1 votes 1 votes If Page size = Frame size is always true and offset in Logical address = offset in physical address holds Then how/why Logical address size can come different than physical address size? Operating System paging operating-system memory-management + – Mrityudoot asked Mar 20 Mrityudoot 237 views answer comment Share Follow See 1 comment See all 1 1 comment reply Biswajit Kumar commented Mar 21 reply Follow Share LA(Logical Address) = p + d where p = no. of pages & d = page offset. PA(Physical Address) = f + d where f = no. of frames & d = frame offset. These Page offset and Frame offset are same as Page Size = Frame Size; But the no. of pages != the no. of frames, Therefore, LA != PA. (as size of PAS != LAS) [But there may be a case where these are equal!] 0 votes 0 votes Please log in or register to add a comment.
Best answer 6 votes 6 votes Logical address is the amount of memory every process assumes to have. Its not true that the entire memory is available to the processes everytime. Its logical to assume that all processes have the same amount of memory. Lets say we have system with 4GB RAM. Every process in the system assumes that it has 4 GB space.Its is worth noting that $Programmer's$ $View$ : Thinks just like the process and doesn't have to care about the amount of memory available to execute a program.So the programmer can complete focus on his logics. $Processor's$ $View$ : It sees the processes and its pages .It is bound to generate a virtual address which is mapped to physical by MMU. There is a very common misconception that the processor generates Main Memory address. Infact it cannot generate any address as its a dumb hardware. It executes instructions which provides the virtual to physical translation by MMU. $Process'$ $View$: In a system with 4GB it is almost never possible that each process can consume 4GB of space that it assumes to have. This is because that much memory isn't even physically available.Lets consider we have 64-bit system with those 4 gigs of RAM.Every process will assume that it can consume 264 B of storage.Although a very large portion of it remains unused.That is why the stack and heap portions of a process' memory map is kept to be floating and has no fixed boundary.U can learn more on these from HPCA 4 . Psy Duck answered Mar 22 • selected Mar 22 by Mrityudoot Psy Duck comment Share Follow See all 0 reply Please log in or register to add a comment.