288 lines
7.2 KiB
TeX
288 lines
7.2 KiB
TeX
\documentclass[12pt]{article}
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\title{Computer Architecture -- Assignment 11 Bonus}
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\author{Claudio Maggioni \and Tommaso Rodolfo Masera}
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\date{}
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\begin{document}
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\maketitle
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\section{Bonus Question 1}
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\subsection{Decimal to Hexadecimal}
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\textbf{All the numbers are going to be first converted to binary and then to
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hexadecimal for an easier calculation}
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\begin{itemize}
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\item[a)] \textbf{\large 11};
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First we find how 11 is expressed in binary which is 1011 and then we convert it to
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scientific base 2 notation: $1011 = 1.\textbf{011 } \times 2^3$.
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The sign bit is 0 since $11>0$ and therefore the number is positive.
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To determine the exponent we take the single precision bias (127) and we add to it
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the amount of times we moved a bit past the floating point: $127 + 3 = 130$.
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We then convert 130 to decimal by successive halving:\\
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\begin{tabular}[h]{l|c}
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\textbf{Fraction} & \textbf{Rest}\\
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\hline \\
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$130/2 = 65$ & 0 \\\\
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$65/2 = 32$ & 1 \\
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$32/2 = 16$ & 0 \\
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$16/2= 8$ & 0 \\
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$8/2 = 4$ & 0 \\
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$4/2 = 2$ & 0 \\
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$2/2 = 1$ & 0 \\
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$1/2 = 0$ & 1 \\
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\end{tabular}
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By reading from most significant bit to least significant bit we get 10000010 for our
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exponent.
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We then finally assemble the floating point binary number with the parts we have
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converted:
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$0\ 10000010\ 01100000000000000000000 = 01000001001100000000000000000000$
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And we convert to hexadecimal by splitting the number in 4 bit groups and converting
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each group to its equivalent hexadecimal digit:
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$0100\ 0001\ 0011\ 0000\ 0000\ 0000\ 0000\ 0000 = 4\ 1\ 3\ 0\ 0\ 0\ 0\ 0\ = 41300000$. \\\\
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\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.}
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\item[b)] \textbf{\large 5/64};
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Sign bit = 0
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$\frac{5}{64} = 0.078125$
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Conversion to binary:\\
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\begin{itemize}
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\item $0.078125 \times 2 = \textbf{0} + 0.15625$;
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\item $0.15625 \times 2 = \textbf{0} + 0.3125$;
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\item $0.3125 \times 2 = \textbf{0} + 0.625$;
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\item $0.625 \times 2 = \textbf{1} + 0.25$;
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\item $0.25 \times 2 = \textbf{0} + 0.5$;
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\item $0.5 \times 2 = \textbf{1} + 0$;
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\end{itemize}
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Result of conversion: $0.000101$.
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Normalization to scientific notation: $1.\textbf{01} \times 2^{-4}$.
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Exponent bits in decimal = $127 - 4 = 123$
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Conversion of exponent:\\
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\begin{tabular}[h]{l|c}
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\textbf{Fraction} & \textbf{Rest}\\
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\hline \\
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$123/2 = 61$ & 1 \\\\
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$61/2 = 30$ & 1 \\
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$30/2 = 15$ & 0 \\
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$15/2= 7$ & 1 \\
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$7/2 = 3$ & 1 \\
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$3/2 = 1$ & 1 \\
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$1/2 = 0$ & 1 \\
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\end{tabular}
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Exponent bits in binary = $01111011$
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Final binary number = $0\ 01111011\ 01000000000000000000000 =$
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$00111101101000000000000000000000$
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Conversion to hexadecimal:
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$0011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000 = 3$ D A $0\ 0\ 0\ 0\ 0 = 3$DA$00000$.
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\item[c)] \textbf{\large -5/64};
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With this being the negative counterpart of the previous number we just need to flip
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the sign bit of $\frac{5}{64}$.
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With that we get: $10111101101000000000000000000000$.
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Which, converted to hexadecimal, gives:
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$1011\ 1101\ 1010\ 0000\ 0000\ 0000\ 0000\ 0000 =$ B D A $0\ 0\ 0\ 0\ 0 =$ BDA$00000$.
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\item[d)] \textbf{\large 6.125};
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Sign bit = 0
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Conversion to binary:
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Integer part = $6_{10} = 110_{2}$
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Fractional part = $0.125_{10}$:
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\begin{itemize}
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\item $0.125 \times 2 = \textbf{0} + 0.25$;
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\item $0.25 \times 2 = \textbf{0} + 0.5$;
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\item $0.5 \times 2 = \textbf{1} + 0$;
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\end{itemize}
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= $001_2$.
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Normalized binary = $110.001 = 1.\textbf{10001} \times 2^2$
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Exponent bits in decimal = $127 + 2 = 129$.
