📚 End of Unit 3: Exam-style case study questions | 单元3结束：考试风格案例研究题
Case study questions in Cambridge International AS and A Level Computer Science (9618) push you beyond simple recall. They present a realistic scenario, often combining several topics from a unit, and ask you to apply your knowledge to analyse, design, or evaluate a technical solution. Unit 3 – Hardware – is rich with opportunities for examiners to weave together logic circuits, input/output devices, storage media, and basic processor concepts into a single, context-driven task. This article unpacks three full-length, exam-style case studies typical of an end-of-unit assessment, complete with step-by-step answers and examiner-style commentary. Use these to sharpen your problem-solving skills and spot the subtle ways marks are allocated.
案例分析题在剑桥国际AS和A Level计算机科学（9618）考试中，绝不仅是对概念的简单复述。这类题目会给出一个真实场景，常常融合同一单元中的多个知识点，要求你运用所学去分析、设计或评价一个技术方案。单元3——硬件——内容非常丰富，出题人很容易将逻辑电路、输入输出设备、存储介质以及基本的处理器概念糅合进一道情境驱动的试题中。本文拆解了三道完整的、与单元结束时测验难度匹配的考试风格案例题，并配有分步解答和考官视角的点评。请用这些题目磨炼你的问题解决能力，并细心体会分数是如何分配给各个要点的。
1. Why Unit 3 Case Studies Matter | 为什么单元三案例题至关重要
Unit 3 builds the foundation for understanding how a computer system physically operates. From logic gates forming the building blocks of combinational circuits, to the role of sensors, actuators, and memory, every concept links to real-world machinery. Examiners design case studies around these links. Instead of asking 'State the purpose of ROM' in isolation, they might describe an embedded vending machine and ask you to justify the choice of memory types, sketch a control circuit using NAND gates, and evaluate the impact of using an SSD for transaction logging. The ability to move fluently between specification points while staying rooted in the given scenario is exactly what distinguishes a high-scoring script.
单元3为理解计算机系统如何物理运行奠定了基础。从构成组合电路基本单元的逻辑门，到传感器、执行器和存储器的作用，每一个概念都关联着真实的机器设备。考官正是围绕这些联系来设计案例题。他们不会孤立地问“简述ROM的用途”，而是可能描述一台嵌入式自动售货机，然后让你论证所选存储器类型的合理性，用与非门画出控制电路，并评估采用固态硬盘记录交易日志的影响。能否在紧扣给定场景的同时，流畅地穿梭于考纲各个知识点之间，正是高分答卷的决胜关键。
2. How Case Study Questions Are Structured | 案例题的典型结构
A typical end-of-unit case study begins with a paragraph or two setting the scene – a home automation project, a small business upgrading its IT, a simple medical device. Then come three to six sub-questions, often increasing in difficulty and mark weight: (a) identify or list (1–2 marks), (b) explain or compare (3–4 marks), (c) design or draw (5–6 marks), and finally (d) evaluate or discuss (5–6 marks). The final part frequently tests your ability to argue a balanced judgement, such as whether a sensor-based system is more reliable than a manual override. Notice how marks climb; invest your time proportionately.
单元结束的典型案例题开头会有一两段背景描述——家庭自动化项目、小企业升级IT系统、一种简易医疗器械等等。接着是3到6个子问题，难度和分值通常递升：(a) 识别或列举 (1–2分)，(b) 解释或比较 (3–4分)，(c) 设计或绘图 (5–6分)，最后 (d) 评估或讨论 (5–6分)。最后一问往往考查你是否能做出平衡的判断，例如“基于传感器的系统是否比人工手动控制更可靠”。注意分数是如何攀升的，要按比例分配你的作答时间。
3. Case Study 1: Smart Corridor Lighting | 案例一：智能走廊照明
Scenario: ect is designing an energy-efficient corridor for a new office block. Two movement sensors – one at the east end (E) and one at the west end (W) – movement. However, during office hours (indicated by a timer signal H = 1 for hours 08:00– f sensor readings. Outside office hours (H = 0), the lights should follow the sensor rule.
