Re: )))))))Vu & Company((((((( help required regarding operating system's assignment....

sis goto insert and select chart

On 6/30/11, mc080404722 Maria Shahzadi <mc080404722@vu.edu.pk> wrote:
> hi
> who is there plz tell me that how to make a chart in ms-word i forgot.. plz
> tell me early..
>
> On Fri, Jul 1, 2011 at 12:22 AM, Mohammad Sajjad
> <ghulamemastan@gmail.com>wrote:
>
>> dekha mera nam hi kafi hai ab samajh i k samjhaon?
>>
>> On 6/30/11, mc090202888 Saba Naureen <mc090202888@vu.edu.pk> wrote:
>> > hmmmmmmmmm samajh aa gai pehle sawal ki.......
>> > how intelligent i am .........
>> >
>> > On Thu, Jun 30, 2011 at 11:39 PM, mc090202888 Saba Naureen <
>> > mc090202888@vu.edu.pk> wrote:
>> >
>> >> A computer system has a 36-bit virtual address space with a page size
>> >> of
>> >> 8K, and 4 bytes per page table entry.
>> >>
>> >> 1. How many pages are in the virtual address space?
>> >> 2. What is the maximum size of addressable physical memory in this
>> >> system?
>> >> 3. If the average process size is 8GB, would you use a one-level,
>> >> two-level, or three-level page table? Why?
>> >> 4. Compute the average size of a page table in question 3 above.
>> >>
>> >> Solution:
>> >>
>> >> 1. A 36 bit address can address 2^36 bytes in a byte addressable
>> >> machine. Since the size of a page 8K bytes (2^13), the number of
>> >> addressable
>> >> pages is 2^36 / >2^13 = 2^23
>> >> 2. With 4 byte entries in the page table we can reference 2^32
>> >> pages.
>> >> Since each page is 2^13 B long, the maximum addressable physical
>> memory
>> >> size
>> >> is 2^32 * 2^13 = 2^45 B (assuming no protection bits are used).
>> >> 3. 8 GB = 2^33 B
>> >>
>> >> We need to analyze memory and time requirements of paging schemes in
>> order
>> >> to make a decision. Average process size is considered in the
>> calculations
>> >> below.
>> >>
>> >> *1 Level Paging*
>> >> Since we have 2^23 pages in each virtual address space, and we use 4
>> bytes
>> >> per page table entry, the size of the page table will be 2^23 * 2^2 =
>> >> 2^25.
>> >> This is 1/256 of the process' own memory space, so it is quite costly.
>> (32
>> >> MB)
>> >>
>> >> *2 Level Paging*
>> >> The address would be divided up as 12 | 11 | 13 since we want page
>> >> table
>> >> pages to fit into one page and we also want to divide the bits roughly
>> >> equally.
>> >>
>> >> Since the process' size is 8GB = 2^33 B, I assume what this means is
>> that
>> >> the total size of all the distinct pages that the process accesses is
>> 2^33
>> >> B. Hence, this process accesses 2^33 / 2^13 = 2^20 pages. The bottom
>> level
>> >> of the page table then holds 2^20 references. We know the size of each
>> >> bottom level chunk of the page table is 2^11 entries. So we need 2^20 /
>> >> 2^11
>> >> = 2^9 of those bottom level chunks.
>> >>
>> >> The total size of the page table is then:
>> >>
>> >> //size of the outer page table
>> >>
>> >> //total size of the inner pages
>> >>
>> >> 1 * 2^12 * 4
>> >>
>> >> + 2^9 * 2^11 * 4
>> >>
>> >> = 2^20 * ( 2^-6 + 4) ~4MB
>> >>
>> >> *3 Level Paging*
>> >> For 3 level paging we can divide up the address as follows:
>> >> 8 | 8 | 7 | 13
