Chapter 3, Problem 3.40E

Program Plan Intro

Given numbers: $1.666015625\times {10}^0$, $1.9760\times {10}^4$ and $-1.9744\times {10}^4$.

Binary Multiplication:

Binary multiplication follows the following rules:

$\begin{array}{l}0x0=0\\ 0x1=0\\ 1x0=0\\ 1x1=1\end{array}$

Given Information:

A, B and C are stored in 16-bit half precision format.

Explanation of Solution

Compare: $1.666015625\times {10}^0\times (1.9760\times {10}^4-1.9744\times {10}^4)$ and $(1.666015625\times {10}^0\times 1.9760\times {10}^4)+(1.666015625\times {10}^0\times -1.9744\times {10}^4)$

For part (1):

Here, $1.666015625\times {10}^0$ is represented as “A”, $1.9760\times {10}^4$ is represented as “B” and $-1.9744\times {10}^4$  is represented as “C”.

Converting the above values to binary:

$1.666015625\times {10}^0=1.1010101010\times 2^0$

$1.9760\times {10}^4=1.0011010011\times 2^{14}$

$-1.9744\times {10}^4=-1.0011010010\times 2^{14}$

The exponents match and hence shifting is not necessary.

Exponent=$0+4=4$

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of $B+C$ is represented below:

$\begin{array}{l}(B)1.0011010011\\ (C)-1.0011010010\\ ----------\\ (B+C)0.0000000001\times 2^{14}\\ (B+C)1.0000000000\times 2^4\end{array}$

The value of $B+C$ is $1.0000000000\times 2^4$.

The fraction “$A\times (B+C)$” is represented below:

$\begin{array}{l}(A)1.1010101010\\ (B+C)\times 1.0000000000\\ -------\\ 11010101010\\ ----------\\ A\times (B+C)1.10101010100000000000Guard=0,Round=0,Sticky=0:Noround\end{array}$

After rounding up to even, the value of “$A\times (B+C)$” is $1.1010101010\times 2^4$.

For part (2):

Here, $1.666015625\times {10}^0$ is represented as “A”, $1.9760\times {10}^4$ is represented as “B” and $-1.9744\times {10}^4$  is represented as “C”.

Converting the above values to binary:

$1.666015625\times {10}^0=1.1010101010\times 2^0$

$1.9760\times {10}^4=1.0011010011\times 2^{14}$

$-1.9744\times {10}^4=-1.0011010010\times 2^{14}$

The exponents match and hence shifting is not necessary.

Exponent=$0+14=14$

Here, the sign of the result is positive as all the numbers are positive and hence their product is positive. It is assumed that it has “1” guard, “1” round bit and “1” sticky bit and round to the nearest even.

The fraction of $A\times B$ is represented below:

$\begin{array}{l}(A)1.1010101010\\ (B)\times 1.0011010011\\ ---------\\ 11010101010\\ 11010101010\\ 11010101010\\ 11010101010\\ 11010101010\\ ---------------\\ 10.0000001001100001111Normalize,add1to\exp onent\end{array}$

Here, the guard is “1”, Round is “1”, Sticky is “1: Round” and the value of $A\times B$ is $10.0000001001100001111$.

Hence, the final value of $A\times B$ is $1.0000000101\times 2^{15}$.

The fraction “$A\times B+A\times C$” is represented below:

$\begin{array}{l}(A)1.1010101010\\ (C)\times 1.0011010010\\ ----------\\ 11010101010\\ 11010101010\\ 11010101010\\ 11010101010\\ 11010101010\\ -------------\\ 10.000000011111011010\\ Normalizeandadd1to\exp onent\\ A\times C-1.0000000100\times 2^{15}\\ A\times B1.0000000101\times 2^{15}\\ A\times C-1.0000000100\times 2^{15}\\ ------------\\ A\times B+A\times C.0000000001\times 2^{15}\\ A\times B+A\times C1.0000000000\times 2^5\\ \end{array}$

Hence, the value of $A\times B+A\times C$ is $1.0000000000\times 2^5$.

The value of both the subparts is not equal as $A\times (B+C)$ is $A\times (B+C)=1.1010101010\times 2^4=26.65625$ whereas the value of

$(A\times B)+(A\times C)=1.0000000000\times 2^5=32$.