End of Unit 5: Exam-style Case Study Questions | 单元五结课：考试风格案例研究题

📚 End of Unit 5: Exam-style Case Study Questions | 单元五结课：考试风格案例研究题

Unit 5 of the Cambridge International AS & A Level Computer Science course focuses on system software, including operating systems, language translators, and utility programs. Exam-style case study questions often ask you to apply these ideas to a realistic scenario, such as an embedded device or a network server. This article provides a structured set of case studies with model answers and marking guidance to help you prepare for the end-of-unit assessment.

剑桥国际 AS & A Level 计算机科学课程的第五单元聚焦系统软件，包括操作系统、语言翻译器和实用程序。考试风格的案例研究题通常会要求你把这些概念应用到实际场景中，例如嵌入式设备或网络服务器。本文提供一组结构化的案例研究，并附上模型答案和评分指导，帮助你为单元结束评估做好准备。

1. Overview of Unit 5 Assessment | 单元五考核概览

In the Cambridge 9618 syllabus, Unit 5 covers the role of the operating system, interrupt handling, process scheduling, memory management, high- and low-level languages, assemblers, compilers, interpreters, linkers, loaders, and utility software. Case study questions typically allocate 8-12 marks and expect you to identify features, explain mechanisms, and evaluate trade-offs in context.

在剑桥 9618 大纲中，第五单元涵盖操作系统的作用、中断处理、进程调度、内存管理、高低级语言、汇编器、编译器、解释器、链接器、加载器以及实用工具软件。案例研究题通常占 8-12 分，要求你识别特征、解释机制，并结合情境评价权衡。

2. Case Study 1: Operating System Functions in a Library Kiosk | 案例研究1：图书馆自助终端中的操作系统功能

A public library has installed a self-service kiosk that scans books, updates the loan database, and prints receipts. The device has a single processor, 512 MB of RAM, and no secondary storage except a small flash chip. Identify four functions of the operating system that the kiosk relies on, and explain how each supports the kiosk’s operation.

一家公共图书馆安装了一台自助服务终端，用来扫描图书、更新借阅数据库并打印收据。该设备只有一个处理器、512 MB 内存，除了一块小闪存芯片外没有辅助存储器。请指出该终端依赖的操作系统的四项功能，并解释每项功能如何支持终端的运行。

Model answer points: memory management to allocate space for the scanning program and database buffers; device drivers to communicate with the barcode scanner and printer; file management to store configuration data on the flash chip; and interrupt handling to respond to key presses or scanner input without busy waiting.

模型答案要点：内存管理为扫描程序和数据库缓冲区分配空间；设备驱动程序与条码扫描器和打印机通信；文件管理在闪存芯片上存储配置数据；中断处理响应按键或扫描输入，无需忙等待。

3. Case Study 2: Language Translation for a Scientific Calculator | 案例研究2：科学计算器的语言翻译

A manufacturer writes the firmware of a scientific calculator in C, but the processor only understands machine code. The company uses a cross-compiler on a development PC. Explain why a compiler is used instead of an interpreter, and describe one advantage of cross-compilation for this embedded device.

一家制造商用 C 语言编写科学计算器的固件，但处理器只理解机器码。公司在开发 PC 上使用交叉编译器。请解释为什么使用编译器而不是解释器，并描述交叉编译对这种嵌入式设备的一个优点。

A compiler translates the entire C source code into machine code before execution, so the calculator does not need to carry a language translator at run time. This saves memory and improves execution speed. Cross-compilation allows the firmware to be built on a powerful PC even though the target device has limited resources.

编译器在执行之前将整个 C 源代码翻译成机器码，因此计算器在运行时无需携带语言翻译器。这节省了内存并提高了执行速度。交叉编译允许在功能强大的 PC 上构建固件，即使目标设备资源有限。

4. Case Study 3: Interrupt Handling in a Smart Thermostat | 案例研究3：智能恒温器中的中断处理

A smart thermostat uses a temperature sensor that sends an interrupt every 60 seconds when a new reading is ready. The main processor also runs a display update loop. Explain the steps the processor takes when the sensor interrupt is received, and state why interrupts are preferable to polling in this battery-powered device.

智能恒温器使用温度传感器，每 60 秒当新读数准备好时发送一次中断。主处理器还运行显示更新循环。请解释处理器收到传感器中断时采取的步骤，并说明为什么在这种电池供电设备中中断优于轮询。

Steps: the current instruction finishes; the processor saves the program counter and status register on the stack; it identifies the interrupt source and jumps to the interrupt service routine; the ISR reads the temperature and stores it; then the processor restores the saved context and resumes the display loop. Interrupts save power because the processor does not need to check the sensor repeatedly; it can sleep or perform other tasks until an event occurs.

步骤：当前指令执行完毕；处理器将程序计数器和状态寄存器保存到栈中；识别中断源并跳转到中断服务程序；ISR 读取温度并存储；然后处理器恢复保存的上下文并继续显示循环。中断节省电量，因为处理器无需反复检查传感器；它可以休眠或执行其他任务，直到事件发生。

5. Case Study 4: Scheduling Algorithms in a Web Server | 案例研究4：网络服务器中的调度算法

A web server handles three types of jobs: short HTTP requests from browsers, long video transcoding jobs, and interactive terminal sessions. The administrator can choose between round robin, shortest job first (SJF), and multilevel feedback queue. Compare the suitability of each algorithm for this workload.

一台网络服务器处理三类作业：来自浏览器的短 HTTP 请求、长视频转码作业以及交互式终端会话。管理员可以在轮转法、最短作业优先（SJF）和多级反馈队列之间选择。比较每种算法对该工作负载的适用性。

Round robin gives all jobs equal time slices, so short HTTP requests are handled promptly but long transcoding jobs are interrupted often, causing overhead. SJF minimises average waiting time and is good for many short requests, but long jobs may suffer starvation. Multilevel feedback queue separates jobs into priority queues; interactive sessions get high priority, short requests get quick service, and long jobs run in lower-priority queues when the CPU is free, which is the best fit for this mixed workload.

轮转法给所有作业相等的时间片，因此短 HTTP 请求能得到及时处理，但长转码作业会频繁中断，产生开销。SJF 使平均等待时间最小，适合大量短请求，但长作业可能饥饿。多级反馈队列将作业分入不同优先级队列；交互式会话获得高优先级，短请求得到快速服务，长作业在 CPU 空闲时的低优先级队列中运行，最适合这种混合工作负载。

表格总结：Algorithm 算法，Strengths 优点，Weaknesses 缺点。

6. Case Study 5:

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