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Question

Consider the machine with 64 MB Physical Memory and a 34 bit Virtual Address Space. If the page size is 4KB, the appropriate sizes of conventional and inverted page table sizes are:

a) 4M, 4K

b)4K, 4M

c)4M, 16K

d)16K, 4M.

soln: is Option C. but my ans is (7MB, 28KB) 

Answer 1

Is my ans correct ?

Comments

Is there any other way of doing this? What about the conventional page table size is that correct?

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while calculating the conventional size of the page table, we know that the page table will have a lot of page entries and each page table entry will contain frame number(this thing will tell to the CPU that which page is settled in which frame in main memory) along with that. there are some protection bits, invalid/invalid, present/absent bit, the dirty bit, this information is also stored in each page table entry.

Since there is no other information given in question itself that about these bits, but we have a physical address by which we can calculate at least how many frames are available in main memory. On that basis, we will multiply the number of pages with the size of page table entry(here numbers of bits required to represent frame number).

So the size of page table = 2²² * 14 bits

                                        = 2²² * 2 Bytes                   ∴ 14 bits nearly equal to 2 Bytes

                                        = 2²³ Bytes

                                        = 8 MB

The concept of an inverted page table was given because there is a headache of storing lot page tables because each process will have a 1-page table. All the page need not be in the main memory at the same time, so it will be a good idea to store page numbers in the frame of main memory, I  mean we will cut down the main memory in the frames and now each frame will contain page numbers of process and along with this information there is some process id information should also available in each frame table entry so that CPU could distinguish same Pages of different processes.

So the size of inverted page table  = Number of frames * size of the inverted page table entry

Here, size of inverted page table entry is not given, so I m assuming we are storing only page numbers with no process id's but it is not a good way.

size of the inverted page table = 2¹⁴ * 22 bits

                                                  = 2¹⁴ * 3 Bytes

                                                  =  48 KB

 

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Inverted page table size= No of frames * no of bits required to represent each page

no of pages= PAS/ page size = 2^x

no of bits required to represent each page = x

here........x = 22 bits......3B approx

therefore...... Inverted page table size= 2^14 * 3 = 48KB

Answer 2

Page size = $2^{12}$

PAS = $2^{26}$

VAS = $2^{34}$

No of pages $= 2^{34}/2^{12} = 4M$

No of frames $= 2^{26}/2^{12} = 16K$

Page table entry has frame number, no of bits required to address frames $= log 2^{14} = 14 bits$

So Page Table size $= 4M * 14bits = 7MB$

Inverted Page table entry has frame number, no of inverted page table entry = no of frames.

So Page Table size $= 16k * 14bits = 28KB$