Chapter 3, Problem 3.37E

Program Plan Intro

Given numbers: $3.41796875\times {10}^{-3}$, $6.34765625\times {10}^{-3}$, and $1.05625\times {10}^2$.

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: $3.41796875\times {10}^{-3}\times (6.34765625\times {10}^{-3}\times 1.05625\times {10}^2)$ and $(3.41796875\times {10}^{-3}\times 6.34765625\times {10}^{-3})\times 1.05625\times {10}^2$

For: $3.41796875\times {10}^{-3}\times (6.34765625\times {10}^{-3}\times 1.05625\times {10}^2)$

Here, $3.41796875\times {10}^{-3}$ is represented as “A”, $6.34765625\times {10}^{-3}$ is represented as “B” and $1.05625\times {10}^2$ is represented as “C”.

Converting the above values to binary:

$3.41796875\times {10}^{-3}=1.1100000000\times 2^{-9}$

$6.34765625\times {10}^{-3}=1.0001000000\times 2^{-8}$

$1.05625\times {10}^2=1.1010011010\times 2^6$

The exponent for $A\times B$ can be represented as:

Exponent=$-8+6=2$

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 $B\times C$ is represented below:

$\begin{array}{l}(B)1.0001000000\\ (C)\times 1.1010011010\\ ----------\\ 10001000000\\ 10001000000\\ 10001000000\\ 10001000000\\ 10001000000\\ 10001000000\\ ------------\\ 1.110000001110100000000\\ 1.110000001110100000000Guard51,Round50,Sticky51:Round\end{array}$

The value of $B\times C$ is $1.1100000100\times 2^{-2}$.

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

$\begin{array}{l}(A)1.1100000000\\ (B\times C)\times 1.1100000100\\ ---------\\ 11100000000\\ 11100000000\\ 11100000000\\ ------------\\ 11.00010001110000000000Normalize,add1to\exp onent\\ 1.100010011100000000000Guard=1,Round=0,Sticky=0:\\ \\ \end{array}$

Hence after rounding up to even, the value of “$A\times (B\times C)$” is $1.1000100100\times 2^{-10}$.

For: $(3.41796875\times {10}^{-3}\times 6.34765625\times {10}^{-3})\times 1.05625\times {10}^2$

Here, $3.41796875\times {10}^{-3}$ is represented as “A”, $6.34765625\times {10}^{-3}$ is represented as “B” and $1.05625\times {10}^2$ is represented as “C”.

Converting the above values to binary:

$3.41796875\times {10}^{-3}=1.1100000000\times 2^{-9}$

$6.34765625\times {10}^{-3}=1.0001000000\times 2^{-8}$

$1.05625\times {10}^2=1.1010011010\times 2^6$

The exponent can be represented as:

Exponent=$-9-8=-17$

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 the multiplication is represented as:

$\begin{array}{l}(A)1.1100000000\\ (B)\times 1.0001000000\\ ---------\\ 11100000000\\ 11100000000\\ ----------\\ 1.11011100000000000000\\ A\times B1.11011100000000000000\\ Guard=0,Round=0,Sticky=0,NoRound\\ \end{array}$

Hence, the result of the product “$A\times B$” is “$1.1101110000\times 2^{-17}$”.

There is underflow as A and B are both small and their product does not fit into the 16-bit floating point format being used and hence cannot be represented.

As in 2^nd case the value of $(A\times B)\times C$ cannot be found whereas in 1^st case the value of “$A\times (B\times C)$” is $1.1000100100\times 2^{-10}$ and hence they are not equal.