Chapter 2, Problem 2.85HW

A.

Program Plan Intro

IEEE floating-point representation:

The IEEE floating-point standard denotes a number in a form  ${\text{V = (-1)}}^{\text{S}}\text{ × }{\text{M × 2}}^{\text{E}}$

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 “$\text{2 - }\in$” or between “0” and “$\text{1 - }\in$”.
• 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 $\text{exp}\text{=}{\text{e}}_{\text{k-1}}\mathrm{..........}{\text{e}}_{\text{1}}{\text{e}}_{\text{0}}$ converts the exponent “E”.
• The n-bit fraction field $\text{frac}\text{=}{\text{f}}_{\text{n-1}}\mathrm{..........}{\text{f}}_{\text{1}}{\text{f}}_{\text{0}}$ 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.

    

• “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.

    

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, $\text{E}\text{=}\text{e}\text{-}\text{bias}$
    • Here, “e” represents unsigned number containing bit representation ${\text{e}}_{\text{k-1}}\mathrm{..........}{\text{e}}_{\text{1}}{\text{e}}_{\text{0}}$ and bias value is ${\text{2}}^{\text{k-1}}\text{-}\text{1}$.
  • The fraction field “frac” is interpreted as representing the fractional value “f”.
  • The formula for significand “M” is “1 + f”.

    

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

    

• 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”.

    

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

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 “$\text{0b1}\text{.1111111}\mathrm{....}$”
• The value of “f” will be “$\text{0b0}\text{.111111111}\mathrm{....}$” (Here “n” bits “1”)

Then value of “E”, E = n.

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

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 “$\text{0000}\mathrm{.....}$”, it can be represented by “$\text{0b0}\text{.00000}$”
• 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, $\text{E = e - bias = 1 - bias}$
  • Computing value of “M” by “M = 1 + f”.
• From the given representation “$\text{0000}\mathrm{.....}$”, value of “f” is “0”. So, M = 1 + f = 1 + 0 = 1. So, value of “M” is “1” it can be write as “$\text{0b1}\text{.00000}$”
• Therefore, the value of “M” must be “$\text{0b1}\text{.00000}$” and the value of “f” must be “$\text{0b0}\text{.00000}$”.

Now compute the value of “V” by using ${\text{V = (-1)}}^{\text{S}}\text{ × }{\text{M × 2}}^{\text{E}}$

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

$\begin{array}{c}{\text{V = (-1)}}^{\text{S}}\text{ × }{\text{M × 2}}^{\text{E}}\\ \text{=}{\text{(-1)}}^{\text{0}}\text{ × }{\text{1 × 2}}^{\text{1-bias}}\\ \text{=}\text{1}\end{array}$