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Key Facts: Engineer Computer System Exam

80 items

Official written exam paper length (4 subjects × 20 items)

Q-Net Engineer Computer System qualification specification (jmCd 0269)

120 minutes

Written CBT examination duration (30 minutes per subject)

Q-Net 필기시험 접수안내 (기사·산업기사 과목별 30분)

40 / 60

Written passing threshold: 40% subject floor (과락) and 60% overall average

National Technical Qualifications Act Enforcement Rules

KRW 19,400

Written examination fee on Q-Net portal

Q-Net portal jmCd=0269, checked 2026

KRW 22,600

Practical examination fee on Q-Net portal

Q-Net portal jmCd=0269, checked 2026

2 years

Written pass exemption validity period

National Technical Qualifications Act Enforcement Decree Article 21

컴퓨터시스템기사 is South Korea's unified Engineer-grade qualification for computer architecture and systems engineering (jmCd 0269, MOEL / HRD Korea). The 2026 written exam tests 80 four-option MCQs across 4 subjects in 120 minutes (40% subject floor, 60% average), followed by a 2-hour written-answer practical in 컴퓨터시스템 실무. Registration fees are KRW 19,400 written and KRW 22,600 practical. OpenExamPrep provides an independent 100-item English study bank.

Sample Engineer Computer System Practice Questions

Try these sample questions to review concepts for the Engineer Computer System exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1In operating system process management, which process state transition occurs when the CPU scheduler assigns a processor core to a process currently residing in the ready queue?
A.Dispatch (디스패치): Ready to Running
B.Timeout (시간 초과): Running to Ready
C.Block / Wait (대기): Running to Blocked
D.Wakeup (깨움): Blocked to Ready
Explanation: The transition from the Ready state (준비 상태) to the Running state (실행 상태) is performed by the operating system dispatcher and is formally called Dispatch (디스패치). During this transition, the OS saves the context of the outgoing process, loads the registers and program counter of the newly selected process, switches to user mode, and jumps to the proper location in the program.
2Which data structure is maintained by the operating system kernel for every active process to store its execution context, register values, and scheduling metadata?
A.Process Control Block (PCB, 프로세스 제어 블록)
B.Translation Lookaside Buffer (TLB, 변환 참조 버퍼)
C.File Allocation Table (FAT, 파일 할당 테이블)
D.Interrupt Vector Table (IVT, 인터럽트 벡터 테이블)
Explanation: The Process Control Block (PCB, 프로세스 제어 블록) is the kernel data structure that stores all information needed to manage an individual process. It contains the Process ID (PID), process state, Program Counter (PC), CPU registers, CPU scheduling priority, memory management pointers (base/limit registers or page table pointers), accounting info, and list of open file descriptors.
3Four processes (P1, P2, P3, P4) arrive simultaneously at time t = 0 with CPU burst times of 6 ms, 8 ms, 7 ms, and 3 ms, respectively. What is the average waiting time under non-preemptive Shortest Job First (SJF, 최단 작업 우선) scheduling?
A.7.0 ms
B.9.5 ms
C.13.0 ms
D.16.5 ms
Explanation: Under non-preemptive SJF, jobs are scheduled in ascending order of burst length: P4 (3 ms), P1 (6 ms), P3 (7 ms), and P2 (8 ms). The start times (which equal waiting times because all arrive at t=0) are: P4 waits 0 ms; P1 starts at 3 ms (waits 3 ms); P3 starts at 3 + 6 = 9 ms (waits 9 ms); P2 starts at 9 + 7 = 16 ms (waits 16 ms). The average waiting time is (0 + 3 + 9 + 16) / 4 = 28 / 4 = 7.0 ms.
4In a Round Robin (RR) CPU scheduling system, the time quantum is configured to 4 ms, and each context switch incurs an operating system overhead of 1 ms. Assuming processes continuously consume their full allotted time slices, what is the effective CPU utilization devoted to user processes?
A.80%
B.75%
C.20%
D.25%
Explanation: In each Round Robin scheduling cycle where a process consumes its full quantum, the CPU spends 4 ms executing useful process code followed by 1 ms performing the context switch to the next process. The total elapsed period is 4 ms + 1 ms = 5 ms. Therefore, effective CPU utilization is (useful computation time) / (total time) = 4 / 5 = 0.80, or 80%.
5Which mechanism is used in a Multilevel Feedback Queue (MLFQ, 다단계 피드백 큐) scheduler to prevent long-running, CPU-bound processes from suffering indefinite starvation?
A.Aging (에이징): periodically boosting the priority of processes that wait too long in lower-priority queues
B.Preemptive priority inversion by disabling hardware timer interrupts
C.Compaction of memory segments across priority bands
D.Static queue binding where processes are permanently fixed to their initial queue
