Chapter 1.7, Problem 4QE

Explanation of Solution

The largest value represented using floating point:

An excess notation system using bit patterns of length three is:

Bit Pattern	Value represented
$111$	$3$
$110$	$2$
$101$	$1$
$100$	$0$
$011$	$-1$
$010$	$-2$
$001$	$-3$
$000$	$-4$

• The sign bit has to be $0$ for maximum value.

$\underset{\_}{0}\_\_\_\_\_\_\_$

• The exponent should have maximum positive value so that radix moves by maximum distance to the right.
• So, the value of exponent from table 1 is $111$ which has value of $3$.

$\underset{\_}{0}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\text{\_}\text{\_}\text{\_}\text{\_}$

• The elements of mantissa should be all $1$ for maximum value. Therefore, the mantissa has to be $.1111$.

$\underset{\_}{0}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}$

• Move the radix by $3$ points to the right of mantissa $.1111$.

$\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}.\underset{\_}{1}$

• The value of $\underset{\_}{1}\underset{\_}{1}\underset{\_}{1}.\underset{\_}{1}$ is:

$\begin{array}{c}{\left(111.1\right)}_2={\left(1\times 2^2+1\times 2^1+1\times 2^0+\frac{1}{2}\right)}_{10}\\ ={\left(7\frac{1}{2}\right)}_{10}\end{array}$

Thus, the largest value represented using floating point is “$7\frac{1}{2}$”