📝 EXAMPLE of ARM ARCHITECTURE (9 MCQs)

📖 Explanation: Two‑level paging is used for 4‑KB and 16‑KB pages in the ARM architecture, so the correct choice is A. One‑level paging applies only to 1‑MB and 16‑MB sections, making B incorrect. Direct mapping and segmentation are not part of ARM's virtual memory scheme, so C and D are wrong. In a real smartphone, the processor must translate many small pages quickly, and the two‑level scheme provides the needed granularity and speed.

📖 Explanation: ARM focuses on design excellence, allowing it to innovate rapidly while partners handle costly fabrication. Option A is partly true but not the main reason; licensing does not directly lower design costs. Patent restrictions (C) are not the driving factor, and ARM does face competition, so D is inaccurate. By licensing, ARM can spread its architecture across many devices, such as smartphones and tablets, enabling widespread adoption without owning fabs.

📖 Explanation: The micro TLB supplies the base of the 1‑MB section (0x12000000). For a 4‑KB page, the inner index (0x34) is multiplied by the page size (0x1000) giving 0x34000. Adding the offset 0x56 yields 0x34056, which combined with the base gives 0x12034056, so A is correct. Option B mistakenly places the inner index before the offset, C mixes offset and index, and D adds extra digits. This calculation mirrors how an operating system translates virtual addresses to physical memory in a real ARM‑based device.

📖 Explanation: ASID means Address Space Identifier, a tag that allows the micro TLB to distinguish entries belonging to different processes, preventing unnecessary flushes. The other options describe unrelated hardware terms; B refers to a signaling device, C is a design concept, and D describes a type of integrated circuit, none of which match the TLB functionality. In multitasking smartphones, ASIDs let the processor keep separate virtual‑to‑physical mappings for each app, improving performance.

📖 Explanation: Using 1‑MB sections cuts the number of page‑table entries dramatically, which speeds up look‑ups (B). However, allocating memory in 1‑MB chunks can waste space when processes need only a few kilobytes, leading to internal fragmentation (A). Therefore the correct answer combines both effects, C. Options A and B isolate one benefit, and D ignores the impact. In a tablet, this trade‑off means the system may run faster but could consume more RAM than necessary for small workloads.

📖 Explanation: Having distinct instruction and data micro TLBs enables the processor to fetch an instruction address and resolve a data address at the same time, improving pipeline throughput, which is why A is correct. Reducing power (B) is a secondary benefit but not the primary design goal. Simplifying hardware (C) would be easier with a unified TLB, not more complex. Different ASID handling (D) is unnecessary because both TLBs can share the same ASID scheme. Real‑world ARM cores benefit from parallel translation when executing code and accessing memory concurrently.

📖 Explanation: The average latency is calculated by weighting the hit and miss cases. Hits occur 70 % of the time at 2 cycles each (0.7 × 2 = 1.4 cycles). Misses occur 30 % of the time at 120 cycles each (0.3 × 120 = 36 cycles). Adding them yields 1.4 + 36 = 37.4 cycles, so answer B is correct. Option A overestimates, C underestimates, and D is unrelated. This figure reflects how TLB performance directly influences overall execution time in real ARM‑based applications.

📖 Explanation: A 1‑MB section covers 256 times the memory of a 4‑KB page (since 1 MB = 256 × 4 KB). Therefore, the page table needs one entry for each section rather than for each page, cutting the entry count by a factor of 256, which makes A correct. Options B and C contradict this reduction, and D suggests no effect, which is inaccurate. In devices with limited RAM, a smaller page table can free valuable memory, illustrating the practical benefit of section‑based mapping.

📖 Explanation: With 16‑KB pages, the offset occupies 14 bits. ARM’s outer page table typically maps 1‑MB regions (2^20 bytes). Dividing the total virtual space (4 GB = 2^32 bytes) by 1 MB (2^20) yields 2^(32‑20) = 2^12 = 4096 entries, so C is correct. Option A would correspond to 16‑MB regions, B to 4‑MB, and D to 0.5‑MB regions, none matching the architecture’s layout. Knowing the outer table size helps system designers allocate appropriate memory for page‑table structures in real ARM‑based platforms.