Why are bus widths in multiples of eight?

Answer 1

Because the data the bus carries is in Binary form. Each byte in binary is a group of 8 bits, so making the bus widths multiples of eight allows for the transportation of the Binary data. Prior to the introduction of the 8 bit byte and the EBCDIC character set by IBM in 1964 for their new System/360 series of computers, the standard bus widths used on computers were: 12, 18, 36, 72, and 144 (with a small minority that had 24, 48, or 60) because the standard binary character size was 6 bits and these bus widths are all multiples of 6.

You always want the computer to be able to read/write whole characters in a single memory operation to optimize efficiency regardless of the number of bits in the character.

Some of the newest computers use a 256 bit wide bus! Even wider busses are likely in the future.

Answer 2

because 1 byte is 8 bits, and a bus is set up to hold whole bytes. In order to transfer 1 byte, you need 8 connections. (very old technology used an 8 bit bus)

as technology improved, 16 bit, and very quickly 32 bit bus came to be used. Now with today's technology we're using 64 bit bus, a 64 bit bus can carry 8 bytes of data at a time.

Answer 3

Everything on a computer from memory and chips to programing is a multiple of 8 because computers typically have an 8 bit chip, (formally 4 bits). To make calculations efficient, processes have to be multiples of 8 binary digits. If a 8 bit chips divides binary data by 7 it will have large decimal numbers left over which takes up extra memory and processing time. Multiples of 8 return clean integers.

If you had a 9 bit processor everything would be in multiples of 9.

So basically everything is a multiple of 8 to make calculations efficient on an 8 bit processor.

http://en.wikipedia.org/wiki/8-bit

http://en.wikipedia.org/wiki/Byte

http://www.vision-street.co.uk

Answer 4

In actuality, the bus width is in multiples of 1000, but binary. In binary, there are only the digits 0 & 1, so, 0=0, 1=1, 2=10, 3=11, 4=100, 5=101, 6=110, 7=111, and 8=1000.

The next step is 10000 binary (or 16), then 100000 (32) and so on.

This is also where we get the computer architecture term of 32 bit, 64 bit, and (in a few more years) 128 bit computers

32 bit has 32 BInary digiTs (100,000,000,000,000,000,000,000,000,000,000 binary or 4,294,967,296) which is running into the limitation that it can only work with 4294967296 addresses, which is not really enough for newer operating systems and software.

64 bit has a theoretical limit of 18,446,744,073,709,552,000 (or 18 exabytes) so should have some room left for expansion

Answer 5

They are not. The're in multiple of 2^x.