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Effective memory access time

Discussion in 'Operating System' started by rahul roy, May 4, 2010.

  1. rahul roy

    rahul roy Banned

    May 4, 2010
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    An operating system allocates memory in units of 1 KB pages. The address space of a process can be up to 64 MB in size; however, at any point of time, a process can be allocated at most 16 MB of physical memory. In addition the kernel uses 65 KB of physical memory to
    store page table entries of the current process. The OS also uses a translation-lookaside buffer (TLB) to cache page table entries. You are also given the following information:
    a) size of a page table entry is 4 bytes,
    b) TLB hit ratio is 90%,
    c) time for a TLB lookup is negligible,
    d) time for a memory read is 100 nanoseconds,
    e) time to a read a page from the swapping device into physical memory is 10 milliseconds.
    Calculate the effective memory access time for a process whose address space is 20 MB? Assume that memory accesses are random and distributed uniformly over the entire address space.
    Anybody can solve this problem ? plz reply.:cuss:

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