Questions: Memory Management Fundamentals

Question 1

A program running on a modern OS generates the address 0x4000. At what point is this logical address translated to a physical address?

• A. When the program is compiled
• B. When the program is loaded into memory
• C. At execution time, by the MMU hardware
• D. At execution time, by the OS kernel in software

Answer: (C)

Modern systems use execution-time binding via hardware (the MMU). This allows the OS to relocate processes freely in physical memory without recompiling or reloading. The MMU — not the OS kernel software — performs the translation on every memory access; the kernel only configures the MMU's mapping tables.

Question 2

Two processes can legally have identical logical address values (e.g., both use address 0x1000) without interfering with each other.

• T. True
• F. False

Answer: True

Because logical addresses are translated to physical addresses by the MMU using per-process mapping tables, two processes can each have a logical address 0x1000 that maps to completely different physical locations. This is one of the key benefits of execution-time binding — it enables process isolation.

Question 3

Why is execution-time address binding preferred over compile-time or load-time binding in modern operating systems?

Model answer: Execution-time binding allows the OS to move a process to a different physical memory location at any time, because the MMU translates addresses on every access. Compile-time and load-time binding fix physical addresses early, preventing relocation and making multiprogramming and memory isolation much harder.

If addresses are baked in at compile or load time, the OS cannot freely place or relocate processes — every program would need exclusive access to a fixed memory region. Execution-time binding via the MMU decouples logical layout (what the program sees) from physical layout (where RAM is actually used), enabling flexible multiprogramming and process protection.