diff options
| -rwxr-xr-x | render.py | 1 | ||||
| -rw-r--r-- | sem6/dig/m5/ex2.vhdl | 2 | ||||
| -rw-r--r-- | sem6/dig/mpc2/opgaver.tex | 80 | ||||
| -rw-r--r-- | sem6/prob/eksamnen/notes.tex | 47 | ||||
| -rw-r--r-- | sem6/prob/m2/noter.tex | 48 | ||||
| -rw-r--r-- | sem6/prob/m2/opgaver.md | 30 | ||||
| -rw-r--r-- | sem6/prob/m3/noter.md | 40 |
7 files changed, 134 insertions, 114 deletions
@@ -14,6 +14,7 @@ tex_template = """\\documentclass[12pt]{article} \\usepackage{float} \\usepackage{amsthm} \\usepackage{booktabs} +\\usepackage{siunitx} \\usepackage{tikz} \\usetikzlibrary{automata, positioning, arrows} diff --git a/sem6/dig/m5/ex2.vhdl b/sem6/dig/m5/ex2.vhdl index bed016a..86f22a9 100644 --- a/sem6/dig/m5/ex2.vhdl +++ b/sem6/dig/m5/ex2.vhdl @@ -17,7 +17,7 @@ architecture impl of ex2 is begin output_int <= std_logic_vector(value); output <= output_int; - leds <= output_int(7 downto 0); + leds <= output_int(23 downto 16); process (clk) begin diff --git a/sem6/dig/mpc2/opgaver.tex b/sem6/dig/mpc2/opgaver.tex index 6d0826e..74aba7d 100644 --- a/sem6/dig/mpc2/opgaver.tex +++ b/sem6/dig/mpc2/opgaver.tex @@ -1,8 +1,28 @@ \title{Opgaver til Microprocessors 2} \date{2021-03-24} +Har fundet ud af at jeg har lavet de forkerte opgaver \texttt{:-(}. + \section{Problem 4.1} +\begin{opg} + What are the four steps CPUs use to execute instructions +\end{opg} + +\paragraph{Fetch} comes first, where the instruction is fetched from memory. +This is taken from where the instruction pointer is pointing. + +\paragraph{Decode} instruction, where it most likely requires multiple microcode instructions. + +Whether to \textbf{Access memory} is determined in the decoding step. +If this is required, this memory must be fetched from memory. + +\paragraph{Execute} the instruction using the fetched memory and register values. + +\paragraph{Repeat} from the beginning with a new fetch. + +\section{Problem 4.2} + \emph{In Fig. 4-6, the B bus register is encoded in a 4-bit field, but the C bus is represented as a bit map. Why?} @@ -11,13 +31,13 @@ Therefore one cannot take the shortcut with a 4-bit field, as that would only al One cannot present 1 and 2 at the same time in 4-bit field, as that would activate register 3. -\section{Problem 4.5} +\section{Problem 4.4} {\itshape Suppose that in the example of Fig. 4-14(a) the statement - \begin{verbatim} - k = 5; - \end{verbatim} +\begin{verbatim} + k = 5; +\end{verbatim} is added after the if statement. What would the new assembly code be? Assume that the compiler is an optimizing compiler. } @@ -38,13 +58,46 @@ Well k is set either way, so one can invert the if. ISTORE k \end{verbatim} +\section{Problem 4.4 Moodle} + +\begin{opg} + Give two different IJVM translations for the following IJVM code: +\begin{verbatim} + i = j + m + 8; +\end{verbatim} +\end{opg} + +Dette kan man gøre ved at load $j$ og $m$ fra stacken og add dem. +Derefter kan push 8 og add den. +Herefter gemmer man i $i$. + +\begin{verbatim} + ILOAD j + ILOAD m + IADD + BIPUSH 8 + IADD + ISTORE i +\end{verbatim} + +En anden måde er at push alle ting først også add flere gange efter hinnanden. + +\begin{verbatim} + ILOAD j + ILOAD m + BIPUSH 8 + IADD + IADD + ISTORE i +\end{verbatim} + \section{Problem 4.9} {\itshape How long does a 2.5-GHz Mic-1 take to execute the Java statement - \begin{verbatim} - i = j + k - \end{verbatim} +\begin{verbatim} + i = j + k; +\end{verbatim} Give your answer in nanoseconds } @@ -59,13 +112,12 @@ First we "compile" the java statement :-). Then we can add how many microinstructions each takes (\textbf{bold} number) multiplied with how many times it is used. -\begin{equation} - \underbrace{\mathbf 1 \cdot 4}_{MAIN} + \underbrace{\mathbf 5 \cdot 2}_{ILOAD} + \underbrace{\mathbf 3}_{IADD} + \underbrace{\mathbf 6}_{ISTORE} = 23 -\end{equation} +\[ +\underbrace{\mathbf 1 \cdot 4}_{\mathrm{MAIN}} + \underbrace{\mathbf 5 \cdot 2}_{\mathrm{ILOAD}} + \underbrace{\mathbf 3}_{\mathrm{IADD}} + \underbrace{\mathbf 6}_{\mathrm{ISTORE}} = 23 +\] Then we can multiply with the nanoseconds a single instruction takes -\begin{equation} - \frac 1 {2.5 \cdot 10^9} \cdot 23 = 9.2 \cdot 10^{-9}\,, -\end{equation} +\[ + \frac 1 {\SI{2.5e9}{Hz}} \cdot 23 = \SI{9.2e-9}{s}\,, +\] which is 9.2 Nanoseconds. - diff --git a/sem6/prob/eksamnen/notes.tex b/sem6/prob/eksamnen/notes.tex deleted file mode 100644 index 4dfee30..0000000 --- a/sem6/prob/eksamnen/notes.tex +++ /dev/null @@ -1,47 +0,0 @@ -\title{Eksamnens Noter} - - -The universal set or sample space is the set everything, and is denoted $S$. -Therefore the probability of hitting $S$ is $P(S) = 1$. - -This is the first of 3 axioms repeated below. - -\begin{enumerate} - \item For any event $A$, $P(A) \geq 0$. - \item The probability of hitting sample space is always 1, $P(S) = 1$. - \item If events $A_1, A_2, ...$ are \textbf{disjoint} event, then - \begin{equation} - P(A_1 \cup A_2 ...) = P(A_1) + P(A_2)\,. - \end{equation} -\end{enumerate} - -The last axiom requires that the events $A_n$ are disjoint. -If they aren't one should subtract the part they have in common. -This is called the \emph{Inclusion-Exclusion Principle}. - -\begin{principle} - The \emph{Inclusion-Exclusion Principle} is defined as - \begin{equation} - P(A \cup B) = P(A) + P(B) - P(A \cap B)\,. - \end{equation} - Definition with 3 events can be found in the in the book. -\end{principle} - -\section{Counting} - -The probability of a event $A$ can be found by -\begin{equation} - P(A) = \frac {|A|} {|S|}\,. -\end{equation} -It is therefore required to count how many elements are in $S$ and $A$. -The most simple method is the \emph{multiplication principle}. - -\begin{principle}[Multiplication principle] - Let there be $r$ random experiments, where the $k$'th experiment has $n_k$ outcomes. - Then there are - \begin{equation} - n_1 \cdot n_2 \cdot ... \cdot n_r - \end{equation} - possible outcomes over all $r$ experiments. -\end{principle} - diff --git a/sem6/prob/m2/noter.tex b/sem6/prob/m2/noter.tex index 3eb2e4f..c35f52b 100644 --- a/sem6/prob/m2/noter.tex +++ b/sem6/prob/m2/noter.tex @@ -1,6 +1,5 @@ \title{Noter til probability m2} -\section{Random Variables} Her mapper man fra et sample space S til en variabel. Her kalder man variablen et stort tal R eller sådan noget. @@ -15,7 +14,7 @@ P(X = x) = 0 $$ -\subsection{Cumulative Distribution Function} +\section{Cumulative