>> >>
>> >> Again using the same reasoning as above we need 2^20/2^7 = 2^13 level 3
>> >> page table chunks. Each level 2 page table chunk references 2^8 level 3
>> >> page
>> >> table chunks. So we need 2^13/2^8 = 2^5 level-2 tables. And, of course,
>> >> one
>> >> level-1 table.
>> >>
>> >> The total size of the page table is then:
>> >>
>> >> //size of the outer page table
>> >>
>> >> //total size of the level 2 tables
>> >>
>> >> //total size of innermost tables
>> >>
>> >> 1 * 2^8 * 4
>> >>
>> >> 2^5 * 2^8 *4
>> >>
>> >> 2^13 * 2^7 * 4
>> >>
>> >> ~4MB
>> >>
>> >> As easily seen, 2-level and 3-level paging require much less space then
>> >> level 1 paging scheme. And since our address space is not large enough,
>> >> 3-level paging does not perform any better than 2 level paging. Due to
>> the
>> >> cost of memory accesses, choosing a 2 level paging scheme for this
>> process
>> >> is much more logical.
>> >>
>> >> 1. Calculations are done in answer no. 3
>> >>
>> >>
>> >>
>> >>
>> >> On Thu, Jun 30, 2011 at 11:37 PM, mc090202888 Saba Naureen <
>> >> mc090202888@vu.edu.pk> wrote:
>> >>
>> >>> A computer system has a 32-bit virtual address space with a page size
>> of
>> >>> 4K and 8
>> >>> bytes per page table entry.
>> >>> 1. How many pages can be in the virtual address space?
>> >>> 2. What is the maximum size of addressable physical memory in this
>> >>> system?
>> >>>
>> >>> is question ki exact ulat example mili hai..... net se
>> >>>
>> >>>
>> >>> On Thu, Jun 30, 2011 at 11:33 PM, Mohammad Sajjad <
>> >>> ghulamemastan@gmail.com> wrote:
>> >>>
>> >>>> knsa hai swal ji?
>> >>>>
>> >>>> On 6/30/11, mc090202888 Saba Naureen <mc090202888@vu.edu.pk> wrote:
>> >>>> > bhaiya assignment ke sawal hi nahi hal ho rahe..... kiya
>> >>>> > karan......
>> >>>> >
>> >>>> > On Thu, Jun 30, 2011 at 11:28 PM, Mohammad Sajjad
>> >>>> > <ghulamemastan@gmail.com>wrote:
>> >>>> >
>> >>>> >> yes what help u need?
>> >>>> >>
>> >>>> >> On 6/30/11, mc090202888 Saba Naureen <mc090202888@vu.edu.pk>
>> wrote:
>> >>>> >> > yar kisi ne is stupid assignment ko try kiya hai????
>> >>>> >> >
>> >>>> >> > --
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>> >>>> >> >
>> >>>> >> >
>> >>>> >>
>> >>>> >>
>> >>>> >> --
>> >>>> >> ~Ghulam-e-Mastan~
>> >>>> >> الله محمد چار یار حاجی خواجہ قطب فرید
>> >>>> >> ( حق فرید یا فرید ( رحمتہ الله علیہ
>> >>>> >>
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>> >>>> >>
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>> >>>> >
>> >>>>
>> >>>>
>> >>>> --
>> >>>> ~Ghulam-e-Mastan~
>> >>>> الله محمد چار یار حاجی خواجہ قطب فرید
>> >>>> ( حق فرید یا فرید ( رحمتہ الله علیہ
>> >>>>
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>> >>>>
>> >>>>
>> >>>
>> >>
>> >
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>> >
>> >
>>
>>
>> --
>> ~Ghulam-e-Mastan~
>> الله محمد چار یار حاجی خواجہ قطب فرید
>> ( حق فرید یا فرید ( رحمتہ الله علیہ
>>
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>>
>>
>
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--
~Ghulam-e-Mastan~
الله محمد چار یار حاجی خواجہ قطب فرید
( حق فرید یا فرید ( رحمتہ الله علیہ

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