Question

Suppose that you use 1 bit for the sign, 3 bits for the exponent, and 14 bits for the mantissa and that you represent numbers in normalized notation. Assume that setting exponent bits to all 0s or 1s are special cases.

What is the smallest number that you can represent, assuming the same bias? Again, truncate the result to the third decimal place.

Answer 1

According to IEEE 754 standard the floating number can be represented in terms of sign,exponent and mantissa, and according to the question given the space for the sign bit be 1 , size for exponent bit will be 3 and for mantissa it will be 14.

Now first we calculate the value of bias which can be calculated by using formula 2^k-1-1 where k represents the number of bits in the exponent. in given question the value of k is 3 therefore on substituting the value of k=3 we get the value of bias as--

=>2^(3-1)-1 =>2²-1 =>4-1 =>3

Since we have to calculate the smallest floating number therefore the value of the sign bit will be 0 as it will represent the positive number. now for the smallest float number the value of the mantissa will be 000....upto 14 bits.

Now for the smallest floating number the value of the exponent bits should also be smallest therefore the value of exponent will be 0 but it will not be possible because all zero's represents the reserved bits therefore we will take next smallest value which will be 1.

We also know that the real value of the exponent bit is the summation of the bias and the real exponent, since for the smallest value of float number this summation should be 1 therefore the value of the real exponent will be calculated as follows--

Bias + real exponent =1 ........(for smallest value of float number) => 3+ real exponent = 1 => real exponent = -2

Hence the smallest float number will become 1.00000000000000 *2^-2 which is equivalent to 0.0100000000000000, on truncating the result upto three decimal place the number will become 0.010 which is the smallest floating number to represent in the given format.