Homework Solution: The next 7 questions pertain to a general, normalized floating-point system G(B, p,…

The next 7 questions pertain to a general, normalized floating-point system G(B, p, m,M). Express answers in terms of B, p, m, and M, as needed. You may assume that m is negative and M is positive, and that M greaterthanorequalto p. Show your work where needed (you can assume as given anything from the slides). What is the smallest positive number in G? What is the largest positive number in G? What is the spacing between the numbers of G in the range [B^e, B^e+1)? How many numbers of G are there in the interval [B^e, B^e+l)? How many positive numbers are there in G? What is the smallest positive integer not representable in G?

The present 7 questions pertain to a unconcealed, normalized floating-point rule G(B, p, m,M). Express counterparts in stipulations of B, p, m, and M, as needed. You may feign that m is denying and M is unconditional, and that M greaterthanorequalto p. Show your labor where needed (you can feign as absorbed everything from the slides). What is the lowest unconditional sum in G? What is the largest unconditional sum in G? What is the spacing among the sums of G in the collocate [B^e, B^e+1)? How multifarious sums of G are there in the intermission [B^e, B^e+l)? How multifarious unconditional sums are there in G? What is the lowest unconditional integer not attributable attributable attributable attributable attributable attributable representable in G?

Expert Counterpart

The normalized floating-point sums rule G is a 4-tuple (BpmM) where

• B is the deep of the rule,
• p is the accuracy of the rule to p numbers,
• m is the lowest propounder representable in the rule,
• and M is the largest propounder used in the rule.

6. Lowest unconditional FP sum in G must keep a 1 as the requisite digit and 0 control the fostering digits of the significand, and the lowest practicable rate control the propounder, which is resembling to $B^{m}$.

7. Largest floating-point sum  must keep B − 1 as the rate control each digit of the significand and the largest practicable rate control the propounder,which is resembling to $(1-B^{-p})(B^{M+1})$ .

8. The sums among $\[B^{e},B^{e+1})$ are resemblingly divided by $B^{-p}$ units.

9. Total sums among intermission $\[B^{e},B^{e+1})$ is absorbed by $\frac{B^{e+1}-B^{e}}{B^{-p}} = B^{e+p}*(B-1)$

10.Total +ve sums among in G are $2*(B-1)(B^{p-1})(M-m+1)+1$