# Simple Compression question

1. Apr 6, 2005

### justagirl

What am I doing wrong in this problem?

You need to record an entire collection of your favorite music. It consists of 39 hours of music. Assuming you are using MP3 encoding (MPEG Audio Level III at about 128kb per second), how many CDs will you need? (hint: a CD can store up to 700 MB of information.)

I did: 128 kb/s * 3600 * 19 = 8755200 kb (to be coded)
Compression ratio: 8 (?)

8755200 * 8 = 70041600 kb (to be coded on CD)

One CD can store: (700 MB) = (700) * (2^10) kB * 8 = 5734400 kbits

# of CDs needed = 8755200 / 5734400 = 13 CDs.

Thanks!

2. Apr 7, 2005

### Cliff_J

Where did the 19 in the first line come from? Or is it 19hrs instead of 39hrs?

In the CD figuring line, you need 2^20 for a MB and then you'll come up with a realistic number of bits that can be stored on a CDROM.

Note sure where you're getting your 8 from but I'll explain mine.

Ok, you need to be really careful about lower-case b and upper case B since they mean different things in the world of computers. It seems you have a bit of understanding on this but I'll re-iterate to make sure.

b = bit = 0 or 1
B = byte = 8 bits

Its far worse than the typical truth-strecthing where manufacturers use base 10 numbers (1MB = 1,000,000 Bytes) and software uses base 2 numbers (1MB = 1,048,576 Bytes).

The difference between bits/bytes becomes most obvious when you use networks where the bits is a dominant measure, like a modem connection that is 44kb/s (kilo-bits with little b) where you download files at maybe 5kB/s (kilo-Bytes with big B) as reported in the software.

I've double-checked my math with a few MP3 files I have on my computer, compressed them myself off my CDs so I assume they are accurate.

128 kb/s * 3600 seconds = 460,800kb/hr / 8 bits = 57,600kB/hr

I come up with just a shade more than 12 hours per 700 MB capacity CDROM. I'll let you complete your own math for your example.