Explanation: Aging (에이징) is the standard technique in priority scheduling and MLFQ to avoid starvation. If a low-priority, CPU-bound process waits in a lower-priority queue for a configured threshold without receiving CPU time, the operating system kernel increments its priority or moves it into a higher-priority queue, guaranteeing eventual execution.
6To provide a correct software solution for the Critical Section Problem (임계 구역 문제), which three classical requirements must be satisfied?
A.Mutual Exclusion (상호 배제), Progress (진행), and Bounded Waiting (한계 대기)
B.Atomicity (원자성), Consistency (일관성), and Durability (지속성)
C.Hold and Wait (점유와 대기), Circular Wait (환형 대기), and No Preemption (비선점)
D.Paging (페이징), Segmentation (세그멘테이션), and Swapping (스와핑)
Explanation: Dijkstra and subsequent operating system literature establish three fundamental criteria for any valid critical section solution: 1) Mutual Exclusion (no two processes can be executing in their critical sections simultaneously), 2) Progress (if no process is in its critical section, only processes wishing to enter can participate in deciding who enters next, and this decision cannot be postponed indefinitely), and 3) Bounded Waiting (there must be a bound on the number of times other processes are allowed to enter after a process has requested entry).
7In Peterson's algorithm for two-process mutual exclusion, process P0 sets `flag[0] = true` and `turn = 1`. What condition does P0 check in its busy-wait while loop before entering the critical section?
A.while (flag[1] && turn == 1);
B.while (flag[0] && turn == 0);
C.while (!flag[1] || turn == 0);
D.while (flag[0] == flag[1]);
Explanation: In Peterson's algorithm, when process P0 wants to enter, it announces its intent by setting `flag[0] = true` and generously yields precedence by assigning `turn = 1`. P0 then spins while process P1 is also interested (`flag[1] == true`) and it is currently P1's turn (`turn == 1`). Once P1 finishes its critical section and clears `flag[1]`, or if P1 was never interested, P0 exits the loop and enters the critical section.
8What atomic operation occurs when a process executes the `wait()` (or `P()`) primitive on a counting semaphore initialized to integer value S?
A.It decrements S by 1; if S < 0, the calling process blocks and is placed in the semaphore wait queue
B.It increments S by 1; if S > 0, it awakens a blocked process from the semaphore wait queue
C.It reads S without modification and context-switches to the kernel scheduler
D.It resets S to 0 and broadcasts a signal to all threads associated with a condition variable
Explanation: The classical `wait()` (or `P()`, from Dutch 'proberen') operation decrements the semaphore counter `S`. If the resulting value is negative (or if S was 0 prior to decrement in non-negative implementations), the calling process cannot acquire the resource, suspends execution, and is placed into the semaphore's blocked queue. Conversely, `signal()` (`V()`) increments S and wakes up a blocked process.
9Which of the following is NOT one of the four Coffman conditions necessary for a system deadlock (교착 상태) to occur?
A.Preemptive resource reclamation (선점 자원 회수)
B.Mutual exclusion (상호 배제)
C.Hold and wait (점유와 대기)
D.Circular wait (환형 대기)
Explanation: Deadlock requires 'No preemption' (비선점), meaning resources cannot be forcibly confiscated from a process holding them; they must be released voluntarily. If the system allows 'Preemptive resource reclamation' (선점 자원 회수), deadlocks can be systematically broken or prevented by preemption.
10A system running the Banker's Algorithm (은행원 알고리즘) has 5 processes (P0–P4) and 3 resource types A, B, C with total vector (10, 5, 7). Current Allocation is P0:(0,1,0), P1:(2,0,0), P2:(3,0,2), P3:(2,1,1), P4:(0,0,2). Maximum Demand is P0:(7,5,3), P1:(3,2,2), P2:(9,0,2), P3:(2,2,2), P4:(4,3,3). Which sequence represents a valid safe execution order?
A.<P1, P3, P0, P2, P4>
B.<P0, P1, P2, P3, P4>
C.<P2, P4, P1, P3, P0>
D.<P4, P2, P3, P1, P0>
Explanation: Total allocated resources sum to A: 0+2+3+2+0=7, B: 1+0+0+1+0=2, C: 0+0+2+1+2=5. Available = (10, 5, 7) - (7, 2, 5) = (3, 3, 2). Need matrices (Max - Allocation) are: P0:(7,4,3), P1:(1,2,2), P2:(6,0,0), P3:(0,1,1), P4:(4,3,1). Comparing Need with Available (3,3,2): P1 can execute because (1,2,2) <= (3,3,2). After P1 completes, Available becomes (3,3,2) + (2,0,0) = (5,3,2). Then P3 can execute: (0,1,1) <= (5,3,2), releasing (2,1,1) to make Available (7,4,3). Then P0 can execute: (7,4,3) <= (7,4,3), releasing (0,1,0) -> (7,5,3). Then P2: (6,0,0) <= (7,5,3), releasing (3,0,2) -> (10,5,5). Finally P4 finishes. Thus, <P1, P3, P0, P2, P4> is a valid safe sequence.

