ISRO2008-23

Question

A computer uses $8$ digit mantissa and $2$ digit exponent. If $a=0.052$ and $b=28E+11$ then $b+a-b$ will :

• A. result in an overflow error
• B. result in an underflow error
• C. be $0$
• D. be $5.28E$+$11$

Comments

Related Terminologies:-

Overflow is when the absolute value of the number is too high for the computer to represent it.

Underflow is when the absolute value of the number is too close to zero for the computer to represent it.

Answer 1

$a = 0.052$, mantissa $=0.52$, exponent $= -1$.

$b = 28 E +11$, mantissa $=0.28$, exponent $= 13$.

Now, to do $b + a$, we have to make the exponent same. Lets do this by rewriting $b$ as shown here. 

So, $b = 28000000000000 E -1$. So, $b+a = 28000000000000.52E -1$. 

Using 8 digit mantissa, this will be truncated to 0.28000000 with exponent as 13 which is $28 E +11$. So, now $b-b = 0$. 

Comments

Sir here Least Significant digits were truncated. Does it has any relation with how the number is stored in memory (Little Endian or Big Endian)

---

No, it does not. The floating point representation takes care of the endianness.

---

@Arjun sir,
the content for converting b is actually requiring logging in information. can you please write how to convert b

Answer 2

In standard form: a=0.052 = 0.52 E-1; and  b=28E+11=0.28 E+13

To add b+a, Small exponent number (a) is shifted to (13- -1 =14 ) places to right side

0.0000000000000052E+13.

But computer uses only 8 digit mantissa. Digits beyond 8^th position will be discarded.

So a = 0.00000000E+13 = 0.0 E+13

Hence b+a = (0.28E+13) + (0.0E+13 ) = 0.28E+13

Then b+a-b = (0.28E+13)- (0.28E+13) =0

Answer 3

Answer is C i.e, 0

Explanation: 

Addition will be performed first. b+a evaluate to 'b' as a significant digits of 'a' will be lost when it is converted to exponent 11.So b+a-b is b-b which is 0.

In other words,

'a' value is very less compare to 'b'.Once you perform exponent of 11 of 'b' then 'a' value becomes negligible.It means 'a' value can't change the world of 'b' with that power.So we consider 'a' is zero/nothing and 'b' is left. So 'b' - 'b' becomes 0.

Answer 4

Let's break down the question first: 

A computer uses 8 digit mantissa and 2 digit exponent

"digit" means we're talking about decimal number system, and not binary(terminology: bits). So, we need not convert any given number to binary.

 

a=0.052 and b=28E+11

What does this E + 11 mean? It means $* 10^{11}$, or, 11 zeroes after the number 28.

---

 

Now let's solve this:

a = 0.052

b = 2800000000000

 

Standard forms:

a = 0.52 E-1

b = 0.28 E+13

 

To do a+b, or any operation; we make the exponents equal.

a = 0.0000000000000052 E+13 //Added 14 zeroes after decimal

b= 0.28 E+13

Now, as mantissa is only realised upto 8 digits, a is seen as

a = 0.00000000 E+13

a is as good as 0, so b+a = b+0 = b.

 

So, b+a-b = b+0-b = b-b = 0.

---

 

Solving by making exponents equal the other way around:-

 

a = 0.52 E-1

b = 28000000000000.0 E-1

b+a = 28000000000000.52 E-1

Now this term minus b:-

28000000000000.52 E-1 - 28000000000000.0 E-1

=00000000000000.52 E-1

 

Standard form:

0.0000000000000052 E+13

Mantissa is realised only until 8 digits

So, 0.00000000 E+13

=0

---