Year 11 Eduqas Computer Science: Formulas & Theorems Quick Reference | Year 11 Eduqas 计算机：公式定理速查手册

📚 Year 11 Eduqas Computer Science: Formulas & Theorems Quick Reference | Year 11 Eduqas 计算机：公式定理速查手册

This handbook compiles every key formula, conversion and Boolean law you will need for the Year 11 Eduqas Computer Science specification. It is designed as a rapid revision tool, covering data representation, compression, logic, algorithms and error detection. Each formula is paired with a worked explanation so you can apply it confidently in an exam.

本手册汇编了 Year 11 Eduqas 计算机科学课程中所有关键的公式、转换方法和布尔定律。它是一份快速复习工具，涵盖数据表示、压缩、逻辑、算法和错误检测。每个公式都配有详细解释，帮助你在考试中自信应用。

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1. Units of Data | 数据单位

Data is measured from the smallest unit, a bit (binary digit, 0 or 1). Eight bits form one byte, which is typically the storage capacity needed for a single ASCII character. For larger quantities we use powers of 1024 because computer memory is addressable in binary.

数据以最小单位比特（二进制位，0 或 1）度量。8 个比特构成一个字节，这通常是一个 ASCII 字符所需的存储容量。对于更大的量我们使用 1024 的幂次，因为计算机内存是按二进制寻址的。

1 byte = 8 bits
1 KB = 1024 bytes (2¹⁰)
1 MB = 1024 KB (2²⁰ bytes)
1 GB = 1024 MB (2³⁰ bytes)
1 TB = 1024 GB (2⁴⁰ bytes)

1 字节 = 8 位
1 KB = 1024 字节 (2¹⁰)
1 MB = 1024 KB (2²⁰ 字节)
1 GB = 1024 MB (2³⁰ 字节)
1 TB = 1024 GB (2⁴⁰ 字节)

1024 = 2048 KB) and divide when going from smaller to larger (e.g. 4096 MB ÷ 1024 = 4 GB).

在单位之间转换时，从大单位转为小单位用乘法（例如 2 MB = 2 × 1024 = 2048 KB），从小单位转为大单位用除法（例如 4096 MB ÷ 1024 = 4 GB）。

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2. Binary & Hexadecimal Conversions | 二进制与十六进制转换

Binary numbers use base 2; each digit (bit) represents a power of 2. To convert a binary number to decimal, sum the products of each bit and its place value, where the rightmost bit is 2⁰.

二进制数以 2 为基数；每个数字（位）代表 2 的幂次。要将二进制数转换为十进制，将每个位与其位值相乘后求和，最右边的位是 2⁰。

Decimal = d_n-1×2^n-1 + d_n-2×2^n-2 + ... + d₀×2⁰

十进制 = d_n-1×2^n-1 + d_n-2×2^n-2 + ... + d₀×2⁰

corresponds to a nibble (4 bits). To convert binary to hex, group bits in fours from the right; each nibble becomes one hex digit. The reverse applies for hex to binary.

十六进制（基数 16）使用符号 0-9 和 A-F（A=10，F=15）。一个十六进制数字对应一个半字节（4 位）。将二进制转换为十六进制时，从右向左每四位一组，每个半字节转换为一个十六进制数字。十六进制转二进制则相反。

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3. Text File Size | 文本文件大小

The storage required for plain text is the number of characters multiplied by the bits per character. Standard ASCII uses 7 bits (usually stored as 8 bits, one byte). Extended ASCII uses one full byte per character. Unicode, encoded as UTF-8, uses 8 to 32 bits depending on the character.

纯文本所需的存储容量等于字符数乘以每个字符的位数。标准 ASCII 使用 7 位（通常存为 8 位，即 1 字节）。扩展 ASCII 每个字符使用整整一个字节。Unicode（UTF-8 编码）根据字符使用 8 到 32 位。

Text file size (bits) = number of characters × bits per character

文本文件大小（位）= 字符数 × 每字符位数

For a quick estimate, assume 1 byte per character for ASCII and prepare to justify your choice. If higher characters are needed, Unicode may demand 2 or 4 bytes per character.

