TheSameAs

This is a lesson idea for Maths or Upper Primary teachers, and came out of the fact that 25 million people's records would actually take up an awful lot of space - is the two discs claim actually true?

In the first part of the lesson, give the children the list of items stolen, and talk about how long each string would be - try testing this with the longest name in the class for example. Here is the list of data taken for each person below:

• First name
• Last name
• Address 1
• Address 2
• Town
• County
• Postcode
• National Insurance number
• Dob
• Child benefit number
• Bank sort code
• Bank account number

By my estimates, you would need, as a minimum, 175 bytes for each record, although I personally think it is best if you have a wide range of estimates, as it forces individual calculations!

The class would need to then mulitply that by the number of records stolen, which is estimated at 25 million. Although there are of course various ways of multiplying these two, calculating 175 by 25 and then increasing that by a million is probably the one with least headaches!

Next would be to find out how much data in bytes you could fit onto one disc. According to this site, this is calculated as follows:

"The amount of information that can be written is determined by the disc’s recording capacity as well as the physical and logical formats used.
Each of the five main CD physical formats devotes a different amount of space to user data (audio = 2,352 bytes/block, CD-ROM Mode 1 = 2,048 bytes/block, CD-ROM Mode 2 = 2,336/bytes/block, XA Form 1 = 2,048 bytes/block, XA Form 2 = 2,324 bytes/block). For any given data format disc capacity can be calculated by multiplying the appropriate user data area size by the CD data transfer rate of 75 blocks per second by 60 seconds by the minute size of disc. For example, a 80 minute disc written in CD-ROM Mode 1 format: user data area of 2048 bytes/block x 75 blocks/second = 153,600 bytes/second x 60 seconds = 9,216,000 bytes/minute x 80 minutes = 737,280,000 bytes. This rounds to roughly 700 MB (dividing by 1,024 to convert into KB and again by 1,024 to convert into MB). It should be noted, however, that in the real world capacity can vary slightly among discs from different media manufacturers."

For my class, we simply assumed that the data would be compared to the calculation of 737,280,000 bytes, as worked out above, but teachers of older pupils might want to calcluate the available byte space themselves.

It was then simply a case of comparing our estimate of 4,375,000,000 bytes to the disc storage size of 737,280,000 bytes! the discs clearly must have been DVD discs, or there were more discs than they originally said.

To calculate a DVD-ROM disc, you can make a crude estimation by knowing that the standard DVD-ROM is 4.7GB and simply considering how many 700MB discs of data you could fit into 4.7GB (roughly 6.7 worth).

Multiplying our original number of 737,280,000 bytes (bytes on a CD-ROM) by 6.7 (700MB discs on a DVD-ROM), gives us 4,947,148,800 bytes, slightly over our estimate of the data size.

This exercise assumes no accurate knowledge of how the data was organised, or even if our variables are correct. I am happy to update this feature, or produce a worksheet if there is any interest!