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Demand paging system

Considering the equation the data doesnt match ??? Any help or comments in this regard!!
Consider a pure demand paging system with a paging disk that has
An average access and transfer time of 20 milliseconds. Addresses
Are translated through a page table in main memory, with an access
Time of 50 nanoseconds per memory access. Thus, each memory
Reference through the page table takes two accesses. To improve this time, we have added cache memory that reduces access time to one memory reference, if page table entry is in the cache. Assume that cache access time is negligible.
Assume that 80% of the accesses are in the cache, of the remaining,
10% cause page faults, and 50% of the replaced pages are dirty.
Write down the expression / equation for the effective memory
Access time for the computer system.
I am Linux 9.0 user.

Comments

  • ABDUL HAYEEABDUL HAYEE Member Posts: 81
    : Considering the equation the data doesnt match ??? Any help or comments in this regard!!
    : Consider a pure demand paging system with a paging disk that has
    : An average access and transfer time of 20 milliseconds. Addresses
    : Are translated through a page table in main memory, with an access
    : Time of 50 nanoseconds per memory access. Thus, each memory
    : Reference through the page table takes two accesses. To improve this time, we have added cache memory that reduces access time to one memory reference, if page table entry is in the cache. Assume that cache access time is negligible.
    : Assume that 80% of the accesses are in the cache, of the remaining,
    : 10% cause page faults, and 50% of the replaced pages are dirty.
    : Write down the expression / equation for the effective memory
    : Access time for the computer system.
    : I am Linux 9.0 user.


    Hello ,
    Send me the result you got.
    then i can verify if it was correct .
    Bye,
    :

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