Layout of the NSA’s Utah Data Storage Facility. Image: Wikimedia.

The recent revelations of NSA warrantless data collection is of interest mostly because one has to wonder why so many people seem surprised about this. It wasn’t a matter of reading tea leaves or anything, it was right out there in the open for anyone who was paying attention.

One of the big “clues” has been the construction of a massive NSA data storage facility in Utah for all the stuff they collect. Last week I got to thinking about the question whether the NSA collected more than just “metadata” such as who originates, and receives data, when, where, etc., but not actual content as they so piously insisted. Of course, we now know that they do collect content, but I arrived at that conclusion last week.

How, you ask?

Sometimes all it takes is a little back-of-the-envelope calculation to see if numbers don’t add up or, what they in fact add up to. I asked myself, how the proposed size of the new data collection facility comports with the claim the NSA only collects only metadata.

My first step is to do some data collecting of my own. The amount of metadata collected from say, a phone call (number calling, number receiving, duration of the call, etc.) doesn’t really need much data. This paragraph is more than long enough to accommodate it. (218 characters, excluding spaces)

But let’s allow one (1) kilobyte or 1000 characters for each record. That should be more than enough. Some quick research on the web gave me the following statistics about phone and other forms of traffic in the US:

• Phone calls per day: 3 billion
• Text messages: 4.1 billion
• Emails: 249 billion

That comes to a total of 256.1 billion records per day. Just in case we missed other items (credit card purchases, chat sessions, VOIP calls, etc.) that get scooped up, let’s double it to 512 billion, or 5.12 x 10¹¹.

Since each record uses 10³ bytes, the total information collected in a single day using our model comes to 5.12 x 10¹⁴ bytes.

Now, let’s look at that Utah storage facility. A government web site remarks that the storage capacity is measured in “zettabytes” but does not say how many. A zetta- of something is 10²¹, or 10 followed by 21 zeros.

We can now ask, “How many days would it take to fill one zettabyte of data at the current rate?” Divide the daily take by the total capacity:

5.12 x 10¹⁴ / 10²¹ = 5.12 x 10⁷

This comes to 51,200,000 days, or in excess of 140,000 years. Ah, but the flow of electronic traffic is going to grow, isn’t it? Let’s look at this another way. How many times can the rate of traffic double before we hit one zettabyte?

(5.12 x 10¹⁴) * 2^t = 10²¹

Where t is the number of doublings.  Divide through by the daily rate:

2^t = 10²¹/5.12 x 10¹⁴

2^t = 10²¹/5.12 x 10¹⁴

2^t = .195 x 10⁷ or 1.95 x 10⁶

Take the log of both sides:

tlog(2) = 6 (roughly)

t = 6/log(2) = 19.9

This means the rate can double about 20 times. So, if the amount of traffic doubles every 5 years, it will take 100 years to fill just one zettabyte. If it doubles every year, it will take just under 20 years. That’s a huge amount of storage. Something tells me that (a) the facility will hold more than one zettabyte and (b) the numbers point to storing much, much more than just metadata.

Of course, as I mentioned, we now have some confirmation that this is the case, but with a little math and the willingness to play with some numbers, one can peek behind the curtain just a bit.

Call it a case of using open source intelligence, but I prefer to think of it as an answer to that whiney question from bored high school students in math classes of, “Why are we learning this?”