How many pages will be required to convert a logical address of 24 bits into physical address during mapping if the size of each page is 2 K?
Answers
Answer:
Paging is a memory management scheme that eliminates the need for contiguous allocation of physical memory. This scheme permits the physical address space of a process to be non – contiguous.
Logical Address or Virtual Address (represented in bits): An address generated by the CPU
Logical Address Space or Virtual Address Space( represented in words or bytes): The set of all logical addresses generated by a program
Physical Address (represented in bits): An address actually available on memory unit
Physical Address Space (represented in words or bytes): The set of all physical addresses corresponding to the logical addresses
Example:
If Logical Address = 31 bit, then Logical Address Space = 231 words = 2 G words (1 G = 230)
If Logical Address Space = 128 M words = 27 * 220 words, then Logical Address = log2 227 = 27 bits
If Physical Address = 22 bit, then Physical Address Space = 222 words = 4 M words (1 M = 220)
If Physical Address Space = 16 M words = 24 * 220 words, then Physical Address = log2 224 = 24 bits
The mapping from virtual to physical address is done by the memory management unit (MMU) which is a hardware device and this mapping is known as paging technique.
The Physical Address Space is conceptually divided into a number of fixed-size blocks, called frames.
The Logical address Space is also splitted into fixed-size blocks, called pages.
Page Size = Frame Size
Let us consider an example:
Physical Address = 12 bits, then Physical Address Space = 4 K words
Logical Address = 13 bits, then Logical Address Space = 8 K words
Page size = frame size = 1 K words (assumption)
size
Address generated by CPU is divided into
Page number(p): Number of bits required to represent the pages in Logical Address Space or Page number
Page offset(d): Number of bits required to represent particular word in a page or page size of Logical Address Space or word number of a page or page offset.
Physical Address is divided into
Frame number(f): Number of bits required to represent the frame of Physical Address Space or Frame number.
Frame offset(d): Number of bits required to represent particular word in a frame or frame size of Physical Address Space or word number of a frame or frame offset.
The hardware implementation of page table can be done by using dedicated registers. But the usage of register for the page table is satisfactory only if page table is small. If page table contain large number of entries then we can use TLB(translation Look-aside buffer), a special, small, fast look up hardware cache.
The TLB is associative, high speed memory.
Each entry in TLB consists of two parts: a tag and a value.
When this memory is used, then an item is compared with all tags simultaneously.If the item is found, then corresponding value is returned.
Size1
Main memory access time = m
If page table are kept in main memory,
Effective access time = m(for page table) + m(for particular page in page table)
save2
Questions asked in GATE on Paging:
GATE CS 2001 Question 46
This article has been contributed by Vikash Kumar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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Article Tags : Operating Systems memory-management
Practice Tags : Operating Systems