📚 A-Level AQA Computer Science Formula Handbook | A-Level AQA 计算机公式汇总手册
This handbook compiles the essential formulas you need to master for the AQA A-Level Computer Science specification. Using these formulas correctly will help you solve calculation questions on data representation, file sizes, network transmission, CPU performance and more. Bookmark this page for quick revision.
本手册汇总是您在 AQA A-Level 计算机科学考试中需要掌握的关键公式。正确运用这些公式将帮助您解决有关数据表示、文件大小、网络传输和 CPU 性能等方面的计算题。请收藏本页以便快速复习。
1. Image File Size | 图像文件大小
The size of a bitmap image file (ignoring metadata) is determined by its pixel dimensions, colour depth, and unit conversions. Larger resolution or higher bit depth increases file size proportionally.
位图图像文件的大小（忽略元数据）由其像素尺寸、颜色深度和单位换算决定。分辨率越高或颜色深度越大，文件大小会成比例增加。
Image Size (bits) = Width (px) × Height (px) × Bit Depth (bits per pixel)
图像大小（位）= 宽度（像素）× 高度（像素）× 颜色深度（位/像素）
To convert to bytes, divide the total bits by 8. For kilobytes or megabytes, divide by 1024 (using binary IEC prefixes) or by 1000 depending on context. AQA often expects the binary interpretation: 1 KiB = 2¹⁰ bytes.
要转换为字节，需将总位数除以 8。转换为千字节或兆字节时，根据上下文除以 1024（二进制 IEC 前缀）或 1000。AQA 考试通常期望使用二进制换算法，即 1 KiB = 2¹⁰ 字节。
2. Sound File Size | 声音文件大小
An uncompressed audio file’s size is calculated from its sample rate, sample resolution, number of channels, and duration. This formula ignores any file header or metadata.
未压缩音频文件的大小由采样率、采样分辨率、声道数和时长计算得出。此公式忽略文件头或元数据。
Sound Size (bits) = Sample Rate (Hz) × Sample Resolution (bits) × Number of Channels × Duration (s)
声音大小（位）= 采样率（Hz）× 采样分辨率（位）× 声道数 × 时长（秒）
To obtain the size in bytes, divide by 8. For example, CD-quality stereo audio with a 44.1 kHz sample rate, 16-bit resolution, and 2 channels produces 44 100 × 16 × 2 = 1 411 200 bits per second (about 176.4 KB/s).
要得到以字节为单位的大小，需除以 8。例如，CD 品质的立体声音频采用 44.1 kHz 采样率、16 位分辨率和 2 声道，每秒产生 44 100 × 16 × 2 = 1 411 200 位（约 176.4 KB/秒）。
3. Text File Size | 文本文件大小
The size of a plain text file depends on the number of characters and the encoding scheme used. Common encodings include ASCII (7 or 8 bits per character) and Unicode (UTF-16, typically 16 bits).
纯文本文件的大小取决于字符数和所使用的编码方案。常见编码包括 ASCII（每字符 7 或 8 位）和 Unicode（UTF-16，通常为 16 位）。
Text Size (bits) = Number of Characters × Bits per Character
文本大小（位）= 字符数 × 每字符位数
Divide by 8 to get the size in bytes. Always check which encoding is specified in the question – a space is also a character.
要想转换为字节，除以 8 即可。务必检查题目中指定的编码方式 – 空格也算一个字符。
4. Data Transfer Time | 数据传输时间
The time required to send a file over a network is found by dividing the total data size by the transfer rate. Ensure both values use the same unit of bits or bytes before calculating.
通过网络发送文件所需的时间等于数据总大小除以传输速率。计算前请确保两者的单位统一为位或字节。
Transfer Time (s) = Data Size (bits) / Transfer Rate (bps)
传输时间（秒）= 数据大小（位）/ 传输速率（bps）
If data size is given in bytes, multiply it by 8 first. For very large files, be prepared to express the time in minutes or hours.
如果数据大小以字节给出，请先乘以 8。对于非常大的文件，需要将时间表示为分钟或小时。
5. CPU Performance Metrics | CPU 性能指标
Two linked formulas describe CPU execution time. The first uses clock frequency, the second uses clock cycle time. Both give the total seconds a program takes to run.
两个相互关联的公式描述了 CPU 执行时间。第一个使用时钟频率，第二个使用时钟周期时间。两者均能计算程序运行所需的总秒数。
Execution Time (s) = (Instruction Count × CPI) / Clock Rate (Hz)
执行时间（秒）=（指令数 × CPI）/ 时钟频率（Hz）
Execution Time (s) = Instruction Count × CPI × Clock Cycle Time (s)
执行时间（秒）= 指令数 × CPI × 时钟周期时间（秒）
CPI stands for Cycles Per Instruction. Clock Cycle Time is the reciprocal of Clock Rate (1/f). Reducing CPI, improving clock rate, or reducing instruction count all shorten execution time.
CPI 表示每条指令所需时钟周期数。时钟周期时间是时钟频率的倒数（1/f）。降低 CPI、提升时钟频率或减少指令数都能缩短执行时间。
6. Data Unit Conversions | 数据单位换算
Understanding units of digital information is vital. The table below shows the most common binary and decimal prefixes. For file size and memory questions, AQA typically uses the binary IEC prefixes (e.g., KiB, MiB).
理解数字信息单位至关重要。下表显示了最常见的二进制和十进制前缀。在文件大小和内存相关题目中，AQA 通常采用二进制 IEC 前缀（如 KiB、MiB）。
| Unit (单位) | Abbreviation (缩写) | Equivalent Bytes (字节数) |
|---|---|---|
| 1 bit | b | 1/8 byte |
| 1 nibble | – | 4 bits = 0.5 byte |
| 1 byte | B | 1 byte = 8 bits |
| 1 kibibyte | KiB | 2¹⁰ bytes = 1024 bytes |
| 1 mebibyte | MiB | 2²⁰ bytes = 1 048 576 bytes |
| 1 gibibyte | GiB | 2³⁰ bytes = 1 073 741 824 bytes |
| 1 kilobyte (decimal) | KB | 10³ bytes = 1000 bytes |
Pay close attention to whether the question uses decimal or binary multipliers, and always show your working.
请仔细分辨题目使用的是十进制还是二进制乘数，并始终展示计算过程。
7. Boolean Algebra Laws | 布尔代数定律
These laws help simplify Boolean expressions and logic circuits. AQA candidates should be able to apply them in truth tables, gate transformations, and expression reduction.
这些定律有助于化简布尔表达式和逻辑电路。AQA 考生应能在真值表、门电路转换和表达式化简中加以应用。
| Law (定律) | Expression (表达式) | 中文描述 |
|---|---|---|
| Commutative | A ∧ B = B ∧ A A ∨ B = B ∨ A |
交换律：与/或运算的次序可交换。 |
| Associative | A ∧ (B ∧ C) = (A ∧ B) ∧ C A ∨ (B ∨ C) = (A ∨ B) ∨ C |
结合律：与/或运算的分组方式不影响结果。 |
| Distributive | A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C) |
更多咨询请联系16621398022（同微信）
CommentsMore posts |
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply