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

@Abhrajyoti00

Why 9.M ?

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@Arjun @Habibkhan sir can you pls tell why is it taken 9.M at time of Max and 1.M at time of min either both should be 9.M or both should be 1.M @Habibkhan you earlier mentioned that BCD normalized pattern is  $(-1)^s * 1.M * 10^{e-bias}$ so max sould be $1.9999* 10^{e-bias}$ or either min should be 9.0000 * $10^{-50}$ pls clearify this ?

---

There is no implied bit mentioned!

So, shouldn't the answer be

\[
\text{Maximum value} = 0.9999 \times 10^{(99 - 50)}
\]

\[
\text{Minimum value} = 0.1000 \times 10^{(00 - 50)}
\]

Answer 2

 

The "Why": Is Sawaal Mein "Trap" Kya Hai?

Sabse pehla aur sabse zaroori kaam hai iske "trap" ko pehchaanna.
This is NOT IEEE 754!

Yeh ek poori tarah se custom-made, hypothetical floating point system hai. Iska matlab humein IEEE 754 ke (S|E|M, binary mantissa, bias=127) saare niyam bhool kar, is question ke naye niyam-kanoon se khelna padega.

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Concept Deep Dive: Decoding the Custom BCD Format

Chaliye, is naye system ke niyam-kanoon ko decode karte hain.

• The Base: Question mein "BCD" (Binary Coded Decimal) likha hai. Jaise hi aap BCD dekhein, samajh jao ki hum Decimal (Base-10) ki duniya mein baat kar rahe hain, Binary (Base-2) ki nahi.

• The Formula: Iska general formula hoga:
  Value = (Mantissa)₁₀ * 10^(Actual Exponent)

Ab hum Mantissa aur Exponent, dono ke "aukaat" (range) ko nikaalenge.

1. Decoding the Exponent

• Format: "2 BCD digit exponent". BCD matlab har digit 4 bit ka. Toh 2 digit = 8 bits.

• Range of Stored Exponent: 2 BCD digits se hum kaun-kaun se decimal number bana sakte hain? 00 se lekar 99 tak.

• Bias: Question kehta hai "excess-50". Iska matlab Bias = 50.

• Actual Exponent ki Range:

  • Actual Exp = Stored Exp - Bias

  • Minimum Actual Exp: 00 (stored) - 50 = -50.

  • Maximum Actual Exp: 99 (stored) - 50 = +49.

2. Decoding the Mantissa

• Format: "4 BCD digit mantissa". 4 digit = 16 bits.

• Normalization: Question kehta hai "in normalized form". IEEE 754 mein iska matlab 1.M hota tha. BCD/Decimal duniya mein iska matlab hota hai ki mantissa ko hamesha 0.something ki form mein likha jaata hai aur pehla digit zero nahi ho sakta.

  • Mantissa = 0.d₁d₂d₃d₄ where d₁ ≠ 0.

• Range of Mantissa:

  • Minimum Normalized Mantissa: Pehla digit 1 hona chahiye, baaki 0. → 0.1000.

  • Maximum Normalized Mantissa: Saare digits 9 hone chahiye. → 0.9999.

---

Let's Find X (The Minimum Positive Number)

The "How": Sabse chhota number banane ke liye, humein Mantissa ko minimum aur Exponent ko minimum rakhna padega.

• Minimum Normalized Mantissa: 0.1000

• Minimum Actual Exponent: -50

• Assemble X:

  • X = 0.1000 * 10⁻⁵⁰

---

Let's Find Y (The Maximum Positive Number)

The "How": Sabse bada number banane ke liye, humein Mantissa ko maximum aur Exponent ko maximum rakhna padega.

• Maximum Normalized Mantissa: 0.9999

• Maximum Actual Exponent: +49

• Assemble Y:

  • Y = 0.9999 * 10⁴⁹

---

The Final Calculation: X * Y

Ab bas in dono ko multiply karna hai.

• X * Y = (0.1000 * 10⁻⁵⁰) * (0.9999 * 10⁴⁹)

• Powers ko ek saath, values ko ek saath:
  = (0.1 * 0.9999) * (10⁻⁵⁰ * 10⁴⁹)

• = (0.09999) * (10⁻¹)

• = 0.009999

The Last Step: Rounding Off
Question kehta hai "Rounded to one decimal place".

• 0.009999 ko one decimal place tak round karne par answer aata hai 0.0.

Answer 3

Given:

Bias $= q = 50$

Mantissa / Significand digits $= 4$

Exponent digits $= 2$

Base $= 10$

Exponent Range: $0 \leq E \leq 99$

Mantissa Range: $0.1000 \leq m \leq 0.9999$
 

Because, Normalized form requires the most significant digit to be non-zero.

Maximum value:

Exponent = 99

Mantissa = 0.9999

\[
\text{Number} = 10^{(99 - 50)} \times (0.9999)
\]

Minimum value:

Exponent = 00

Mantissa = 0.1000

\[
\text{Number} = 10^{(00 - 50)} \times (0.1000)
\]

Reference :

COA - William Stallings 10th edition (Problems 10.22)

Answer 4

The trick is to treat it as it's own floating point system...