Chapter13.3: Floating-point numbers,Representation and Manipulation

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Format

What are the effect of decreasing the number of bits allocated to Mantissa and increasing the exponent

  1. Reduction in precision
    • As the number of bits in mantissa has decreased.
  2. Increasing in range
    • As the number of bits in exponent has increased.

The denary number 513 cannot be stored accurately as normalised floating-point number in this computer system: (10 bits for mantissa, 6 bits for exponent)

Explain reasons for this:

  • Require more than 10 bits/11bits to store; the maximum number that can be stored is 511
  • The denary 513 in binary is 1000000001 // Normalised: 0.1000000001
  • Results in overflow

Describe an alteration to the way floating-point numbers are stored to enable this number to be stored accurately with the same total number of bits:

  • The number of bits for mantissa must be increased
  • 11 bits for mantissa and 5 bits for exponent
  • Exponent too large to fit 4 bits as two’s complement number
  • Exponent will turn negative
  • … therefore the binary point moves the wrong way
  • Value will be approximately +0.029(296875)

Explain the trade-off between either using a large number of bits for the mantissa, or a large number of bits for the exponent

  • The trade-off is between range and precision
  • Any increase in the number of bits for the mantissa means fewer number of bits for the exponent
  • More bits used in mantissa would result in better precision
  • More bits used in exponent would result in larger range

Conversion

Calculate the normalised binary number for -3.75. Show your working

  • -3.75 = 100.01000 // -4 + 1/4 // -4 + 0.25
  • 100.01000 becomes 1.0001000 Exponent=+2
  • Answer: Mantissa=1.0001000 Exponent=0010

Calculate the normalised floating-point representation of +1.5625 in this system (12bit-mantissa, 4bit-exponent). Show your working

  • Correct conversion to binary: 01.1001
  • Correct calculation of the exponent: 1
  • Answer: Mantissa=0110 0100 0000 | Exponent= 0001

Normalisation

Why binary/floating-point numbers are stored in normalized form

  • To store the maximum range of numbers in the minimum number of bits
  • Normalisation minimizes the number of leading zeros/ones represented
  • Maximizing the number of significant bits // maximizing the number of precision/accuracy with given number of bits
  • Enable large/small numbers to be stored with accuracy
  • Avoids the possibility that many numbers have multiple representation
  • --
  • There will be a unique representation for a number
  • The format will ensure it will be represented with greatest possible accuracy
  • Multiplication is performed more accurately

Problems that can occur when a floating-pointer number is not normalised

  • Lost of precision
  • Redundant leading zeros in the mantissa
  • Lost of the least-significant bits (bits on the right-hand end)
  • Multiple representation of a single number

Approximation & Rounding errors

State why some binary representation can lead to rounding errors

  • There’s no exact binary conversion for some numbers
  • More bits are needed to store the number
  • 0.2 and 0.4 cannot be represented exactly in binary, there is a rounding error
  • 0.2 has been represented by a number just greater than 0.2
  • This is similar for 0.4
  • Therefore, multiplying these two representations together increases the difference
  • Difference after calculation is significant enough to be seen
  • 0.1 cannot be represented exactly in binary, there is a rounding error
  • 0.1 is represented by a value just less than 0.1
  • The loop keeps adding this approximate value to the counter
  • Until all accumulated small difference become significant enough to be seen