Chapter 3, Problem 3.31E

Program Plan Intro

Given numbers: $8.625\times {10}^1$ and $-4.875\times {10}^0$

Given Information:

A and B are stored in 16-bit half precision format.

Division:

$Dividend=Quotient\times Dividend+Remainder$

Booth’s Algorithm:

In order to divide two binary numbers, Booth algorithm is used.

Binary Divisions:

The binary number division follows the multiplication rules:

$\begin{array}{l}0x0=0\\ 0x1=0\\ 1x0=0\\ 1x1=1\end{array}$

For example:

“$1011/11$” can be calculated as:

$\begin{array}{l}11)1011(11\\ -11\\ ---\\ 101\\ -11\\ ---\\ 10<--\mathrm{Remainder}(R)\end{array}$

Explanation of Solution

Calculate: $8.625\times {10}^1/-4.875\times {10}^0$

Converting the above values to binary:

$8.625\times {10}^1=1.0101100100\times 2^6$

$-4.875=-1.0011100000\times 2^2$

Here, the exponent can be calculated as:

Exponent= $6-2=4,4+15=19(10011)$

Here, the sign is negative as one number is negative and the other is positive and hence their division yields a negative sign. It is assumed that it has a guard, a round bit, and a sticky bit.

The fraction generated by $8.625\times {10}^1/-4.875\times {10}^0$ is represented as:

$\begin{array}{l}10011100000.\left|10101100100.0000000000000000\right|1.00011011000100111\\ -10011100000.\\ -----------\\ 10000100.0000\\ -1001110.0000\\ ------------\\ 1100110.00000\\ -100111.00000\\ -------------\\ 1111.0000000\\ -1001.1100000\\ ---------\\ 101.01000000\\ -100.11100000\\ ----------\\ 000.011000000000\\ -.010011100000\\ -----------\\ .000100100000000\\ -.000010011100000\\ ------------\\ .00000011011000000\\ -.00000010011100000\\ --------------\\ .00000000110000000\end{array}$

The result of the division is “1.000110110001001111” has a Guard =0, Sticky=1: No Round as it is assumed that it has a guard a sticky bit and it has fixed sign.

Converting the above result into decimal:

$-1.0001101100\times 2^4=1101000001101100=10001.101100=-17.6875$

The original division in decimal results into:

$86.25/-4.875=-17.692307692307$

There was underflow as some information was lost as the result did not fit into the available 10-bit field and the answer differed from decimal by “$\text{.00480769230}$”.