# Represent (124.0625)10 using IEEE 754 single and double precision representations – IGNOU MCA Assignment 2015 – 16

By | August 29, 2015

MASTER OF COMPUTER APPLICATIONS
Course Code : MCS-012
Course Title : Computer Organisation and Assembly Language Programming
Assignment Number : MCA (2)/012/Assign /2015-16
Maximum Marks : 100
Weightage : 25%

Explain the double precision floating point IEEE 754 representation. Represent the number (124.0625)10 using IEEE 754 single precision and double precision representations.

Single Precision

S stands for Sign (white color)

E stands for Exponent (yellow color)

N stands for Number (also called Mantissa or Significand) (green color)

0this Sign bit (1 bit)

1st to 8this Exponent bits (7 bits)

9thto 31this Exponent bits (24 bits)

Double Precision

S stands for Sign (white color)

E stands for Exponent (yellow color)

N stands for Number (also called Mantissa or Significand) (green color)

0this Sign bit (1 bit)

1st to 11this Exponent bits (10 bits)

12thto 63this Exponent bits (53 bits)

First of all to represent we have (63.125)10 using IEEE 754

Since number is a positive number

Sign bit is: 0

 0

Now let’s work on Mantissa part

Convert 124 into binary form i.e 1111100

Remaining is fraction part 0.0625

Now we multiply fraction value with 2 till we get whole number 1 at the end.

We reach 1 so we stop

Write number from top to bottom

We get 0.0625 = 0001

Number (124.0625)10 = (1111100.0001)2

Convert Binary number to Exponent Form

1111100.0001 Will be 1.1111000001 * 26

Take 1111000001 as Significand

Significand number = (1111000001)2

Significand bits are:

 1 1 1 1 0 0 0 0 0 1

Fill the Significand in the beginning

Rest all fills it with 0’s

We will have

Significand bits are:

 1 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0

+6 is the Exponent

So we have to subtract it with 127

Exponent = 127+6=132

Exponent number (132)10 = (10000100)2

Exponent bits are:

 0 0 0 0 1 0 0

Note: we have to discard overflow bit since we have only 7 bits

For Double Precision Only Exponent

+6 is the Exponent

So we have to subtract it with 1023

Exponent = 1023+6=1029

Exponent number (1029)10 = (10000000101)2

Exponent bits are:

 0 0 0 0 0 0 0 1 0 1

Note: we have to discard overflow bit since we have only 11 bits

Significand bits are:

 1 1 1 1 0 0 0 0 0 1

Significand will only have a change in terms of Zero’s which are added at the end.

For single precision 23-10=13 Zero’s

For double precision 53-10=43 Zero’s