done bonus question 1

Homework 11/bonus.tex

@@ -0,0 +1,280 @@
+\documentclass[12pt]{article}
+\title{Computer Architecture -- Assignment 11 Bonus}
+\author{Claudio Maggioni \and Tommaso Rodolfo Masera}
+\date{}
+
+\begin{document}
+\maketitle
+
+\section{Bonus Question 1}
+
+\subsection{Decimal to Hexadecimal}
+
+\textbf{All the numbers are going to be first converted to binary and then to
+ hexadecimal for an easier calculation}
+
+\begin{itemize}
+
+\item[a)] \textbf{\large 11};
+
+First we find how 11 is expressed in binary which is 1011 and then we convert it to
+scientific base 2 notation: $1011 = 1.\textbf{011 } \times 2^3$.
+
+The sign bit is 0 since $11>0$ and therefore the number is positive.
+
+To determine the exponent we take the single precision bias (127) and we add to it
+the amount of times we moved a bit past the floating point: $127 + 3 = 130$.
+
+We then convert 130 to decimal by successive halving:\\
+\begin{tabular}[h]{l|c}
+\textbf{Fraction} & \textbf{Rest}\\
+\hline \\
+$130/2 = 65$ & 0 \\\\
+$65/2 = 32$ & 1 \\
+$32/2 = 16$ & 0 \\
+$16/2= 8$ & 0 \\
+$8/2 = 4$ & 0 \\
+$4/2 = 2$ & 0 \\
+$2/2 = 1$ & 0 \\
+$1/2 = 0$ & 1 \\
+\end{tabular}
+
+By reading from most significant bit to least significant bit we get 10000010 for our
+exponent.
+
+We then finally assemble the floating point binary number with the parts we have
+converted:
+
+$0\ 10000010\ 01100000000000000000000 = 01000001001100000000000000000000$
+
+And we convert to hexadecimal by splitting the number in 4 bit groups and converting
+each group to its equivalent hexadecimal digit:
+
+$0100\ 0001\ 0011\ 0000\ 0000\ 0000\ 0000\ 0000 = 4\ 1\ 3\ 0\ 0\ 0\ 0\ 0\ = 41300000$. \\\\
+
+\textbf{From now on the passages are going to include only the calculations to avoid verbosity. Nonetheless, the same reasoning will be applied for each conversion.}
+
+\item[b)] \textbf{\large 5/64};
+
+Sign bit = 0
+
+$\frac{5}{64} = 0.078125$
+
+Conversion to binary:\\
+\begin{itemize}
+\item $0.078125 \times 2 = \textbf{0} + 0.15625$;
+\item $0.15625 \times 2 = \textbf{0} + 0.3125$;
+\item $0.3125 \times 2 = \textbf{0} +  0.625$;
+\item $0.625 \times 2 = \textbf{1} + 0.25$;
+\item $0.25 \times 2 = \textbf{0} + 0.5$;
+\item $0.5 \times 2 = \textbf{1} + 0$;
+\end{itemize}
+
+Result of conversion: $0.000101$.
+
+Normalization to scientific notation: $1.\textbf{01} \times 2^{-4}$.
+
+Exponent bits in decimal = $127 - 4 = 123$
+
+Conversion of exponent:\\
+\begin{tabular}[h]{l|c}
+\textbf{Fraction} & \textbf{Rest}\\
+\hline \\
+$123/2 = 61$ & 1 \\\\
+$61/2 = 30$ & 1 \\
+$30/2 = 15$ & 0 \\
+$15/2= 7$ & 1 \\
+$7/2 = 3$ & 1 \\
+$3/2 = 1$ & 1 \\
+$1/2 = 0$ & 1 \\
+\end{tabular}
+
+Exponent bits in binary = $01111011$
+
+Final binary number = $0\ 01111011\ 01000000000000000000000 =$
+
+ $00111101101000000000000000000000$
+
+Conversion to hexadecimal:
+
+$0011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000 = 3$ D A $0\ 0\ 0\ 0\ 0 = 3$DA$00000$.
+
+\item[c)] \textbf{\large -5/64};
+
+With this being the negative counterpart of the previous number we just need to flip
+the sign bit of $\frac{5}{64}$.
+
+With that we get: $10111101101000000000000000000000$.
+
+Which, converted to hexadecimal, gives:
+
+$1011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000 =$ B D A $0\ 0\ 0\ 0\ 0 =$ BDA$00000$.
+
+\item[d)] \textbf{\large 6.125};
+
+Sign bit = 0
+
+Conversion to binary:
+ 
+ Integer part = $6_{10} = 110_{2}$
+ 
+ Fractional part = $0.125_{10}$:
+ \begin{itemize}
+\item $0.125 \times 2 = \textbf{0} + 0.25$;
+\item $0.25 \times 2 = \textbf{0} + 0.5$;
+\item $0.5 \times 2 = \textbf{1} + 0$;
+ \end{itemize}
+ = $001_2$.
+ 
+Normalized binary = $110.001 = 1.\textbf{10001} \times 2^2$
+ 
+Exponent bits in decimal = $127 + 2 = 129$.
