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

acha ji ab koi help kr bhi day..... saray nakhray kr rahe hain.......

On Fri, Jul 1, 2011 at 1:28 AM, mc080203417 Iffat Arif <mc080203417@vu.edu.pk> wrote:

insert pictur then chart option

On Fri, Jul 1, 2011 at 12:34 AM, 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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>>>>
>>>>
>>>> --
>>>> ~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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