COMP2421 Assignment 1

MA Mingyu (Derek) | 14110562D | Sep 17, 2017
derek.ma | derek.ma@connect.polyu.hk

Question 1

Convert the binary number to decimal number:
\[
01110001_{two} = 1*2^6+1*2^5+1*2^4+1*2^0 = 113_{ten}
\]. For 8-bit word, complement to \(2_{ten}^8 = 256_{ten}\):
\[
256_{ten} - 113_{ten} = 143_{ten}
\]. Thus the value of 2's complement of \(01110001_{two}\) in decimal is \(143\).

Question 2

According to the straightforward approach for making two's complement, take the bitwise complement of the value and then add 1:

Bitwise complement of \(10011101_{two}\) is: \(01100010_{two}\) ;
then add 1: \(01100010_{two} + 1 = 01100011_{two}\).

Here, we get the 2's complement value of the given value in the question in binary, then convert it to decimal:

\[
01100011_{two} = 1*2^6+1*2^5+1*2^1+1*2^0 = 99_{ten}
\]. Thus, the value of 2's complement of \(10011101_{two}\) in decimal is \(99\).

Question 3

The original real number is \(111.111_{ten}\).

Get binary number for \(111_{ten}\):
\(111 = 55*2+1\)
\(55 = 27*2+1\)
\(27 = 13*2+1\)
\(13 = 6*2+1\)
\(6 = 3*2+0\)
\(3 = 1*2+1\)
\(1 = 0*2+1\)

\[111_{ten} = 1101111_{two}\]

Get binary number for \(.111_{ten}\):
\(0.111*2 = 0.222\)
\(0.222*2 = 0.444\)
\(0.444*2 = 0.888\)
\(0.888*2 = 1.776\)
\(0.776*2 = 1.552\)
\(0.552*2 = 1.104\)
\(0.104*2 = 0.208\)
\(0.208*2 = 0.416\)
...

\[.111_{ten} = .00011100011010100111111011111001110..._{two}\]

So we can know:
\[111.111_{ten} = 1101111.00011100..._{two} = (-1)^0 *1.10111100011100011..._{two}*2^6\];
\(S=0\);
\(Fraction=101111000111000110101001111...\);
\(e=6\);
\(Exponent = e+bias = 6+127 = 133_{ten} = 10000101_{two}\).

So the single-precision float is:
0 10000101 10111100011100011010100.

After optional rounding process, it can be:
0 10000101 10111100011100011010101.

Question 4

Original single-precision float is:

\[1 01111111 01010101010101010101010\]
, so we can get:
\[S = 1\];
\[
\begin{align}
Fraction &= .01010101010101010101010_{two} \\
&= 2^{-2}+2^{-4}+2^{-6}+2^{-8}+2^{-10}+2^{-12}+2^{-14}+2^{-16}+2^{-18}+2^{-20}+2^{-22} \\
&= 0.333333254_{ten}
\end{align}
\];
\[Exponent = 01111111_{two} = 127_{ten}\]
. Then calculate value of \(e\) by subtracting the bias:
\[e = Exponent - 127 = 127-127 = 0\].

Finally, the real number is:
\[
\begin{align}
x &= (-1)^S * (1+Fraction)*2^{e} \\
&= (-1)^1* 1.333333254 * 2^0 \\
&= - 1.333333254_{ten}
\end{align}
\]

Question 5

Represent pi using single-precision

The original real number is \(3.1415926535_{ten}\).

Get binary number for \(3_{ten}\):
\(3 = 1*2+1\)
\(1 = 0*2+1\)

\[3_{ten} = 11_{two}\]

Get binary number for \(0.1415926535_{ten}\):
\(.1415926535*2 = .283185307\)
\(.283185307*2 = .566370614\)
\(.566370614*2 = 1.132741228\)
\(.132741228*2 = .265482456\)
\(.265482456*2 = .530964912\)
\(.530964912*2 = 1.061929824\)
\(.061929824*2 = .123859648\)
...

\[.1415926535_{ten} = 0.001001000011111101101010100..._{two}\]

So we can know:
\[
\begin{align}
3.1415926535_{ten} &= 11.001001000011111101101..._{two} \\
&= (-1)^0 *1.1001001000011111101101..._{two}*2^1
\end{align}
\];
\(S=0\);
\(Fraction=1001001000011111101101010100010000010001011...\);
\(e=1\);
\(Exponent = e+bias = 1+127 = 128_{ten} = 10000000_{two}\).

So the single-precision float for pi with 10 decimal places is:
0 10000000 10010010000111111011010
; while after rounding the last digit of the fraction, it's:
0 10000000 10010010000111111011011.

Evaluate accuracy

Accuracy for not rounded single-precision float

Convert the above single-precision float back to a real number:
\(.10010010000111111011010_{two} = 0.5707962512969970703125\)
\[
\begin{align}
x &= (-1)^0*1.5707962512969970703125*2^{128-127} \\
&= 3.14159250259
\end{align}
\]

The single-precision value without rounding is: \(3.14159250259\). Then calculate the accuracy:

\[
\begin{align}
Error &= |single-precision value - actual value| * 100\% / actual value \\
&= \frac{|3.14159250259 - 3.1415926535|}{3.1415926535} *100\% \\
&= \frac{1.5091e-7}{3.1415926535} *100\% \\
&= 0.0000048036145 \%
\end{align}
\].

Thus we can know the accuracy for using single-precision to express Pi value without rounding with 10 decimal places: the error is \(0.0000048036145\%\), the accuracy is \(99.9999951964\%\).

Accuracy for rounded single-precision float

Convert the above rounded single-precision float back to a real number:

\(.10010010000111111011011_{two} = 0.57079637050628662109375\)
\[
\begin{align}
x &= (-1)^0*1.57079637050628662109375*2^{128-127} \\
&= 3.1415927410125732421875
\end{align}
\].

The single-precision value is: \(3.1415927410125732421875\). Then calculate the accuracy:

\[
\begin{align}
Error &= |single-precision value - actual value| * 100\% / actual value \\
&= \frac{|3.1415927410125732421875 - 3.1415926535|}{3.1415926535} *100\% \\
&= \frac{.0000000875125732}{3.1415926535} *100\% \\
&= 0.00000278561172\%
\end{align}
\].

Thus we can know the accuracy for using single-precision to express Pi value with 10 decimal places: the error is \(0.00000278561172\%\), the accuracy is \(99.9999972144\%\).

Notes: we can found that the accuracy after rounding is actually better than the solution without rounding.