Tag Archives: Addressability

1. Computer Systems, Revision

Calculations in computer systems

I was sure I had this here already but I can’t find it.

Here are the main formula we used in the computer systems unit

  • Largest number with n bits = 2n-1
  • No of memory locations/different colours = 2n
  • Coloumn heading in a binary number at n bit =2n-1
  • Size of bitmap = no. of Horizontal pixels X no. of Vertical pixels X Colour Depth
  • No. of pixels = DPI X size in inches
  • Total storage capacity = no. of memory locations X location size.

You should also be able to convert from bits, Bytes, Kilobytes, Megabytes, Gigabytes & Terabytes.

  • Bit= 1/0 (Binary digIT)
  • 1 Byte = 8 Bits
  • 1 Kilobytes= 1024 Bytes  (210)
  • 1 Megabytes= 1024 KB  (220)
  • 1 Gigabytes = 1024 MB  (230)
  • 1 Terabytes = 1024 GB  (240)

For colour depth

  • 8 bits = 256 colours (28)
  • 16 bits = High Colour 65,536 colours (216)
  • 24 bits = True Colour 16,777,216 colours (224)
1. Computer Systems, Computing Higher

Buses and Addressability

Today’s lesson was more a revision of topics that went before. However, we did learn a few important things.

  • To get the processors attention peripherals can make use of
    • Polling its a little like the teacher not letting anyone talk and asking everyone in turn if they need help
    • Interrupts this is like someone putting up their hand and the teacher stopping to deal with their problem before continuing.
  • The word size of the computer is the number of bits that can be manipulated as a single unit by the processor.
    • An ideal computer has a data bus that is the same size as its memory locations
  • The address bus determines the number of memory locations, however the data bus determines the size of each location. So to work out the amount of addressable memory, we must multiply the number of addresses by their size.
    • Total Addressable Memory = (2^address bus width) * Data bus width
    • IE a machine with a 16 bit Data Bus and 32 bit address bus would have
    • (2^32)*16 bits of accessible storage
    • or 8GB – Do the math yourself to prove it.