page size (logical address space) = 512 = 2 power 9
so 9 bits

base address = 10 bits

so combining them 10+ 9 = 19 bits

On Sat, Sep 10, 2011 at 7:18 PM, rohit kumar <[email protected]> wrote:

> please explain ...
>
>
> On Sat, Sep 10, 2011 at 5:10 PM, bharatkumar bagana <
> [email protected]> wrote:
>
>> physical address:19 bits(10+9)
>> logical address : 15 bits(6+9)
>>
>>
>> On Sat, Sep 10, 2011 at 6:48 AM, ravi maggon <[email protected]>wrote:
>>
>>> we have page table having 64 entries of 10 bits each. and page size is of
>>> 512 bytes.
>>> tell the no of bits required for physical and logical address.
>>>
>>> --
>>>
>>> Regards
>>> Ravi Maggon
>>> Final Year, B.E. CSE
>>> Thapar University
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