About the Engineer Computer System Exam

Engineer Computer System (컴퓨터시스템기사) is South Korea's unified Engineer-grade national technical qualification in computer hardware and system software engineering, administered by the Human Resources Development Service of Korea (HRD Korea / Q-Net) under the Ministry of Employment and Labor (MOEL). Newly unified in 2026 from the former Engineer Computer (전자계산기기사) and Engineer Computer System Application (전자계산기조직응용기사) qualifications under the National Technical Qualifications Act Enforcement Rules, the credential validates comprehensive competence in computer architecture, operating systems, systems programming, and digital communications. The written examination tests 80 four-option items across four subjects (20 items each), followed by a 2-hour written-answer practical examination in 컴퓨터시스템 실무. This study bank provides an independent English-language MCQ study adaptation for written preparation and is not an official translation or practical examination simulation.

Exam sponsor: HRD Korea (Q-Net) / Ministry of Employment and Labor. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Written paper covering 운영체제 및 시스템소프트웨어, 컴퓨터구조, 컴퓨터프로그래밍, and 디지털회로 및 데이터통신 with 20 items and 30 minutes each (80 items / 120 minutes total), followed by a 2-hour, 100-point written-answer practical examination in 컴퓨터시스템 실무. OpenExamPrep study weights cover all four written subjects with 25 questions each.

Time Limit

120 minutes written; 2 hours practical

Passing Score

Written: 40+ per subject floor and 60 average; practical: 60/100

Exam / Certification Fees

KRW 19,400 written / KRW 22,600 practical (Q-Net, 2026)

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

25%

Operating Systems and System Software (운영체제 및 시스템소프트웨어)

Process lifecycle, thread architectures, CPU scheduling algorithms (FCFS, SJF, SRTF, Round Robin, Multilevel Feedback Queue), process synchronization primitives (mutex locks, semaphores, monitors), deadlock characterization and handling (prevention, avoidance with Banker's algorithm, detection, recovery), memory management architectures (paging, segmentation, page tables, TLB effective access time calculations), virtual memory page replacement algorithms (FIFO, LRU, LFU, Optimal, Clock algorithm, working set, thrashing), file system organization (FAT, inode, ext4, NTFS, contiguous/indexed allocation), I/O subsystems and disk scheduling (SSTF, SCAN, C-SCAN), and system software toolchains (macro processors, assemblers, linkers, loaders, runtime loaders).

25%

Computer Architecture (컴퓨터구조)

CPU architecture and instruction execution cycles (fetch, decode, execute, interrupt), micro-operations and control unit design (hardwired vs microprogrammed), instruction formats and addressing modes (immediate, direct, indirect, register, displacement, PC-relative), RISC vs CISC architectures, instruction-level parallelism and pipelining (MIPS 5-stage pipeline, pipeline speedup calculations, throughput, structural hazards, data hazards and forwarding, branch hazards and branch prediction), cache memory organization (direct-mapped, fully associative, set-associative, tag/index/offset address partitioning), cache replacement policies, cache write policies (write-through, write-back, write-allocate), average memory access time (AMAT) calculations for multi-level caches, cache coherence protocols (MESI, MOESI), virtual memory hardware interfaces, and system bus architectures and I/O interfacing (programmed I/O, interrupt-driven I/O, direct memory access DMA, PCIe, USB, bus arbitration).

25%

Computer Programming (컴퓨터프로그래밍)

C systems programming language fundamentals (data types, operators, control structures, storage classes, variable scope and lifetime), pointer mechanics and memory models (pointer arithmetic, arrays and multi-dimensional array address calculations, function pointers, dynamic memory allocation with malloc/calloc/realloc/free, memory leaks, dangling pointers), structs, unions, bit-fields, and alignment/padding calculations, preprocessor macros and header guards, linear data structures (arrays, singly/doubly linked lists, circular lists, stacks, queues, circular queues, deques), non-linear data structures (binary trees, binary search trees, AVL trees, heaps and priority queues, hash tables with chaining and open addressing, graphs and adjacency representations), algorithm design and complexity analysis (asymptotic Big-O notation, Master theorem, recursion relations), fundamental sorting and searching algorithms (bubble, insertion, selection, quicksort, mergesort, heapsort, binary search), graph algorithms (BFS, DFS, Dijkstra, Prim, Kruskal), and embedded systems programming fundamentals (memory-mapped I/O, volatile qualifier, interrupt service routines ISR, hardware timer configuration).

