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In a logical address space (LA) of 8 pages with 1024 words each, the total logical address space is 8,192 words. If this is mapped into a memory of 32 frames, each frame can hold 256 words (since 8,192 words / 32 frames = 256 words per frame). This means not all frames will be utilized for the given pages, as only 8 pages are needed. Consequently, there are additional frames available in memory that can accommodate future pages or other processes.

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Consider a logical address space with 4 pages of 1024 bytes per pageeach mapped onto a physical memory of 64 frames. how many bits are there in logical address?

4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits


Consider a logical address space of eight pages of 1024words each mapped onto a physical memory of 32 frames how many bits are there in the logical address?

As was given for a 4 Page, 1024 words & 64 frames (shown below) 4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits The answer for this problem is 13. 8 pages -> 2^3 bits 1024 bytes -> 2^10 bits 32 frames -> 2^5 bits Therefore: Logical memory = 3+10=13 bits (Page + Word) Physical memory = 10 + 5 =15 bits (Word + Frame)


How many bits are in logical addresss If a logical address space of eight pages of 1024 words each mapped onto a physical memory of 32 frames?

a.Logical address will have3 bits to specify the page number (for 8 pages) .10 bits to specify the offset into each page (210 =1024 words) = 13 bits.b.For (25) 11 32 frames of 1024 words each (Page size = Frame size)We have 5 + 10 = 15 bits.


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