Chapter 1

\[\begin{align}S = \frac{1}{(1-\alpha)+\alpha/k} && \text{(Amdahl’s Law)} \end{align}\]

Practice Problem 1.1

1.1.png

A. It takes 25 hours to travel from Boise to Minneapolis and travelling through Montana takes 20 hours. Thus \(\alpha = \frac{\text{20hr}}{\text{25hr}} = 0.6\). The average speed improved from \(\text{100km/hr}\) to \(\text{150km/hr}\). Thus, \(k=\frac{\text{150km/hr}}{\text{100km/hr}}=1.5\). By Amdahl’s law, \(S=\frac{1}{(1-0.6 + 0.6/1.5)}=1.25\).

B. If we rearrange Amdhal’s law for \(k\), we get \[k=\frac{\alpha S}{1+S(\alpha-1)}\]

Let \(v\) represent the speed at which the vehicle travels through Montana. Then, \[\alpha = \frac{\text{1500km}/v + (\text{1000km}/ \text{100km/hr})}{\text{25hr}} \]. Given \(S = 1.67\), we have \[\frac{v}{100} = \frac{1.67(\frac{\text{10hr}\cdot v + \text{1500km}}{\text{25hr}\cdot v})} {1 + 1.67(\frac{\text{(10hr)}\cdot v + 1500km}{\text{25hr}\cdot v}-1)}\]. Thus, \(v=\text{305.56km/hr}\)

Practice Problem 1.2

Given \(\alpha=0.9\) and \(S=4\), we have to improve 90% of this machine by a factor of \(k=\frac{(0.9)(0.4)}{1+4\cdot (0.9-1)} = 6\)

Chapter 2

Practice Problem 2.1

A. \( (\text{0010 0101 1011 1001 1101 0010})_{2} \)

B. \( (\text{AE49})_{16}\)

C. \( (\text{1010 1000 1011 0011 1101})_2 \)

D. \( (\text{322D96})_{16} \)

Practice Problem 2.2

\(n\) \(2^n\)(decimal) \(2^n\)(hexadecimal)
5 32 0x20
23 8388608 0x800000
15 32768 0x8000
13 8192 0x2000
12 4096 0x1000
6 64 0x40
8 256 0x100

Practice Problem 2.3

Decimal Binary Hexadecimal
0 0000 0000 0x00
158 1001 1110 0x9E
76 0100 1100 0x4C
145 1001 0001 0x91
174 1010 1110 0xAE
60 0011 1100 0x3C
241 1111 0001 0xF1

Practice Problem 2.4

A. 0x6061

B. 0x603C

C. 0x607c

D. 0x9c

Practice Problem 2.5

A. Little endian: 0x78 Big endian: 0x12 B. Little endian: 0x7856 Big endian: 0x1234 C. Little endian: 0x785634 Big endian: 0x123456

Practice Problem 2.6

A. \(\begin{align} \text{0x0027C8F8 = 0000 0000 0010 0111 1100 1000 1111 1000} \\ \text{0x4A1F23E0 = 0100 1010 0001 1111 0010 0011 1110 0000}\end{align}\)

B. \(\begin{align}\text{0x0027c8f8 = 0000 0000 0010} \textbf{ 0111 1100 1000 1111 1000} \\ \text{0x4A1F23E0} \gg \text{2 = 0001 0010 1000} \textbf{ 0111 1100 1000 1111 1000} && \text{(20 bits)}\end{align}\)
C. See 2.6 B

Practice Problem 2.7

\(\text{0x7271706F6E6D}\)

Practice Problem 2.8

Operation Result
\(a\) \([01001110]\)
\(b\) \([11100001]\)
\(\sim a\) \([10010110]\)
\(\sim b\) \([10101010]\)
\(a\& b\) \([01000001]\)
\(a | b \) \([01111101]\)
\(a^\widehat{} b\) \([00111100]\)

Practice Problem 2.9

A.

Colour Complement
Black \(111\)
Blue \(110\)
Green \(101\)
Cyan \(100\)
Red \(011\)
Magenta \(010\)
Yellow \(001\)
White \(000\)

B.

\(\begin{align} \text{Blue} | \text{Green} &= \text{Cyan} \\ \text{Yellow} \& \text{Cyan} &= \text{Blue} \\ \text{Red}^\widehat{} \text{Magenta} &= \text{Blue} \end{align} \)

Practice Problem 2.10

Step *x * y
Initially a b
Step 1 a a^b
Step 2 a^a^b = b a^b
Step 3 b b^a^b = a

Practice Problem 2.11

A. \(\lceil{\frac{2k+1}{2}}\rceil\)

B. a[first] and a[last] point to the same variable in the last iteration when cnt is odd. The function inplace_swap xor these two variables however since they point to the same number, the operation evaluates to zero.

C.

    for (first = 0, last=cnt-1;
        first < last;
        first++, last--)

Practice Problem 2.12

A. x &= 0xFF

B. x ^= ~0xFF

C. x |= 0xFF

Practice Problem 2.13

/* Declarations of functions implementing operations bis and bic */
int bis(int x, int m);
int bic(int x, int m);
/* Compute x|y using only calls to functions bis and bic */
int bool_or(int x, int y) {
    int result = bis(x, y);
    return result;
}
/* Compute x^y using only calls to functions bis and bic */
int bool_xor(int x, int y) {
    int result = /* left to reader to solve (I couldn't do it myself) */;
    return result;
}

Practice Problem 2.14

\(\begin{align} a &= 0101\ 0101 = 0x55\\ b &= 0100\ 0110 = 0x46\end{align}\)

Expression Value Expression Value      
a & b 68 a&&b 0x01      
a b 87 a   b 0x01
~a ~b -71 !a   !b 0x00
a & !b -71 a&&~b 0x01      

Skipped to Problem 2.45 due to lack of interest and knowing most of the content.

Practice Problem 2.45

Fractional value Binary representation Decimal representation
\(\frac{1}{8}\) 0.001 0.125
\(\frac{3}{4}\) 0.11 0.75
\(\frac{5}{16}\) 0.0011 0.3125
… … …

You get the point

Practice Problem 2.46

A.

\(\begin{align}0.1 &= 0.0001\ 1001\ 1001\ 1001\ 1001\ 1001\ [1001]\\x &= 0.0001\ 1001\ 1001\ 1001\ 1001\ 0100\\0.1-x &= 0.0000\ 0000\ 0000\ 0000\ 0000\ 0101\ [1001]\end{align}\)

B. \(\begin{align} 0.1-x &= 2^{-22} + 2^{-24} + [(2^{-25} + 2^{-28}) + (2^{-29} + 2^{-32}) + …] \\ &= 2^{-22} + 2^{-24} + \sum_{i=0}^{\infty} (2^{-25-4i} + 2^{-28-4i})\\ &= 2^{-22} + 2^{-24} + \frac{1}{5} \cdot 2^{-32} & \text{Simplify the sum and apply geometric series formula}\\ &= 2^{-22}(1+2^{-2}+\frac{2^{-10}}{5})\\ &= 2^{-22}(\frac{6401}{5120}) \\ &= 2.9806979\cdot 10^{-7}\end{align}\)

C.

\(\text{100hr} = 3600000 \cdot \text{0.1s} = 3600000\cdot (0.000110011[0011]\cdots)_2\\ \) \(3600000\cdot 0.1\text{s} - 3600000\cdot (x)\text{s} = 3600000(0.1\text{s}-(x)s)=3600000\cdot 2.9806979\cdot 10^{-7}\text{s} = 1.073051244\text{s}\)

D.