+
+\begin{tabular}[h]{l|c}
+\textbf{Fraction} & \textbf{Rest}\\
+\hline \\
+$129/2 = 64$ & 1 \\\\
+$64/2 = 32$ & 0  \\
+$32/2 = 16$ & 0 \\
+$16/2= 8$ & 0 \\
+$8/2 = 4$ & 0 \\
+$4/2 = 2$ & 0 \\
+$2/2 = 1$ & 0 \\
+$1/2 = 0$ & 1 \\
+\end{tabular}
+
+Exponent bits in binary = $10000001$
+
+Final binary number = $0\ 10000001\ 10001000000000000000000 =$
+
+ $01000000110001000000000000000000$.
+ 
+Conversion to hexadecimal:
+$0100\ 0000\ 1100\ 0100\ 0000\ 0000\ 0000\ 0000 = 4\ 0$ C $4\ 0\ 0\ 0\ 0 = 40$C$40000$.
+ 
+\end{itemize}
+
+\subsection{Hexadecimal to Decimal}
+
+\textbf{All the numbers are going to be first converted to binary and then to
+decimal for an easier calculation}
+
+\begin{itemize}
+
+\item[a)] \textbf{\large 42E48000$_{16}$}
+
+Conversion to binary:
+
+$42$E$48000 = 0100\ 0010\ 1110\ 0100\ 1000\ 0000\ 0000\ 0000 =$
+
+$ 0\ 10000101\ 11001001000000000000000 = $
+
+$ 01000010111001001000000000000000$
+
+From this we can observe that the sign bit is 0 (i.e. positive).
+
+Conversion of exponent from binary to decimal:
+
+$10000101 = 2^7 + 2^2 + 2^0 = 128 + 4 + 1 = 133$
+
+To find how many bits have been moved beyond the floating point we now find the
+exponent for our notation:
+
+$ 133 - 127 = 6$
+
+And we then take the mantissa to multiply it by $2^6$ (where 6 is the number we found through the last passage) to get the final binary number to convert to decimal:
+
+$1.11001001000000000000000 \times 2^6 = 1.11001001 \times 2^6 = 1110010.01$
+
+Conversion to decimal:
+
+$1110010.01 = 2^6 + 2^5 + 2^4 + 2 + 2^{-2} = 64 + 32 + 16 + 2 + 0.25 = 114.25$
+
+\item[b)] \textbf{\large 3F880000$_{16}$}
+
+Conversion to binary:
+
+$3$F$880000 = 0011\ 1111\ 1000\ 1000\ 0000\ 0000\ 0000\ 0000\ =$
+
+$0\ 01111111\ 00010000000000000000000 =$
+
+$00111111100010000000000000000000$
+
+Sign bit = 0
+
+Conversion of exponent from binary to decimal:
+
+$01111111 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$
+
+Power of 2 = $127 - 127 = 0$
+
+Final binary number:
+
+$1.0001 \times 2^0 = 1.0001$
+
+Conversion to decimal:
+
+$1.0001 = 2^0 + 2^{-4} = 1 + 0.0625 = 1.0625$
+
+\item[c)] \textbf{\large 00800000$_{16}$}
+
+Conversion to binary:
+
+$00800000 = 0000\ 0000\ 1000\ 0000\ 0000\ 0000\ 0000\ 0000 =$
+
+$0\ 00000001\ 000\ 00000000000000000000 =$
+
+$00000000100000000000000000000000$
+
+Sign bit = 0
+
+Conversion of exponent from binary to decimal:
+
+$00000001 = 2^0 = 1$
+
+Power of 2 = $1 - 127 = -126$
+
+ Final binary number:
+ 
+ $1.00000000000000000000 \times 2^{-126}$
+ 
+ Conversion to decimal:
+ 
+$1.00000000000000000000 \times 2^{-126} = 2^{-126} = 1.1754944 \times 10^{-38}$
+
+\item[d)] \textbf{\large C7F00000$_{16}$}
+
+Conversion to binary:
+
+C$7$F$00000 = 1100\ 0111\ 1111\ 0000\ 0000\ 0000\ 0000\ 0000 =$
+
+$1\ 10001111\ 11100000000000000000000 =$
+
+$11000111111100000000000000000000$
+
+Sign bit = 1
+
+Conversion of exponent from binary to decimal:
+
+$10001111 = 2^7 + 2^3 + 2^2 + 2^1 + 2^0 = 128 + 8 + 4 + 2 + 1 = 143$
+
+Power of 2 = $143 - 127 = 16$
+
+Final binary number:
+
+$1.11100000000000000000000 \times 2^{16} = 1.111 \times 2^16 =$
+
+$11110000000000000.0$
+
+Conversion to decimal:
+
+$11110000000000000.0 = 2^16 + 2^15 + 2^14 + 2^13 = 65536 + 32768 + 16384 +
+8192 = 122880 $ and, since the sign bit is 1 (i.e. negative), $= -122880$.
+
+\end{itemize}
+
+\section{Bonus Question 2}
+
+
+
+\end{document}