场景：一名建筑师正在为一栋新办公楼设计节能走廊。走廊东端（E）和西端（W）各装有一个运动传感器，用于检测是否有人经过。每当至少一个传感器检测到移动时，天花板照明（L）应被点亮。然而，在办公时间段内（由定时器信号H表示，H = 1代表08:00–18:00），无论传感器读数如何，照明必须保持常亮。在非办公时间（H = 0），照明则须按照传感器规则工作。
(a) State the logic expression for L in terms of E, W and H when H = 0. (1 mark)
(a) 写出当H = 0时，L关于E、W和H的逻辑表达式。(1分)
(b) Draw the truth table for all combinations of E, W and H, and complete the output column for L. (3 marks)
(b) 画出E、W和H所有组合的真值表，并写出输出L对应的值。(3分)
(c) Using only two-input NAND gates, design a circuit that implements the logic above. You may use a K-map to simplify the expression first. (6 marks)
(c) 仅使用二输入与非门，设计一个实现上述逻辑的电路。你可以先通过卡诺图化简表达式。(6分)
egardless of H, E and W. Without drawing a new circuit, explain how you would modify the original design to accommodate this requirement. (2 marks)
(d) 建筑师还要求增加一个手动维修开关（M），当其激活时（M = 1）应强制点亮照明，而无需顾及H、E和W的值。不画新电路图，请解释你会如何修改原设计以满足这一要求。(2分)
Answers and commentary | 答案与点评
(a) When H = 0, the lights are on if at least one sensor detects movement: L = E OR W. Stated simply as 'E OR W' earns the mark. Many students overcomplicate this initial step; remember that part (a) is designed to get you into the context.
(a) 当H = 0时，至少一个传感器检测到移动即开灯：L = E OR W。直接回答“E OR W”即可得分。很多学生在这第一步就想得过于复杂；记住，(a)问的目的就是让你快速进入情境。
(b) Truth table:
(b) 真值表：
| E | W | H | L |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
The output is 1 for every row except the first. A blank truth table is usually provided in the exam; check that you fill every cell. (3 marks for all eight rows correct; deduct one mark per error.)
除第一行外，所有行的输出都为1。考试中通常会提供一个空白真值表，务必填满每一格。(8行全对得3分，每处错误扣1分。)
(c) From the truth table we obtain an unsimplified sum-of-products expression. However, a quick K-map reveals that L simplifies to H + (E OR W). In Boolean algebra: L = H + E + W alent to three inputs OR-ed. A two-level NAND realisation for OR: first NAND each input with itself (forming NOT), then NAND those outputs together? ngle three-input OR cannot be directly built from two-input NANDs without using De Morgan: H + E + W = (H' · E' · W')' tes (each from a NAND with tied inputs), feed the inverted signals into a three-input NAND? But we only have two-input NANDs; we can cascade: first compute (H' NAND E') = X, then X is H'·E' inverted, and then NAND that with W'? ee NAND gates used as inverters (connect both inputs of a NAND to the signal) to produce H', E', W'. Then feed H', E' into a two-input NAND, producing (H' NAND E') = (H'·E')' need to OR that with W. We can do another stage: produce a NOT of (H+E) using an inverter? No, we want (H+E)+W. We can combine by using another NAND: let S = (H' NAND E'), which is (H+E). We want S OR W. Convert to NAND form: S OR W = (S' · W')' . So we take S through an inverter to get S', W through an inverter for W' NAND with tied inputs. So total gates: three inverters (NANDs) for H, E, W; one NAND for H' and E'; one inverter from that output; one NAND for that inverted output and W'? Actually, let's design stepwise. A robust NAND-only circuit: L = ( (H NAND H) NAND (E NAND E) ) NAND (W NAND W) gives (A NAND B) where A = (H' NAND E')?, not correct. I'll provide a canonical NAND-only solution: L = NAND( NAND(H,H) , NAND( NAND(E,E) , NAND(W,W) ) ) yields L = H + E + W; using De Morgan, L = ( (H NAND H) NAND (E NAND E) NAND (W NAND W) )' which requires a three-input NAND, so we break it: first compute X = (E' NAND W')? We'll accept a clear explanation: use three NANDs as inverters to generate H', E', W'; feed E' and W' into a NAND to give (E+W); feed that output and H' into another NAND with its output inverted? to recognising the simple OR relationship and successfully translating it into NAND gates.
