.

A few weeks into the quarter, everyone gets assigned a random group of 5. This is your “peer review” group. (Your partner will be in his/her own peer review group). Since 340 doesn’t seem to feature any TA feedback, this is what the class gives you instead: feedback from other students who probably know the same or less than you do about what you’re working on.

The real kicker? You have to bring their feedback into your document and either act on it or explain why you chose not to. The feedback we got was generally useless: at best it was people pointing out UI bugs we already knew about, at worst it was a hot take dashed out 2 minutes before the feedback was due that suggested little to no reading comprehension on the reviewer’s part. At least we didn’t have to all meet at the same time like we did in 361.

Our CS340 project

We made a museum ticketing system. This project was a bit large for the time given but I enjoyed working on it.

In our app, you can add museums, add exhibits to museums, “sell” tickets to guests with a variety of exhibit entitlements, create new guests, add orders to existing guests, search for tickets by ID, date or transaction, view transactions, refund exhibit entitlements from tickets, and refund tickets themselves. You can also rename exhibits and museums, and update guest info.

We used Bootstrap on a node.js/Express/MySQL stack. (I’d have preferred to use an actual front-end framework but my partner was much newer to all this so we went with a nice big bowl of JQuery spaghetti instead.)

Here’s a few screenshots from our completed web app:

Our app’s default page – you can change which museum to view tickets and exhibits for.
Here’s the “sell tickets” flow, where the user chooses which guest types to create tickets for as well as a visit date.
The user can manage which extra exhibits each guest ticket should have access to.
Review your order before proceeding…
Enter guest info and complete the purchase! The ticket info is added to the database.
The Transactions page shows all pages and provides links to individual tickets.
In the Museums page the user views all museums and can choose to rename them.

Here’s one of the queries from the project that I wrote:

app.get('/get-transactions', function(request, response) {
var context = {};
var queryStr = "SELECT tr.transaction_id, tr.trans_date, tr.trans_time, g.fname, g.lname, tr.pymt_type, GROUP_CONCAT(ti.ticket_id) AS ticket_ids, GROUP_CONCAT(ti.admission_type) AS guest_types "+
"FROM transactions tr "+
"JOIN guests g ON tr.guest_id = g.guest_id "+
"JOIN tickets ti ON tr.transaction_id = ti.transaction_id "+
"WHERE tr.museum_id = ? "+
"GROUP BY tr.transaction_id;";
pool.query(queryStr, [], function(err, result, fields) {
if (err) {

context.transactions = result;

GROUP_CONCAT is a neat trick we used to get a string of ticket IDs and admissions types back from the db (which you can then parse on the front-end).

My favorite parts of this class were when I got to do something new and exciting in SQL.

CS340: What’s missing?

I wish the class had covered any (or all) of these topics:

  • Non-relational databases
  • Input sanitization
  • Security
  • Stored procedures
  • Thread locks
  • Transactions
  • Advanced SQL (this class never goes further than a SELECT within a SELECT)
  • Best practices

And this is just “what I know I don’t know”.

If I were to redesign this course, I would give students a defined project (“Make a theme park ride ticketing system”) and provide a functioning front-end so the class can super deep dive into database-specific topics instead. I know databases don’t exist in isolation, but the sheer amount of front-end work it required to interact meaningfully with our data greatly overshadowed the database work.

A few final tips for CS340

The course is disorganized and the assignments never fall into a predictable “rhythm”, so double-check everything. Every week it’s something different: this week you turn in a PDF, the next week a .zip. Sometimes you turn it into Canvas, sometimes you post it to your “peer review” discussion group. Does it count for you and your partner or just you? It varies week to week. Is there a quiz that opened last week due this week? These aren’t difficult things to figure out, but they add a lot of “overhead” and I saw more than a few people going “OMG I thought that submission counted for both of us!!” in the Slack chat.

Take 290 first. Easily the best thing you can do for yourself to ensure success in CS340. The last 290 assignment will have you make routes for interacting with a database. You can use that work as boilerplate for your CS340 project and save yourself a ton of grief. Can you imagine taking 290 and 340 at the same time and being stuck on the same problem for both classes? Take 290 first! It should be a hard pre-req for 340.

Skip the lectures. They’re a mess. I mean, I won’t tell you how to live your life, but the lectures in this class are worthless. They won’t help you build your site, they’re thin on examples, and sometimes the topic they introduce was actually needed for the homework due last week so… yeah. I’m a diehard “watch the lectures no matter what” person and I gave up on them. It’s like they’re from a previous version of the course or something.

Don’t reinvent the wheel. Use Bootstrap (or similar) to make your front-end look nice, choose a project idea that lends itself to lots of pairings (customers to orders, people to tickets, etc).

Overall, I was disappointed by CS340. I’ve weathered other not-so-great courses and found the good in them, but this one was just a whole lotta making a bigger website than we made in 290 and writing a few SQL queries on the side. My database/SQL skills didn’t grow much in this course and I’m bummed that it didn’t live up to my expectations. My partner was awesome, though, so there was that. :)