Question

Consider a 8 bit floating point system that is designed to hold positive numbers only; four bits are assigned to mantissa, three bits are assigned to exponential and one bit is assigned to the sign of exponential. What is the absolute error (not the percentage relative error) when this system is used to represent zero?

Answer 1

To find the solution to this question we will have to first understand how a number is stored in the floating point format.

In the floating point format the first bit is the sign bit, which is for positive number and for negative number.

The exponent and the mantissa part of a floating point number is found out by finding the binary representation of the number in its normal for. For eg. If the number is then its binary form is and in its normal for it is : $1.01\times2^2$ .

Here, is the mantissa and is the expontent.

While storeing it as a floating point, the exponent is stored in the biased form and in the mantissa the first bit i.e of the mantisa is not included in it (it is hidden i.e it is always considered that there is a before the bits of the mantissa). Biased form of the exponent means that a bias is added to the exponent and then the resultant sum is stored in its binary form. This bias is equal to $2^{k-1}-1$ , where is the number of bits reserved for exponent.This makes the range of the exponent from
-(2k-1 - 1) to 2k-1

In this case,

No. of bits reserved for exponent, k = 3

No. of bits reserved for mantisa = 4

Here it is said that the floating point number is designed to store only positive number, so we will ignore the sign bit

The bias will be $2^{3-1}-1=2^2-1=3$ . And the lowest possible value of the exponent will be $-(2^{3-1}-1)=-3$ .

While representing in a floating point format, there is a problem. The problem is that the first bit of the mantisa is always considered to be . So, we can't actually store absolute in this format, but in place of it we can store a very small number. And this is the error in storeing

The smallest possible number (positive) that we can store will have the smallest value of exponent i.e. and the mantisa will be all . So the representation of will actually be  $1.0000 \times 2^{-3}=0.001$.

In decimal will be  .

The absolute error when the system is used to represent zero is