Consider a 16-bit floating-point representation based on the IEEE floating-point

Question

Consider a 16-bit floating-point representation based on the IEEE floating-point format (the format we dis- cussed in the class), with one sign bit, seven exponent bits (k = 7), and eight fraction bits (n = 8). The exponent bias is then 271 1 = 63.

(a) {2 mark} What is the smallest denormalized number that can be encoded in this format?

(b) {2 mark} What is the smallest value > 7.25 that can be encoded in this format?

(c) {2 mark} The number 7.1 cannot be represented accurately using the floating-point representation. What is the closest decimal number to 7.1 that is representable in the 16-bit floating-point representation?

(d) {2 mark} What is the number with hex representation 80C0? (i.e., decode 80C0, where 80C0 represents a floating point)

Explanation / Answer

Solution:

The give floating point representation contains 1 bit as sign, 7 bit as exponent and 8 bit as fraction bits or Mantissa

so smallest number represented will be

in normalized form this number will be 1.00000000*2^-63

in decimal, the number will be 2.1684043e-19.

b)

we will get the smallest value greater than 7.25 by forst representing 7.25 in the given format

7.25 in binary is 111.01,

in normalized format 1.1101*2^0010

Exponent= Bias + Actual exponent= 63+2= 65 in binary 1000001

now if we put 1 at LSB that will be the smallest number greater than 7.25.

1.11010001* 2^0010

= 1.81640625*4= 7.265625

c)

7.1 in binary is 111.0001100110011001101 since fraction bits will be more than 8 bits that is why it is not valid.

so the closest number will be 7.0

whose binary equivalent is 111

d)

80C0 in binary is

1000 0000 1100 0000,

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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