Consider a logical address space of eight pages of 1024 words each, mapped into a physical memory of 32 frames. How many bits are there in the logical address and in the physical address, respectively?

Question

Arrange the following layers of MS DOS operating system starting from inner most to outer most.
(A). ROM BIOS Device drivers
(B). Resident System Program
(C). MS DOS Device Drivers
(D). Application Program

Answer 1

To determine the bit counts for logical and physical addresses: Logical Address: 8 pages require log2(8) = 3 bits for the page number. With 1024 words per page, log2(1024) = 10 bits are needed for the word offset. The total logical address is 3 + 10 = 13 bits. Physical Address: 32 frames require log2(32) = 5 bits for the frame number. As each frame holds 1024 words, 10 bits are for the word offset. The total physical address is 5 + 10 = 15 bits.

Consider a logical address space of eight pages of 1024 words each, mapped into a physical memory of 32 frames. How many bits are there in the logical address and in the physical address, respectively?

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The Correct Option is B

Solution and Explanation

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