3.32 Let A=3.984375*10^-1, B=3.4375*10^-1, C= 1.771*10³ be the three numbers in scientific notations in decimal. Calculate (A+B) + C by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit floating point format and in decimal.

First, convert the normalized scientific notations from decimal to binary as follows:

A=3.984375*10^-1 =0.3984375 = 1.*2^-2 (Fill in the fraction of 10 binary bits)

B=3.4375*10^-1 = 0.34375 = 1. *2^-2 (Fill in the fraction of 10 binary bits)

C= 1.771*10³ = 1771 = 1.*2¹⁰ (Fill in the fraction of 10 binary bits)

Next, align the binary point of A and B, compute A+B, and then get the normalized scientific notations of the sum A+B = 1. *2^-1 (Fill in the fraction of 10 binary bits)

Next, align the binary point of (A+B) and C, compute (A+B)+C, and then get the normalized scientific notations of the sum (A+B)+C as follows.

(A+B)+C = (0.000000000010111110000)* 2¹⁰ + 1.1011101011*2¹⁰ = (0.000000000010111110000 + 1.1011101011 )* 2¹⁰=(1.1011101011 10 111110000)* 2¹⁰ =1.1011101100* 2¹⁰ (Round up to have 10-bit fraction).

So, (A+B)+C is equal to a decimal value of , and its 16-bit half precision format is (Fill in 16 binary bits).

3.33 Now calculate A + (B+C) by hand, assuming each of the values are stored in the 16-bit half precision format described in Exercise 3.27 (and also described in the text). Assume 1 guard, 1 round bit, and 1 sticky bit, and round to the nearest even. Show all the steps, and write your answer in both the 16-bit floating point format and in decimal.

A+(B+C) = 1.1001100000*2^-2 + (1.0110000000*2^-2 + 1.1011101011*2¹⁰) =1.1001100000*2^-2 + (0.00000 00000 01 0110000000*2¹⁰ + 1.1011101011*2¹⁰) = 1.1001100000*2^-2 + (1.1011101011 01 0110000000*2¹⁰) =1.1001100000*2^-2 + (1.1011101011 *2¹⁰ ) (Round down to 10-bit fraction). After aligning the binary point, we have A+)B+C) = 0.000000000001 1001100000*2¹⁰ + (1.1011101011 *2¹⁰) = (0.0000000000 01 1001100000 + 1.1011101011) *2¹⁰ = (1.1011101011 01 1001100000) *2¹⁰ (round down to get 10-bit fraction) = 1. *2¹⁰ (10 binary bit).

So, A+(B+C) is equal to a decimal value of , and its 16-bit half precision format is (Fill in 16 binary bits)

Is (A+B)+C equal to A+(B+C) (y/n)? (Fill in either y or n)