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Thread: CS604-Assignment 3

  1. #1
    Junior Member Minnow's Avatar
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    CS604-Assignment 5

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    Calculate the following

    1. Calculate the swap out + swap in time of a process which has 2MB of size having transfer rate of 10MB/sec and whose average latency is 20ms. [5 marks]

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    2. The logical address space of the process is 16 pages and the size of each page is 2 Kb. There are 128 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer the following questions: [2.5 * 4 = 10 marks]

    Calculate the size of Logical Address?
    Find the size of the Logical Memory?
    Calculate the size of Physical Address?
    Find the size of the Physical Memory?

    3. Consider a 64-bit linear address having 16k page size. Calculate maximum number of pages in a process address space, number of bits needed for d and for p. [5 marks]
    Last edited by Minnow; 07-21-2010 at 08:19 PM.

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    Senior Member viki's Avatar
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    please mention the duedate of it
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    mention the duedate of assignment is 21 july(today)

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    ok thanks. we are trying our best to provide you solution

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    IDEA SOLUTION

    (1) Calculate the swap out + swap in time of a process which has 2MB of size having transfer rate of 10MB/sec and whose average latency is 20ms.

    (Answer)
    swap out time = process size / transfer rate
    net swap out time= latency time swap out time
    and then
    swap out + swap in = net swap out time + net swap out time
    by using this method my answer is
    swap out + swap in = 440 ms


    (2) The logical address space of the process is 16 pages and the size of each page is 2 Kb. There are 128 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer the following questions:

    • Calculate the size of Logical Address?
    • Find the size of the Logical Memory?
    • Calculate the size of Physical Address?
    • Find the size of the Physical Memory?

    (Answer)
    Size of logical address= |p| + |d|=15 bits
    Size of logical memory= No. of pages * page size= 64
    Size of physical address= |f|+ |d|=18 bits
    Size of physical memory= No. of frames * size of each frame= 256


    (3) Consider a 64-bit linear address having 16k page size. Calculate maximum number of pages in a process address space, number of bits needed for d and for p.

    (Answer)
    Logical address = 64-bit
    Page size = 16 k

    Maximum pages in a process address space = 2^16 / 16k = 4
    No. of bits needed for d =|d| = log 2 4k = 12 bits
    No. of bits needed for p = |p|= 16 – 12 = 3 bits

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    That's good you provide the solution for just idea.

  7. #7
    Senior Member viki's Avatar
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    hmmmm that is really smart of u ......thankx
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    Junior Member Minnow's Avatar
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    Thanks alot

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    you are welcome dear. Just post your assignment on very first day.

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