Consider the finite floating point number system F with x0di

Consider the finite floating point number system F with x=±0.did2x10°, 1 di 9, 0 d2 9, -2 e xEF 2. 2 for any 0

Solution

(a)Excluding 0 number of floating point numbers are: 9 choices for d1 and 10 choices for d2 and 5 choices for e, all three can be chosen independently. Hence ,9*5*10=450

Including 0 gives 451 total floating point numbers in F

(b) Largest positive number is obtained by giving largest possible value to d1,d2 and e

So we get: 0.99*10^2=99

Smallest positive number is obtained by giving smallest possible values to d1,d2 and e

So we get:0.10*10^(-2)=0.001

(c)

Machine precision is the smallest number e such that the difference between 1 and 1 + e is nonzero

In this system:1=0.1*10^1

Adding 0.1 to 1 gives: 1.1 =0.11*10^-1

Largest of the number smaller than 0.1 is 0.09. Adding to 1 gives:1.09=0.10*10^-1

Hence adding anything smaller than 0.1 to 1 leaves 1 unchanged.

So machine precision is:0.1

d)

a=(98+5)-40

First system computes number within the brackets ie 98+5=103

But largest number in the system is 99. Hence the system ignore the 1 at the hundreths place and reads the number as 03=.03*10^2

a=.3*10^1-.4*10^2=-.37*10^2

For a reliable computation we can rearrange the brackets as follows:

a=98+(5-40)=98-35=63=.63*10^2