快速估算时，假设 ASCII 每个字符 1 字节，并准备说明你的依据。如果需要高位字符，Unicode 可能需要每个字符 2 或 4 字节。

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4. Image File Size | 图像文件大小

A bitmap image' els) and the colour depth (bits per pixel). Each pixel stores a colour value that takes up 'colour depth' bits.

位图图像的原始大小取决于分辨率（宽 × 高，单位像素）和颜色深度（每像素位数）。每个像素存储一个颜色值，占用“颜色深度”位。

Image file size (bits) = width × height × colour depth

图像文件大小（位）= 宽 × 高 × 颜色深度

-bit colour (true colour) requires 300×200×24 = 1,440,000 bits = 180,000 bytes ≈ 175.8 KB.

要将位转换为字节除以 8。例如，一张 300×200 的 24 位真彩色图像需要 300×200×24 = 1,440,000 位 = 180,000 字节 ≈ 175.8 KB。

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5. Sound File Size | 声音文件大小

Uncompressed sound file size is calculated from the sample rate (Hz), bit depth, number of channels and duration. Each second of audio contains 'sample rate' samples, each stored with 'bit depth' bits.

未压缩的声音文件大小由采样率（Hz）、位深度、声道数和时长计算得出。每秒钟音频包含“采样率”个采样点，每个采样点以“位深度”位存储。

Sound file size (bits) = sample rate × bit depth × duration (seconds) × number of channels

声音文件大小（位）= 采样率 × 位深度 × 时长（秒）× 声道数

tereo, 180 seconds) gives: 44100×16×180×2 = 254,016,000 bits = 31,752,000 bytes ≈ 30.3 MB.

要得到字节数，将结果除以 8。CD 质量（44100 Hz, 16 位, 立体声, 180 秒）计算为：44100×16×180×2 = 254,016,000 位 = 31,752,000 字节 ≈ 30.3 MB。

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6. Data Transfer Time | 数据传输时间

The time needed to transfer data over a network is determined by the size of the file and the transfer rate (bit rate). Always ensure that both quantities are expressed in matching units (bits or bytes).

通过网络传输数据所需的时间由文件大小和传输速率（比特率）决定。务必确保两个量使用一致的单位（位或字节）。

Transfer time (seconds) = data size (bits) / transfer rate (bps)

传输时间（秒）= 数据大小（位） / 传输速率（bps）

,886,080 bits) over a 100 Mbps connection (100,000,000 bps) takes about 0.84 seconds in ideal conditions.

如果文件大小以字节给出，先乘以 8 得到位数。例如，一个 10 MB 的文件（10×1024×1024×8 = 83,886,080 位）通过 100 Mbps 的连接传输（100,000,000 bps），理想情况下大约需要 0.84 秒。

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7. Compression Ratio | 压缩比

Compression ratio measures how much a file size is reduced. It compares the size before compression to the size after compression. A higher ratio means greater compression.

压缩比衡量文件大小缩小的程度。它比较压缩前与压缩后的大小。比值越高表示压缩越强。

Compression ratio = uncompressed size / compressed size

压缩比 = 未压缩大小 / 压缩后大小

You may also be asked for percentage compression reduction. The formula is ((uncompressed – compressed) / uncompressed) × 100% a file that shrinks from 200 KB to 50 KB, compression ratio is 4:1 and the reduction is 75%.

你可能还需要计算压缩的百分比减少量。公式为 ((未压缩 – 压缩后) / 未压缩) × 100%。如果一个文件从 200 KB 缩小到 50 KB，压缩比为 4:1，减幅为 75%。

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8. Boolean Algebra Laws | 布尔代数定律

Boolean algebra uses variables that can only be 0 or 1. The basic operations are AND (conjunction, ·), OR (disjunction

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