Computer Systems: Program... -Access
Computer Systems: Program... -Access
3rd Edition
ISBN: 9780134071923
Author: Bryant
Publisher: PEARSON
Question
Chapter 2, Problem 2.85HW

A.

Program Plan Intro

IEEE floating-point representation:

The IEEE floating-point standard denotes a number in a form  V = (-1)S × M × 2E

From the above form,

  • The sign is denoted by “s”. It is used to define whether the number is in negative or positive.
    • If the number is positive, then “s” is “0”.
    • If the number is negative, then “s” is “1”.
  • The significand is denoted by “M”. It is a fractional binary number.
    • The number ranges either between “1” and “2 - ” or between “0” and “1 - ”.
  • The exponent is denoted by “E”. Its weights the value by a power of 2.

In floating-point representation, the bit is denoted by three fields such as sign, exponent and fraction field.

  • The single sign bit “s” directly converts the sign
    “s”.
  • The k-bit exponent field exp=ek-1..........e1e0 converts the exponent “E”.
  • The n-bit fraction field frac=fn-1..........f1f0 converts the significant “M”.
  • There are two formats are used for floating-point bit representation. They are “32-bit” format and “64-bit” format.
  • “32-bit” format:
    • It is the single precision format.
    • In this format, “1” bit for sign field, “8” bit for exponent field and “23” bits for fraction field.

      Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  1

  • “64-bit” format:
    • It is the double precision format.
    • In this format, “1” bit for sign field, “11” bit for exponent field and “52” bits for fraction field.

      Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  2

There are three types of cases occurs based on the single precision format. It is occur when the value encoded by a given bit representation can be divided into three different cases.

  • Case 1: Normalized value
    • This case occurs when the bit of “exp” is neither all zeros or nor all ones.
      • Numeric value for all zeros is “0”.
      • Numeric value for all ones is “255”.
    • In this case, the formula for exponent value, E=e-bias
      • Here, “e” represents unsigned number containing bit representation ek-1..........e1e0 and bias value is 2k-1-1.
    • The fraction field “frac” is interpreted as representing the fractional value “f”.
    • The formula for significand “M” is “1 + f”.

      Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  3

  • Case 2: Denormalized value
    • This case occurs when the exponent field is all zeros.
    • The formula for exponent value is “E=1-bias”.
    • Here the formula for significand “M” is “M = f”.

      Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  4

  • Case 3: Special values
    • This case occurs in two formats such as “infinity” and “NaN”.
    • When the exponent field is all ones and the fraction field is all zeros, then the resulting value is represented by “infinity”.

      Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  5

    • When the exponent field is all ones and the fraction field is not all zeros, then the resulting value is represented by “NaN”.

Computer Systems: Program... -Access, Chapter 2, Problem 2.85HW , additional homework tip  6

B.

Explanation of Solution

For largest odd integer:

Here, consider “bias >> n”.

Consider the value for the largest odd integer is

  • The value of “M” must be “0b1.1111111....
  • The value of “f” will be “0b0.111111111....” (Here “n” bits “1”)

Then value of “E”, E = n.

Now V = 0b0.111111111....{Here (n + 1) bits 1} which implies 2n+1-1

C.

Explanation of Solution

For reciprocal of the smallest positive normalized value:

From the smallest positive normalized value, the bit for exponent bit and fraction bit is given below:

  • For exponent field, place “1” in the least significant bit and remaining are all zeros. So, exponent becomes “0b1”.
  • For fraction field, all bits are zeros. So, it becomes “0000.....”, it can be represented by “0b0.00000
  • From the normalized case, the formula for “E” and “M” is given below.
    • Computing value of “E” by “E = e – bias”.
      • From the given binary representation “0b1”, decimal value of “e” is “1”.
        • Hence, E = e - bias = 1 - bias
    • Computing value of “M” by “M = 1 + f”.
  • From the given representation “0000.....”, value of “f” is “0”. So, M = 1 + f = 1 + 0 = 1. So, value of “M” is “1” it can be write as “0b1.00000
  • Therefore, the value of “M” must be “0b1.00000” and the value of “f” must be “0b0.00000”.

Now compute the value of “V” by using V = (-1)S × M × 2E

From the given question, the value is in positive. Hence, value of “s” is “0”.

V = (-1)S × M × 2E=(-1)0 × 1 × 21-bias=1

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Chapter 2 Solutions

Computer Systems: Program... -Access

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