GATE CSE 1989 | Question: 1-vi

Question

Consider an excess -$50$ representation for floating point numbers with $4$ BCD digit mantissa and $2$ BCD digit exponent in normalised form. The minimum and maximum positive numbers that can be represented are __________ and _____________ respectively.

Comments

floating point standard not given.

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No floating point standard in 1989 :)

Answer 1

In binary we have normalized number of the form $(-1)^{S} \times 1.M \times 2^{E-\text{ Bias}}$ where,

• $S:$ sign bit 
• $M :$ Mantissa
• $E:$ Exponent

Similarly for for $\textsf{Binary Coded Decimal (BCD)}$ numbers the normalized number representation will be $:(-1)^{S} \times 1.M \times 10^{E-\text{ Bias}}$

Here bias is given to be excess - $50$ meaning that we need to subtract $50$ from the base exponent field to get the actual exponent.

So, maximum mantissa value with $4\;\text{BCD}$ digits $= 9999$

Maximum base exponent value with $2\;\text{BCD}$ digits $=  99$

So, maximum actual exponent value possible with $2\;\text{BCD}$ digits $= 99\;\text{- Bias} $

$\qquad \qquad \qquad =  99 - 50$

$\qquad \qquad \qquad  =  49$

So, the magnitude of the largest positive number $=  9.9999 \times 10^{49}$

Similarly, 

To get a minimum positive number, we have to set mantissa $= 0$ and exponent field  $= 0$

So, doing that we get exponent  $=  0 - 50  =  -50$

So, magnitude of minimum positive number $=  1.0000 \times 10^{-50}$

Therefore, maximum positive number $ = 9.9999 \times 10^{49}$

Minimum positive number $= 1.0000 \times 10^{-50}$

Comments

@Arjun Sir please edit 3^rd last line, power of 10 should be -50.

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@Habibkhan, @Arjun sir, thank you! I’m sure we are not going to find such a thorough explanation and discussion that too for a 1989 question!

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@Abhrajyoti00

Why 9.M ?