\title{Opgaver til Microprocessors 2}
\date{2021-03-24}

\section{Problem 4.1}

\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?}

It is often wanted to save the result from the ALU in multiple destination registers.
Therefore one cannot take the shortcut with a 4-bit field, as that would only allow one save at the time.

One cannot present 1 and 2 at the same time in 4-bit field, as that would activate register 3.

\section{Problem 4.5}

{\itshape
    Suppose that in the example of Fig. 4-14(a) the statement
    \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.
}

Well k is set either way, so one can invert the if.

\begin{verbatim}
    ILOAD j
    ILOAD k
    IADD
    BIPUSH 3
    IF_ICMPEQ L1
    ILOAD j
    BIPUSH 1
    ISUB
    ISTORE j
    L1: BIPUSH 5
    ISTORE k
\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}
    Give your answer in nanoseconds
}

First we "compile" the java statement :-).

\begin{verbatim}
    ILOAD j
    ILOAD k
    IADD
    ISTORE i
\end{verbatim}

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}

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}
which is 9.2 Nanoseconds.