Question

Calculate the sum of $2.6125\times {10}^1$and$4.150390625\times {10}^{-1}$ by hand, assuming A and B are stored in the 16-bit half precision described in Exercise 3.27. Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps

Short answer

The required result will be$2.6546875\times {10}^1$.

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 26.125 is equal to 11010.001 in the binary number system.

The decimal number 0.4150390625 is equal to 0.011010100111 in the binary number system.

Determine the calculation

Given numbers are $2.6125\times {10}^1$and $4.150390625\times {10}^1$.

By doing the simplification of the number $2.6125\times {10}^1$

$\begin{array}{rcl}2.6125\times {10}^1& =& 26.125\times {10}^0\\ & =& 26.125\times 1\\ & =& 26.125\\ & =& 11010.001(inbinary)\\ & =& 1.1010001000\times 2^4\end{array}$

By doing the simplification of the number $4.150390625\times {10}^1$

$\begin{array}{rcl}4.150390625\times {10}^1& =& 0.4150390625\times {10}^0\\ & =& 0.4150390625\times 1\\ & =& 0.4150390625\\ & =& .011010100111(inbinary)\\ & =& 1.1010100111\times 2^{-2}\end{array}$

For the addition,

$$\begin{gathered} 1.101000100000 \\ 1.0000011010100111 \\ ------------- \\ 1.101010001010(As,thelastbitsismorethanhalfofthelistsignificantbit. \\ hencethevalueisroundedup) \end{gathered}$$

Hence, for the simplification of the result-

$$\begin{gathered} 1.1010100011\times 2^4 \\ =11010.100011\times 2^0 \\ =26.546875(indecimals) \\ =2.6546875\times {10}^1 \end{gathered}$$