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# 第一次期中考重點

## Chap 1

### Performance

:::success
<font class="red">效能是執行時間的倒數</font>

$$
\text{Performance}_x\ =\ \frac{1}{\text{Execution time}_x}
$$

:::info
**Example**:
Computer A runs a program in 10 sec, computer B runs the **same program** in 15 sec. **How much faster is A than B?**

**Ans.**

$$
\rm\frac{Performance_A}{Performance_B}\ =\ \frac{Execution\ time_B}{Execution\ time_A}\ =\ \frac{15}{10}\ =\ 1.5
$$

So, **A is 1.5 times faster than B**, or **B is 1.5 times slower than A**.
:::

:::success
- **Clock cycle** (tick, clock tick, clock period, clock, cycle)
- The time for **one clock period** (ps)
- **Clock rate**
- The **inverse** of the clock period (Hz)
:::

### CPU performance and its factors

:::warning
- **CPU execution time** for a program
= <font class="blue">**CPU clock cycles for a program $\times$ Clock cycle time**</font>
= <font class="blue">**CPU clock cycles for a program / Clock rate**</font>

:::info
**Example**: Improving performance
**Q.**
- A program runs in 10 seconds on computer A, which has a 2 GHz clock.
- We want to build a computer B that will run this program in 6 seconds.
- To increase the clock rate is possible, but will require 1.2 times as many clock cycles as computer A.
- **What clock rate should computer B have?**

**A.**
1.
$$
\begin{split}
\rm{CPU\ Time_A} &= \rm{CPU\ Clock\ Cycles_A} / {Clock\ Rate_A} \\
10 &= \rm{CPU\ Clock\ Cycles_A} / (2 \times 10^9) \\
\rm CPU\ Clock\ &\rm Cycles_A = 20 \times 10^9 Cycles
\end{split}
$$
2.
$$
\begin{split}
\rm{CPU\ Time_B} &= 1.2 \times \rm{CPU\ Clock \ Cycles_A} / {Clock\ Rate_B} \\
6 &= 1.2\times (20 \times 10^9) / \rm{Clock\ Rate_B} \\
&\rm{Clock\ Rate_B} = 4GHz
\end{split}
$$

- ***To run the program in 6 seconds, B must have twice the clock rate of A***.
:::

### Instruction performance

The execution time must depend on the number of instructions in a program:

:::success
- **CPU clock cycles required for a program**
= <font class="blue">**Instructions for a program $\times$ Average clock cycles per instruction**</font>
- **Clock cycles per instruction (CPI)**
= <font class="blue">**The <font style="background-color: lightblue">average</font> number of clock cycles each instruction takes to execute**</font>

:::info
**Example**: Using the performance equation
**Q.**
- **Computer A and B have the same ISA**
- Computer A: clock cycle time 250 ps, CPI 2.0 for **same** program
- Computer B: clock cycle time 500 ps, CPI 1.2 for the **same** program
- Which computer is faster for this program and by how much?

**A.**
- Assume the program has $I$ instruction
- CPU clock cycles A = $I$ $\times$ 2.0, CPU clock cycles B = $I$ $\times$ 1.2
- CPU timeA = $I$ $\times$ 2.0 $\times$ 250 = 500 $\times$ $I$ ps **faster !**
CPU timeB = $I$ $\times$ 1.2 $\times$ 500 = 600 $\times$ $I$ ps
- ***Computer A is 1.2 times as fast as computer B for this program***
:::

### The classic CPU performance equation

:::warning
- Instruction count
- The number of instructions executed by the program
- **CPU time**
= <font class="blue">**Instruction count $\times$ CPI $\times$ Clock cycle time**</font>
= <font class="blue">**Instruction count $\times$ CPI / Clock rate**</font>

:::info
**Example**: Comparing code segments
**Q.**
- A compiler designer is trying to decide between two code sequence for a particular computer
- Which code sequence executes the most instructions?
- Which will be faster?
- What is the CPI for each sequence?


**A.**
- The number of executed instructions
- Sequence 1: 2 + 1 + 2 = 5
- Sequence 2: 4 + 1 + 1 = 6 **more !**
- Execution time
- CPU clock cycles~1~ = (2 $\times$ 1) + (1 $\times$ 2) + (2 $\times$ 3) = 10 cycles
- CPU clock cycles~2~ = (4 $\times$ 1) + (1 $\times$ 2) + (1 $\times$ 3) = 9 cycles **faster !**
- CPI
- CPI~1~ = CPU clock cycles~1~ / Instruction count~1~ = 10/5 = 2
- CPI~2~ = CPU clock cycles~2~ / Instruction count~2~ = 9/6 = 1.5
:::

## Chap 2

- 定點數的進位轉換
- 浮點數加減乘除運算
- IEEE 754 的運算
- 定點數的乘除法 (表格)