I need to find the total number of positive numbers that this number system can

Question

I need to find the total number of positive numbers that this number system can hold when beta=2, n=4 and M=3, just like question question 2 part A of the picture above.

I figured that 2^4 would describe the different numbers created through the mantissa and then the choice of 7 different exponents (-3 to +3) would leave a possible 2^4*7=112 different positive numbers, but I have no idea if this is the correct logic behind this question.

2. The floating point representation of a real number has the form r 0 did Be, where di 0,0 di KB-1 -MS e S M. a) Suppose B 2, n 4, M 3. How many different nonzero positive numbers can be represented in this system? b) Find fl(V3), the floating point representation of V3 in this system. Apply the usual rules for rounding numbers up and down. Then express the final result in decimal form. How large is the roundoff error? c) In class we saw that Imax 7.5, r 0.0625 (these are the largest and smallest positive numbers that can be represented in this system). Suppose we switch to a system with m 5. Find the new values of rmax, rmin.

Explanation / Answer

2.

a) Given n=4 so for d1d2d3d4 we will have 16 combinations from 0000 to 1111(as the base b is given as 2).

Also mentioned that we need to find non zero postive numbers so 0000 will be eliminated as it leads to 0.

Now we have 15 combinations left from 0001 to 1111 and in each combination we can represent 7 numbers as

the values of e = -3 to 3 (given M =3). Total 15*7=105 non zero positive numbers can be represented.

ex:- for 0001 we have (0.0001)2 .2-3 to (0.0001)2 .2+3

   .   

   .

   .

   for 1111 we have (0.1111)2 .2-3  to (0.1111)2 .2+3

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