Distribution Function} Her måler man prob for at ens random er mindre end et bestemt tal. @@ -33,7 +32,7 @@ Ved discrete random variables vil denne være en slags trappe. Kan sige at den er \emph{continues from the right} eftersom man har $\leq$ i definition. -\subsection{Probability Mass Function} +\section{Probability Mass Function} Works only for discrete random variables. Is defines as the probability that $X = a$: @@ -49,15 +48,17 @@ F(a) = \sum_{all x \leq a} p(a) $$ -\subsection{Probability Density Function} +\section{Probability Density Function} Her finder man P i et evigt lille interval: -Is the derivative of the CDF. +In the following formula PDF is $f$. +\begin{equation*} + \begin{split} + F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\ + f(a) = \frac{d}{da} F(a) + \end{split} +\end{equation*} -$$ -F(a) = P(X \in (-\infty,a]) = \int_{-\infty}^a f(x) dx \\ -f(a) = \frac{d}{da} F(a) -$$ The following must be true: @@ -65,7 +66,7 @@ $$ \int_{-\infty}^{\infty} f(x) dx = 1 $$ -\subsection{Multiple Random Variables} +\section{Multiple Random Variables} Have multiple random variables, which can be or is not correlated. Can define the joined CDF: @@ -81,15 +82,28 @@ F_X(x) = P(X \leq x) = P(X \leq, Y < \infty) = F(x, \infty) $$ One can not go from marginal to the joined, as they do not contain enough information. -This is only possible if X and Y are \emph{independent}. -$$ -F_{XY}(x,y) = F_X(x) \cdot F_Y(x) \\ -p(x,y) = p_X(x) \cdot p_Y(y) \\ -f(x,y) = f_X(x) \cdot f_Y(y) -$$ +However if two random variables, and $A$ and $B$ are two sets of real numbers: +\[ + P(X \in A, Y \in B) = P(X \in A) P(Y \in B)\,. +\] +% This is only possible if X and Y are \emph{independent}. +% \begin{align*} +% F_{XY}(x,y) &= F_X(x) \cdot F_Y(x) \\ +% p(x,y) &= p_X(x) \cdot p_Y(y) \\ +% f(x,y) &= f_X(x) \cdot f_Y(y) +% \end{align*} + +\section{Conditional PDF} + +If $X$ and $Y$ have a joint PDF, then the conditional PDF of X given that $Y=y$ is +\[ + F_{X|Y}(x|y) = \frac {f(x, y)} {f_Y(y)} +\] + +There is also one for PMF not listed here. -\subsection{Joined PMF} +\section{Joined PMF} $$ P_{XY}(x,y) = P(X = x, Y = y) diff --git a/sem6/prob/m2/opgaver.md b/sem6/prob/m2/opgaver.md index 0ce9c77..601aa86 100644 --- a/sem6/prob/m2/opgaver.md +++ b/sem6/prob/m2/opgaver.md @@ -5,16 +5,16 @@ Man kan sige at chances for at en kvinde kommer først er 50%. $$ -P(1) = \frac 1 2 +P(1) = \frac 1 2 = 0.5 $$ -Herefter kræver det at en mand for først og en kvinde får næste. +Herefter kræver det at en mand får først og en kvinde får næste. -$$ -P(2) = \frac 5 10 \cdot \frac 5 9 \\ -P(3) = \frac 5 10 \cdot \frac 4 9 \frac 5 8 \\ -P(4) = \frac 5 10 \cdot \frac 4 9 \frac 3 8 \cdot \frac 5 8 -$$ +\begin{align*} + P(2) &= \frac 5 {10} \cdot \frac 5 9 = 0.2778\\ + P(3) &= \frac 5 {10} \cdot \frac 4 9 \cdot \frac 5 8 = 0.1389 \\ + P(4) &= \frac 5 {10} \cdot \frac 4 9 \cdot \frac 3 8 \cdot \frac 5 7 = 0.0595 +\end{align*} osv. @@ -52,22 +52,22 @@ $$ Først skal man finde $\lambda$. -$$ - \int_{0}^{\infty} \lambda e^{- \frac x {100}} \mathrm{dx} = 1 \\ - \left[ - \lambda 100 \cdot e^{- \frac x {100}}\right]_{0}^{\infty} = 1 \\ - \lambda \cdot 100 = 1 \\ - \lambda = \frac 1 {100} -$$ +\begin{align*} + \int_{0}^{\infty} \lambda e^{- \frac x {100}} \mathrm{dx} &= 1 \\ + \left[ - \lambda 100 \cdot e^{- \frac x {100}}\right]_{0}^{\infty} &= 1 \\ + \lambda \cdot 100 &= 1 \\ + \lambda &= \frac 1 {100} +\end{align*} Nu kan man sætte 50 til 150 ind. $$ - P(50 < x \leq 150) = \int_{50}^{150} f(x) \mathrm{dx} = - e^{- \frac {150} {100}} + e^{ - \frac {50} {100}} = 0.3834 + P(50 < x \leq 150) = \int_{50}^{150} f(x) \mathrm{dx} = \left[ - 100 \cdot \frac 1 {100} \cdot e^{- \frac x {100}} \right]_{50}^{150} = -e^{- \frac{150} {100}} + e^{ - \frac {50} {100}} \approx 0.3834 $$ Derefter kan vi tage fra 0 til 100. $$ - P(x < 100) = \int_{0}^{100} f(x) \mathrm{dx} = - e^{- \frac {100} {100}} = - \frac 1 e + P(x < 100) = \int_{0}^{100} f(x) \mathrm{dx} = - e^{- \frac {100} {100}} + 1 = 1 - \frac 1 e \approx 0.6321 $$ diff --git a/sem6/prob/m3/noter.md b/sem6/prob/m3/noter.md index 23990ef..c784a6a 100644 --- a/sem6/prob/m3/noter.md +++ b/sem6/prob/m3/noter.md @@ -19,7 +19,7 @@ $$ E[X] = \int_{\infty}^{\infty} x f(x) \mathrm{dx} $$ -Can also calculate expectation distribution function: +Can also calculate expectation distribution function, however this can only be used if all values are non-negative: $$ E[X] = \sum_{k=0}^{\infty} P(X > k) \\ @@ -55,7 +55,7 @@ $$ E[Z] = \sum_{i} \sum_{j} g(x_i, y_j) \cdot p(x_i, y_j) $$ -If discrete just use integrals instead. +If continues just use integrals instead. $$ E[Z] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x, y) \cdot f(x, y) \mathrm{dxdy} @@ -69,27 +69,27 @@ $$ If $X$ and $Y$ are **independent** the following is true: -$$ - E[g_1(X) \cdot g_2(Y)] = E[g_1(X)] \cdot E[g_2(Y)] \\ - E[X \cdot Y] = E[X] \cdot E[Y] -$$ +\begin{align*} + E[g_1(X) \cdot g_2(Y)] &= E[g_1(X)] \cdot E[g_2(Y)] \\ + E[X \cdot Y] &= E[X] \cdot E[Y] +\end{align*} ## Variance Describes the mean of the distance between outcomes and the overall mean. Good way to describe the spread of the random variable. -$$ -Var(X) = E[(X - E[X])^2] \\ -Var(X) = E[X^2] - E[X]^2 -$$ +\begin{align*} + Var(X) &= E[(X - E[X])^2] \\ + Var(X) &= E[X^2] - E[X]^2 +\end{align*} If there is no power of two, it will be mean minus mean, which wont work. One can define the *standard deviation* to bring back the unit from squared. $$ - Std(X) = \sqrt{ (Var(X)) } + Std(X) = \sqrt{ Var(X) } $$ A rule for variance: @@ -110,21 +110,21 @@ If X and Y are independent the Cov part disappears. ## Covariance -$$ - Cov(X,Y) = E[(X - E[X]) \cdot (Y - E[Y])] \\ - Cov(X,Y) = E[XY] - E[X] \cdot E[Y] -$$ +\begin{align*} + Cov(X,Y) &= E[(X - E[X]) \cdot (Y - E[Y])] \\ + Cov(X,Y) &= E[XY] - E[X] \cdot E[Y] +\end{align*} Shows whether two variables vary together, can be both positive and negative. If it is possible $X$ and $Y$ are varying from the average together. Some rules below: -$$ - Cov(X, X) = Var(X) \\ - Cov(a X, Y) = a Cov(X, Y) \\ - Cov(X + Y, Z) = Voc(X, Z) + Cov(Y, Z) -$$ +\begin{align*} + Cov(X, X) &= Var(X) \\ + Cov(a X, Y) &= a Cov(X, Y) \\ + Cov(X + Y, Z) &= Cov(X, Z) + Cov(Y, Z) +\end{align*} If X and Y are independent, then covariance is zero (X and Y are *uncorrelated*). X and Y can be uncorrelated and not be independent. |