ISRO CSE 2014 | Question: 8

Question

In the standard $\text{IEEE}$ $754$ single precision floating point representation, there is $1$ bit for sign, $23$ bits for fraction and $8$ bits for exponent. What is the precision in terms of the number of decimal digits?

• A. $5$            
• B. $6$                    
• C. $7$                   
• D. $8$

Answer 1

Option C

Explanation

IEEE-754 Format   $(-1)^{S}  1.M^{E-127}$
$S \to \text{sign}$
$M \to \text{Mantissa}$
$E \to \text{Exponent}$

Precision is represented  by   ' 1.M '  where M is 23 bits and in total 24 bits are used for representing a precised number .

With 24 bits we can represent  $2^{24}$ numbers $(0 - 2^{24} -1).$  What is that value in decimal ?

Largest number that can be represented in any base $x$ is $x^{\text{no. of digits}} -1$ (starting from 0) . So,

$\text{base X}^{\text{no. of digits}} = \text{baseY}^{\text{no. of digits}}$

binary is base 2 and decimal is base 10

So, $2^{24} = 10^x$ . We need to find $x$

taking log  on both sides,

$\log_{2}2^{^{24}}$ = $\log_{2}10^{^{x}}$

$24=x \log_{2}10$

$x=7.22$

so we need minimum 7 digits in decimal

Comments

Sir  will it not require 8 decimal digits, because answer is 7.22. So we cant represent 24 bits in 7 digits completely, hence we need 8 digits.. Please correct me where I am wrong

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I have the same query as above why 7 and not 8?

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@anchit and Aakanchha , 7 is correct... To represent a "n" digit decimal number , let's say we need minimum "x" number of bits..then maximum decimal number with "n" decimal digits = 10ⁿ  - 1 and maximum decimal number with "x" binary bits = 2^x - 1...So, 10ⁿ -1 <= 2^x - 1 implies 10ⁿ <= 2^x ...Now , we verify this formula , let's say we have a decimal no 99(maximum 2 digit decimal number)..Since to represent 99 , we need 7 binary bits..Now , according to formula , since 99 has 2 decimal digits... So , 10²   <= 2^x ..So , x >= 2*lg10 ...So, x >= 2*3.32 ...So, x >= 6.67..so x = 7 ..So it proves that we need minimum 7 bits  to represent 99...We can verify it by other examples also..Now , In this question , we have 24 binary bits ..So, x = 24..and we have to find number of decimal digits ..So , 10ⁿ <= 2²⁴..So , n*lg 10 <= 24 ...n*3.32 <= 24.... n <= 24/3.32 ..So, n <= 7.22..So ,maximum possible value of n = 7... So , 8 is not possible here..

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sir,why here we consider 24 bit for precision calculation....as we know the precision is determined by the mantissa part so here the calculation for the inequality  should be  2^23 -1<=10^x-1.....

please correct me if i had done anything wrong here...

Answer 2

Precision is the number of digits we can represent accurately.

$1.2 \times 10^8$ - precision is just 2 and rest we have 8 zeroes which might not be accurate.

So, in IEEE 754 floating point representation we have 23 precision bits and we also have one implied bit before '.' making it 24 precision bits. With 24 bits we can represent $\lfloor 24 \log 2 \rfloor = 7$ decimal digits.

Comments

@arjun sir :from where this formula come from.please comment!

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baseX^{no. of digits} = baseY^{no. of digits}

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@Arjun Sir, @Mangilal Saraswat The maximum value that can be represented 24 precision bits = 2^24-1 . So, the no.of decimal digits = log(2^24) = 24 log(2).

Is this correct understanding? 

Answer 3

In IEEE 754 single precision format, the precision is 24 bits or 7 decimal digits. ($24 *log_{10}(2) ≈ 7.22$)

In IEEE 754 double precision format, the precision is 53 bits or 15 decimal digits. ($53 *log_{10}(2) ≈ 15.95$)

Notice that in bits, the precision is $|Mantissa| + 1$

Answer 4

Mantissa bits are precision bits. there are 23 M bits

2^23 = 8388608

this is 7 digits in decimal .

hence the precision in decimal is of 7 digits