(c) 从真值表可得未化简的积之和表达式。但通过卡诺图很快发现，L可化简为H + (E OR W)。写成布尔代数即为L = H + E + W（逻辑或）。由于限定使用二输入与非门，需将表达式转换为纯与非形式。函数 L = H + E + W 相当于三个输入相或。利用德·摩根定律：H + E + W = (H' · E' · W')'。因此，先用三个与非门分别接成反相器产生H'、E'、W'，然后将这三个信号送入一个三输入与非门——但我们只有二输入门，所以需要级联：用二输入与非门先计算E' NAND W'，其输出等于(E+W)；再将此输出与H'经过类似的门组合，必要时再加一次反相。典型A-Level解答会配合简洁的电路图，这里我们勾勒出思路：总共需要五到六个与非门。关键在于识别出简单的或逻辑，并成功转换为与非实现。(6分)
(d) The new input M forces L = 1 when M = 1. This is simply an OR condition: L = M + (previous expression). To modify the NAND-only circuit, take the existing output (which gives H + E + W) and OR it with M. In NAND logic, OR-ing M can be achieved by connecting M (inverted if needed) and the old output into additional NAND structures. The simplest modification: treat M as an additional input to the final stage, just as we did with H, E, W. In other words, we can regenerate the expression as L = H + E + W + M, and rebuild the circuit identically, adding one more inverter and adjusting the cascade. Since the question says 'without drawing a new circuit', a textual description earns credit: 'Extend the K-map to four variables (H, E, W, M); the SOP simplifies to H+E+W+M. Add a NAND inverter for M and widen the final combining NAND stage (e.g., use a four-input NAND or extra cascade) to accept the additional term.' (2 marks)
(d) 新输入M要求在M=1时强制L=1。这显然是一个或的逻辑：L = M + (先前的逻辑表达式)。要修改现有纯与非电路，可以把M看作额外的输入，并入最后一级。最简单的做法是将逻辑函数扩展为L = H + E + W + M，然后如法炮制，增加一个反相器处理M，并扩大最终的与非合并级（可选用四输入与非门或再级联一个二输入门）。由于题目要求“不画新电路图”，文字说明也可得分：“将卡诺图扩展为四个变量(H, E, W, M)，SOP化简得H+E+W+M。增加一个M的反相器，并在最终与非组合级中加入这个新项。” (2分)
4. Case Study 2: School Computer Laboratory Refresh | 案例二：学校计算机实验室设备更新
Scenario: Greenwood International School plans to replace its 10-year-old desktop computers in the main ICT laboratory. The laboratory hosts 30 workstations used for programming, video editing, and online examinations. The school's IT manager must choose between upgrading to mid-range desktops with traditional HDDs or higher-cost machines with SSDs. Additionally, each workstation currently has 4 GiB of RAM; the new software demands at least 8 GiB, with some multimedia applications recommending 16 GiB. Budget is limited, but reliability during online exams is critical.
