Question

Calculate $\left(1.666015625\times {10}^0\times 1.9760\times {10}^4\right)+\left(1.666015625\times {10}^0\times -1.9744\times {10}^4\right)$ by hand, assuming each of the values is 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.

Short answer

The value of $\left(1.666015625\times {10}^0\times 1.9760\times {10}^4\right)+\left(1.666015625\times {10}^0\times -1.9744\times {10}^4\right)$is as follows:

1.0000000000 x 2⁵

The decimal value is 32

Step-by-step solution

Determine the 16-bit half-precision format.

The 16-bit half-precision format will have the 1 sign bit, 5 exponent bit, and 10 fraction bits.

If the exponent is large to represent in the exponent field, then it is called overflow.

Guard bit is the first two extra bits in the right of the intermediate calculations that help in rounding.

Round is the nearest number that can be fit into the floating format.

A sticky bit is used to round in addition to guard and to set the round when there are bits present right to the round bit.

Determine the value of  1.666015625×100×1.9760×104+1.666015625×100×1.9744×104

Convert each value to the half-precision format as follows:

Consider 1.666015625 x 10⁰ as A , The exponent is zero, so simply convert the number to binary.

$$\begin{gathered} A=1.666015625\times {10}^0 \\ =1.1010101010\times 2^0 \end{gathered}$$

Consider 1.9760 x 10⁴as B. The exponent is 4, so place the decimal point four digits right as 19760.0 ., Now convert to binary.

$$\begin{gathered} B=1.9760\times {10}^4 \\ =19760.0 \\ =1.0011010011\times 2^{14} \end{gathered}$$

The floating-point is adjusted to the power of 2.

Consider -1.9744 x 10⁴ as C . The exponent is 4, so place the decimal point four digits right as -19744.0., Now convert to binary.

$$\begin{gathered} C=-1.9744\times {10}^4 \\ =19744.0 \\ =-1.0011010010\times 2^{14} \end{gathered}$$

The floating-point is adjusted to the power of 2.

The fraction part is calculated as follows:

$A\times B=1.1010101010\times 1.0011010011$

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$$\begin{gathered} 000000000011010101010 \\ 00000000011010101010 \\ 00000011010101010 \\ 000011010101010 \\ 00011010101010 \\ 11010101010 \end{gathered}$$

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$10.0000001001100001111$

Normalize the above value and add 1 to the exponent.

A x B = 1.0000001001100001111

Guard bit=1,

Round bit=1,

Sticky bit=1

Then, AxB = 1.000000101 x 2¹⁵

Now, Calculate : A x C

$A\times C=1.1010101010\times 1.0011010010$

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$$\begin{gathered} 00000000011010101010 \\ 00000011010101010 \\ 000011010101010 \\ 00011010101010 \\ 11010101010 \end{gathered}$$

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$10.0000000111110111010$

Normalize the above value and add 1 to the exponent

$A\times C=1.00000001111101110100$

Guard bit=1,

Round bit=1,

Sticky bit=1

$A\times C=-1.0000000100\times {10}^{15}$

Finally, Add and

$$\begin{gathered} A\times B=1.0000000101\times {10}^{15} \\ A\times C=-1.0000000100\times {10}^{15} \\ \left(A\times B\right)+\left(A\times C\right)=.0000000001\times {10}^{15} \end{gathered}$$

Normalize the above value, then the floating-point value of $\left(1.666015625\times {10}^0\times 1.9760\times {10}^4\right)+\left(1.666015625\times {10}^0\times 1.9744\times {10}^4\right)$

is 1.0000000000 x 2⁵

The decimal value is 32