The size of the logical address space of the process = 32 x 2 bytes= 64 bytes= 2

Question

The size of the logical address space of the process = 32 x 2 bytes= 64 bytes= 26x23 = 29m = 9 bits The size of the physical address space of the main memory = 32 x 2 bytes= 64 bytes The number of bits required for the page number P =m- n bits = 9 - 4 =5 bits Page size = 2 bytes = 21 x 23 = 24 So, n=4 bits The number of bits required for offset (d) = n=4 bits The number of entries in the page table = number of pages = 32 With reference to the RESULTS above, assume that there is no concept of virtual memory involved here as well as no possibility of preemptive scheduling and moreover the size of each process is same. Keeping all this in mind, what can yon infer from the results? Just explain in one or two lines.

Explanation / Answer

Givent he physical address space is 64 bytes Each page size is 2 bytes. Therefore number of pages = (physical address space / pagesize)                                       =64/2                                          =32 pages Therefore 5 bits are required to represent 32 pages. Now each page size is 2bytes = 16bits Therefore 4 bits (offset) are required to represent all theaddresses in the page. Therefore the logical address has 9 bits ( page number + offset inpage = 5+4) Now we will look into the page table and find the relevant locationof the page in physical memory, location of page in physical memory + offset in page gives usthe physical address

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