场景：绿林国际学校计划更换其主ICT实验室中已使用十年的台式计算机。该实验室设有30台工作站，用于编程、视频编辑和在线考试。校IT经理必须选择究竟是升级为搭配传统HDD的中档台式机，还是成本更高的固态硬盘机器。此外，目前每台工作站内存为4 GiB，新软件至少需要8 GiB，某些多媒体应用则推荐16 GiB。预算有限，但在线考试期间的可靠性至关重要。
(a) Compare the roles of RAM and secondary storage in a workstation during an online examination. (3 marks)
(a) 比较在一次在线考试中，RAM与辅助存储器在工作站中分别扮演的角色。(3分)
(b) Explain two advantages of SSDs over HDDs that are directly relevant to the school's video-editing classes. (4 marks)
(b) 解释SSD相比于HDD的两项优势，并说明它们与学校视频编辑课程的直接关联。(4分)
(c) The IT manager decides to purchase 30 desktops, each with a 256 GiB SSD and 8 GiB of RAM. However, a teacher argues that for the same total cost, the school could get 35 desktops with HDDs. Discuss the trade-off between performance and quantity, making reference to the specific uses of the laboratory. (6 marks)
(c) IT经理决定购买30台台式机，每台配备256 GiB固态硬盘和8 GiB内存。但有教师提出，用同样的总成本学校可以买到35台搭载HDD的机器。请讨论性能与数量之间的权衡，并具体联系实验室的用途。(6分)
Answers and commentary | 答案与点评
(a) During an online exam, RAM holds the currently running examination software, the web browser, and any active data, providing fast read/write access so the CPU can execute instructions smoothly. Secondary storage (the SSD/HDD) stores the operating system, the exam platform's local cache, and any autosave backups. If RAM is insufficient, the system may slow down as it resorts to virtual memory, risking timeouts – a crucial reliability concern.
(a) 在线考试期间，RAM保存着当前运行的考试软件、浏览器以及任何活动数据，提供快速读写访问，以便CPU顺畅执行指令。辅助存储器（SSD或HDD）则存放操作系统、考试平台的本地缓存和任何自动保存的备份。如果RAM不足，系统将被迫使用虚拟内存，可能导致速度变慢甚至超时——这是一个重大的可靠性隐患。
(b) First, SSDs offer much faster read/write speeds, dramatically reducing the time needed to load large video files and render previews; this directly improves students' workflow during editing sessions. Second, SSDs have no moving parts, making them more resistant to physical shock and less prone to mechanical failure – important in a classroom where machines may be accidentally jolted. The durability ensures that video projects are not lost to a head crash.
(b) 第一，固态硬盘提供高得多的读写速度，能显著缩短加载大型视频文件和渲染预览的时间，直接提升学生在编辑课堂中的创作流程。第二，固态硬盘没有活动部件，更抗物理冲撞，不易发生机械故障——这在经常可能意外碰撞的教室环境中至关重要。其耐久性可确保视频项目不会因磁头碰撞而丢失。
(c) This question requires a balanced evaluation. Opting for 30 SSD machines offers higher per-machine performance: quicker boot times, faster application loading, and smoother multitasking – essential for video editing and maintaining exam stability. However, 35 HDD machines would increase capacity, allowing more simultaneous users or smaller class sizes; beneficial if the school's timetable often sees oversubscribed lab slots. Yet HDDs' slower speed could frustrate students during editing and increase the risk of exam lags. The trade-off centres on quality versus quantity. Given that online examination reliability is 'critical' and multimedia work demands speed, the performance gain justifies the smaller quantity. A stronger school might add a few HDD machines for basic tasks, but for this lab's specific purposes, 30 SSDs represent better value. Clear references to the scenario requirements earn top marks.
(c) 此题需要进行平衡评估。选择30台固态硬盘机器，每台性能更高：启动更快、应用程序加载更迅速、多任务处理更流畅——这对视频编辑和确保考试稳定性至关重要。而35台HDD机器则增加了容量，可容纳更多同时上机的学生或支持小班教学；若学校课表常出现实验室满员的情况，则这一优势明显。不过，HDD的低速会让学生在编辑时感到沮丧，并增加考试迟滞的风险。本质上，这是质量与数量的取舍。题干明确指出在线考试可靠性“至关重要”，多媒体工作又需要速度，因此性能提升足以证明数量减少是合理的。资金充裕的学校或可为简单任务增配几台HDD设备，但就该实验室的特定用途而言，30台固态硬盘设备性价比更高。清晰引用场景要求是获得满分的关键。
5. Case Study 3: Digital Dice Embedded System | 案例三：数字骰子嵌入式系统
Scenario: A hobbyist constructs a 'digital dice' device using a microcontroller, seven LEDs arranged as a standard die face, and a push button. When the button is pressed, a pseudo-random number between 1 and 6 is generated by the microcontroller and displayed by lighting the appropriate LEDs. The device uses a crystal oscillator running at 4 MHz. Each press triggers
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