25%

Digital Circuits and Data Communications (디지털회로 및 데이터통신)

Boolean algebra theorems (De Morgan's laws, duality, consensus theorem), Boolean function minimization using algebraic manipulation and Karnaugh maps (SOP, POS, don't-care conditions, prime implicants), combinational logic building blocks (half adders, full adders, carry-lookahead adders, subtractors, decoders, encoders, multiplexers, demultiplexers, parity generators), sequential logic circuits (latches, SR, D, JK, and T flip-flops, master-slave flip-flops, excitation tables, setup/hold times), registers and counters (shift registers, synchronous and asynchronous ripple counters, ring and Johnson counters, modulus calculations), finite state machine synthesis (Mealy vs Moore models, state diagram, state reduction, state assignment), digital transmission media and signal characteristics (attenuation, distortion, Nyquist and Shannon channel capacity), OSI 7-layer and TCP/IP protocol architectures, data link layer protocols (framing, flow control with stop-and-wait and sliding window, error detection with parity, checksum, CRC polynomial division, ARQ error correction), and network/transport layers (IPv4 addressing, subnet masking, CIDR host range calculations, IPv6, routing protocols RIP, OSPF, BGP, TCP 3-way handshake and congestion control, UDP).

Preparing for the Engineer Computer System Exam

What You Need to Know

  • Passing score: Written: 40+ per subject floor and 60 average; practical: 60/100
  • Assessment: Written paper covering 운영체제 및 시스템소프트웨어, 컴퓨터구조, 컴퓨터프로그래밍, and 디지털회로 및 데이터통신 with 20 items and 30 minutes each (80 items / 120 minutes total), followed by a 2-hour, 100-point written-answer practical examination in 컴퓨터시스템 실무. OpenExamPrep study weights cover all four written subjects with 25 questions each.
  • Time limit: 120 minutes written; 2 hours practical
  • Exam / certification fees: KRW 19,400 written / KRW 22,600 practical (Q-Net, 2026) Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

Engineer Computer System: Suggested Study Strategy

1Master high-yield calculation problems across all 4 subjects: Expect numerical questions such as Effective Memory Access Time (EMAT) with TLB, multi-level cache AMAT, pipeline speedup and branch stall CPI, Karnaugh map minimization, and IPv4 CIDR subnetting.
2Safeguard against the 40-point per-subject disqualification rule (과락): With 20 questions per subject, scoring fewer than 8 questions correct in any single subject results in an automatic fail regardless of how high your overall score is.
3Familiarize yourself with Korean computer engineering terms alongside English concepts: Key exam terms include 문맥교환 (context switch), 선점 (preemption), 교착상태 (deadlock), 파이프라인 해저드 (pipeline hazard), 주소지정방식 (addressing mode), and 부호확장 (sign extension).
4Treat this English bank as conceptual preparation for both written CBT and the practical paper: The practical examination (컴퓨터시스템 실무) tests written-answer problem solving, C code tracing, and architecture calculations in Korean.

Frequently Asked Questions

What is Engineer Computer System (컴퓨터시스템기사)?

It is an Engineer-grade National Technical Qualification administered by HRD Korea on Q-Net under the Ministry of Employment and Labor (MOEL). It was newly unified in 2026 under the National Technical Qualifications Act Enforcement Rules from the former Engineer Computer (전자계산기기사) and Engineer Computer System Application (전자계산기조직응용기사) qualifications to cover both computer hardware systems and core system software.

How is the official exam structured in 2026?

The written stage is a 120-minute CBT consisting of 80 four-option multiple-choice questions across four subjects: Operating Systems and System Software, Computer Architecture, Computer Programming, and Digital Circuits and Data Communications (20 items each). Candidates who achieve a score of at least 40% in each subject and an overall average of 60% advance to a 2-hour written-answer practical examination (필답형) in 컴퓨터시스템 실무 (Computer System Practice).

What language is the official exam, and what is this study bank?

The official Q-Net examination is administered in Korean at designated CBT and written testing centers across South Korea. This page provides an independent English-language multiple-choice study adaptation designed to master core concepts, calculation methods, and bilingual Korean terminology. It is not an official government translation, not a live CBT platform simulation, and not a substitute for practical exam preparation.

What are the 2026 exam registration fees and timing?

Official Q-Net fees for Engineer Computer System (jmCd 0269) are KRW 19,400 for the written CBT paper and KRW 22,600 for the practical examination (checked 2026). The written exam allocates 30 minutes per subject (120 minutes total for 4 subjects), and the practical examination lasts 2 hours.

How long is a written exam pass valid towards the practical examination?

Under Article 21 of the Enforcement Decree of the National Technical Qualifications Act (국가기술자격법 시행령 제21조), candidates who pass the written paper receive a 2-year exemption from repeating the written examination, calculated from the official pass announcement date.