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a.
Logical address will have
3 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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Related Questions

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


What Bridges filter network frames based on?

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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)


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What device is used to move frames between a source and a destination on a MAC address?

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