Question

Calculate $\left(3.984375\times {10}^{-1}+3.4375\times {10}^{-1}\right)+1.711\times {10}^3$by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit floating point format and in decimal.

Short answer

The required result will be 1772.

Step-by-step solution

Define the concept.

The decimal number system uses base 10 and the binary number system uses base 2.

For example, “12” is also equivalent to the value of the binary number system “1100”.

The decimal number $3.984375\times {10}^{-1}$is equal to $1.1001100000\times 2^{-2}$in the binary number system.

The decimal number $3.4375\times {10}^{-1}$is equal to $-1.0110000000\times 2^{-2}$in the binary number system.

The decimal number$1.771\times {10}^3$is equal to$-1.1011101011\times 2^{10}$in the binary number system.

Determine the calculation

Given numbers are $3.984375\times {10}^{-1}$, $3.4375\times {10}^{-1}$and $1.771\times {10}^3$.

Let, $x=3.984375\times {10}^{-1}$,

$y=3.4375\times {10}^{-1}$

and $z=1.771\times {10}^3$

role="math" localid="1654942319721" $\begin{array}{rcl}x& =& 1.1001100000\times 2^{-2}\\ y& =& +1.0110000000\times 2^{-2}\\ & & --------------\\ \left(x+y\right)& =& 10.1111100000\left(normalized\right)\\ 10.1111100000& =& 1.01111100000\times 2^{-1}\\ (x+y)& =& 1.0111110000\\ z& =& +1.1011101011[.000000000010111110000]\\ & & ---------------\\ \left(x+y\right)+z& =& 1.1011101011101[Roundedup]\\ 1.1011101100\times 2^{10}& =& 0110101011101100=1772[Indecimal]\end{array}$

Hence, the result will be 1772.