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\begin{tabular}[h]{l|c}
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\textbf{Fraction} & \textbf{Rest}\\
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\hline \\
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$129/2 = 64$ & 1 \\\\
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$64/2 = 32$ & 0 \\
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$32/2 = 16$ & 0 \\
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$16/2= 8$ & 0 \\
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$8/2 = 4$ & 0 \\
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$4/2 = 2$ & 0 \\
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$2/2 = 1$ & 0 \\
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$1/2 = 0$ & 1 \\
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\end{tabular}
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Exponent bits in binary = $10000001$
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Final binary number = $0\ 10000001\ 10001000000000000000000 =$
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$01000000110001000000000000000000$.
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Conversion to hexadecimal:
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$0100\ 0000\ 1100\ 0100\ 0000\ 0000\ 0000\ 0000 = 4\ 0$ C $4\ 0\ 0\ 0\ 0 = 40$C$40000$.
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\end{itemize}
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\subsection{Hexadecimal to Decimal}
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\textbf{All the numbers are going to be first converted to binary and then to
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decimal for an easier calculation}
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\begin{itemize}
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\item[a)] \textbf{\large 42E48000$_{16}$}
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Conversion to binary:
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$42$E$48000 = 0100\ 0010\ 1110\ 0100\ 1000\ 0000\ 0000\ 0000 =$
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$ 0\ 10000101\ 11001001000000000000000 = $
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$ 01000010111001001000000000000000$
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From this we can observe that the sign bit is 0 (i.e. positive).
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Conversion of exponent from binary to decimal:
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$10000101 = 2^7 + 2^2 + 2^0 = 128 + 4 + 1 = 133$
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To find how many bits have been moved beyond the floating point we now find the
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exponent for our notation:
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$ 133 - 127 = 6$
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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:
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$1.11001001000000000000000 \times 2^6 = 1.11001001 \times 2^6 = 1110010.01$
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Conversion to decimal:
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$1110010.01 = 2^6 + 2^5 + 2^4 + 2 + 2^{-2} = 64 + 32 + 16 + 2 + 0.25 = 114.25$
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\item[b)] \textbf{\large 3F880000$_{16}$}
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Conversion to binary:
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$3$F$880000 = 0011\ 1111\ 1000\ 1000\ 0000\ 0000\ 0000\ 0000\ =$
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$0\ 01111111\ 00010000000000000000000 =$
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$00111111100010000000000000000000$
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Sign bit = 0
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Conversion of exponent from binary to decimal:
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$01111111 = 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2 + 1 = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$
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Power of 2 = $127 - 127 = 0$
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Final binary number:
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$1.0001 \times 2^0 = 1.0001$
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Conversion to decimal:
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$1.0001 = 2^0 + 2^{-4} = 1 + 0.0625 = 1.0625$
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\item[c)] \textbf{\large 00800000$_{16}$}
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Conversion to binary:
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$00800000 = 0000\ 0000\ 1000\ 0000\ 0000\ 0000\ 0000\ 0000 =$
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$0\ 00000001\ 000\ 00000000000000000000 =$
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$00000000100000000000000000000000$
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Sign bit = 0
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Conversion of exponent from binary to decimal:
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$00000001 = 2^0 = 1$
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Power of 2 = $1 - 127 = -126$
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Final binary number:
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$1.00000000000000000000 \times 2^{-126}$
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Conversion to decimal:
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$1.00000000000000000000 \times 2^{-126} = 2^{-126} = 1.1754944 \times 10^{-38}$
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\item[d)] \textbf{\large C7F00000$_{16}$}
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Conversion to binary:
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C$7$F$00000 = 1100\ 0111\ 1111\ 0000\ 0000\ 0000\ 0000\ 0000 =$
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$1\ 10001111\ 11100000000000000000000 =$
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$11000111111100000000000000000000$
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Sign bit = 1
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Conversion of exponent from binary to decimal:
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$10001111 = 2^7 + 2^3 + 2^2 + 2^1 + 2^0 = 128 + 8 + 4 + 2 + 1 = 143$
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Power of 2 = $143 - 127 = 16$
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Final binary number:
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$1.11100000000000000000000 \times 2^{16} = 1.111 \times 2^16 =$
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$11110000000000000.0$
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Conversion to decimal:
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$11110000000000000.0 = 2^16 + 2^15 + 2^14 + 2^13 = 65536 + 32768 + 16384 +
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8192 = 122880 $ and, since the sign bit is 1 (i.e. negative), $= -122880$.
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\end{itemize}
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\section{Bonus Question 2}
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Loss of significance can occur via subtraction of values that are close to each other so
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that most of the significant digits get cancelled.
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For example let x $= 1.5 \times 2^0$
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and y $=\ 1.0 \times 2^{26}$ such that the sequence of operations $r =(x+y) - y $ results in r = 0.0 which is incorrect as the actual value should be 1.5.
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This is due to catastrophic cancellation which occurred during the subtraction.
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This kind of occurrences can't be identified unless data is further examined leaving no
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way of checking that 0.0 is the wrong result.
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\end{document} |